Introduction to the Op Amp Outline Basic Concept What About Other

Outline

Introduction to the Op Amp 1 Voltage Ampliﬁers

Phyllis R. Nelson 2 Operational ampliﬁer

Cal Poly Pomona 3 Ideal op amp ECE 207 - Winter 2015

4 Applications of the ideal op amp model

Phyllis R. Nelson (Cal Poly Pomona) Introduction to the Op Amp ECE 207 - Winter 2015 1 / 22 Phyllis R. Nelson (Cal Poly Pomona) Introduction to the Op Amp ECE 207 - Winter 2015 2 / 22 Basic concept What about other dependent sources?

Voltage-controlled current source

Amplify: increase, magnify, intensify measures the voltage vd between two nodes

outputs a current Gvd What circuit elements from ECE 109 can amplify?

Voltage-controlled voltage source RS Output power: measures the voltage between two nodes 2 + Po = (Avd) RL Is outputs that voltage times a gain V + v Av R Input power: S − d d L Pi = vdIS - Current-controlled current source measures a branch current outputs that current times a gain vd = ? IS = ?

Phyllis R. Nelson (Cal Poly Pomona) Introduction to the Op Amp ECE 207 - Winter 2015 3 / 22 Phyllis R. Nelson (Cal Poly Pomona) Introduction to the Op Amp ECE 207 - Winter 2015 4 / 22 Measuring voltages Back to the example . . .

RS Imeter R1 + Is + 2 I I2 = Is Imeter V v R Av R P = (Av ) R s − S − d i d L o d L V + R I s − 2 2 V = R I = R I R2 2 2 6 2 s -

Ri VS vd = VS = VS if Ri RS If Imeter = 0, the measurement changes the voltage! Ri + Rs 1 + RS ' 6 Ri VS The circuit that measures a voltage (a voltmeter) IS = 0 as Ri Pi = vdIS 0 as Ri Ri + Rs → → ∞ → → ∞ is connected between two nodes (in parallel) needs to have an internal resistance large compared to the equivalent This circuit delivers power to the load while drawing very little power from resistance of the circuit it is measuring the source. an ideal voltmeter acts as an open circuit Power ampliﬁcation!

Phyllis R. Nelson (Cal Poly Pomona) Introduction to the Op Amp ECE 207 - Winter 2015 5 / 22 Phyllis R. Nelson (Cal Poly Pomona) Introduction to the Op Amp ECE 207 - Winter 2015 6 / 22 Measuring currents Voltage ampliﬁer example (text Example 5.1)

10 kΩ 20 kΩ V R1 o A = 2 105 × VS VS − 50 Ω Is I = = + s V R1 + Rmeter 6 RS V + v s − s − d 2MΩ V + v V Av o d + o − d = 0 + Av + d − 20 kΩ 50 Ω v V v v V − d − s + − d + − d − o = 0 10 kΩ 2 MΩ 20 kΩ If Vmeter = 0, the measurement changes the current! 6 From the ﬁrst node equation

The circuit that measures a current (an ammeter) 20 kΩ + 50 Ω 6 vd = Vo = (5.0125 10− )Vo = FVo is connected in series A(20 kΩ) 50 Ω × − needs to have an internal resistance small compared to the equivalent 20 kΩ Vo 6 V = = (5.000 10− )V ' A(20 kΩ) o A × o resistance of the circuit it is measuring an ideal ammeter acts as a short circuit

Phyllis R. Nelson (Cal Poly Pomona) Introduction to the Op Amp ECE 207 - Winter 2015 7 / 22 Phyllis R. Nelson (Cal Poly Pomona) Introduction to the Op Amp ECE 207 - Winter 2015 8 / 22 Voltage ampliﬁer example (continued) Voltage ampliﬁer example: doing the numbers 10 kΩ 20 kΩ Vo Vo 2 5 = − − 50 Ω A = 2 10 Vs 1 + F 20 kΩ × Rp + 6 V vd Ω vd = (5.0125 10− )Vo = FVo s − 2M × + + 6 20 kΩ 5 Av vd Vs vd vd Vo F = 5.0125 10− R = 6.645 kΩ F = 1.5088 10− d − − − + − +− − = 0 × p R × 10 kΩ 2 MΩ 20 kΩ p

From the second node equation Vo vd Vo Vs = 1.999 97 + = − Vs − 10 kΩ 2 MΩ 20 kΩ 20 kΩ 10 kΩ k k This result is 20 kΩ very close to Rp = 10 kΩ 2 MΩ 20 kΩ = 6.645 kΩ 10 kΩ 20 kΩ = The model parameters of the k k ' k 3 20 kΩ/10 kΩ = 2 − − nonideal dependent source 1 20 kΩ almost independent of Vo − 2 don’t matter. WHY? = 10 kΩ = − 10 kΩ = − A V 1 F 20 kΩ 20 kΩ s 20 kΩ + R 1 + F 1 + F p Rp Rp the 2 MΩ and 50 Ω resistors Phyllis R. Nelson (Cal Poly Pomona) Introduction to the Op Amp ECE 207 - Winter 2015 9 / 22 Phyllis R. Nelson (Cal Poly Pomona) Introduction to the Op Amp ECE 207 - Winter 2015 10 / 22 A closer look A closer look (continued) 10 kΩ 20 kΩ V o 10 kΩ 20 kΩ Vo − 50 Ω V 20 kΩ + 50 Ω v o 0 for large A + v vd = Vo − d Vs − d 2MΩ A(20 kΩ) 50 Ω ' A ' + v + + − Vs − d 2MΩ Av Vo Power supplied by V d − 0 for A s + Av + ' A → → ∞ d − V 2 P = s s 10 kΩ If vd is small and the resistance where it is measured is large, then the current through the “voltmeter” resistor can be neglected. Power in 20 kΩ resistor

