Operational Amplifier (Op-Amp) Basics
SYRACUSE UNIVERSITY
ELE231 Electrical Fundamentals I
Operational Amplifier (Op-Amp) Basics by Akhan Almagambetov, [email protected] Office Hours: Wed. 1:15-3:15 p.m., Thurs. 12:30-2:30 p.m in CST 3-204 Recitation Session # 7 Week of October 6th, 2008
Perhaps one of the most interesting practical applications of a dependent source is the operational amplifier (referred to as the op-amp). These devices are typically used to amplify small signals, which makes them ideal candidates for use in instrumentation. A common op-amp device that you will see in laboratory is the 741 general-purpose amplifier. It is one of the cheapest (costs only about 25¢ retail and less than a penny to manufacture) and the most readily-available amplifier on the market today. The op-amp is represented on circuit diagrams by the following symbol:
v+
I+
+ LM741
v--
vOUT
-I--
Figure 1: Circuit symbol for the operational amplifier with labeled currents and voltages 1
Operational Amplifier Basics There are a few basic points that you need to keep in mind when analyzing op-amp circuits: • Op amps have a very high input impedance • Op amps generally tend to have very high gain (output vs. input) • They are commonly used to amplify small signals (i.e. instrumentation) Figure 2, below, contains a model of the operational amplifier. The op-amp is the first device we see that contains some sort of a controlled voltage source. As we can observe from the diagram, the value of the controlled voltage source is dependent on the input voltage (across the input resistor, RIN ). +15V
+ non-inverting output
positive rails (supply)
+ vIN
ROUT
RIN +
– inverting output
--
–
vOUT output
AvIN negative rails (supply)
ground
--15V
Figure 2: Simple operational amplifier model Since we know (from class discussions) that the op-amp is a quirky device, there exist a few relationships of currents and voltages that will make your life easier. They are as follows (look on Figure 1 to see all of the voltages and currents that will be discussed below): • v− = v+ : Voltages at the input terminals of an op-amp are always equal. This also makes the voltage drop between the positive and negative terminals equal to zero. • I+ = I− = 0 : Currents going into the op-amp are always zero. • RIN → ∞ : Input resistance of an op-amp is close to ∞ (in other words, really BIG).
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Operational Amplifier Basics
Inverting Configuration In this configuration of the operational amplifier, negative gain is achieved. The inverting configuration is perhaps the most common configuration of the operational amplifier. As you will see from the sections that follow, you can make the op-amp do a number of neat things just by wiring components around it (if these components happen to be strictly resistors, this practice is called op-amp biasing ). I2
I1
Rf
v--
R1
vOUT -+
+
vS
vOUT
v+
--
Figure 3: Inverting op-amp configuration What can we see from the figure above? First, we note that the input voltage at the positive input terminal, v+ , is simply equal to zero (since it’s shorted directly to ground). From the op-amp voltage relationships, we know that the input terminal voltages are equal, which will give us the following: (1)
v+ = v− = 0
Now, all we have to do is nodal analysis! Remember that every op-amp circuit is simply a nodal analysis problem, so try to get the method down now, before we move on to more complex circuits. Let’s first find the currents flowing through the two resistors, I1 and I2 . By Ohm’s Law, we have:
I1 =
I2 =
vS − v− R1
v− − vOU T Rf
→
→ 3
I1 =
I2 =
vS R1 −vOU T Rf
(2)
Operational Amplifier Basics
Inverting Configuration (cont’d) Immediately, we can set the v− terms to zero, from equation (1). Being aware of the fact that currents entering an op-amp are equal to zero (I+ = I− = 0) and by using KCL (currents entering a node must equal the currents leaving), we can set the two currents calculated above equal to each other:
I1 = I2 + I− But since we know that I− = 0, we can say that: I1
=
vS R1
=
I2 −vOU T Rf
(3)
After all of the terms have been rearranged, the output voltage of the op-amp circuit can be expressed as a function of the input (source) voltage: Rf vOU T = vS − R1
(4)
Relevant question: does switching the (+) and (-) terminals on the op-amp affect the solution? Why or why not (from KCL equations and properties of the op-amp, above)? Answer: it really doesn’t.. although the correct notation would be to connect the positive terminal of the op-amp to the source voltage in the case of a non-inverting op-amp and to connect the negative terminal to the source in case of an inverting configuration. The only thing that affects the voltage-in to voltage-out relationships is resistor configuration, as you will observe from the next two examples. To make this concept a little clearer, observe the way we analyze the differential amplifier.
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Operational Amplifier Basics
Non-inverting Configuration This particular biasing scheme will result in positive gain. Note that the only thing that changed from the inverting configuration is the overall location of the resistors. If in the inverting configuration the resistors were hooked up to the source voltage, here they are grounded . Also note the correct schematic representation for this configuration: the positive terminal of the op-amp is connected to the source (versus the negative terminal being connected to the voltage source in the inverting configuration). v+
vOUT
+
+
-v--
R2
I2
vS
vOUT R1 I1
--
Figure 4: Non-inverting op-amp configuration From the diagram above, we can see that the voltage at the positive terminal of the opamp (v+ ) is equal to the source voltage, vS . Looking back to our basic op-amp voltage relationships, we can determine the following: (5)
v+ = v− = vS
Once again, we perform simple nodal analysis to express currents I1 and I2 in terms of what we know already:
I2 =
vOU T − vA R2
I1 =
vS − 0 R1
→
→ 5
I2 =
vOU T − vS R2
I1 =
vS R1
(6)
Operational Amplifier Basics
Non-inverting Configuration (cont’d) Since we already know that I+ = I− = 0 and I1 = I2 + I− , we can say that: I1
=
vS R1
=
I2 vOU T −vS R2
(7)
By rearranging the terms of the above equation, we should get the following relationship, which would define the value of the output voltage as a function of the input voltage, vS , and resistors R1 and R2 :
vOU T
R2 = vS 1 + R1
(8)
Does this make sense? Why, sure it does! Based on equation (8), the value of the output voltage (vOU T ) can never be less than zero (the ratio of the “feedback” resistor (R2 ) over the “ground” resistor (R1 ) will always be positive.. and you’re adding “+1” to boot! ). In the inverting configuration, the value of the output voltage will never be greater than zero (and rightfully so), since the ratio of the “feedback” resistor (Rf ) over the “source” (R1 ) resistor (which is normally positive) is negated.
