CE2004 II‐1 : Operational Amplifiers (Op-Amps)
Weisi Lin Email: [email protected] School of Computer Science and Engineering Nanyang Technological University Singapore
CE 2004 – Circuits & Signal Analysis 1 Major Topics of Part 2
• Principles of Op‐Amps • General Signals & Systems • Laplace Transform & frequency‐based system analysis Important Concepts, Logic & Methodology (Principles of Op-Amps) Introduction to Op-Amps (operational amplifiers) Concept of Negative Feedback Two Commonly-used Circuits with Negative Feedback oNon-Inverting Amplifiers oInverting Amplifiers Recap/Summary
CE 2004 – Circuits & Signal Analysis 3 We are starting to talk about amplifiers.
What is an ideal amplifier?
Considerations: Voltage gain Input impedance Output impedance
CE 2004 – Circuits & Signal Analysis 4 First op amps built in 1934 Invented by a Bell engineer named Harry Black* Didn’t gain the name operational amplifier until the computer age began a decade or so later Vacuum Tube Op-Amps Vacuum Tube Op-Amps Used in WWII to help to strike military targets - Buffers, summers, differentiators, inverters Took ±300V to ± 100V to power Transistors replaced vacuum tubes in 1950’s Integrated circuits (ICs) invented in the 1960’s, op amps were among the first chips to be designed.
*In 1934，Harry Black commuted from his home in New York City to work at Bell Labs in New Jersey by way of a railroad/ferry. The ferry ride relaxed Harry enabling him to do some conceptual thinking…
CE 2004 – Circuits & Signal Analysis 5 Introduction to Operational Amplifiers (Op-Amps) Op-amp is a high gain amplifier having (nearly ideal): - Very high voltage gain (104 to 106) - Very high input impedance (typically a few more megohms) explanations in the next - Low output impedance (less than 100 ohm). pages. An op-amp consists of multi-stage transistor amplifier fabricated in an integrated circuit (IC) form.
CE 2004 – Circuits & Signal Analysis 6 Internal Circuitry of the Standard 741 Op-Amp
Positive DC power supply
An IC consisting of transistors resistors capacitors
Negative DC power supply
CE 2004 – Circuits & Signal Analysis 7 A voltage source followed by an amplifier
(very different from Vs)
Loading effect if minimization: input impedance of an i.e, amplifier should be much With the same Rs, greater (infinite for an ideal case) than the (similar to V ) CE 2004 – Circuits & Signal Analysis s source impedance. 8 An amplifier followed by another amplifier
For amplifier #1:
Loading effect minimization: input impedance of Amplif#2 similar analysis as in should be much greater the previous page (infinite for an ideal case) than output impedance of Amplif#1. if Zi2 >>Zo1 CE 2004 – Circuits & Signal Analysis 9 Typical Uses of Op-Amps: Voltage amplifiers, Oscillators, Filter circuits, Instrumentation circuits, and so on.
Op-Amp Equivalent Circuit
V (see next CC page) As represented in the previous pages V+
ZOUT ZIN Vout
V- Zout
Vo=AoVin VEE Vin Zin ~ AdVd Vout
CE 2004 – Circuits & Signal Analysis 10 Input port: Modeled as a resistance of value ZIN . (Very large, typically 106 ).
The input v is known as non-inverting input and v is the inverting input.
Output port: Modeled as a dependent voltage source, vo in series with a resistance ZOUT .( is very small, ZOUT typically 100 ).
Two DC power supplies: VCC (+ve) and VEE (‐ve). The use of positive and negative voltage allows the output of the op‐amp to swing in both directions.
CE 2004 – Circuits & Signal Analysis 11 Input-Output Relationship: Vout Ao (v v ) ‐Ao is the internal voltage gain (open loop voltage gain) of the op‐amp.
4 6 ‐ Typically, Ao will have values 10 to10 (very high) Output Saturation Limits vo can never exceed the op‐amp’s power supply voltages, VCC and (saturation limits of VEE vo ) .
VEE Ao(v v) VCC
CE 2004 – Circuits & Signal Analysis 12 Voltage transfer characteristics of an op-amp
Vout Positive saturation VCC= +15V VCC
Linear, i.e., Vo=Ao(Vin)
Vin=(V+)–(V-)
V V = -15V Negative saturation EE EE Linear region
. The slope of linear region is big (Why?)
CE 2004 – Circuits & Signal Analysis 13 Op-Amps
1.The saturation limits ofvo are equal to the op-amp supply voltagesVCC andVEE. 2.The gainAo is so high(Ao ) that a very small nonzero value of differential input(v v ) drivesvo to saturation (non linear operation). 3.(v v ) positive vo will saturate at its positive saturation limit VCC . 4.(v v ) negative vo will saturate at its negative saturation limit VEE.
