Op-Amp Circuits: Part 5
M. B. Patil [email protected] www.ee.iitb.ac.in/~sequel
Department of Electrical Engineering Indian Institute of Technology Bombay
M. B. Patil, IIT Bombay R2
Vi R1
Vo
RL
R2 R1 V+ = Vi + Vo (3) R1 + R2 R1 + R2
V V V V i ↑ → + ↑ → o ↑ → + ↑ Eq. 3 Eq. 1 Eq. 3
We now have a positive feedback situation.
As a result, Vo rises (or falls) indefinitely,
Vi V Vo V limited finally by saturation. ↑ → − ↑ → ↓ → − ↓ Eq. 2 Eq. 1 Eq. 2 The circuit reaches a stable equilibrium.
Feedback: inverting amplifier
R2
Vi R1
Vo
RL
Vo = AV(V+ V ) (1) − − Since the Op Amp has a high input resistance,
iR1 = iR2, and we get,
R2 R1 V = Vi + Vo (2) − R1 + R2 R1 + R2
M. B. Patil, IIT Bombay R2
Vi R1
Vo
RL
R2 R1 V+ = Vi + Vo (3) R1 + R2 R1 + R2
V V V V i ↑ → + ↑ → o ↑ → + ↑ Eq. 3 Eq. 1 Eq. 3
We now have a positive feedback situation.
As a result, Vo rises (or falls) indefinitely, limited finally by saturation.
The circuit reaches a stable equilibrium.
Feedback: inverting amplifier
R2
Vi R1
Vo
RL
Vo = AV(V+ V ) (1) − − Since the Op Amp has a high input resistance,
iR1 = iR2, and we get,
R2 R1 V = Vi + Vo (2) − R1 + R2 R1 + R2
Vi V Vo V ↑ → − ↑ → ↓ → − ↓ Eq. 2 Eq. 1 Eq. 2
M. B. Patil, IIT Bombay R2
Vi R1
Vo
RL
R2 R1 V+ = Vi + Vo (3) R1 + R2 R1 + R2
V V V V i ↑ → + ↑ → o ↑ → + ↑ Eq. 3 Eq. 1 Eq. 3
We now have a positive feedback situation.
As a result, Vo rises (or falls) indefinitely, limited finally by saturation.
Feedback: inverting amplifier
R2
Vi R1
Vo
RL
Vo = AV(V+ V ) (1) − − Since the Op Amp has a high input resistance,
iR1 = iR2, and we get,
R2 R1 V = Vi + Vo (2) − R1 + R2 R1 + R2
Vi V Vo V ↑ → − ↑ → ↓ → − ↓ Eq. 2 Eq. 1 Eq. 2 The circuit reaches a stable equilibrium.
M. B. Patil, IIT Bombay R2 R1 V+ = Vi + Vo (3) R1 + R2 R1 + R2
V V V V i ↑ → + ↑ → o ↑ → + ↑ Eq. 3 Eq. 1 Eq. 3
We now have a positive feedback situation.
As a result, Vo rises (or falls) indefinitely, limited finally by saturation.
Feedback: inverting amplifier
R2 R2
Vi Vi R1 R1
Vo Vo
RL RL
Vo = AV(V+ V ) (1) − − Since the Op Amp has a high input resistance,
iR1 = iR2, and we get,
R2 R1 V = Vi + Vo (2) − R1 + R2 R1 + R2
Vi V Vo V ↑ → − ↑ → ↓ → − ↓ Eq. 2 Eq. 1 Eq. 2 The circuit reaches a stable equilibrium.
M. B. Patil, IIT Bombay V V V V i ↑ → + ↑ → o ↑ → + ↑ Eq. 3 Eq. 1 Eq. 3
We now have a positive feedback situation.
As a result, Vo rises (or falls) indefinitely, limited finally by saturation.
Feedback: inverting amplifier
R2 R2
Vi Vi R1 R1
Vo Vo
RL RL
R2 R1 Vo = AV(V+ V ) (1) V+ = Vi + Vo (3) − − R1 + R2 R1 + R2 Since the Op Amp has a high input resistance,
iR1 = iR2, and we get,
R2 R1 V = Vi + Vo (2) − R1 + R2 R1 + R2
Vi V Vo V ↑ → − ↑ → ↓ → − ↓ Eq. 2 Eq. 1 Eq. 2 The circuit reaches a stable equilibrium.
M. B. Patil, IIT Bombay We now have a positive feedback situation.
As a result, Vo rises (or falls) indefinitely, limited finally by saturation.
Feedback: inverting amplifier
R2 R2
Vi Vi R1 R1
Vo Vo
RL RL
R2 R1 Vo = AV(V+ V ) (1) V+ = Vi + Vo (3) − − R1 + R2 R1 + R2 Since the Op Amp has a high input resistance, Vi V+ Vo V+ iR1 = iR2, and we get, ↑ → ↑ → ↑ → ↑ Eq. 3 Eq. 1 Eq. 3
R2 R1 V = Vi + Vo (2) − R1 + R2 R1 + R2
Vi V Vo V ↑ → − ↑ → ↓ → − ↓ Eq. 2 Eq. 1 Eq. 2 The circuit reaches a stable equilibrium.
M. B. Patil, IIT Bombay Feedback: inverting amplifier
R2 R2
Vi Vi R1 R1
Vo Vo
RL RL
R2 R1 Vo = AV(V+ V ) (1) V+ = Vi + Vo (3) − − R1 + R2 R1 + R2 Since the Op Amp has a high input resistance, Vi V+ Vo V+ iR1 = iR2, and we get, ↑ → ↑ → ↑ → ↑ Eq. 3 Eq. 1 Eq. 3
R2 R1 V = Vi + Vo (2) We now have a positive feedback situation. − R1 + R2 R1 + R2 As a result, Vo rises (or falls) indefinitely,
Vi V Vo V limited finally by saturation. ↑ → − ↑ → ↓ → − ↓ Eq. 2 Eq. 1 Eq. 2 The circuit reaches a stable equilibrium.
M. B. Patil, IIT Bombay R2
R1
Vo Vi RL
R1 V+ = Vo (3) R1 + R2
V V V V i ↑ → o ↓ → + ↓ → o ↓ Eq. 1 Eq. 3 Eq. 1
We now have a positive feedback situation.
As a result, Vo rises (or falls) indefinitely,
Vi Vo V Vo limited finally by saturation. ↑ → ↑ → − ↑ → ↓ Eq. 1 Eq. 2 Eq. 1 The circuit reaches a stable equilibrium.
Feedback: non-inverting amplifier
R2
R1
Vo Vi RL
Vo = AV(V+ V ) (1) − − Since the Op Amp has a high input resistance,
iR1 = iR2, and we get,
R1 V = Vo (2) − R1 + R2
M. B. Patil, IIT Bombay R2
R1
Vo Vi RL
R1 V+ = Vo (3) R1 + R2
V V V V i ↑ → o ↓ → + ↓ → o ↓ Eq. 1 Eq. 3 Eq. 1
We now have a positive feedback situation.
As a result, Vo rises (or falls) indefinitely, limited finally by saturation.
The circuit reaches a stable equilibrium.
Feedback: non-inverting amplifier
R2
R1
Vo Vi RL
Vo = AV(V+ V ) (1) − − Since the Op Amp has a high input resistance,
iR1 = iR2, and we get,
R1 V = Vo (2) − R1 + R2
Vi Vo V Vo ↑ → ↑ → − ↑ → ↓ Eq. 1 Eq. 2 Eq. 1
M. B. Patil, IIT Bombay R2
R1
Vo Vi RL
R1 V+ = Vo (3) R1 + R2
V V V V i ↑ → o ↓ → + ↓ → o ↓ Eq. 1 Eq. 3 Eq. 1
We now have a positive feedback situation.
As a result, Vo rises (or falls) indefinitely, limited finally by saturation.
Feedback: non-inverting amplifier
R2
R1
Vo Vi RL
Vo = AV(V+ V ) (1) − − Since the Op Amp has a high input resistance,
iR1 = iR2, and we get,
R1 V = Vo (2) − R1 + R2
Vi Vo V Vo ↑ → ↑ → − ↑ → ↓ Eq. 1 Eq. 2 Eq. 1 The circuit reaches a stable equilibrium.
M. B. Patil, IIT Bombay R1 V+ = Vo (3) R1 + R2
V V V V i ↑ → o ↓ → + ↓ → o ↓ Eq. 1 Eq. 3 Eq. 1
We now have a positive feedback situation.
As a result, Vo rises (or falls) indefinitely, limited finally by saturation.
Feedback: non-inverting amplifier
R2 R2
R1 R1
Vo Vo Vi Vi RL RL
Vo = AV(V+ V ) (1) − − Since the Op Amp has a high input resistance,
iR1 = iR2, and we get,
R1 V = Vo (2) − R1 + R2
Vi Vo V Vo ↑ → ↑ → − ↑ → ↓ Eq. 1 Eq. 2 Eq. 1 The circuit reaches a stable equilibrium.
M. B. Patil, IIT Bombay V V V V i ↑ → o ↓ → + ↓ → o ↓ Eq. 1 Eq. 3 Eq. 1
We now have a positive feedback situation.
As a result, Vo rises (or falls) indefinitely, limited finally by saturation.
Feedback: non-inverting amplifier
R2 R2
R1 R1
Vo Vo Vi Vi RL RL
R1 Vo = AV(V+ V ) (1) V+ = Vo (3) − − R1 + R2 Since the Op Amp has a high input resistance,
iR1 = iR2, and we get,
R1 V = Vo (2) − R1 + R2
Vi Vo V Vo ↑ → ↑ → − ↑ → ↓ Eq. 1 Eq. 2 Eq. 1 The circuit reaches a stable equilibrium.
M. B. Patil, IIT Bombay We now have a positive feedback situation.
As a result, Vo rises (or falls) indefinitely, limited finally by saturation.
Feedback: non-inverting amplifier
R2 R2
R1 R1
Vo Vo Vi Vi RL RL
R1 Vo = AV(V+ V ) (1) V+ = Vo (3) − − R1 + R2 Since the Op Amp has a high input resistance, Vi Vo V+ Vo iR1 = iR2, and we get, ↑ → ↓ → ↓ → ↓ Eq. 1 Eq. 3 Eq. 1
R1 V = Vo (2) − R1 + R2
Vi Vo V Vo ↑ → ↑ → − ↑ → ↓ Eq. 1 Eq. 2 Eq. 1 The circuit reaches a stable equilibrium.
M. B. Patil, IIT Bombay Feedback: non-inverting amplifier
R2 R2
R1 R1
Vo Vo Vi Vi RL RL
R1 Vo = AV(V+ V ) (1) V+ = Vo (3) − − R1 + R2 Since the Op Amp has a high input resistance, Vi Vo V+ Vo iR1 = iR2, and we get, ↑ → ↓ → ↓ → ↓ Eq. 1 Eq. 3 Eq. 1
R1 V = Vo (2) We now have a positive feedback situation. − R1 + R2 As a result, Vo rises (or falls) indefinitely,
Vi Vo V Vo limited finally by saturation. ↑ → ↑ → − ↑ → ↓ Eq. 1 Eq. 2 Eq. 1 The circuit reaches a stable equilibrium.
M. B. Patil, IIT Bombay * Because of positive feedback, both of these circuits are unstable.
* The output at any time is only limited by saturation of the op-amp, i.e., Vo = Vsat. ± * Of what use is a circuit that is stuck at Vo = Vsat? It turns out that these circuits are actually useful! ± Let us see how.
Feedback
Inverting amplifier with + Non-inverting amplifier with + ←→ − ←→ −
R2 R2
Vi R1 R1 Vo Vo
Vi RL RL
M. B. Patil, IIT Bombay * The output at any time is only limited by saturation of the op-amp, i.e., Vo = Vsat. ± * Of what use is a circuit that is stuck at Vo = Vsat? It turns out that these circuits are actually useful! ± Let us see how.
Feedback
Inverting amplifier with + Non-inverting amplifier with + ←→ − ←→ −
R2 R2
Vi R1 R1 Vo Vo
Vi RL RL
* Because of positive feedback, both of these circuits are unstable.