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Operational Amplifier Basics
Differential Amplifier Ah, the differential amplifier.. It’s by far one of the most common op-amp applications (and something that you will build in lab using the LM741 chips). The differential amplifier has weird resistor biasing that’s different from what you’ve seen so far.
[?]
Can you tell which configuration (studied so far) is represented in the figure below?
(That was a rhetorical question.. there is something that looks like an inverting configuration on the top, but there’s also some funky stuff on the bottom, which makes this circuit a whole different ball game when it comes to analysis. In other words, you can’t apply any of the previous voltage-in voltage-out relationships to this circuit.) IR4
vIN-vIN+
IR1
IR2
R1
R2
R4
v--
vOUT --
IO
+
v+
IOUT ROUT
R3 IR3
+
vOUT
--
Figure 5: Differential amplifier Despite the weirdness of the above circuit, we can still apply our good ol’ nodal analysis and break it down. So let’s do just that. Since the non-inverting terminal of the op-amp is grounded, we’ll start by solving for the currents that enter and leave the v+ node. The currents that we would be solving for are IR2 and IR3 . 7
Operational Amplifier Basics
Differential Amplifier (cont’d) Non-inverting terminal (Please note the current direction when writing KCL equations. You need to subtract the node that the current is going toward from the node that the current is leaving, in order to have your equations work out in the end). IR2 =
vIN + − v+ R2
IR3 =
v+ − 0 v+ = R3 R3
(9)
Since we know that the current entering the op-amp is equal to zero, we can simply set the two currents equal (IR2 = IR3 ), as seen below: (vIN + − v+ )
1 1 = (v+ ) R2 R3
By rearraging the terms above, we get the final relationship that ties the node voltage, v+ , to the input voltage, vIN + .
1 1 + R2 R3
v+ =
1 R2
+
vIN +
!
1 R3
v+ =
1 R2
(10)
vIN +
1 R3
Inverting terminal We apply the same procedure to the inverting terminal of the op-amp: IR1 =
vIN − − v− R1
(vIN − − v− )
1 1 + R1 R4
v− =
v− =
1 R1
v− − vOU T R4
(11)
1 1 = (v− − vOU T ) R1 R4
IR4 =
1 R1
vIN − +
vIN − + 1 R1
8
+
1 R4 1 R4
1 R4
vOU T
vOU T
(12)
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Operational Amplifier Basics
Differential Amplifier (cont’d, 2) The final step involves expressing the output voltage, vOU T , as a function of both input voltages, vIN + and vIN − . This value is called the transfer function. In order to find the transfer function, we simply set equations (10) and (12) equal to each other, based on the property of the op-amp, where the input voltages are always equal (v− = v+ ).
1 R2
+
!
1 R3 1 R3
vIN + =
1 R1
vIN − + 1 R1
+
1 R4 1 R4
vOU T
By simple terms manipulation, we get the following equation, which is our transfer function: 1 R3
vOU T =
vOU T =
1 + R1 R1 4 1 + R1 R2 3
!
vIN + −
1 R1
vIN −
1 R4
(R2 (R1 + R4 )) vIN + (R4 ) vIN − − R1 (R2 + R3 ) R1
Simplified to, vOU T =
R4 R2 + R3
R4 R4 + 1 vIN + − vIN − R1 R1
(13)
From here, you can find the output current, IOU T , given the output resistance, ROU T . You would then be able to calculate the current on the output terminal of the operational amplifier, IO , by subtracting IR4 from IOU T .
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Operational Amplifier Basics
Current-to-Voltage Converter As its name suggests, the current-to-voltage converter simply converts the input current to a voltage. R
IIN
--
vREF
+
vOUT
Figure 6: Current-to-voltage converter There are simple relationships that one needs to keep in mind when analyzing circuits that contain current-to-voltage converters. They are as follows: v = IIN R and vOU T = vREF − v Based on the two equations above, the output voltage is dependent on the input current, the value of the “feedback” resistor, as well as the reference voltage (if this terminal is shorted to ground, then your reference voltage will be zero): vOU T = vREF − IIN R
(14)
What does the circuit above look like? Think back to a few lectures ago.. what is one of the weird conceptual devices that you’ve seen in lecture, which requires two equations to determine its output? That’s right, a dependent source! In this case, the circuit represents a current-controlled voltage source (CCVS). In other words, not only do the op-amps contain dependent sources, but they can also be biased to model them! Isn’t that neat?
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Operational Amplifier Basics
Voltage Follower (Unity-gain amplifier) This is one of the simplest op-amp-based devices (in terms of analysis).
-vIN
vOUT
+
Figure 7: Voltage follower This device has a gain of 1 (in the ideal case) and does not contribute any lengthy circuit analysis equations or steps to existing circuits. Several things to keep in mind for this type of configuration: • very high input resistance • typically used as a “buffer” device to prevent excessive loading (you will see this device as the first stage on more complex circuits involving operational amplifiers) • gain of A → 1 Based on the above information, the following holds true:
vOU T = vIN
(15)
This device is used as a “buffer” to provide isolation to more costly circuit components. Since op-amps are relatively cheap (which might not be the case for the rest of the circuit), they are easy to replace and therefore stand “in the direct line of fire” from voltage surges and similar phenomena, which might be potentially hazardous to circuit components.