VCC
V+
ZOUT ZIN Vout
V-
VEE CE 2004 – Circuits & Signal Analysis 14 5.For linear operation, the differential input voltage (v v ) must be close to zero (very small). 6.The internal input resistance Z IN . Hence the input currents (in +ve and –ve terminals) are assumed to be zero. 7.The internal output resistance ZOUT 0. Hence Vout Vo.
Q: What is the main concern in using op-amp as a linear device?
A: Saturates for a small Vin. Q: How do we get rid of this problem, and also obtain the desired voltage gain? VCC A: Negative Feedback
V+
ZOUT ZIN Vout
V-
V CE 2004 – Circuits & Signal Analysis EE 15 Harold Stephen Black (April 14, 1898 – December 11, 1983) invented the negative feedback amplifier while he was a passenger on the Lackawanna Ferry (from Hoboken Terminal to Manhattan) on his way to work at Bell Laboratories (located in Manhattan instead of New Jersey in 1927) on August 2, 1927 (US Patent 2,102,671, issued in 1937)— obviously he was thinking while taking ferry.
His invention is considered as an important breakthrough of the 20th century in electronics, since it has a wide area of applications. On August 8, 1928, Black submitted his invention to the U. S. Patent Office, which took more than 9 years to issue the patent. The patent office was inundated with fraudulent “perpetual motion” (motion of bodies that continues indefinitely) claims, and dismissed Black’s invention at first sight. Black later wrote: "One reason for the delay was that the concept was so contrary to established beliefs that the Patent Office initially did not believe it would work.” CE 2004 – Circuits & Signal Analysis 16 Concept of Negative Feedback
In negative feedback, the output signal (or part of it) is connected back (feedback) to the negative input terminal of an op-amp. Summing Amplifier Node VM VO InputAo Output + - VR output/K (to “‐” terminal) 1/K
Vo (VM VR )Ao (VM Vo / K)Ao
CE 2004 – Circuits & Signal Analysis 17 Ao K Ao Vo 1 VM Ao Vo VM Ao K K
VM Ao Vo K Ao K
K Vo VM Ao K Ao K (As Ao K , VM Ao Ao Ao K Ao)
Vo=K VM
CE 2004 – Circuits & Signal Analysis 18 Negative feedback reduces the closed loop gain to K （this is under the condition that Ao is very high; notice that K is not related to Ao.）
Using resistors (as to be illustrated with non-inverting & inverting Op-Amps next) to obtain appropriate value of K such that the output will not reach the saturation limits.
Thus, negative feedback ensures linear operation of op- amps, and also yield the required voltage gain.
CE 2004 – Circuits & Signal Analysis 19 A question for you to think while we move on…
Why are op-amps designed with very high Ao, while the required voltage gain of amplifiers is much lower in practice (so we use negative feedback to reduce the gain and ensure linear operation)? Why not to design op-amps with a lower Ao?
Hint: thinking along the flexibility provided to users and benefits to manufacturers.
CE 2004 – Circuits & Signal Analysis 20 The Op-amp Golden Rules – with Nagative Feedback Circuits
1.Voltage Rule: Through negative feedback, the output attempts to do whatever is necessary to make the voltage difference between the inputs zero.
In a negative feedback op-amp, V+ -V- ≈ 0 (very small).
2. Current Rule: The inputs draw no current, due to the very high input impedance of the op-amp.
The input current is so low (0.08 microamps for the standard 741 op-amp).
CE 2004 – Circuits & Signal Analysis 21 Linear (not saturated) Op-Amp Circuits
V+ 1. Non-Inverting Amplifier VIN + V is applied to the +ve input VOUT IN V terminal of the op-amp. - _ R1
ApartofVO is fed back to the A way to –ve terminal of the op-amp. R2 materialize the negative feedback How does negative feedback concept presented help? previously.
If a +ve voltage* is applied to VIN :
(V V ) will increase VOU will increase. T
*Similar analysis for a -ve voltage case.
CE 2004 – Circuits & Signal Analysis 22 •A part ofVOUT is fed back to theV terminal. (Determined by the voltage divider resistors R1 and R2).
R2 V VOUT (1) R1 R2 • V increases ( V V ) decreases V OU T decreases.
i.e., due to negative feedback,(V V ) is decreased, and thenVOUT is decreased.
CE 2004 – Circuits & Signal Analysis 23 •Equilibrium condition V OUT forces ( V V ) to a very small value VOUT will not reach the saturation limits. • Op-Amp golden voltage rule: Input difference voltage is very small Approximated as zero (practically, small non-zero).