M. B. Patil, IIT Bombay * Of what use is a circuit that is stuck at Vo = Vsat? It turns out that these circuits are actually useful! ± Let us see how.
Feedback
Inverting amplifier with + Non-inverting amplifier with + ←→ − ←→ −
R2 R2
Vi R1 R1 Vo Vo
Vi RL RL
* Because of positive feedback, both of these circuits are unstable.
* The output at any time is only limited by saturation of the op-amp, i.e., Vo = Vsat. ±
M. B. Patil, IIT Bombay Feedback
Inverting amplifier with + Non-inverting amplifier with + ←→ − ←→ −
R2 R2
Vi R1 R1 Vo Vo
Vi RL RL
* Because of positive feedback, both of these circuits are unstable.
* The output at any time is only limited by saturation of the op-amp, i.e., Vo = Vsat. ± * Of what use is a circuit that is stuck at Vo = Vsat? It turns out that these circuits are actually useful! ± Let us see how.
M. B. Patil, IIT Bombay Consider Vi = 5 V . R1 1 k Case (i): Vo = +Vsat = +10 V → V+ = Vo = 10 V = 1 V . R1 + R2 1 k + 9 k
(V+ − V−) = (1 − 5) = −4 V → Vo = −Vsat .
This is inconsistent with our assumption (Vo = +Vsat).
R1 1 k Case (ii): Vo = −Vsat = −10 V → V+ = Vo = × (−10 V) = −1 V . R1 + R2 1 k + 9 k
(V+ − V−) = (−1 − 5) = −6 V → Vo = −Vsat (consistent)
If we move to the right (increasing Vi ), the same situation applies, i.e., Vo = −Vsat.
10 Vsat
5
0
V+ −5
Vo −10 V − sat V (V) −10 −5 0 5 10 i
Inverting Schmitt trigger
9 k
R2 1 k
R1 Vo
Vi RL
Vsat = 10 V
Because of positive feedback, Vo can only be +Vsat (if V+ > V−) or −Vsat (if V+ < V−).
M. B. Patil, IIT Bombay R1 1 k Case (i): Vo = +Vsat = +10 V → V+ = Vo = 10 V = 1 V . R1 + R2 1 k + 9 k
(V+ − V−) = (1 − 5) = −4 V → Vo = −Vsat .
This is inconsistent with our assumption (Vo = +Vsat).
R1 1 k Case (ii): Vo = −Vsat = −10 V → V+ = Vo = × (−10 V) = −1 V . R1 + R2 1 k + 9 k
(V+ − V−) = (−1 − 5) = −6 V → Vo = −Vsat (consistent)
If we move to the right (increasing Vi ), the same situation applies, i.e., Vo = −Vsat.
10 Vsat
5
0
V+ −5
Vo −10 V − sat V (V) −10 −5 0 5 10 i
Inverting Schmitt trigger
9 k
R2 1 k
R1 Vo
Vi RL
Vsat = 10 V
Because of positive feedback, Vo can only be +Vsat (if V+ > V−) or −Vsat (if V+ < V−).
Consider Vi = 5 V .
M. B. Patil, IIT Bombay This is inconsistent with our assumption (Vo = +Vsat).
R1 1 k Case (ii): Vo = −Vsat = −10 V → V+ = Vo = × (−10 V) = −1 V . R1 + R2 1 k + 9 k
(V+ − V−) = (−1 − 5) = −6 V → Vo = −Vsat (consistent)
If we move to the right (increasing Vi ), the same situation applies, i.e., Vo = −Vsat.
10 Vsat
5
0
V+ −5
Vo −10 V − sat V (V) −10 −5 0 5 10 i
Inverting Schmitt trigger
9 k
R2 1 k
R1 Vo
Vi RL
Vsat = 10 V
Because of positive feedback, Vo can only be +Vsat (if V+ > V−) or −Vsat (if V+ < V−).
Consider Vi = 5 V . R1 1 k Case (i): Vo = +Vsat = +10 V → V+ = Vo = 10 V = 1 V . R1 + R2 1 k + 9 k
(V+ − V−) = (1 − 5) = −4 V → Vo = −Vsat .
M. B. Patil, IIT Bombay R1 1 k Case (ii): Vo = −Vsat = −10 V → V+ = Vo = × (−10 V) = −1 V . R1 + R2 1 k + 9 k
(V+ − V−) = (−1 − 5) = −6 V → Vo = −Vsat (consistent)
If we move to the right (increasing Vi ), the same situation applies, i.e., Vo = −Vsat.
10 Vsat
5
0
V+ −5
Vo −10 V − sat V (V) −10 −5 0 5 10 i
Inverting Schmitt trigger
9 k
R2 1 k
R1 Vo
Vi RL
Vsat = 10 V
Because of positive feedback, Vo can only be +Vsat (if V+ > V−) or −Vsat (if V+ < V−).
Consider Vi = 5 V . R1 1 k Case (i): Vo = +Vsat = +10 V → V+ = Vo = 10 V = 1 V . R1 + R2 1 k + 9 k
(V+ − V−) = (1 − 5) = −4 V → Vo = −Vsat .
This is inconsistent with our assumption (Vo = +Vsat).
M. B. Patil, IIT Bombay If we move to the right (increasing Vi ), the same situation applies, i.e., Vo = −Vsat.
10 Vsat
5
0
V+ −5
Vo −10 V − sat V (V) −10 −5 0 5 10 i
Inverting Schmitt trigger
9 k
R2 1 k
R1 Vo
Vi RL
Vsat = 10 V
Because of positive feedback, Vo can only be +Vsat (if V+ > V−) or −Vsat (if V+ < V−).
Consider Vi = 5 V . R1 1 k Case (i): Vo = +Vsat = +10 V → V+ = Vo = 10 V = 1 V . R1 + R2 1 k + 9 k
(V+ − V−) = (1 − 5) = −4 V → Vo = −Vsat .
This is inconsistent with our assumption (Vo = +Vsat).
R1 1 k Case (ii): Vo = −Vsat = −10 V → V+ = Vo = × (−10 V) = −1 V . R1 + R2 1 k + 9 k
(V+ − V−) = (−1 − 5) = −6 V → Vo = −Vsat (consistent)
M. B. Patil, IIT Bombay If we move to the right (increasing Vi ), the same situation applies, i.e., Vo = −Vsat.
Inverting Schmitt trigger
10 Vsat 9 k 5 R2 1 k 0 R1 Vo V+ Vi −5 RL
Vsat = 10 V Vo −10 V − sat V (V) −10 −5 0 5 10 i
Because of positive feedback, Vo can only be +Vsat (if V+ > V−) or −Vsat (if V+ < V−).
Consider Vi = 5 V . R1 1 k Case (i): Vo = +Vsat = +10 V → V+ = Vo = 10 V = 1 V . R1 + R2 1 k + 9 k
(V+ − V−) = (1 − 5) = −4 V → Vo = −Vsat .
This is inconsistent with our assumption (Vo = +Vsat).
R1 1 k Case (ii): Vo = −Vsat = −10 V → V+ = Vo = × (−10 V) = −1 V . R1 + R2 1 k + 9 k
(V+ − V−) = (−1 − 5) = −6 V → Vo = −Vsat (consistent)
M. B. Patil, IIT Bombay Inverting Schmitt trigger
10 Vsat 9 k 5 R2 1 k 0 R1 Vo V+ Vi −5 RL
Vsat = 10 V Vo −10 V − sat V (V) −10 −5 0 5 10 i
Because of positive feedback, Vo can only be +Vsat (if V+ > V−) or −Vsat (if V+ < V−).
Consider Vi = 5 V . R1 1 k Case (i): Vo = +Vsat = +10 V → V+ = Vo = 10 V = 1 V . R1 + R2 1 k + 9 k
(V+ − V−) = (1 − 5) = −4 V → Vo = −Vsat .
This is inconsistent with our assumption (Vo = +Vsat).
R1 1 k Case (ii): Vo = −Vsat = −10 V → V+ = Vo = × (−10 V) = −1 V . R1 + R2 1 k + 9 k
(V+ − V−) = (−1 − 5) = −6 V → Vo = −Vsat (consistent)
If we move to the right (increasing Vi ), the same situation applies, i.e., Vo = −Vsat.
M. B. Patil, IIT Bombay Inverting Schmitt trigger
10 Vsat 9 k 5 R2 1 k 0 R1 Vo V+ Vi −5 RL
Vsat = 10 V Vo −10 V − sat V (V) −10 −5 0 5 10 i
Because of positive feedback, Vo can only be +Vsat (if V+ > V−) or −Vsat (if V+ < V−).
Consider Vi = 5 V . R1 1 k Case (i): Vo = +Vsat = +10 V → V+ = Vo = 10 V = 1 V . R1 + R2 1 k + 9 k
(V+ − V−) = (1 − 5) = −4 V → Vo = −Vsat .
This is inconsistent with our assumption (Vo = +Vsat).
R1 1 k Case (ii): Vo = −Vsat = −10 V → V+ = Vo = × (−10 V) = −1 V . R1 + R2 1 k + 9 k
(V+ − V−) = (−1 − 5) = −6 V → Vo = −Vsat (consistent)
If we move to the right (increasing Vi ), the same situation applies, i.e., Vo = −Vsat. M. B. Patil, IIT Bombay Consider decreasing values of Vi .
R1 1 k V+ = Vo = (−Vsat) = −1 V . R1 + R2 10 k As long as Vi = V− > V+ = −1 V , Vo remains at −Vsat.
When Vi < V+ = −1 V , Vo changes sign, i.e., Vo = +Vsat.
R1 V+ now becomes (+Vsat) = +1 V . R1 + R2
Decreasing Vi further makes no difference to Vo (since Vi = V− < V+ = +1 V holds).
Now, the threshold at which Vo flips is Vi = +1 V .
Inverting Schmitt trigger
10 Vsat 9 k 5 R2 1 k 0 R1 Vo V+ Vi −5 RL
Vsat = 10 V Vo −10 V − sat V (V) −10 −5 0 5 10 i
M. B. Patil, IIT Bombay R1 1 k V+ = Vo = (−Vsat) = −1 V . R1 + R2 10 k As long as Vi = V− > V+ = −1 V , Vo remains at −Vsat.
When Vi < V+ = −1 V , Vo changes sign, i.e., Vo = +Vsat.
R1 V+ now becomes (+Vsat) = +1 V . R1 + R2
Decreasing Vi further makes no difference to Vo (since Vi = V− < V+ = +1 V holds).
Now, the threshold at which Vo flips is Vi = +1 V .
Inverting Schmitt trigger
10 Vsat 9 k 5 R2 1 k 0 R1 Vo V+ Vi −5 RL
Vsat = 10 V Vo −10 V − sat V (V) −10 −5 0 5 10 i
Consider decreasing values of Vi .
M. B. Patil, IIT Bombay As long as Vi = V− > V+ = −1 V , Vo remains at −Vsat.
When Vi < V+ = −1 V , Vo changes sign, i.e., Vo = +Vsat.
R1 V+ now becomes (+Vsat) = +1 V . R1 + R2
Decreasing Vi further makes no difference to Vo (since Vi = V− < V+ = +1 V holds).
Now, the threshold at which Vo flips is Vi = +1 V .
Inverting Schmitt trigger
10 Vsat 9 k 5 R2 1 k 0 R1 Vo V+ Vi −5 RL
Vsat = 10 V Vo −10 V − sat V (V) −10 −5 0 5 10 i
Consider decreasing values of Vi .
R1 1 k V+ = Vo = (−Vsat) = −1 V . R1 + R2 10 k
M. B. Patil, IIT Bombay When Vi < V+ = −1 V , Vo changes sign, i.e., Vo = +Vsat.
R1 V+ now becomes (+Vsat) = +1 V . R1 + R2
Decreasing Vi further makes no difference to Vo (since Vi = V− < V+ = +1 V holds).
Now, the threshold at which Vo flips is Vi = +1 V .