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LM741
Soldering Information N-Package (10 seconds)
AbsoluteJ-Maximum Ratings or H-Package (10 seconds) (Note 2)
260˚C
260˚C
260˚C
300˚C
300˚C
300˚C
If Military/Aerospace specified devices are required, M-Package please contact the National Vapor PhaseSemiconductor (60 seconds) Sales Office/ 215˚C 215˚C 215˚C Distributors for availability and specifications. Infrared (15 seconds) 215˚C 215˚C 215˚C (Note 7) See AN-450 “Surface Mounting Methods and Their Effect on Product Reliability” for other methods of soldering LM741A LM741 LM741C
Operational Amplifier Basics surface Voltage mount devices. Supply ESD (Note 8) 3) PowerTolerance Dissipation (Note
± 22V
± 22V
± 18V
400V 500 mW
400V 500 mW
400V 500 mW
± 15V
± 30V ± 15V
± 30V ± 15V
± 30V Differential Input Voltage Non-ideal Operational Amplifiers Electrical Characteristics (Note 5) Input Voltage (Note 4)
Output Short Circuit Duration Parameter Conditions Operating Temperature Range
ContinuousLM741A −55˚C toMin +125˚C Typ
ContinuousLM741 −55˚C +125˚C Max toMin Typ
Continuous LM741C 0˚C to +70˚C Max Min Typ
Units Max
This last Input section covers non-ideal operational amplifiers. Simply put, these are amplifiers Storage Temperature Range −65˚C to +150˚C −65˚C to +150˚C −65˚C to +150˚C Offset Voltage T = 25˚C A
Junction Temperature 150˚C 150˚C ≤ 10 kΩ 5.0 100˚C 2.0 6.0 mV that do not conform to theRR usual “ideal” characteristics and1.0may exhibit slightly different Soldering Information ≤ 50Ω 0.8 3.0 mV S S
N-Package (10 260˚C that you260˚C T seconds) ≤ T ≤“real T behavior (these op-amps are the thing” will see in 260˚C laboratory). The input AMIN
A
AMAX
J- or H-Package seconds) RS (10 ≤ 50Ω
300˚C
4.0 300˚C
300˚C
mV
215˚C
15 215˚C
215˚C
µV/˚C
M-Package Ris≤ not 10 kΩ infinite and the gain of the device 6.0 depends7.5 mV type and resistance of these op-amps on the S
Vapor Phase (60 seconds) Average Input Offset
the manufacturer. Usually, non-ideal op-amp are specified in the datasheets Infrared (15 seconds) 215˚C characteristics 215˚C 215˚C Voltage Drift See AN-450 “Surface Mounting Input Offset Voltage V =Methods 20V and Their 10 Effect on Product Reliability” 15 for other methods15of that comeAdjustment with the device. T = 25˚C, soldering Range A
S
±
±
±
Average Input Offset
Electrical Characteristics (Note 5) T = 25˚C, R ≥ 2 kΩ
Current Drift Voltage Gain Large Signal Input BiasParameter Current
Conditions ± 20V, T VAS = 25˚C VO = ± 15V
Input Resistance Offset Voltage
T TA ≤VT VAMIN = ± 10V S = ±≤15V, OAMAX TA = 25˚C ± 20V ≤ TA V ≤ST= , A AMIN25˚C, AMAX
A
T TA ≤VT ±≤15V, VA = ± 10V AMIN S = 25˚C OAMAX R ≤ 50Ω ±≤5V, ± 2V T TA V≤OT=AMAX VAMIN S = S
Output Voltage Swing Average Input Offset
R ≤ 10 kΩ ± 20V VSS =
Voltage Drift
RL ≥ 2 kΩ T ± 15V VS = ± 20V VAS = 25˚C,
Input Offset Voltage www.national.com Adjustment Range
Input Offset Current Output Short Circuit Average Current Input Offset Current Drift Common-Mode Input BiasRatio Current Rejection
Min
30 80 50 LM741A Min Typ 0.210 Max
Min 50
1.0
0.3
Typ
20 200 20 200 85 500 400VLM741C 300 LM741 Typ
Max
Min
LM741 80 500 Typ 1.5 200 Max
Min 20
nA nA Units
Max
nA/˚C
LM741C 80 500 Typ 0.8 200 Max
Units nA V/mV
Typ
L
R ≤ 10 TkΩ TAMIN S A ≤ TAMAX, L ≥ 2≤kΩ, R ≤ 50Ω ± 20V VSS = 20V, VO = ± 15V Input Voltage Range
0.5 Max
Min
mV
LM741
Electrical Characteristics (Note 5) (Continued) 3.0 surface mount devices. Input Offset Current TA = 25˚C 30 ESD Tolerance (Note 400V LM741A 70 400V TAMIN8)≤ Conditions TA ≤ TAMAX Parameter
±
6.0
0.5 0.8
32
2.0 1.0
0.3 5.0
2.0 2.0
3.0
±15 12
25 10
4.0
± 16 ± 15 ± 10
15
± 12
± 13
RL ≥ 10 kΩ T 25˚C RAL = ≥2 kΩ TAMIN ≤ T ≤ TAMAX A = 25˚CA
10
TAMIN ≤ TA ≤ TAMAX
10
TAMIN ≤ TA ≤ TAMAX T RAS =≤ 25˚C 10 kΩ, VCM = ± 12V T ≤ T ≤ T = ± 12V RAMIN S ≤ 50Ω,A VCMAMAX TA = 25˚C, V = ± 20V AMIN ≤ TA ≤STAMAX,
7.5
3.0
30
25
70 35 0.5 40
30
80
80 1.0
95 6.0
0.210
86
96
mV µV/˚C V
± 15
2
MΩ mV MΩ mV V/mV V V/mV mV V V/mV
± 13 6.0
RL ≥ 10 kΩ
6.0
µA V/mV
± 12 ± 10
V mV
± 15
± 14 ±20 13
200
85 25
500
± 12 ± 10
± 14 ±20 13
200
25
300
V nA V nA mA nA/˚C mA
70
80 90
500
70
80 90
1.5
500 0.8
nA dB µA dB MΩ
0.3 2.0 0.3 2.0 Figure 8: LM741 data sheet (courtesy of National Semiconductor) T T ≤ , 5V 0.5 MΩ V = ±≤20V toTV = ±
Input Resistance Supply Voltage Rejection Ratio
Input Voltage Range
AMIN S
A
AMAX S
± 20V V RSS = ≤ 50Ω T RAS =≤ 25˚C 10 kΩ
77 ± 12
96 ± 13
±77 12
±96 13
dB V dB
Figure 8, Transient above,Response has a dataT sheet the LM741 IC from National Semiconductor. As you ≤ T Unity ≤ for T Gain V = 25˚C, AMIN A
A
AMAX
Time 0.25 0.8 0.3 0.3 µs can see, theRise input resistance (listed under “Input Resistance”, rIN = 2.0MΩ), rails (listed Overshoot 6.0 20 5 5 % Bandwidth (Note 6) T = 25˚C 0.437 1.5 under “Input Voltage Range”, ±13V), and the common-mode rejection ratioMHz (also called A
Slew Rate
TA = 25˚C, Unity Gain
0.3
0.7
0.5
0.5
V/µs