V V (2)
Applying (2) in (1): R2 V VOUT VIN (3) R1 R2
Closed loop-gain, A V:
VOUT (R1 R2) R1 Av 1 (4) VIN R2 R2
CE 2004 – Circuits & Signal Analysis 24 Significant Parameters of Non-Inverting Op-Amps
●A V will be always ≥ 1.
●VOUT is always in phase with VIN (i.e., with same polarity, or non-inverting).
CE 2004 – Circuits & Signal Analysis 25 Non-Inverting Op-Amps
since V+=V- Input
impedance: ZIN is very high. Ideal case = .
RL
(to be presented a bit later)
R Output Resistance: V (=(1 1 )V ) is independent of R (no OUT R2 IN L matter what RL is used), so ROUT 0.
CE 2004 – Circuits & Signal Analysis 26 Further questions to think…
We know that an op-amp’s voltage gain is very high; do you think if a different voltage gain value (say, from 200,000 to 250,000) matters to a non-inverting circuit’s output? Discuss if there is any benefit in op-amp manufacturing processes for your conclusion above.
CE 2004 – Circuits & Signal Analysis 27 Example: Design a non-inverting amplifier with a gain of +10: VOUT (R1 R2) R1 Av 1 10 VIN R2 R2 R1 9 R1 9R2 ; we choose R1 9k and R2 1k R2
Example: A sensor signal Vin in the range of 1mV–50mVhastobeamplifiedwithanon- inverting op-amp. If the op-amp power supply is ±15V, find the maximum allowable gain without
causing distortion at output signal Vout. Hint: To avoid output signal distortion,
Vout =(1+R1/R2)x(Vin) ≤ 15Volts (where Vin = 50 mV and (1+R1/R2) = gain) (Answer: 300)
CE 2004 – Circuits & Signal Analysis 28 A variation of the non-inverting Op-Amp circuit as a summer:
10R VCC V1
Vout V2 R 10R VEE
R
Superposition Theorem can be used to find the relationship of Vout and inputs (you can practise it and then use it in the tutorials).
CE 2004 – Circuits & Signal Analysis 29 Additional Example
Calculate all voltages and currents in the circuit, complete with arrows for current direction and polarity markings for voltage polarity.
CE 2004 – Circuits & Signal Analysis 30 Additional Example 1 (cont’d)
AV = 1+22k/47k = 1.468
Vout=3.2Vx1.468=4.698V
IR2=IR1 =Vout/(22k+47k)=68.09A
VR2= IR2 x22K=1.498V
VR1= IR1 x47K=3.2V
(it is verified that V+=V‐) No current
CE 2004 – Circuits & Signal Analysis 31 Determine both the input and output voltage in this circuit:
Vout = 2mA(5k+18k)=46 V Av = 1+18k/5k=4.6 Vin = 46V/4.6=10 V
CE 2004 – Circuits & Signal Analysis 32 Predict how the operation of this operational amplifier circuit will be affected as a result of the following faults. Consider each fault independently (i.e. one at a time, no multiple faults):
• Resistor R1 is open;
• S/C across resistor R2.
CE 2004 – Circuits & Signal Analysis 33 (Solution)
• Resistor R1 is open: Output saturates positively
• S/C across resistor R2: Output saturates positively
CE 2004 – Circuits & Signal Analysis 34
V+ VIN +
VOUT V- _
R1
2. Unity Gain Amplifier R2
V+ For the Non-Inverting VIN + amplifier, if R1 0 (short VOUT V- _ circuit) and R 2 very large ( , i.e. open circuit):
R1 Av VOUT /VIN 1 1 R2
Hence, VOUT VIN Unity gain amplifier or Voltage buffer.
CE 2004 – Circuits & Signal Analysis 35 Why we need an amplifier that seems to do nothing? A unity gain amplifier serves as a buffer circuit, which provides a means of: Isolating an input signal from a load by using a stage having unity voltage gain, with very high input impedance and low output impedance (previously shown), and without phase or polarity inversion, as to be explained again in the next page.
CE 2004 – Circuits & Signal Analysis 36 Acting as an ideal circuit thus minimizing the loading effect (for circuits before & after it)
(for the next stage)
very high input impedance and low output impedance.
CE 2004 – Circuits & Signal Analysis 37 Rf 2. Inverting Amplifier R1 V- - R f , which connects VIN V toV terminal, V OUT V OUT provides the negative + + feedback connection.
Applying op-amp voltage rule:(V+) – (V-) 0 V+ = V-
Thus, V- will be at the same voltage of V+, i.e. ground potential (0V)!