Inverting Schmitt trigger
10 Vsat 9 k 5 R2 1 k 0 R1 Vo V+ Vi −5 RL
Vsat = 10 V Vo −10 V − sat V (V) −10 −5 0 5 10 i
Consider decreasing values of Vi .
R1 1 k V+ = Vo = (−Vsat) = −1 V . R1 + R2 10 k As long as Vi = V− > V+ = −1 V , Vo remains at −Vsat.
M. B. Patil, IIT Bombay When Vi < V+ = −1 V , Vo changes sign, i.e., Vo = +Vsat.
R1 V+ now becomes (+Vsat) = +1 V . R1 + R2
Decreasing Vi further makes no difference to Vo (since Vi = V− < V+ = +1 V holds).
Now, the threshold at which Vo flips is Vi = +1 V .
Inverting Schmitt trigger
10 Vsat 9 k 5 R2 1 k 0 R1 Vo V+ Vi −5 RL
Vsat = 10 V Vo −10 V − sat V (V) −10 −5 0 5 10 i
Consider decreasing values of Vi .
R1 1 k V+ = Vo = (−Vsat) = −1 V . R1 + R2 10 k As long as Vi = V− > V+ = −1 V , Vo remains at −Vsat.
M. B. Patil, IIT Bombay R1 V+ now becomes (+Vsat) = +1 V . R1 + R2
Decreasing Vi further makes no difference to Vo (since Vi = V− < V+ = +1 V holds).
Now, the threshold at which Vo flips is Vi = +1 V .
Inverting Schmitt trigger
10 Vsat 9 k 5 R2 1 k 0 R1 Vo V+ Vi −5 RL
Vsat = 10 V Vo −10 V − sat V (V) −10 −5 0 5 10 i
Consider decreasing values of Vi .
R1 1 k V+ = Vo = (−Vsat) = −1 V . R1 + R2 10 k As long as Vi = V− > V+ = −1 V , Vo remains at −Vsat.
When Vi < V+ = −1 V , Vo changes sign, i.e., Vo = +Vsat.
M. B. Patil, IIT Bombay R1 V+ now becomes (+Vsat) = +1 V . R1 + R2
Decreasing Vi further makes no difference to Vo (since Vi = V− < V+ = +1 V holds).
Now, the threshold at which Vo flips is Vi = +1 V .
Inverting Schmitt trigger
10 Vsat 9 k 5 R2 1 k 0 R1 Vo V+ Vi −5 RL
Vsat = 10 V Vo −10 V − sat V (V) −10 −5 0 5 10 i
Consider decreasing values of Vi .
R1 1 k V+ = Vo = (−Vsat) = −1 V . R1 + R2 10 k As long as Vi = V− > V+ = −1 V , Vo remains at −Vsat.
When Vi < V+ = −1 V , Vo changes sign, i.e., Vo = +Vsat.
M. B. Patil, IIT Bombay Decreasing Vi further makes no difference to Vo (since Vi = V− < V+ = +1 V holds).
Now, the threshold at which Vo flips is Vi = +1 V .
Inverting Schmitt trigger
10 Vsat 9 k 5 R2 1 k 0 R1 Vo V+ Vi −5 RL
Vsat = 10 V Vo −10 V − sat V (V) −10 −5 0 5 10 i
Consider decreasing values of Vi .
R1 1 k V+ = Vo = (−Vsat) = −1 V . R1 + R2 10 k As long as Vi = V− > V+ = −1 V , Vo remains at −Vsat.
When Vi < V+ = −1 V , Vo changes sign, i.e., Vo = +Vsat.
R1 V+ now becomes (+Vsat) = +1 V . R1 + R2
M. B. Patil, IIT Bombay Now, the threshold at which Vo flips is Vi = +1 V .
Inverting Schmitt trigger
10 Vsat 9 k 5 R2 1 k 0 R1 Vo V+ Vi −5 RL
Vsat = 10 V Vo −10 V − sat V (V) −10 −5 0 5 10 i
Consider decreasing values of Vi .
R1 1 k V+ = Vo = (−Vsat) = −1 V . R1 + R2 10 k As long as Vi = V− > V+ = −1 V , Vo remains at −Vsat.
When Vi < V+ = −1 V , Vo changes sign, i.e., Vo = +Vsat.
R1 V+ now becomes (+Vsat) = +1 V . R1 + R2
Decreasing Vi further makes no difference to Vo (since Vi = V− < V+ = +1 V holds).
M. B. Patil, IIT Bombay Now, the threshold at which Vo flips is Vi = +1 V .
Inverting Schmitt trigger
10 Vsat 9 k 5 R2 1 k 0 R1 Vo V+ Vi −5 RL
Vsat = 10 V Vo −10 V − sat V (V) −10 −5 0 5 10 i
Consider decreasing values of Vi .
R1 1 k V+ = Vo = (−Vsat) = −1 V . R1 + R2 10 k As long as Vi = V− > V+ = −1 V , Vo remains at −Vsat.
When Vi < V+ = −1 V , Vo changes sign, i.e., Vo = +Vsat.
R1 V+ now becomes (+Vsat) = +1 V . R1 + R2
Decreasing Vi further makes no difference to Vo (since Vi = V− < V+ = +1 V holds).
M. B. Patil, IIT Bombay Inverting Schmitt trigger
10 Vsat 9 k 5 R2 1 k 0 R1 Vo V+ Vi −5 RL
Vsat = 10 V Vo −10 V − sat V (V) −10 −5 0 5 10 i
Consider decreasing values of Vi .
R1 1 k V+ = Vo = (−Vsat) = −1 V . R1 + R2 10 k As long as Vi = V− > V+ = −1 V , Vo remains at −Vsat.
When Vi < V+ = −1 V , Vo changes sign, i.e., Vo = +Vsat.
R1 V+ now becomes (+Vsat) = +1 V . R1 + R2
Decreasing Vi further makes no difference to Vo (since Vi = V− < V+ = +1 V holds).
Now, the threshold at which Vo flips is Vi = +1 V .
M. B. Patil, IIT Bombay Inverting Schmitt trigger
10 Vsat 9 k 5 R2 1 k 0 R1 Vo V+ Vi −5 RL
Vsat = 10 V Vo −10 V − sat V (V) −10 −5 0 5 10 i
Consider decreasing values of Vi .
R1 1 k V+ = Vo = (−Vsat) = −1 V . R1 + R2 10 k As long as Vi = V− > V+ = −1 V , Vo remains at −Vsat.
When Vi < V+ = −1 V , Vo changes sign, i.e., Vo = +Vsat.
R1 V+ now becomes (+Vsat) = +1 V . R1 + R2
Decreasing Vi further makes no difference to Vo (since Vi = V− < V+ = +1 V holds).
Now, the threshold at which Vo flips is Vi = +1 V . M. B. Patil, IIT Bombay R1 * The threshold values (or “tripping points”), VTH and VTL, are given by Vsat. ± R1 + R2
* The tripping point (whether VTH or VTL) depends on where we are on the Vo axis. In that sense, the circuit has a memory.
* ∆VT = V V is called the “hysteresis width.” TH − TL
Inverting Schmitt trigger
VTL VTH
10 Vsat
9 k Vo 5 R2 1 k 0 R1 Vo
Vi −5 RL
Vsat = 10 V −10 V − sat −10 −5 0 5 10
Vi (V)
M. B. Patil, IIT Bombay * The tripping point (whether VTH or VTL) depends on where we are on the Vo axis. In that sense, the circuit has a memory.
* ∆VT = V V is called the “hysteresis width.” TH − TL
Inverting Schmitt trigger
VTL VTH
10 Vsat
9 k Vo 5 R2 1 k 0 R1 Vo
Vi −5 RL
Vsat = 10 V −10 V − sat −10 −5 0 5 10
Vi (V) R1 * The threshold values (or “tripping points”), VTH and VTL, are given by Vsat. ± R1 + R2
M. B. Patil, IIT Bombay * ∆VT = V V is called the “hysteresis width.” TH − TL
Inverting Schmitt trigger
VTL VTH
10 Vsat
9 k Vo 5 R2 1 k 0 R1 Vo
Vi −5 RL
Vsat = 10 V −10 V − sat −10 −5 0 5 10
Vi (V) R1 * The threshold values (or “tripping points”), VTH and VTL, are given by Vsat. ± R1 + R2
* The tripping point (whether VTH or VTL) depends on where we are on the Vo axis. In that sense, the circuit has a memory.
M. B. Patil, IIT Bombay Inverting Schmitt trigger
VTL VTH
10 Vsat
9 k Vo 5 R2 1 k 0 R1 Vo
Vi −5 RL
Vsat = 10 V −10 V − sat −10 −5 0 5 10
Vi (V) R1 * The threshold values (or “tripping points”), VTH and VTL, are given by Vsat. ± R1 + R2
* The tripping point (whether VTH or VTL) depends on where we are on the Vo axis. In that sense, the circuit has a memory.
* ∆VT = V V is called the “hysteresis width.” TH − TL M. B. Patil, IIT Bombay Consider Vi = 5 V .
Case (i): Vo = −Vsat = −10 V
R2 R1 9 k 1 k → V+ = Vi + Vo = × 5 + × (−10) = 4.5 − 1 = 3.5 V . R1 + R2 R1 + R2 10 k 10 k
(V+ − V−) = (3.5 − 0) = 3.5 V → Vo = +Vsat .
This is inconsistent with our assumption (Vo = −Vsat). 9 k 1 k Case (ii): Vo = +Vsat = +10 V → V+ = × 5 + × 10 = 4.5 + 1 = 5.5 V . 10 k 10 k (V+ − V−) = (5.5 − 0) = 5.5 V → Vo = +Vsat (consistent)
If we move to the right (increasing Vi ), the same situation applies, i.e., Vo = +Vsat.
10 Vsat Vo
5 V+
0
−5
−10 V − sat V (V) −10 −5 0 5 10 i
Non-inverting Schmitt trigger
9 k
R2 1 k Vi R1 Vo
RL
Vsat = 10 V
Because of positive feedback, Vo can only be +Vsat (for V+ > V−) or −Vsat (for V+ < V−).
M. B. Patil, IIT Bombay Case (i): Vo = −Vsat = −10 V
R2 R1 9 k 1 k → V+ = Vi + Vo = × 5 + × (−10) = 4.5 − 1 = 3.5 V . R1 + R2 R1 + R2 10 k 10 k
(V+ − V−) = (3.5 − 0) = 3.5 V → Vo = +Vsat .
This is inconsistent with our assumption (Vo = −Vsat). 9 k 1 k Case (ii): Vo = +Vsat = +10 V → V+ = × 5 + × 10 = 4.5 + 1 = 5.5 V . 10 k 10 k (V+ − V−) = (5.5 − 0) = 5.5 V → Vo = +Vsat (consistent)
If we move to the right (increasing Vi ), the same situation applies, i.e., Vo = +Vsat.
10 Vsat Vo
5 V+
0
−5
−10 V − sat V (V) −10 −5 0 5 10 i
Non-inverting Schmitt trigger
9 k
R2 1 k Vi R1 Vo
RL
Vsat = 10 V
Because of positive feedback, Vo can only be +Vsat (for V+ > V−) or −Vsat (for V+ < V−).
Consider Vi = 5 V .
M. B. Patil, IIT Bombay This is inconsistent with our assumption (Vo = −Vsat). 9 k 1 k Case (ii): Vo = +Vsat = +10 V → V+ = × 5 + × 10 = 4.5 + 1 = 5.5 V . 10 k 10 k (V+ − V−) = (5.5 − 0) = 5.5 V → Vo = +Vsat (consistent)
If we move to the right (increasing Vi ), the same situation applies, i.e., Vo = +Vsat.
10 Vsat Vo
5 V+
0
−5
−10 V − sat V (V) −10 −5 0 5 10 i
Non-inverting Schmitt trigger
9 k
R2 1 k Vi R1 Vo
RL
Vsat = 10 V
Because of positive feedback, Vo can only be +Vsat (for V+ > V−) or −Vsat (for V+ < V−).
Consider Vi = 5 V .