2 2 If vd is nearly zero, the current through the 10 kΩ resistor is approximately Vs Vs 20 kΩ P = (20 kΩ) = = 2Ps V /10 kΩ. This current must ﬂow through the 20 kΩ resistor and 10 kΩ 10 kΩ 10 kΩ s dependent source. Also, 50 Ω 20 kΩ. the dependent source supplies power! ⇒ V V = s 20 kΩ = 2V 0 − 10 kΩ − S Phyllis R. Nelson (Cal Poly Pomona) Introduction to the Op Amp ECE 207 - Winter 2015 11 / 22 Phyllis R. Nelson (Cal Poly Pomona) Introduction to the Op Amp ECE 207 - Winter 2015 12 / 22 Summary Outline

R1 R2 R V V 2 V o o ' − R s 1 r 1 Voltage Ampliﬁers − o v 0 I 0 d ' ri ' V + v r s − d i If + Av + 2 Operational ampliﬁer d − A is large r R ,R i 1 2 r R 3 Ideal op amp o 2

The nonideal dependent source circuit shown in 4 Applications of the ideal op amp model the dashed box is a model of an operational ampliﬁer (op amp).

Phyllis R. Nelson (Cal Poly Pomona) Introduction to the Op Amp ECE 207 - Winter 2015 13 / 22 Phyllis R. Nelson (Cal Poly Pomona) Introduction to the Op Amp ECE 207 - Winter 2015 14 / 22 Operational ampliﬁer symbol Limits on output voltage

The op amp requires external DC voltage sources (V+ and V ) to V − + power the internal dependent source. − Vm V V + V o These voltages limit the maximum possible values of the output voltage. p + V Minus Input V o V − V V m − + + − Vm Pos. saturation Vo V + − r Output p o V− ri vd V o vd + A + Av Linear (slope ) − d v = V V Neg. saturation Plus Input d p − m + V− Vp The power connections are often not shown, but they 5 V− For V+ = V = 10 V and A = 1 10 , the linear region is are required. | −| × 100 µV v 100 µV − . d . Phyllis R. Nelson (Cal Poly Pomona) Introduction to the Op Amp ECE 207 - Winter 2015 15 / 22 Phyllis R. Nelson (Cal Poly Pomona) Introduction to the Op Amp ECE 207 - Winter 2015 16 / 22 Op amp summary Outline

ro Vm Vo Desired properties: − r is large 1 Voltage Ampliﬁers v r + Av i d i − d Input currents 0 + ' A is large Vp 2 Operational ampliﬁer ro is small

3 Ideal op amp In our example, large A and small ro let Vo be determined by the external circuit. 4 resulted in very small vd Applications of the ideal op amp model

If vd is outside a very narrow range near 0 V, Vo is determined by positive or negative saturation.

Phyllis R. Nelson (Cal Poly Pomona) Introduction to the Op Amp ECE 207 - Winter 2015 17 / 22 Phyllis R. Nelson (Cal Poly Pomona) Introduction to the Op Amp ECE 207 - Winter 2015 18 / 22 The ideal op amp Test the ideal op amp model

Vo no current into minus V+ V+ 20 kΩ terminal V − Pos. saturation m V V o 10 kΩ Vs Vm Vm Vo p + − − = − V− Vo 10 kΩ 20 kΩ v + + d Vs −

vd = 0 Vm = Vp = 0 v = V V Neg. saturation ⇒ d p − m (virtual ground) V−

Rules: V 20 kΩ o = = 2 no currrents ﬂow in the inputs Vs −10 kΩ − vd = 0 V (Vm = Vp) Inverting voltage ampliﬁer V is not saturated voltage gain 2 o −

Phyllis R. Nelson (Cal Poly Pomona) Introduction to the Op Amp ECE 207 - Winter 2015 19 / 22 Phyllis R. Nelson (Cal Poly Pomona) Introduction to the Op Amp ECE 207 - Winter 2015 20 / 22 Outline Useful op amp circuits

1 Voltage Amplifiers Noninverting amplifier Voltage follower 2 Operational amplifier Summing amplifer Difference amplifier 3 Ideal op amp Instrumentation amplifier ... 4 Applications of the ideal op amp model

Phyllis R. Nelson (Cal Poly Pomona) Introduction to the Op Amp ECE 207 - Winter 2015 21 / 22 Phyllis R. Nelson (Cal Poly Pomona) Introduction to the Op Amp ECE 207 - Winter 2015 22 / 22