V- is at Virtual Ground (i.e. not a real ground) – to be continued in the next page.
CE 2004 – Circuits & Signal Analysis 38 At the amplifier input there exists a virtual short circuit or virtual ground! (not a real ground). Remember, it is only a “virtual” ground, i.e., there is no
“real” short circuit between V- and ground (see the Internal Circuitry of the Standard 741 Op-Amp given earlier) There is no current through the amplifier input to ground. Current only goes throughR1 andR f . The virtual ground concept permits a simple Virtual s/c solution to determine the overall gain (otherwise difficult to solve due to the feedback signal mixed with input signal), as shown as the next pages. CE 2004 – Circuits & Signal Analysis 39 op-amp
Zin
CE 2004 – Circuits & Signal Analysis 40 Inverting op‐amp equivalent circuit
CE 2004 – Circuits & Signal Analysis 41 Vo I = VIN / R1 I R f
N VIN / R1 = - Vo/Rf
Vo / VIN = - Rf / R1
CE 2004 – Circuits & Signal Analysis 42 Significant Parameters of Inverting Op-Amps
o VOUT is 180 out of phase to VIN (inverting amplifier):
Rf
R1 0 - V+IN - VOUT +
Rf
R1 0 - V- IN V+OUT +
CE 2004 – Circuits & Signal Analysis 43 Significant Parameters of Inverting Op-Amps (cont’d)
The gain can be made less than unity (i.e., an amplifier that attenuate input)—different from a non-inverting one. Input impedance: unlike the non-inverting amplifier, the input current of the inverting amplifier is non-zero (although the input current of an op- amp itself is always close to zero):
I=(VIN-0)/R1=VIN/R1, so the input impedance is finite: Z IN VIN / I R1 (you will use this in the lab).
Output impedance: very close to zero (like the non-inverting amplifier), because V 0 I R V R / R V OUT f I N f 1 ; i.e., OUT is independent of RL (i.e.,
VOUT does not change when RL changes). Rf I: not zero R1 - VIN i=0 VOUT + RL CE 2004 – Circuits & Signal Analysis 44 Example 1:
Fig. 1
What input voltage results in an output of 2V in the circuit of Fig. 1? R f 1M Vo VI VI 2V R1 20k
VI = ‐ 2V/50= ‐ 40 mV
CE 2004 – Circuits & Signal Analysis 45 Another Example: The input signal Vin of the op-amp circuit in the figure varies from 0.1V to 0.5V. If the op-amp power supply is
±15V, find the maximum allowable RF without causing distortion at output signal Vout. Rf 6.8kΩ V- - VIN V V OUT + +
Hint: To prevent output signal distortion,
|Vout |= (RF/R1)x(Vin) ≤ 15Volts where Vin = 0.5 V and R1 = 6.8kΩ (Answer: RF ≤ 204kΩ)
CE 2004 – Circuits & Signal Analysis 46 A further question to think…
Do the 2 circuits below have different voltage gain?
Rf Rf
R1 Vin R V 1 out Vin Vout
R2
CE 2004 – Circuits & Signal Analysis 47 • What happens if two op‐amp circuits are used together: – An inverting circuit followed by a non- inverting one – A non-inverting circuit followed by an inverting one – a summer followed by a non- inverting one – An inverting circuit followed by a summer –…
CE 2004 – Circuits & Signal Analysis 48 Recap/Summary
Op-amp Golden Rules with Nagative Feedback
1.Voltage Rule: the voltage difference between two inputs is zero.
2. Current Rule: two inputs draw no current.
CE 2004 – Circuits & Signal Analysis 49 Recap/Summary (cont’d) Two commonly-used circuits
Non-Inverting Inverting
V+ VIN + Rf V V OUT - _ R1 V V- - R1 IN VOUT V+ +
R2
CE 2004 – Circuits & Signal Analysis 50 Recap/Summary (cont’d) Non-Inverting Inverting
Voltage gain 1+R1/R2>1 -Rf/R1<0 Input impedance Very high R1 Output impedance Very small Very small Additional remarks can form a unity- can form an gain amplifier; with inverter; a good minimum loading voltage supplier effect; a good voltage supplier Both are with negative feedback (to V- terminal).
CE 2004 – Circuits & Signal Analysis 51 With op-amps, how to build 1. An amplifier of a positive voltage gain 2. An isolation amplifier (to provide isolation of one part of a circuit from another) 3.A voltage inverter 4.A voltage summer 5.An Amplifier to minimize loading effect of the previous stage 6.A good voltage supplier
CE 2004 – Circuits & Signal Analysis 52