Case (i): Vo = −Vsat = −10 V
R2 R1 9 k 1 k → V+ = Vi + Vo = × 5 + × (−10) = 4.5 − 1 = 3.5 V . R1 + R2 R1 + R2 10 k 10 k
(V+ − V−) = (3.5 − 0) = 3.5 V → Vo = +Vsat .
M. B. Patil, IIT Bombay 9 k 1 k Case (ii): Vo = +Vsat = +10 V → V+ = × 5 + × 10 = 4.5 + 1 = 5.5 V . 10 k 10 k (V+ − V−) = (5.5 − 0) = 5.5 V → Vo = +Vsat (consistent)
If we move to the right (increasing Vi ), the same situation applies, i.e., Vo = +Vsat.
10 Vsat Vo
5 V+
0
−5
−10 V − sat V (V) −10 −5 0 5 10 i
Non-inverting Schmitt trigger
9 k
R2 1 k Vi R1 Vo
RL
Vsat = 10 V
Because of positive feedback, Vo can only be +Vsat (for V+ > V−) or −Vsat (for V+ < V−).
Consider Vi = 5 V .
Case (i): Vo = −Vsat = −10 V
R2 R1 9 k 1 k → V+ = Vi + Vo = × 5 + × (−10) = 4.5 − 1 = 3.5 V . R1 + R2 R1 + R2 10 k 10 k
(V+ − V−) = (3.5 − 0) = 3.5 V → Vo = +Vsat .
This is inconsistent with our assumption (Vo = −Vsat).
M. B. Patil, IIT Bombay If we move to the right (increasing Vi ), the same situation applies, i.e., Vo = +Vsat.
Non-inverting Schmitt trigger 10 Vsat Vo 9 k 5 V+ R2 1 k V i 0 R1 Vo
−5 RL
Vsat = 10 V −10 V − sat V (V) −10 −5 0 5 10 i
Because of positive feedback, Vo can only be +Vsat (for V+ > V−) or −Vsat (for V+ < V−).
Consider Vi = 5 V .
Case (i): Vo = −Vsat = −10 V
R2 R1 9 k 1 k → V+ = Vi + Vo = × 5 + × (−10) = 4.5 − 1 = 3.5 V . R1 + R2 R1 + R2 10 k 10 k
(V+ − V−) = (3.5 − 0) = 3.5 V → Vo = +Vsat .
This is inconsistent with our assumption (Vo = −Vsat). 9 k 1 k Case (ii): Vo = +Vsat = +10 V → V+ = × 5 + × 10 = 4.5 + 1 = 5.5 V . 10 k 10 k (V+ − V−) = (5.5 − 0) = 5.5 V → Vo = +Vsat (consistent)
M. B. Patil, IIT Bombay If we move to the right (increasing Vi ), the same situation applies, i.e., Vo = +Vsat.
Non-inverting Schmitt trigger 10 Vsat Vo 9 k 5 V+ R2 1 k V i 0 R1 Vo
−5 RL
Vsat = 10 V −10 V − sat V (V) −10 −5 0 5 10 i
Because of positive feedback, Vo can only be +Vsat (for V+ > V−) or −Vsat (for V+ < V−).
Consider Vi = 5 V .
Case (i): Vo = −Vsat = −10 V
R2 R1 9 k 1 k → V+ = Vi + Vo = × 5 + × (−10) = 4.5 − 1 = 3.5 V . R1 + R2 R1 + R2 10 k 10 k
(V+ − V−) = (3.5 − 0) = 3.5 V → Vo = +Vsat .
This is inconsistent with our assumption (Vo = −Vsat). 9 k 1 k Case (ii): Vo = +Vsat = +10 V → V+ = × 5 + × 10 = 4.5 + 1 = 5.5 V . 10 k 10 k (V+ − V−) = (5.5 − 0) = 5.5 V → Vo = +Vsat (consistent)
M. B. Patil, IIT Bombay Non-inverting Schmitt trigger 10 Vsat Vo 9 k 5 V+ R2 1 k V i 0 R1 Vo
−5 RL
Vsat = 10 V −10 V − sat V (V) −10 −5 0 5 10 i
Because of positive feedback, Vo can only be +Vsat (for V+ > V−) or −Vsat (for V+ < V−).
Consider Vi = 5 V .
Case (i): Vo = −Vsat = −10 V
R2 R1 9 k 1 k → V+ = Vi + Vo = × 5 + × (−10) = 4.5 − 1 = 3.5 V . R1 + R2 R1 + R2 10 k 10 k
(V+ − V−) = (3.5 − 0) = 3.5 V → Vo = +Vsat .
This is inconsistent with our assumption (Vo = −Vsat). 9 k 1 k Case (ii): Vo = +Vsat = +10 V → V+ = × 5 + × 10 = 4.5 + 1 = 5.5 V . 10 k 10 k (V+ − V−) = (5.5 − 0) = 5.5 V → Vo = +Vsat (consistent)
If we move to the right (increasing Vi ), the same situation applies, i.e., Vo = +Vsat.
M. B. Patil, IIT Bombay Non-inverting Schmitt trigger 10 Vsat Vo 9 k 5 V+ R2 1 k V i 0 R1 Vo
−5 RL
Vsat = 10 V −10 V − sat V (V) −10 −5 0 5 10 i
Because of positive feedback, Vo can only be +Vsat (for V+ > V−) or −Vsat (for V+ < V−).
Consider Vi = 5 V .
Case (i): Vo = −Vsat = −10 V
R2 R1 9 k 1 k → V+ = Vi + Vo = × 5 + × (−10) = 4.5 − 1 = 3.5 V . R1 + R2 R1 + R2 10 k 10 k
(V+ − V−) = (3.5 − 0) = 3.5 V → Vo = +Vsat .
This is inconsistent with our assumption (Vo = −Vsat). 9 k 1 k Case (ii): Vo = +Vsat = +10 V → V+ = × 5 + × 10 = 4.5 + 1 = 5.5 V . 10 k 10 k (V+ − V−) = (5.5 − 0) = 5.5 V → Vo = +Vsat (consistent)
If we move to the right (increasing Vi ), the same situation applies, i.e., Vo = +Vsat. M. B. Patil, IIT Bombay Consider decreasing values of Vi .
R2 R1 9 k 1 k V+ = Vi + Vo = Vi + Vo . R1 + R2 R1 + R2 10 k 10 k
As long as V+ > 0 V , Vo remains at +Vsat.
R1 1 k When V+ = 0 V , i.e., Vi = − Vsat = − 10 V = −1.11 V , Vo changes sign, i.e., Vo becomes −Vsat. R2 9 k R2 R1 9 k 1 k V+ now follows the equation, V+ = Vi + Vo = Vi − Vsat . R1 + R2 R1 + R2 10 k 10 k
Decreasing Vi further makes no difference to Vo (since V+ remains negative). R1 Now, the threshold at which Vo flips is V+ = 0, i.e., Vi = + Vsat = +1.11 V . R2
Non-inverting Schmitt trigger
10 Vsat Vo 9 k 5 V+ R2 1 k V i 0 R1 Vo
−5 RL
Vsat = 10 V −10 V − sat V (V) −10 −5 0 5 10 i
M. B. Patil, IIT Bombay R2 R1 9 k 1 k V+ = Vi + Vo = Vi + Vo . R1 + R2 R1 + R2 10 k 10 k
As long as V+ > 0 V , Vo remains at +Vsat.
R1 1 k When V+ = 0 V , i.e., Vi = − Vsat = − 10 V = −1.11 V , Vo changes sign, i.e., Vo becomes −Vsat. R2 9 k R2 R1 9 k 1 k V+ now follows the equation, V+ = Vi + Vo = Vi − Vsat . R1 + R2 R1 + R2 10 k 10 k
Decreasing Vi further makes no difference to Vo (since V+ remains negative). R1 Now, the threshold at which Vo flips is V+ = 0, i.e., Vi = + Vsat = +1.11 V . R2
Non-inverting Schmitt trigger
10 Vsat Vo 9 k 5 V+ R2 1 k V i 0 R1 Vo
−5 RL
Vsat = 10 V −10 V − sat V (V) −10 −5 0 5 10 i
Consider decreasing values of Vi .
M. B. Patil, IIT Bombay As long as V+ > 0 V , Vo remains at +Vsat.
R1 1 k When V+ = 0 V , i.e., Vi = − Vsat = − 10 V = −1.11 V , Vo changes sign, i.e., Vo becomes −Vsat. R2 9 k R2 R1 9 k 1 k V+ now follows the equation, V+ = Vi + Vo = Vi − Vsat . R1 + R2 R1 + R2 10 k 10 k
Decreasing Vi further makes no difference to Vo (since V+ remains negative). R1 Now, the threshold at which Vo flips is V+ = 0, i.e., Vi = + Vsat = +1.11 V . R2
Non-inverting Schmitt trigger
10 Vsat Vo 9 k 5 V+ R2 1 k V i 0 R1 Vo
−5 RL
Vsat = 10 V −10 V − sat V (V) −10 −5 0 5 10 i
Consider decreasing values of Vi .
R2 R1 9 k 1 k V+ = Vi + Vo = Vi + Vo . R1 + R2 R1 + R2 10 k 10 k
M. B. Patil, IIT Bombay R1 1 k When V+ = 0 V , i.e., Vi = − Vsat = − 10 V = −1.11 V , Vo changes sign, i.e., Vo becomes −Vsat. R2 9 k R2 R1 9 k 1 k V+ now follows the equation, V+ = Vi + Vo = Vi − Vsat . R1 + R2 R1 + R2 10 k 10 k
Decreasing Vi further makes no difference to Vo (since V+ remains negative). R1 Now, the threshold at which Vo flips is V+ = 0, i.e., Vi = + Vsat = +1.11 V . R2
Non-inverting Schmitt trigger
10 Vsat Vo 9 k 5 V+ R2 1 k V i 0 R1 Vo
−5 RL
Vsat = 10 V −10 V − sat V (V) −10 −5 0 5 10 i
Consider decreasing values of Vi .
R2 R1 9 k 1 k V+ = Vi + Vo = Vi + Vo . R1 + R2 R1 + R2 10 k 10 k
As long as V+ > 0 V , Vo remains at +Vsat.
M. B. Patil, IIT Bombay R1 1 k When V+ = 0 V , i.e., Vi = − Vsat = − 10 V = −1.11 V , Vo changes sign, i.e., Vo becomes −Vsat. R2 9 k R2 R1 9 k 1 k V+ now follows the equation, V+ = Vi + Vo = Vi − Vsat . R1 + R2 R1 + R2 10 k 10 k
Decreasing Vi further makes no difference to Vo (since V+ remains negative). R1 Now, the threshold at which Vo flips is V+ = 0, i.e., Vi = + Vsat = +1.11 V . R2
Non-inverting Schmitt trigger
10 Vsat Vo 9 k 5 V+ R2 1 k V i 0 R1 Vo
−5 RL
Vsat = 10 V −10 V − sat V (V) −10 −5 0 5 10 i
Consider decreasing values of Vi .
R2 R1 9 k 1 k V+ = Vi + Vo = Vi + Vo . R1 + R2 R1 + R2 10 k 10 k
As long as V+ > 0 V , Vo remains at +Vsat.
M. B. Patil, IIT Bombay R2 R1 9 k 1 k V+ now follows the equation, V+ = Vi + Vo = Vi − Vsat . R1 + R2 R1 + R2 10 k 10 k
Decreasing Vi further makes no difference to Vo (since V+ remains negative). R1 Now, the threshold at which Vo flips is V+ = 0, i.e., Vi = + Vsat = +1.11 V . R2
Non-inverting Schmitt trigger
10 Vsat Vo 9 k 5 V+ R2 1 k V i 0 R1 Vo
−5 RL
Vsat = 10 V −10 V − sat V (V) −10 −5 0 5 10 i
Consider decreasing values of Vi .
R2 R1 9 k 1 k V+ = Vi + Vo = Vi + Vo . R1 + R2 R1 + R2 10 k 10 k
As long as V+ > 0 V , Vo remains at +Vsat.