www.national.com Supply Current T = 25˚C 1.7 2.8 1.7 2.8 value mA of 90dB. the common mode voltage gain when not 2measured in decibels, CMRR) A
Power Consumption
TA = 25˚C
V = 20V 80 150 mW The values above can be used to generate an approximate model of the op-amp, just like V = ± 15V 50 85 50 85 mW S
±
S
± 20V LM741A the one drawn in Figure 2.V =The circuit can then be analyzed to find the input resistance S
TA = TAMIN
165
mW
vOU T T =T 135 gain of the circuit (A = mW (RIN P U T ), the output resistance (ROU T P U T ), and the vIN ). A
LM741
AMAX
VS = ± 15V
T =T 60 100 mW The next few pages will talk about finding the gain of the total circuit for a non-ideal opT =T 45 75 mW A
AMIN
A
AMAX
Note 2: “Absolute Maximum Ratings” indicate limits beyond which damage device may occur. Operatingcan Ratings indicate conditions for which device is non-ideal amp in a simple inverting configuration (theto the same process be applied tothe any functional, but do not guarantee specific performance limits.
op-amp circuit to find its gain).
3
12
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Operational Amplifier Basics
Non-ideal Operational Amplifiers Calculation of circuit input resistance, RIN P U T Let’s think back to the days of Thévenin equivalent circuits.. If you look at the figure below, what is the first thing that you notice that differs from other circuits that we have Thévenized before? The DEpendent voltage source. Traditional Thévenization rules explicitly state that we can’t simply “kill” dependent sources, so there must be another way of solving for the equivalent resistance of the circuit below. vd
vIN
IIN
vOUT
Rf
R1
rOUT µvd
vIN
+ +
rIN
–
vOUT
−µv −µ d
--
Figure 9: RIN P U T calculation for a non-ideal op-amp One way to solve for the equivalent resistance involves first expressing the equivalent resistance in terms of the total current through and the voltage across the circuit. In other words, by Ohm’s Law, we can express resistance as RIN P U T =
vIN IIN .
In the circuit above, we
have all it takes to find the equivalent resistance: IIN , which is simply the current through the input resistor (and the entire circuit), R1 ; and vIN , which is the voltage across the entire circuit. Based on this information, we can simply solve KCL equations at node vd and we can express total circuit resistance as a function of the input voltage and current:
KCL @ vd vd − vIN vd − (−µvd ) vd + + =0 R1 Rf + rOU T rIN
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(16)
Operational Amplifier Basics
Non-ideal Op-Amps : Calculation of RIN P U T (cont’d) By looking at KCL equation (16), above, what is the only “lone” term that we see? The only thing that breaks a long chain of vd terms, is the vIN term. Since it’s a symbolic variable, we don’t know anything about it and we would rather eliminate it now than have it stick around and cause problems later on. It would be even better, if we can express vIN in terms of vd : vIN − vd = IIN R1 The equation above happens to be your first current in KCL equation (16). We rearrange the terms to get vIN by itself on one side of the equation (for easier substitution): (17)
vIN = vd + IIN R1 Now, we simply substitute equation (17) into (16): vd − (vd + IIN R1 ) vd + µvd vd + + =0 R1 Rf + rOU T rIN
From the above equation, we see that the vd terms go to zero, cancelling out the R1 term on both the top and the bottom of the first part of the KCL. We’re left with the equation below:
IIN =
vd vd (1 + µ) + Rf + rOU T rIN
Now, the vd term seems suspiciously repetitive.. as if it’s trying to tell us something (it is). To get ourselves a little closer to our goal of finding the equivalent resistance, we divide all of the terms of the equation by vd . We’ll also bring the (1 + µ) term on the top into the denominator: IIN 1+µ 1 = + vd (Rf + rOU T ) rIN 1 1 1 + = R +r OU T f RIN rIN
(18)
1+µ
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Operational Amplifier Basics
Non-ideal Op-Amps : Calculation of RIN P U T (cont’d, 2) What does equation (18) look like? Parallel resistor combinations (i.e.
1 Req
=
1 R1
+
1 R2
+ ... +
1 RN ).
To get the appropriate value of the equivalent resistance, we simply convert the above equation (18) into a more recognizable form: RIN =
Rf + rOU T 1+µ
k (RIN )
(19)
[!]
Remember that equation (19) above is not your final resistance value! Analyzing the circuit from node vd completely neglects R1 , which (as we see from the diagram) is one of the biasing resistors in this inverting configuration. We have to manually add the value of R1 to the equivalent resistance that we found from analyzing our KCL equations, as the last step: RIN P U T = R1 +
Rf + rOU T 1+µ
k (RIN )
(20)
A similar process can be applied to the non-inverting op-amp configuration. The value of µ will always be given to you in the non-ideal op-amp model (it will usually be somewhere in the tens of thousands and will be a unitless quantity).