R1 1 k When V+ = 0 V , i.e., Vi = − Vsat = − 10 V = −1.11 V , Vo changes sign, i.e., Vo becomes −Vsat. R2 9 k
M. B. Patil, IIT Bombay R2 R1 9 k 1 k V+ now follows the equation, V+ = Vi + Vo = Vi − Vsat . R1 + R2 R1 + R2 10 k 10 k
Decreasing Vi further makes no difference to Vo (since V+ remains negative). R1 Now, the threshold at which Vo flips is V+ = 0, i.e., Vi = + Vsat = +1.11 V . R2
Non-inverting Schmitt trigger
10 Vsat Vo 9 k 5 V+ R2 1 k V i 0 R1 Vo
−5 RL
Vsat = 10 V −10 V − sat V (V) −10 −5 0 5 10 i
Consider decreasing values of Vi .
R2 R1 9 k 1 k V+ = Vi + Vo = Vi + Vo . R1 + R2 R1 + R2 10 k 10 k
As long as V+ > 0 V , Vo remains at +Vsat.
R1 1 k When V+ = 0 V , i.e., Vi = − Vsat = − 10 V = −1.11 V , Vo changes sign, i.e., Vo becomes −Vsat. R2 9 k
M. B. Patil, IIT Bombay Decreasing Vi further makes no difference to Vo (since V+ remains negative). R1 Now, the threshold at which Vo flips is V+ = 0, i.e., Vi = + Vsat = +1.11 V . R2
Non-inverting Schmitt trigger
10 Vsat Vo 9 k 5 V+ R2 1 k V i 0 R1 Vo
−5 RL
Vsat = 10 V −10 V − sat V (V) −10 −5 0 5 10 i
Consider decreasing values of Vi .
R2 R1 9 k 1 k V+ = Vi + Vo = Vi + Vo . R1 + R2 R1 + R2 10 k 10 k
As long as V+ > 0 V , Vo remains at +Vsat.
R1 1 k When V+ = 0 V , i.e., Vi = − Vsat = − 10 V = −1.11 V , Vo changes sign, i.e., Vo becomes −Vsat. R2 9 k R2 R1 9 k 1 k V+ now follows the equation, V+ = Vi + Vo = Vi − Vsat . R1 + R2 R1 + R2 10 k 10 k
M. B. Patil, IIT Bombay R1 Now, the threshold at which Vo flips is V+ = 0, i.e., Vi = + Vsat = +1.11 V . R2
Non-inverting Schmitt trigger
10 Vsat Vo 9 k 5 V+ R2 1 k V i 0 R1 Vo
−5 RL
Vsat = 10 V −10 V − sat V (V) −10 −5 0 5 10 i
Consider decreasing values of Vi .
R2 R1 9 k 1 k V+ = Vi + Vo = Vi + Vo . R1 + R2 R1 + R2 10 k 10 k
As long as V+ > 0 V , Vo remains at +Vsat.
R1 1 k When V+ = 0 V , i.e., Vi = − Vsat = − 10 V = −1.11 V , Vo changes sign, i.e., Vo becomes −Vsat. R2 9 k R2 R1 9 k 1 k V+ now follows the equation, V+ = Vi + Vo = Vi − Vsat . R1 + R2 R1 + R2 10 k 10 k
Decreasing Vi further makes no difference to Vo (since V+ remains negative).
M. B. Patil, IIT Bombay R1 Now, the threshold at which Vo flips is V+ = 0, i.e., Vi = + Vsat = +1.11 V . R2
Non-inverting Schmitt trigger
10 Vsat Vo 9 k 5 V+ R2 1 k V i 0 R1 Vo
−5 RL
Vsat = 10 V −10 V − sat V (V) −10 −5 0 5 10 i
Consider decreasing values of Vi .
R2 R1 9 k 1 k V+ = Vi + Vo = Vi + Vo . R1 + R2 R1 + R2 10 k 10 k
As long as V+ > 0 V , Vo remains at +Vsat.
R1 1 k When V+ = 0 V , i.e., Vi = − Vsat = − 10 V = −1.11 V , Vo changes sign, i.e., Vo becomes −Vsat. R2 9 k R2 R1 9 k 1 k V+ now follows the equation, V+ = Vi + Vo = Vi − Vsat . R1 + R2 R1 + R2 10 k 10 k
Decreasing Vi further makes no difference to Vo (since V+ remains negative).
M. B. Patil, IIT Bombay Non-inverting Schmitt trigger
10 Vsat Vo 9 k 5 V+ R2 1 k V i 0 R1 Vo
−5 RL
Vsat = 10 V −10 V − sat V (V) −10 −5 0 5 10 i
Consider decreasing values of Vi .
R2 R1 9 k 1 k V+ = Vi + Vo = Vi + Vo . R1 + R2 R1 + R2 10 k 10 k
As long as V+ > 0 V , Vo remains at +Vsat.
R1 1 k When V+ = 0 V , i.e., Vi = − Vsat = − 10 V = −1.11 V , Vo changes sign, i.e., Vo becomes −Vsat. R2 9 k R2 R1 9 k 1 k V+ now follows the equation, V+ = Vi + Vo = Vi − Vsat . R1 + R2 R1 + R2 10 k 10 k
Decreasing Vi further makes no difference to Vo (since V+ remains negative). R1 Now, the threshold at which Vo flips is V+ = 0, i.e., Vi = + Vsat = +1.11 V . R2
M. B. Patil, IIT Bombay Non-inverting Schmitt trigger
10 Vsat Vo 9 k 5 V+ R2 1 k V i 0 R1 Vo
−5 RL
Vsat = 10 V −10 V − sat V (V) −10 −5 0 5 10 i
Consider decreasing values of Vi .
R2 R1 9 k 1 k V+ = Vi + Vo = Vi + Vo . R1 + R2 R1 + R2 10 k 10 k
As long as V+ > 0 V , Vo remains at +Vsat.
R1 1 k When V+ = 0 V , i.e., Vi = − Vsat = − 10 V = −1.11 V , Vo changes sign, i.e., Vo becomes −Vsat. R2 9 k R2 R1 9 k 1 k V+ now follows the equation, V+ = Vi + Vo = Vi − Vsat . R1 + R2 R1 + R2 10 k 10 k
Decreasing Vi further makes no difference to Vo (since V+ remains negative). R1 Now, the threshold at which Vo flips is V+ = 0, i.e., Vi = + Vsat = +1.11 V . R2 M. B. Patil, IIT Bombay R1 * The threshold values VTH and VTL are given by Vsat. ± R2
* As in the inverting Schmitt trigger, this circuit has a memory, i.e., the tripping point (whether VTH or VTL) depends on where we are on the Vo axis.
* ∆VT = V V is called the “hysteresis width.” TH − TL
Non-inverting Schmitt trigger
VTL VTH
10 Vsat 9 k Vo 5 R2 1 k V i 0 R1 Vo
−5 RL
Vsat = 10 V −10 V − sat −10 −5 0 5 10
Vi (V)
M. B. Patil, IIT Bombay * As in the inverting Schmitt trigger, this circuit has a memory, i.e., the tripping point (whether VTH or VTL) depends on where we are on the Vo axis.
* ∆VT = V V is called the “hysteresis width.” TH − TL
Non-inverting Schmitt trigger
VTL VTH
10 Vsat 9 k Vo 5 R2 1 k V i 0 R1 Vo
−5 RL
Vsat = 10 V −10 V − sat −10 −5 0 5 10
Vi (V) R1 * The threshold values VTH and VTL are given by Vsat. ± R2
M. B. Patil, IIT Bombay * ∆VT = V V is called the “hysteresis width.” TH − TL
Non-inverting Schmitt trigger
VTL VTH
10 Vsat 9 k Vo 5 R2 1 k V i 0 R1 Vo
−5 RL
Vsat = 10 V −10 V − sat −10 −5 0 5 10
Vi (V) R1 * The threshold values VTH and VTL are given by Vsat. ± R2
* As in the inverting Schmitt trigger, this circuit has a memory, i.e., the tripping point (whether VTH or VTL) depends on where we are on the Vo axis.
M. B. Patil, IIT Bombay Non-inverting Schmitt trigger
VTL VTH
10 Vsat 9 k Vo 5 R2 1 k V i 0 R1 Vo
−5 RL
Vsat = 10 V −10 V − sat −10 −5 0 5 10
Vi (V) R1 * The threshold values VTH and VTL are given by Vsat. ± R2
* As in the inverting Schmitt trigger, this circuit has a memory, i.e., the tripping point (whether VTH or VTL) depends on where we are on the Vo axis.
* ∆VT = V V is called the “hysteresis width.” TH − TL M. B. Patil, IIT Bombay Schmitt triggers Vo Inverting Vsat
R2
R1 V V Vo o V i TL VTH Vi Vi RL
R1 VTH, VTL = Vsat V ± R1 + R2 − sat
Vo Non−inverting Vsat
R2
Vi R1 Vo Vi Vo V V TL TH Vi
RL
R1 VTH, VTL = Vsat V ± R2 − sat M. B. Patil, IIT Bombay An op-amp in the open-loop configuration serves as a comparator because of its high gain ( 105) in the linear region. ∼ As seen earlier, the width of the linear region, [Vsat ( Vsat)]/AV , is small ( 0.1 mV ), and could be treated as 0. − − ∼
i.e., if V+ > V−, Vo = +Vsat , if V+ < V−, Vo = Vsat . −
Comparators
Vo +Vsat
V+
Vo (V+ V ) − − V − V − sat
M. B. Patil, IIT Bombay As seen earlier, the width of the linear region, [Vsat ( Vsat)]/AV , is small ( 0.1 mV ), and could be treated as 0. − − ∼
i.e., if V+ > V−, Vo = +Vsat , if V+ < V−, Vo = Vsat . −
Comparators
Vo +Vsat
V+
Vo (V+ V ) − − V − V − sat
An op-amp in the open-loop configuration serves as a comparator because of its high gain ( 105) in the linear region. ∼
M. B. Patil, IIT Bombay i.e., if V+ > V−, Vo = +Vsat , if V+ < V−, Vo = Vsat . −
Comparators
Vo +Vsat
V+
Vo (V+ V ) − − V − V − sat
An op-amp in the open-loop configuration serves as a comparator because of its high gain ( 105) in the linear region. ∼ As seen earlier, the width of the linear region, [Vsat ( Vsat)]/AV , is small ( 0.1 mV ), and could be treated as 0. − − ∼
M. B. Patil, IIT Bombay Comparators
Vo +Vsat
V+
Vo (V+ V ) − − V − V − sat
An op-amp in the open-loop configuration serves as a comparator because of its high gain ( 105) in the linear region. ∼ As seen earlier, the width of the linear region, [Vsat ( Vsat)]/AV , is small ( 0.1 mV ), and could be treated as 0. − − ∼
i.e., if V+ > V−, Vo = +Vsat , if V+ < V−, Vo = Vsat . −
M. B. Patil, IIT Bombay A comparator can be used to convert an analog signal into a digital (high/low) signal for further processing with digital circuits. In practice, the input (analog) signal can have noise or electromagnetic pick-up superimposed on it. As a result, erroneous operation of the circuit may result.
Comparators
Vi
t
Vi
Vo Vsat Vo
t
V − sat
M. B. Patil, IIT Bombay In practice, the input (analog) signal can have noise or electromagnetic pick-up superimposed on it. As a result, erroneous operation of the circuit may result.
Comparators
Vi
t
Vi
Vo Vsat Vo
t
V − sat A comparator can be used to convert an analog signal into a digital (high/low) signal for further processing with digital circuits.
M. B. Patil, IIT Bombay Comparators
Vi
t
Vi
Vo Vsat Vo
t
V − sat A comparator can be used to convert an analog signal into a digital (high/low) signal for further processing with digital circuits. In practice, the input (analog) signal can have noise or electromagnetic pick-up superimposed on it. As a result, erroneous operation of the circuit may result. M. B. Patil, IIT Bombay The comparator has produced multiple (spurious) transitions or “bounces,” referred to as “comparator chatter.” A Schmitt trigger can be used to eliminate the chatter.
corrupted expanded input expand view signal
t t
t t
Comparators
original V input i signal
t Vi
Vo
Vsat
Vo
t
V − sat
M. B. Patil, IIT Bombay The comparator has produced multiple (spurious) transitions or “bounces,” referred to as “comparator chatter.” A Schmitt trigger can be used to eliminate the chatter.
expanded expand view
t
t
Comparators
original corrupted V input i input signal signal
t t Vi
Vo
Vsat
Vo
t t
V − sat
M. B. Patil, IIT Bombay The comparator has produced multiple (spurious) transitions or “bounces,” referred to as “comparator chatter.” A Schmitt trigger can be used to eliminate the chatter.
expanded view
t
t
Comparators
original corrupted Vi input input expand signal signal
t t Vi
Vo
Vsat
Vo
t t
V − sat
M. B. Patil, IIT Bombay The comparator has produced multiple (spurious) transitions or “bounces,” referred to as “comparator chatter.” A Schmitt trigger can be used to eliminate the chatter.
Comparators
original corrupted expanded Vi input input expand view signal signal
t t t Vi
Vo
Vsat
Vo
t t t
V − sat
M. B. Patil, IIT Bombay A Schmitt trigger can be used to eliminate the chatter.
Comparators
original corrupted expanded Vi input input expand view signal signal
t t t Vi
Vo
Vsat
Vo
t t t
V − sat The comparator has produced multiple (spurious) transitions or “bounces,” referred to as “comparator chatter.”
M. B. Patil, IIT Bombay Comparators
original corrupted expanded Vi input input expand view signal signal
t t t Vi
Vo
Vsat
Vo
t t t
V − sat The comparator has produced multiple (spurious) transitions or “bounces,” referred to as “comparator chatter.” A Schmitt trigger can be used to eliminate the chatter. M. B. Patil, IIT Bombay * While going from positive to negative values, Vi needs to cross VTL (and not 0 V ) to cause a change in Vo . * In the reverse direction (negative to positive), Vi needs to cross VTH . * The circuit gets rid of spurious transitions, a major advantage over the simple comparator. * The hysteresis width (VTH − VTL) should be designed to be larger than the spurious excursions riding on Vi . i V expanded view expand VTH o V t
VTL sat V sat V −
t
corrupted input Vi signal
t
Vi Vo
Vsat
Vo
t
V − sat
M. B. Patil, IIT Bombay * While going from positive to negative values, Vi needs to cross VTL (and not 0 V ) to cause a change in Vo . * In the reverse direction (negative to positive), Vi needs to cross VTH . * The circuit gets rid of spurious transitions, a major advantage over the simple comparator. * The hysteresis width (VTH − VTL) should be designed to be larger than the spurious excursions riding on Vi . i V expanded view
VTH o V t
VTL sat V sat V −
t
corrupted input Vi signal expand
t
Vi Vo
Vsat
Vo
t
V − sat
M. B. Patil, IIT Bombay * While going from positive to negative values, Vi needs to cross VTL (and not 0 V ) to cause a change in Vo . * In the reverse direction (negative to positive), Vi needs to cross VTH . * The circuit gets rid of spurious transitions, a major advantage over the simple comparator. * The hysteresis width (VTH − VTL) should be designed to be larger than the spurious excursions riding on Vi . i V o V sat V sat V −
corrupted expanded input Vi view signal expand VTH
t t
VTL Vi Vo
Vsat
Vo
t t
V − sat
M. B. Patil, IIT Bombay * In the reverse direction (negative to positive), Vi needs to cross VTH . * The circuit gets rid of spurious transitions, a major advantage over the simple comparator. * The hysteresis width (VTH − VTL) should be designed to be larger than the spurious excursions riding on Vi . i V o V sat V sat V −
corrupted expanded input Vi view signal expand VTH
t t
VTL Vi Vo
Vsat
Vo
t t
V − sat
* While going from positive to negative values, Vi needs to cross VTL (and not 0 V ) to cause a change in Vo .
M. B. Patil, IIT Bombay * In the reverse direction (negative to positive), Vi needs to cross VTH . * The circuit gets rid of spurious transitions, a major advantage over the simple comparator. * The hysteresis width (VTH − VTL) should be designed to be larger than the spurious excursions riding on Vi . i V corrupted expanded input Vi view signal expand VTH o V t t
VTL
Vi Vo sat V sat
Vsat V −
Vo
t t
V − sat
* While going from positive to negative values, Vi needs to cross VTL (and not 0 V ) to cause a change in Vo .
M. B. Patil, IIT Bombay * The circuit gets rid of spurious transitions, a major advantage over the simple comparator. * The hysteresis width (VTH − VTL) should be designed to be larger than the spurious excursions riding on Vi . i V corrupted expanded input Vi view signal expand VTH o V t t
VTL
Vi Vo sat V sat
Vsat V −
Vo
t t
V − sat
* While going from positive to negative values, Vi needs to cross VTL (and not 0 V ) to cause a change in Vo . * In the reverse direction (negative to positive), Vi needs to cross VTH .
M. B. Patil, IIT Bombay * The hysteresis width (VTH − VTL) should be designed to be larger than the spurious excursions riding on Vi . i V corrupted expanded input Vi view signal expand VTH o V t t
VTL
Vi Vo sat V sat
Vsat V −
Vo
t t
V − sat
* While going from positive to negative values, Vi needs to cross VTL (and not 0 V ) to cause a change in Vo . * In the reverse direction (negative to positive), Vi needs to cross VTH . * The circuit gets rid of spurious transitions, a major advantage over the simple comparator.
M. B. Patil, IIT Bombay i V corrupted expanded input Vi view signal expand VTH o V t t
VTL
Vi Vo sat V sat
Vsat V −
Vo
t t
V − sat
* While going from positive to negative values, Vi needs to cross VTL (and not 0 V ) to cause a change in Vo . * In the reverse direction (negative to positive), Vi needs to cross VTH . * The circuit gets rid of spurious transitions, a major advantage over the simple comparator. * The hysteresis width (VTH − VTL) should be designed to be larger than the spurious excursions riding on Vi .
M. B. Patil, IIT Bombay + − * A Schmitt trigger has two states, Vo = L and Vo = L . * With a suitable RC network, it can be made to freely oscillate between L+ and L−. Such a circuit is called an “astable multivibrator” or a “free-running multivibrator.” * An astable multivibrator produces oscillations without an input signal, the frequency being controlled by the component values. * The maximum operating frequency of these oscillators is typically 10 kHz, due to op-amp speed limitations. ∼
Waveform generation using Schmitt triggers
Vo Vo Non−inverting Schmitt trigger L+ Inverting Schmitt trigger L+
Vi Vo Vi Vo VTL VTH Vi VTL VTH Vi
L− L−
M. B. Patil, IIT Bombay * With a suitable RC network, it can be made to freely oscillate between L+ and L−. Such a circuit is called an “astable multivibrator” or a “free-running multivibrator.” * An astable multivibrator produces oscillations without an input signal, the frequency being controlled by the component values. * The maximum operating frequency of these oscillators is typically 10 kHz, due to op-amp speed limitations. ∼
Waveform generation using Schmitt triggers
Vo Vo Non−inverting Schmitt trigger L+ Inverting Schmitt trigger L+
Vi Vo Vi Vo VTL VTH Vi VTL VTH Vi
L− L−
+ − * A Schmitt trigger has two states, Vo = L and Vo = L .
M. B. Patil, IIT Bombay * An astable multivibrator produces oscillations without an input signal, the frequency being controlled by the component values. * The maximum operating frequency of these oscillators is typically 10 kHz, due to op-amp speed limitations. ∼
Waveform generation using Schmitt triggers
Vo Vo Non−inverting Schmitt trigger L+ Inverting Schmitt trigger L+
Vi Vo Vi Vo VTL VTH Vi VTL VTH Vi
L− L−
+ − * A Schmitt trigger has two states, Vo = L and Vo = L . * With a suitable RC network, it can be made to freely oscillate between L+ and L−. Such a circuit is called an “astable multivibrator” or a “free-running multivibrator.”
M. B. Patil, IIT Bombay * The maximum operating frequency of these oscillators is typically 10 kHz, due to op-amp speed limitations. ∼
Waveform generation using Schmitt triggers
Vo Vo Non−inverting Schmitt trigger L+ Inverting Schmitt trigger L+
Vi Vo Vi Vo VTL VTH Vi VTL VTH Vi
L− L−
+ − * A Schmitt trigger has two states, Vo = L and Vo = L . * With a suitable RC network, it can be made to freely oscillate between L+ and L−. Such a circuit is called an “astable multivibrator” or a “free-running multivibrator.” * An astable multivibrator produces oscillations without an input signal, the frequency being controlled by the component values.
M. B. Patil, IIT Bombay Waveform generation using Schmitt triggers
Vo Vo Non−inverting Schmitt trigger L+ Inverting Schmitt trigger L+
Vi Vo Vi Vo VTL VTH Vi VTL VTH Vi
L− L−
+ − * A Schmitt trigger has two states, Vo = L and Vo = L . * With a suitable RC network, it can be made to freely oscillate between L+ and L−. Such a circuit is called an “astable multivibrator” or a “free-running multivibrator.” * An astable multivibrator produces oscillations without an input signal, the frequency being controlled by the component values. * The maximum operating frequency of these oscillators is typically 10 kHz, due to op-amp speed limitations. ∼
M. B. Patil, IIT Bombay + At t = 0, let Vo = L , and Vc = 0 V . The capacitor starts charging toward L+. − When Vc crosses VTH , the output flips. Now, the capacitor starts discharging toward L .
When Vc crosses VTL, the output flips again → oscillations. Note that the circuit oscillates on its own, i.e., without any input. Q: Where is the energy coming from?
Waveform generation using a Schmitt trigger
6 Vo L+ Vo Vi 4
RL 2
R 0 Vc C VTL VTH Vi + L =+5 V −2 L− = 5 V − −4 R = 2 k VTH =+1 V L− C = 1 µF VTL = 1 V −6 − 0 1 2 3 4 5 6 7 8 9 10 t (msec)
M. B. Patil, IIT Bombay The capacitor starts charging toward L+. − When Vc crosses VTH , the output flips. Now, the capacitor starts discharging toward L .
When Vc crosses VTL, the output flips again → oscillations. Note that the circuit oscillates on its own, i.e., without any input. Q: Where is the energy coming from?
Waveform generation using a Schmitt trigger
6 Vo L+ Vo Vi 4
RL 2
R 0 Vc C VTL VTH Vi + L =+5 V −2 L− = 5 V − −4 R = 2 k VTH =+1 V L− C = 1 µF VTL = 1 V −6 − 0 1 2 3 4 5 6 7 8 9 10 t (msec)
+ At t = 0, let Vo = L , and Vc = 0 V .
M. B. Patil, IIT Bombay The capacitor starts charging toward L+. − When Vc crosses VTH , the output flips. Now, the capacitor starts discharging toward L .
When Vc crosses VTL, the output flips again → oscillations. Note that the circuit oscillates on its own, i.e., without any input. Q: Where is the energy coming from?
Waveform generation using a Schmitt trigger
6 Vo L+ Vo Vi 4
RL 2
R 0 Vc C VTL VTH Vi + L =+5 V −2 L− = 5 V − −4 R = 2 k VTH =+1 V L− C = 1 µF VTL = 1 V −6 − 0 1 2 3 4 5 6 7 8 9 10 t (msec)
+ At t = 0, let Vo = L , and Vc = 0 V .
M. B. Patil, IIT Bombay − When Vc crosses VTH , the output flips. Now, the capacitor starts discharging toward L .
When Vc crosses VTL, the output flips again → oscillations. Note that the circuit oscillates on its own, i.e., without any input. Q: Where is the energy coming from?
Waveform generation using a Schmitt trigger
6 Vo L+ Vo Vi 4
RL 2
R 0 Vc C VTL VTH Vi + L =+5 V −2 L− = 5 V − −4 R = 2 k VTH =+1 V L− C = 1 µF VTL = 1 V −6 − 0 1 2 3 4 5 6 7 8 9 10 t (msec)
+ At t = 0, let Vo = L , and Vc = 0 V . The capacitor starts charging toward L+.