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Operational Amplifier Basics
Non-ideal Operational Amplifiers Calculation of circuit output resistance, ROU T P U T vd
vOUT
Rf
R1
rOUT
µvd +
rIN
–
−µv −µ d
Figure 10: ROU T P U T circuit with killed independent source The calculation of output resistance is the same nodal analysis problem with a twist.. we have to kill the independent voltage source (by shorting it), but since we cannot Thévenize circuits with dependent current sources, we have to leave the −µvd dependent source in the circuit (just like we had done in the previous problem). What should we do? vd
vOUT
Rf
R1
rOUT
µvd
rIN
+ –
0
ITEST −µv −µ d
vTEST
Figure 11: ROU T P U T calculation for a non-ideal op-amp 16
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Operational Amplifier Basics
Non-ideal Op-Amps : Calculation of ROU T P U T (cont’d) Once again, our trusty little ohmmeter comes into play. As we see from Figure 11, we simply add a test voltage source vT EST and a 0 or 1Ω resistor in series with the test source. This way, we can find the total current in the circuit. We immediately see that since we’ve added a negligible resistance to the source branch, the voltage at the vOU T node will be the same as vT EST . We therefore can analyze the KCL currents leaving the node vOU T , which will completely eliminate vT EST from our equations (when writing our KCL equations for the node, we can state that the current leaving is equal to IT EST , as shown below:
KCL @ vOU T vOU T + µvd vOU T + = IT EST rOU T R1 + Rf
(21)
Note that almost no current goes through the rIN resistor (if you remember from our discussions during lecture, the input resistance of the op-amp is VERY large), so we can safely assume that node vT EST is connected directly to ground through resistors Rf and R1 .. hence the second term in equation (21). We immediately see that our “lone” term here is vd so we’ll try to express it in terms of what we need to make this equation play nice and give us a value of the output resistance. In particular, we need to express vd in terms of vT EST . Once again, we rely upon the fact that the input resistance of the op-amp is HUGE , therefore having the R1 resistor in parallel with rIN would roughly give you the value of R1 .. this fact turns thevd node into R1 a simple voltage divider calculation (for simplicity’s sake we express R1 +Rf as α, which is approximately equal to α ≈ 0.1 for op-amps): vd =
R1 R1 + Rf
vT EST = αvT EST
(22)
Since we have vd expressed in terms of vT EST , we can now substitute equation (22) into our KCL equation (21) and get the following relationship: vT EST
1 + µα rOU T
+ vT EST
17
1 R1 + Rf
= IT EST
Operational Amplifier Basics
Non-ideal Op-Amps : Calculation of ROU T P U T (cont’d, 2) We now divide through by vT EST and place the (1 + µα) in the denominator of the first term to get an equation that resembles parallel resistor combinations: 1 + µα 1 IT EST + = rOU T R1 + Rf vd 1
rOU T 1+µα
+
1 1 = R1 + Rf ROU T P U T
All we have left is to express the equation above in a more concise form and we’re done! ROU T P U T = (R1 + RF ) k
rOU T 1 + µα
(23)
Now that we have calculated both the input and output resistance of this inverting opamp configuration, we’ll move on to gain calculation!
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Operational Amplifier Basics
Non-ideal Operational Amplifiers Calculation of circuit gain, A =
vIN
IIN
vOU T vIN
vd
R1
vOUT
Rf
+ rOUT
vIN
vOUT
µvd
rIN
+ –
µvd --
Figure 12: Gain calculation for a non-ideal op-amp Compared to the other characteristics (whew!), gain calculation is relatively simple to do.. you will have two nodal equations, which you would need to solve to get vOU T expressed in terms of vIN (I would suggest first expressing vOU T in terms of vd , and then simply substituting an expression for vIN .. otherwise you might get tripped up by the math).
KCL @ vd vd − vIN vd − 0 vd − vOU T + + =0 R1 rIN Rf
(24)
vOU T − vd vOU T − µvd + =0 Rf ROU T
(25)
KCL @ vOU T
We solve the above equations to get vOU T in terms of vd and vOU T . We then get rid of the remaining vd terms by substitution. 19
ELE231 Electrical Fundamentals I
Operational Amplifier (Op-Amp) Basics by Akhan Almagambetov, [email protected] Office Hours: Wed. 1:15-3:15 p.m., Thurs. 12:30-2:30 p.m in CST 3-204 Recitation Session # 7 Week of October 6th, 2008
Perhaps one of the most interesting practical applications of a dependent source is the operational amplifier (referred to as the op-amp). These devices are typically used to amplify small signals, which makes them ideal candidates for use in instrumentation. A common op-amp device that you will see in laboratory is the 741 general-purpose amplifier. It is one of the cheapest (costs only about 25¢ retail and less than a penny to manufacture) and the most readily-available amplifier on the market today. The op-amp is represented on circuit diagrams by the following symbol:
v+
I+
+ LM741
v--
vOUT
-I--
Figure 1: Circuit symbol for the operational amplifier with labeled currents and voltages 1
Operational Amplifier Basics There are a few basic points that you need to keep in mind when analyzing op-amp circuits: • Op amps have a very high input impedance • Op amps generally tend to have very high gain (output vs. input) • They are commonly used to amplify small signals (i.e. instrumentation) Figure 2, below, contains a model of the operational amplifier. The op-amp is the first device we see that contains some sort of a controlled voltage source. As we can observe from the diagram, the value of the controlled voltage source is dependent on the input voltage (across the input resistor, RIN ). +15V
+ non-inverting output
positive rails (supply)
+ vIN
ROUT
RIN +
– inverting output
--
–
vOUT output
AvIN negative rails (supply)
ground
--15V
Figure 2: Simple operational amplifier model Since we know (from class discussions) that the op-amp is a quirky device, there exist a few relationships of currents and voltages that will make your life easier. They are as follows (look on Figure 1 to see all of the voltages and currents that will be discussed below): • v− = v+ : Voltages at the input terminals of an op-amp are always equal. This also makes the voltage drop between the positive and negative terminals equal to zero. • I+ = I− = 0 : Currents going into the op-amp are always zero. • RIN → ∞ : Input resistance of an op-amp is close to ∞ (in other words, really BIG).