M. B. Patil, IIT Bombay − When Vc crosses VTH , the output flips. Now, the capacitor starts discharging toward L .
When Vc crosses VTL, the output flips again → oscillations. Note that the circuit oscillates on its own, i.e., without any input. Q: Where is the energy coming from?
Waveform generation using a Schmitt trigger
6 Vo L+ Vo Vi 4
RL 2
R 0 Vc C VTL VTH Vi + L =+5 V −2 L− = 5 V − −4 R = 2 k VTH =+1 V L− C = 1 µF VTL = 1 V −6 − 0 1 2 3 4 5 6 7 8 9 10 t (msec)
+ At t = 0, let Vo = L , and Vc = 0 V . The capacitor starts charging toward L+.
M. B. Patil, IIT Bombay When Vc crosses VTL, the output flips again → oscillations. Note that the circuit oscillates on its own, i.e., without any input. Q: Where is the energy coming from?
Waveform generation using a Schmitt trigger
6 Vo L+ Vo Vi 4
RL 2
R 0 Vc C VTL VTH Vi + L =+5 V −2 L− = 5 V − −4 R = 2 k VTH =+1 V L− C = 1 µF VTL = 1 V −6 − 0 1 2 3 4 5 6 7 8 9 10 t (msec)
+ At t = 0, let Vo = L , and Vc = 0 V . The capacitor starts charging toward L+. − When Vc crosses VTH , the output flips. Now, the capacitor starts discharging toward L .
M. B. Patil, IIT Bombay When Vc crosses VTL, the output flips again → oscillations. Note that the circuit oscillates on its own, i.e., without any input. Q: Where is the energy coming from?
Waveform generation using a Schmitt trigger
6 Vo L+ Vo Vi 4
RL 2
R 0 Vc C VTL VTH Vi + L =+5 V −2 L− = 5 V − −4 R = 2 k VTH =+1 V L− C = 1 µF VTL = 1 V −6 − 0 1 2 3 4 5 6 7 8 9 10 t (msec)
+ At t = 0, let Vo = L , and Vc = 0 V . The capacitor starts charging toward L+. − When Vc crosses VTH , the output flips. Now, the capacitor starts discharging toward L .
M. B. Patil, IIT Bombay Note that the circuit oscillates on its own, i.e., without any input. Q: Where is the energy coming from?
Waveform generation using a Schmitt trigger
6 Vo L+ Vo Vi 4
RL 2
R 0 Vc C VTL VTH Vi + L =+5 V −2 L− = 5 V − −4 R = 2 k VTH =+1 V L− C = 1 µF VTL = 1 V −6 − 0 1 2 3 4 5 6 7 8 9 10 t (msec)
+ At t = 0, let Vo = L , and Vc = 0 V . The capacitor starts charging toward L+. − When Vc crosses VTH , the output flips. Now, the capacitor starts discharging toward L .
When Vc crosses VTL, the output flips again → oscillations.
M. B. Patil, IIT Bombay Note that the circuit oscillates on its own, i.e., without any input. Q: Where is the energy coming from?
Waveform generation using a Schmitt trigger
6 Vo L+ Vo Vi 4
RL 2
R 0 Vc C VTL VTH Vi + L =+5 V −2 L− = 5 V − −4 R = 2 k VTH =+1 V L− C = 1 µF VTL = 1 V −6 − 0 1 2 3 4 5 6 7 8 9 10 t (msec)
+ At t = 0, let Vo = L , and Vc = 0 V . The capacitor starts charging toward L+. − When Vc crosses VTH , the output flips. Now, the capacitor starts discharging toward L .
When Vc crosses VTL, the output flips again → oscillations.
M. B. Patil, IIT Bombay Waveform generation using a Schmitt trigger
6 Vo L+ Vo Vi 4
RL 2
R 0 Vc C VTL VTH Vi + L =+5 V −2 L− = 5 V − −4 R = 2 k VTH =+1 V L− C = 1 µF VTL = 1 V −6 − 0 1 2 3 4 5 6 7 8 9 10 t (msec)
+ At t = 0, let Vo = L , and Vc = 0 V . The capacitor starts charging toward L+. − When Vc crosses VTH , the output flips. Now, the capacitor starts discharging toward L .
When Vc crosses VTL, the output flips again → oscillations. Note that the circuit oscillates on its own, i.e., without any input. Q: Where is the energy coming from?
M. B. Patil, IIT Bombay Charging: Let Vc (t) = A1 exp(−t/τ) + B1, with τ = RC. + Using Vc (0) = VTL, Vc (∞) = L , find A1 and B1.
At t = t1, Vc = VTH → VTH = A1 exp(−t1/τ) + B1 → find t1.
Discharging: Let Vc (t) = A2 exp(−(t − t1)/τ) + B2. − Using Vc (t1) = VTH , Vc (∞) = L , find A2 and B2.
At t = t2, Vc = VTL → VTL = A2 exp(−(t2 − t1)/τ) + B2 → find (t2 − t1) → t2 → T = t2. + − L + VT If L = L, L = −L, VTH = VT , VTL = −VT , show that T = 2 RC ln . L − VT
Waveform generation using a Schmitt trigger
L+
T Vo Vo Vi V TH Vc RL t R VTL C Vc
L− 0 t1 t2
M. B. Patil, IIT Bombay Discharging: Let Vc (t) = A2 exp(−(t − t1)/τ) + B2. − Using Vc (t1) = VTH , Vc (∞) = L , find A2 and B2.
At t = t2, Vc = VTL → VTL = A2 exp(−(t2 − t1)/τ) + B2 → find (t2 − t1) → t2 → T = t2. + − L + VT If L = L, L = −L, VTH = VT , VTL = −VT , show that T = 2 RC ln . L − VT
Waveform generation using a Schmitt trigger
L+
T Vo Vo Vi V TH Vc RL t R VTL C Vc
L− 0 t1 t2
Charging: Let Vc (t) = A1 exp(−t/τ) + B1, with τ = RC. + Using Vc (0) = VTL, Vc (∞) = L , find A1 and B1.
At t = t1, Vc = VTH → VTH = A1 exp(−t1/τ) + B1 → find t1.
M. B. Patil, IIT Bombay + − L + VT If L = L, L = −L, VTH = VT , VTL = −VT , show that T = 2 RC ln . L − VT
Waveform generation using a Schmitt trigger
L+
T Vo Vo Vi V TH Vc RL t R VTL C Vc
L− 0 t1 t2
Charging: Let Vc (t) = A1 exp(−t/τ) + B1, with τ = RC. + Using Vc (0) = VTL, Vc (∞) = L , find A1 and B1.
At t = t1, Vc = VTH → VTH = A1 exp(−t1/τ) + B1 → find t1.
Discharging: Let Vc (t) = A2 exp(−(t − t1)/τ) + B2. − Using Vc (t1) = VTH , Vc (∞) = L , find A2 and B2.
At t = t2, Vc = VTL → VTL = A2 exp(−(t2 − t1)/τ) + B2 → find (t2 − t1) → t2 → T = t2.
M. B. Patil, IIT Bombay Waveform generation using a Schmitt trigger
L+
T Vo Vo Vi V TH Vc RL t R VTL C Vc
L− 0 t1 t2
Charging: Let Vc (t) = A1 exp(−t/τ) + B1, with τ = RC. + Using Vc (0) = VTL, Vc (∞) = L , find A1 and B1.
At t = t1, Vc = VTH → VTH = A1 exp(−t1/τ) + B1 → find t1.
Discharging: Let Vc (t) = A2 exp(−(t − t1)/τ) + B2. − Using Vc (t1) = VTH , Vc (∞) = L , find A2 and B2.
At t = t2, Vc = VTL → VTL = A2 exp(−(t2 − t1)/τ) + B2 → find (t2 − t1) → t2 → T = t2. + − L + VT If L = L, L = −L, VTH = VT , VTL = −VT , show that T = 2 RC ln . L − VT M. B. Patil, IIT Bombay Note that Op-Amp 411 (slew rate: 10 V /µs) gives sharper waveforms as compared to Op-Amp 741 (slew rate: 0.5 V /µs). SEQUEL files: schmitt osc 741.sqproj, schmitt osc 411.sqproj (Ref: J. M. Fiore, “Op-Amps and linear ICs”)
Waveform generation using a Schmitt trigger
Op−Amp 741 Op−Amp 411 15 Vo Vo 10
5 Vc Vc 0
−5
−10
−15 0 0.2 0.4 0.6 0.8 1.0 0 0.2 0.4 0.6 0.8 1.0 t (msec) t (msec) Vo Vi
RL
R Vc C
M. B. Patil, IIT Bombay SEQUEL files: schmitt osc 741.sqproj, schmitt osc 411.sqproj (Ref: J. M. Fiore, “Op-Amps and linear ICs”)
Waveform generation using a Schmitt trigger
Op−Amp 741 Op−Amp 411 15 Vo Vo 10
5 Vc Vc 0
−5
−10
−15 0 0.2 0.4 0.6 0.8 1.0 0 0.2 0.4 0.6 0.8 1.0 t (msec) t (msec) V V o i Note that Op-Amp 411 (slew rate: 10 V /µs) gives sharper waveforms as compared to Op-Amp 741 (slew rate: 0.5 V /µs). RL
R Vc C
M. B. Patil, IIT Bombay Waveform generation using a Schmitt trigger
Op−Amp 741 Op−Amp 411 15 Vo Vo 10
5 Vc Vc 0
−5
−10
−15 0 0.2 0.4 0.6 0.8 1.0 0 0.2 0.4 0.6 0.8 1.0 t (msec) t (msec) V V o i Note that Op-Amp 411 (slew rate: 10 V /µs) gives sharper waveforms as compared to Op-Amp 741 (slew rate: 0.5 V /µs). RL SEQUEL files: schmitt osc 741.sqproj, schmitt osc 411.sqproj R Vc C (Ref: J. M. Fiore, “Op-Amps and linear ICs”)
M. B. Patil, IIT Bombay 1 Z 1 Z For the integrator, Vo = − Vo dt ≡ − Vo dt 1 RC 2 τ 2 + − Vo2 = L → Vo2 decreases linearly, Vo2 = L → Vo2 increases linearly.
dVo1 Vo2 ∆Vo1 |Vo2| ∆Vo1 = − . If Vo2 is constant, = → ∆t = τ dt τ ∆t τ Vo2
VTH − VTL T = τ . 1 L+
VTH − VTL T = τ . 2 −L−
Waveform generation using a Schmitt trigger
Integrator Schmitt trigger Vo2 L+ L+ Vo2 T1 T2 C VTH Vo1 R Vo2 t VTL VTH Vo1 Vo1 VTL
L− L−
M. B. Patil, IIT Bombay + − Vo2 = L → Vo2 decreases linearly, Vo2 = L → Vo2 increases linearly.
dVo1 Vo2 ∆Vo1 |Vo2| ∆Vo1 = − . If Vo2 is constant, = → ∆t = τ dt τ ∆t τ Vo2
VTH − VTL T = τ . 1 L+
VTH − VTL T = τ . 2 −L−
Waveform generation using a Schmitt trigger
Integrator Schmitt trigger Vo2 L+ L+ Vo2 T1 T2 C VTH Vo1 R Vo2 t VTL VTH Vo1 Vo1 VTL
L− L−
1 Z 1 Z For the integrator, Vo = − Vo dt ≡ − Vo dt 1 RC 2 τ 2
M. B. Patil, IIT Bombay
dVo1 Vo2 ∆Vo1 |Vo2| ∆Vo1 = − . If Vo2 is constant, = → ∆t = τ dt τ ∆t τ Vo2
VTH − VTL T = τ . 1 L+
VTH − VTL T = τ . 2 −L−
Waveform generation using a Schmitt trigger
Integrator Schmitt trigger Vo2 L+ L+ Vo2 T1 T2 C VTH Vo1 R Vo2 t VTL VTH Vo1 Vo1 VTL
L− L−
1 Z 1 Z For the integrator, Vo = − Vo dt ≡ − Vo dt 1 RC 2 τ 2 + − Vo2 = L → Vo2 decreases linearly, Vo2 = L → Vo2 increases linearly.
M. B. Patil, IIT Bombay VTH − VTL T = τ . 1 L+
VTH − VTL T = τ . 2 −L−
Waveform generation using a Schmitt trigger
Integrator Schmitt trigger Vo2 L+ L+ Vo2 T1 T2 C VTH Vo1 R Vo2 t VTL VTH Vo1 Vo1 VTL
L− L−
1 Z 1 Z For the integrator, Vo = − Vo dt ≡ − Vo dt 1 RC 2 τ 2 + − Vo2 = L → Vo2 decreases linearly, Vo2 = L → Vo2 increases linearly.
dVo1 Vo2 ∆Vo1 |Vo2| ∆Vo1 = − . If Vo2 is constant, = → ∆t = τ dt τ ∆t τ Vo2
M. B. Patil, IIT Bombay VTH − VTL T = τ . 2 −L−
Waveform generation using a Schmitt trigger
Integrator Schmitt trigger Vo2 L+ L+ Vo2 T1 T2 C VTH Vo1 R Vo2 t VTL VTH Vo1 Vo1 VTL
L− L−
1 Z 1 Z For the integrator, Vo = − Vo dt ≡ − Vo dt 1 RC 2 τ 2 + − Vo2 = L → Vo2 decreases linearly, Vo2 = L → Vo2 increases linearly.
dVo1 Vo2 ∆Vo1 |Vo2| ∆Vo1 = − . If Vo2 is constant, = → ∆t = τ dt τ ∆t τ Vo2
VTH − VTL T = τ . 1 L+
M. B. Patil, IIT Bombay Waveform generation using a Schmitt trigger
Integrator Schmitt trigger Vo2 L+ L+ Vo2 T1 T2 C VTH Vo1 R Vo2 t VTL VTH Vo1 Vo1 VTL
L− L−
1 Z 1 Z For the integrator, Vo = − Vo dt ≡ − Vo dt 1 RC 2 τ 2 + − Vo2 = L → Vo2 decreases linearly, Vo2 = L → Vo2 increases linearly.
dVo1 Vo2 ∆Vo1 |Vo2| ∆Vo1 = − . If Vo2 is constant, = → ∆t = τ dt τ ∆t τ Vo2
VTH − VTL T = τ . 1 L+
VTH − VTL T = τ . 2 −L−
M. B. Patil, IIT Bombay * When Vo2 = +Vsat, D1 is forward-biased (with a voltage drop of Von), and D2 is reverse-biased. The Zener breakdown voltage (VZ ) is chosen so that D2 operates under breakdown condition. Vo = Von + VZ . → 3 * When Vo = Vsat, D is forward-biased (with a voltage drop of Von), and D is reverse-biased. 2 − 2 1 Vo = Von VZ . → 3 − − * R3 serves to limit the output current for OA2.
SEQUEL file: opamp osc 1.sqproj
Limiting the output voltage
20
Vo2 C 10 R2 Vo3 R R3 Vo1 R1 0 OA1 Vo3 Vo2 OA2 D1 −10
D2
−20 0 0.1 t (msec) Integrator Schmitt trigger Limiter
M. B. Patil, IIT Bombay * When Vo = Vsat, D is forward-biased (with a voltage drop of Von), and D is reverse-biased. 2 − 2 1 Vo = Von VZ . → 3 − − * R3 serves to limit the output current for OA2.
SEQUEL file: opamp osc 1.sqproj
Limiting the output voltage
20
Vo2 C 10 R2 Vo3 R R3 Vo1 R1 0 OA1 Vo3 Vo2 OA2 D1 −10
D2
−20 0 0.1 t (msec) Integrator Schmitt trigger Limiter
* When Vo2 = +Vsat, D1 is forward-biased (with a voltage drop of Von), and D2 is reverse-biased. The Zener breakdown voltage (VZ ) is chosen so that D2 operates under breakdown condition. Vo = Von + VZ . → 3
M. B. Patil, IIT Bombay * R3 serves to limit the output current for OA2.
SEQUEL file: opamp osc 1.sqproj
Limiting the output voltage
20
Vo2 C 10 R2 Vo3 R R3 Vo1 R1 0 OA1 Vo3 Vo2 OA2 D1 −10
D2
−20 0 0.1 t (msec) Integrator Schmitt trigger Limiter
* When Vo2 = +Vsat, D1 is forward-biased (with a voltage drop of Von), and D2 is reverse-biased. The Zener breakdown voltage (VZ ) is chosen so that D2 operates under breakdown condition. Vo = Von + VZ . → 3 * When Vo = Vsat, D is forward-biased (with a voltage drop of Von), and D is reverse-biased. 2 − 2 1 Vo = Von VZ . → 3 − −
M. B. Patil, IIT Bombay SEQUEL file: opamp osc 1.sqproj
Limiting the output voltage
20
Vo2 C 10 R2 Vo3 R R3 Vo1 R1 0 OA1 Vo3 Vo2 OA2 D1 −10
D2
−20 0 0.1 t (msec) Integrator Schmitt trigger Limiter
* When Vo2 = +Vsat, D1 is forward-biased (with a voltage drop of Von), and D2 is reverse-biased. The Zener breakdown voltage (VZ ) is chosen so that D2 operates under breakdown condition. Vo = Von + VZ . → 3 * When Vo = Vsat, D is forward-biased (with a voltage drop of Von), and D is reverse-biased. 2 − 2 1 Vo = Von VZ . → 3 − − * R3 serves to limit the output current for OA2.
M. B. Patil, IIT Bombay Limiting the output voltage
20
Vo2 C 10 R2 Vo3 R R3 Vo1 R1 0 OA1 Vo3 Vo2 OA2 D1 −10
D2
−20 0 0.1 t (msec) Integrator Schmitt trigger Limiter
* When Vo2 = +Vsat, D1 is forward-biased (with a voltage drop of Von), and D2 is reverse-biased. The Zener breakdown voltage (VZ ) is chosen so that D2 operates under breakdown condition. Vo = Von + VZ . → 3 * When Vo = Vsat, D is forward-biased (with a voltage drop of Von), and D is reverse-biased. 2 − 2 1 Vo = Von VZ . → 3 − − * R3 serves to limit the output current for OA2.
SEQUEL file: opamp osc 1.sqproj M. B. Patil, IIT Bombay * Design the circuit to obtain V = 2.5 V and ∆VT = VTH VTL = 0.4 V. 0 − * Verify your design with simulation (and in the lab).
Schmitt trigger
Vo
Vm Vi Vo
R1 R3 VR VTL VTH Vi
R2
V − m V0
A Schmitt trigger circuit is shown in the figure along with its Vo -Vi relationship. Assume that Vsat 14 V for ≈ the op-amp. The reference voltage VR can be adjusted using a pot (not shown in the figure).
M. B. Patil, IIT Bombay * Verify your design with simulation (and in the lab).
Schmitt trigger
Vo
Vm Vi Vo
R1 R3 VR VTL VTH Vi
R2
V − m V0
A Schmitt trigger circuit is shown in the figure along with its Vo -Vi relationship. Assume that Vsat 14 V for ≈ the op-amp. The reference voltage VR can be adjusted using a pot (not shown in the figure).
* Design the circuit to obtain V = 2.5 V and ∆VT = VTH VTL = 0.4 V. 0 −
M. B. Patil, IIT Bombay Schmitt trigger
Vo
Vm Vi Vo
R1 R3 VR VTL VTH Vi
R2
V − m V0
A Schmitt trigger circuit is shown in the figure along with its Vo -Vi relationship. Assume that Vsat 14 V for ≈ the op-amp. The reference voltage VR can be adjusted using a pot (not shown in the figure).
* Design the circuit to obtain V = 2.5 V and ∆VT = VTH VTL = 0.4 V. 0 − * Verify your design with simulation (and in the lab).
M. B. Patil, IIT Bombay (R1 R2) ∆VT = 0.4 V 2Vm k = 0.4 V. → (R R ) + R 1 k 2 3 (R2 R3) V0 = 2.5 V VR k = 2.5 V. → (R R ) + R 2 k 3 1
Schmitt trigger
Vo
Vm Vi Vo
R1 R3 VR VTL VTH Vi
R2
V − m V0
(R2 R3) (R1 R2) V+ = VR k Vm k . (R R ) + R ± (R R ) + R 2 k 3 1 1 k 2 3
M. B. Patil, IIT Bombay (R2 R3) V0 = 2.5 V VR k = 2.5 V. → (R R ) + R 2 k 3 1
Schmitt trigger
Vo
Vm Vi Vo
R1 R3 VR VTL VTH Vi
R2
V − m V0
(R2 R3) (R1 R2) V+ = VR k Vm k . (R R ) + R ± (R R ) + R 2 k 3 1 1 k 2 3 (R1 R2) ∆VT = 0.4 V 2Vm k = 0.4 V. → (R R ) + R 1 k 2 3
M. B. Patil, IIT Bombay Schmitt trigger
Vo
Vm Vi Vo
R1 R3 VR VTL VTH Vi
R2
V − m V0
(R2 R3) (R1 R2) V+ = VR k Vm k . (R R ) + R ± (R R ) + R 2 k 3 1 1 k 2 3 (R1 R2) ∆VT = 0.4 V 2Vm k = 0.4 V. → (R R ) + R 1 k 2 3 (R2 R3) V0 = 2.5 V VR k = 2.5 V. → (R R ) + R 2 k 3 1
M. B. Patil, IIT Bombay (R/2) Let R1 = R2 = 5 k 2Vsat = 0.4 V R3 = 172.5 k. → (R/2) + R3 →
(R2 R3) V0 = 2.5 V VR k = 2.5 V VR = 5.07 V. → (R R ) + R → 2 k 3 1 (SEQUEL file: schmitt 1.sqproj)
Schmitt trigger
Vo
Vm Vi Vo
R1 R3 VR VTL VTH Vi
R2
V − m V0
(R1 R2) ∆VT = 0.4 V 2Vm k = 0.4 V. → (R R ) + R 1 k 2 3
M. B. Patil, IIT Bombay (R2 R3) V0 = 2.5 V VR k = 2.5 V VR = 5.07 V. → (R R ) + R → 2 k 3 1 (SEQUEL file: schmitt 1.sqproj)
Schmitt trigger
Vo
Vm Vi Vo
R1 R3 VR VTL VTH Vi
R2
V − m V0
(R1 R2) ∆VT = 0.4 V 2Vm k = 0.4 V. → (R R ) + R 1 k 2 3 (R/2) Let R1 = R2 = 5 k 2Vsat = 0.4 V R3 = 172.5 k. → (R/2) + R3 →
M. B. Patil, IIT Bombay (SEQUEL file: schmitt 1.sqproj)
Schmitt trigger
Vo
Vm Vi Vo
R1 R3 VR VTL VTH Vi
R2
V − m V0
(R1 R2) ∆VT = 0.4 V 2Vm k = 0.4 V. → (R R ) + R 1 k 2 3 (R/2) Let R1 = R2 = 5 k 2Vsat = 0.4 V R3 = 172.5 k. → (R/2) + R3 →
(R2 R3) V0 = 2.5 V VR k = 2.5 V VR = 5.07 V. → (R R ) + R → 2 k 3 1
M. B. Patil, IIT Bombay Schmitt trigger
Vo
Vm Vi Vo
R1 R3 VR VTL VTH Vi
R2
V − m V0
(R1 R2) ∆VT = 0.4 V 2Vm k = 0.4 V. → (R R ) + R 1 k 2 3 (R/2) Let R1 = R2 = 5 k 2Vsat = 0.4 V R3 = 172.5 k. → (R/2) + R3 →
(R2 R3) V0 = 2.5 V VR k = 2.5 V VR = 5.07 V. → (R R ) + R → 2 k 3 1 (SEQUEL file: schmitt 1.sqproj)
M. B. Patil, IIT Bombay