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Operational Amplifier Basics
Inverting Configuration In this configuration of the operational amplifier, negative gain is achieved. The inverting configuration is perhaps the most common configuration of the operational amplifier. As you will see from the sections that follow, you can make the op-amp do a number of neat things just by wiring components around it (if these components happen to be strictly resistors, this practice is called op-amp biasing ). I2
I1
Rf
v--
R1
vOUT -+
+
vS
vOUT
v+
--
Figure 3: Inverting op-amp configuration What can we see from the figure above? First, we note that the input voltage at the positive input terminal, v+ , is simply equal to zero (since it’s shorted directly to ground). From the op-amp voltage relationships, we know that the input terminal voltages are equal, which will give us the following: (1)
v+ = v− = 0
Now, all we have to do is nodal analysis! Remember that every op-amp circuit is simply a nodal analysis problem, so try to get the method down now, before we move on to more complex circuits. Let’s first find the currents flowing through the two resistors, I1 and I2 . By Ohm’s Law, we have:
I1 =
I2 =
vS − v− R1
v− − vOU T Rf
→
→ 3
I1 =
I2 =
vS R1 −vOU T Rf
(2)
Operational Amplifier Basics
Inverting Configuration (cont’d) Immediately, we can set the v− terms to zero, from equation (1). Being aware of the fact that currents entering an op-amp are equal to zero (I+ = I− = 0) and by using KCL (currents entering a node must equal the currents leaving), we can set the two currents calculated above equal to each other:
I1 = I2 + I− But since we know that I− = 0, we can say that: I1
=
vS R1
=
I2 −vOU T Rf
(3)
After all of the terms have been rearranged, the output voltage of the op-amp circuit can be expressed as a function of the input (source) voltage: Rf vOU T = vS − R1
(4)
Relevant question: does switching the (+) and (-) terminals on the op-amp affect the solution? Why or why not (from KCL equations and properties of the op-amp, above)? Answer: it really doesn’t.. although the correct notation would be to connect the positive terminal of the op-amp to the source voltage in the case of a non-inverting op-amp and to connect the negative terminal to the source in case of an inverting configuration. The only thing that affects the voltage-in to voltage-out relationships is resistor configuration, as you will observe from the next two examples. To make this concept a little clearer, observe the way we analyze the differential amplifier.
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Operational Amplifier Basics
Non-inverting Configuration This particular biasing scheme will result in positive gain. Note that the only thing that changed from the inverting configuration is the overall location of the resistors. If in the inverting configuration the resistors were hooked up to the source voltage, here they are grounded . Also note the correct schematic representation for this configuration: the positive terminal of the op-amp is connected to the source (versus the negative terminal being connected to the voltage source in the inverting configuration). v+
vOUT
+
+
-v--
R2
I2
vS
vOUT R1 I1
--
Figure 4: Non-inverting op-amp configuration From the diagram above, we can see that the voltage at the positive terminal of the opamp (v+ ) is equal to the source voltage, vS . Looking back to our basic op-amp voltage relationships, we can determine the following: (5)
v+ = v− = vS
Once again, we perform simple nodal analysis to express currents I1 and I2 in terms of what we know already:
I2 =
vOU T − vA R2
I1 =
vS − 0 R1
→
→ 5
I2 =
vOU T − vS R2
I1 =
vS R1
(6)
Operational Amplifier Basics
Non-inverting Configuration (cont’d) Since we already know that I+ = I− = 0 and I1 = I2 + I− , we can say that: I1
=
vS R1
=
I2 vOU T −vS R2
(7)
By rearranging the terms of the above equation, we should get the following relationship, which would define the value of the output voltage as a function of the input voltage, vS , and resistors R1 and R2 :
vOU T
R2 = vS 1 + R1
(8)
Does this make sense? Why, sure it does! Based on equation (8), the value of the output voltage (vOU T ) can never be less than zero (the ratio of the “feedback” resistor (R2 ) over the “ground” resistor (R1 ) will always be positive.. and you’re adding “+1” to boot! ). In the inverting configuration, the value of the output voltage will never be greater than zero (and rightfully so), since the ratio of the “feedback” resistor (Rf ) over the “source” (R1 ) resistor (which is normally positive) is negated.
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Operational Amplifier Basics
Differential Amplifier Ah, the differential amplifier.. It’s by far one of the most common op-amp applications (and something that you will build in lab using the LM741 chips). The differential amplifier has weird resistor biasing that’s different from what you’ve seen so far.
[?]
Can you tell which configuration (studied so far) is represented in the figure below?
(That was a rhetorical question.. there is something that looks like an inverting configuration on the top, but there’s also some funky stuff on the bottom, which makes this circuit a whole different ball game when it comes to analysis. In other words, you can’t apply any of the previous voltage-in voltage-out relationships to this circuit.) IR4
vIN-vIN+
IR1
IR2
R1
R2
R4
v--
vOUT --
IO
+
v+
IOUT ROUT
R3 IR3
+
vOUT
--
Figure 5: Differential amplifier Despite the weirdness of the above circuit, we can still apply our good ol’ nodal analysis and break it down. So let’s do just that. Since the non-inverting terminal of the op-amp is grounded, we’ll start by solving for the currents that enter and leave the v+ node. The currents that we would be solving for are IR2 and IR3 . 7
Operational Amplifier Basics
Differential Amplifier (cont’d) Non-inverting terminal (Please note the current direction when writing KCL equations. You need to subtract the node that the current is going toward from the node that the current is leaving, in order to have your equations work out in the end). IR2 =
vIN + − v+ R2
IR3 =
v+ − 0 v+ = R3 R3
(9)
Since we know that the current entering the op-amp is equal to zero, we can simply set the two currents equal (IR2 = IR3 ), as seen below: (vIN + − v+ )
1 1 = (v+ ) R2 R3
By rearraging the terms above, we get the final relationship that ties the node voltage, v+ , to the input voltage, vIN + .
1 1 + R2 R3
v+ =
1 R2
+
vIN +
!
1 R3
v+ =
1 R2
(10)
vIN +
1 R3
Inverting terminal We apply the same procedure to the inverting terminal of the op-amp: IR1 =
vIN − − v− R1
(vIN − − v− )
1 1 + R1 R4
v− =
v− =
1 R1
v− − vOU T R4
(11)
1 1 = (v− − vOU T ) R1 R4
IR4 =
1 R1
vIN − +
vIN − + 1 R1
8
+
1 R4 1 R4
1 R4
vOU T
vOU T
(12)
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Operational Amplifier Basics
Differential Amplifier (cont’d, 2) The final step involves expressing the output voltage, vOU T , as a function of both input voltages, vIN + and vIN − . This value is called the transfer function. In order to find the transfer function, we simply set equations (10) and (12) equal to each other, based on the property of the op-amp, where the input voltages are always equal (v− = v+ ).
1 R2
+
!
1 R3 1 R3
vIN + =
1 R1
vIN − + 1 R1
+
1 R4 1 R4
vOU T
By simple terms manipulation, we get the following equation, which is our transfer function: 1 R3
vOU T =
vOU T =
1 + R1 R1 4 1 + R1 R2 3
!
vIN + −
1 R1
vIN −
1 R4
(R2 (R1 + R4 )) vIN + (R4 ) vIN − − R1 (R2 + R3 ) R1
Simplified to, vOU T =
R4 R2 + R3
R4 R4 + 1 vIN + − vIN − R1 R1
(13)
From here, you can find the output current, IOU T , given the output resistance, ROU T . You would then be able to calculate the current on the output terminal of the operational amplifier, IO , by subtracting IR4 from IOU T .
9
Operational Amplifier Basics
Current-to-Voltage Converter As its name suggests, the current-to-voltage converter simply converts the input current to a voltage. R
IIN
--
vREF
+
vOUT
Figure 6: Current-to-voltage converter There are simple relationships that one needs to keep in mind when analyzing circuits that contain current-to-voltage converters. They are as follows: v = IIN R and vOU T = vREF − v Based on the two equations above, the output voltage is dependent on the input current, the value of the “feedback” resistor, as well as the reference voltage (if this terminal is shorted to ground, then your reference voltage will be zero): vOU T = vREF − IIN R
(14)
What does the circuit above look like? Think back to a few lectures ago.. what is one of the weird conceptual devices that you’ve seen in lecture, which requires two equations to determine its output? That’s right, a dependent source! In this case, the circuit represents a current-controlled voltage source (CCVS). In other words, not only do the op-amps contain dependent sources, but they can also be biased to model them! Isn’t that neat?
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Operational Amplifier Basics
Voltage Follower (Unity-gain amplifier) This is one of the simplest op-amp-based devices (in terms of analysis).
-vIN
vOUT
+
Figure 7: Voltage follower This device has a gain of 1 (in the ideal case) and does not contribute any lengthy circuit analysis equations or steps to existing circuits. Several things to keep in mind for this type of configuration: • very high input resistance • typically used as a “buffer” device to prevent excessive loading (you will see this device as the first stage on more complex circuits involving operational amplifiers) • gain of A → 1 Based on the above information, the following holds true:
vOU T = vIN
(15)
This device is used as a “buffer” to provide isolation to more costly circuit components. Since op-amps are relatively cheap (which might not be the case for the rest of the circuit), they are easy to replace and therefore stand “in the direct line of fire” from voltage surges and similar phenomena, which might be potentially hazardous to circuit components.
11
LM741
Soldering Information N-Package (10 seconds)
AbsoluteJ-Maximum Ratings or H-Package (10 seconds) (Note 2)
260˚C
260˚C
260˚C
300˚C
300˚C
300˚C
If Military/Aerospace specified devices are required, M-Package please contact the National Vapor PhaseSemiconductor (60 seconds) Sales Office/ 215˚C 215˚C 215˚C Distributors for availability and specifications. Infrared (15 seconds) 215˚C 215˚C 215˚C (Note 7) See AN-450 “Surface Mounting Methods and Their Effect on Product Reliability” for other methods of soldering LM741A LM741 LM741C
Operational Amplifier Basics surface Voltage mount devices. Supply ESD (Note 8) 3) PowerTolerance Dissipation (Note
± 22V
± 22V
± 18V
400V 500 mW
400V 500 mW
400V 500 mW
± 15V
± 30V ± 15V
± 30V ± 15V
± 30V Differential Input Voltage Non-ideal Operational Amplifiers Electrical Characteristics (Note 5) Input Voltage (Note 4)
Output Short Circuit Duration Parameter Conditions Operating Temperature Range
ContinuousLM741A −55˚C toMin +125˚C Typ
ContinuousLM741 −55˚C +125˚C Max toMin Typ
Continuous LM741C 0˚C to +70˚C Max Min Typ
Units Max
This last Input section covers non-ideal operational amplifiers. Simply put, these are amplifiers Storage Temperature Range −65˚C to +150˚C −65˚C to +150˚C −65˚C to +150˚C Offset Voltage T = 25˚C A
Junction Temperature 150˚C 150˚C ≤ 10 kΩ 5.0 100˚C 2.0 6.0 mV that do not conform to theRR usual “ideal” characteristics and1.0may exhibit slightly different Soldering Information ≤ 50Ω 0.8 3.0 mV S S
N-Package (10 260˚C that you260˚C T seconds) ≤ T ≤“real T behavior (these op-amps are the thing” will see in 260˚C laboratory). The input AMIN
A
AMAX
J- or H-Package seconds) RS (10 ≤ 50Ω
300˚C
4.0 300˚C
300˚C
mV
215˚C
15 215˚C
215˚C
µV/˚C
M-Package Ris≤ not 10 kΩ infinite and the gain of the device 6.0 depends7.5 mV type and resistance of these op-amps on the S
Vapor Phase (60 seconds) Average Input Offset
the manufacturer. Usually, non-ideal op-amp are specified in the datasheets Infrared (15 seconds) 215˚C characteristics 215˚C 215˚C Voltage Drift See AN-450 “Surface Mounting Input Offset Voltage V =Methods 20V and Their 10 Effect on Product Reliability” 15 for other methods15of that comeAdjustment with the device. T = 25˚C, soldering Range A
S
±
±
±
Average Input Offset
Electrical Characteristics (Note 5) T = 25˚C, R ≥ 2 kΩ
Current Drift Voltage Gain Large Signal Input BiasParameter Current
Conditions ± 20V, T VAS = 25˚C VO = ± 15V
Input Resistance Offset Voltage
T TA ≤VT VAMIN = ± 10V S = ±≤15V, OAMAX TA = 25˚C ± 20V ≤ TA V ≤ST= , A AMIN25˚C, AMAX
A
T TA ≤VT ±≤15V, VA = ± 10V AMIN S = 25˚C OAMAX R ≤ 50Ω ±≤5V, ± 2V T TA V≤OT=AMAX VAMIN S = S
Output Voltage Swing Average Input Offset
R ≤ 10 kΩ ± 20V VSS =
Voltage Drift
RL ≥ 2 kΩ T ± 15V VS = ± 20V VAS = 25˚C,
Input Offset Voltage www.national.com Adjustment Range
Input Offset Current Output Short Circuit Average Current Input Offset Current Drift Common-Mode Input BiasRatio Current Rejection
Min
30 80 50 LM741A Min Typ 0.210 Max
Min 50
1.0
0.3
Typ
20 200 20 200 85 500 400VLM741C 300 LM741 Typ
Max
Min
LM741 80 500 Typ 1.5 200 Max
Min 20
nA nA Units
Max
nA/˚C
LM741C 80 500 Typ 0.8 200 Max
Units nA V/mV
Typ
L
R ≤ 10 TkΩ TAMIN S A ≤ TAMAX, L ≥ 2≤kΩ, R ≤ 50Ω ± 20V VSS = 20V, VO = ± 15V Input Voltage Range
0.5 Max
Min
mV
LM741
Electrical Characteristics (Note 5) (Continued) 3.0 surface mount devices. Input Offset Current TA = 25˚C 30 ESD Tolerance (Note 400V LM741A 70 400V TAMIN8)≤ Conditions TA ≤ TAMAX Parameter
±
6.0
0.5 0.8
32
2.0 1.0
0.3 5.0
2.0 2.0
3.0
±15 12
25 10
4.0
± 16 ± 15 ± 10
15
± 12
± 13
RL ≥ 10 kΩ T 25˚C RAL = ≥2 kΩ TAMIN ≤ T ≤ TAMAX A = 25˚CA
10
TAMIN ≤ TA ≤ TAMAX
10
TAMIN ≤ TA ≤ TAMAX T RAS =≤ 25˚C 10 kΩ, VCM = ± 12V T ≤ T ≤ T = ± 12V RAMIN S ≤ 50Ω,A VCMAMAX TA = 25˚C, V = ± 20V AMIN ≤ TA ≤STAMAX,
7.5
3.0
30
25
70 35 0.5 40
30
80
80 1.0
95 6.0
0.210
86
96
mV µV/˚C V
± 15
2
MΩ mV MΩ mV V/mV V V/mV mV V V/mV
± 13 6.0
RL ≥ 10 kΩ
6.0
µA V/mV
± 12 ± 10
V mV
± 15
± 14 ±20 13
200
85 25
500
± 12 ± 10
± 14 ±20 13
200
25
300
V nA V nA mA nA/˚C mA
70
80 90
500
70
80 90
1.5
500 0.8
nA dB µA dB MΩ
0.3 2.0 0.3 2.0 Figure 8: LM741 data sheet (courtesy of National Semiconductor) T T ≤ , 5V 0.5 MΩ V = ±≤20V toTV = ±
Input Resistance Supply Voltage Rejection Ratio
Input Voltage Range
AMIN S
A
AMAX S
± 20V V RSS = ≤ 50Ω T RAS =≤ 25˚C 10 kΩ
77 ± 12
96 ± 13
±77 12
±96 13
dB V dB
Figure 8, Transient above,Response has a dataT sheet the LM741 IC from National Semiconductor. As you ≤ T Unity ≤ for T Gain V = 25˚C, AMIN A
A
AMAX
Time 0.25 0.8 0.3 0.3 µs can see, theRise input resistance (listed under “Input Resistance”, rIN = 2.0MΩ), rails (listed Overshoot 6.0 20 5 5 % Bandwidth (Note 6) T = 25˚C 0.437 1.5 under “Input Voltage Range”, ±13V), and the common-mode rejection ratioMHz (also called A
Slew Rate
TA = 25˚C, Unity Gain
0.3
0.7
0.5
0.5
V/µs
www.national.com Supply Current T = 25˚C 1.7 2.8 1.7 2.8 value mA of 90dB. the common mode voltage gain when not 2measured in decibels, CMRR) A
Power Consumption
TA = 25˚C
V = 20V 80 150 mW The values above can be used to generate an approximate model of the op-amp, just like V = ± 15V 50 85 50 85 mW S
±
S
± 20V LM741A the one drawn in Figure 2.V =The circuit can then be analyzed to find the input resistance S
TA = TAMIN
165
mW
vOU T T =T 135 gain of the circuit (A = mW (RIN P U T ), the output resistance (ROU T P U T ), and the vIN ). A
LM741
AMAX
VS = ± 15V
T =T 60 100 mW The next few pages will talk about finding the gain of the total circuit for a non-ideal opT =T 45 75 mW A
AMIN
A
AMAX
Note 2: “Absolute Maximum Ratings” indicate limits beyond which damage device may occur. Operatingcan Ratings indicate conditions for which device is non-ideal amp in a simple inverting configuration (theto the same process be applied tothe any functional, but do not guarantee specific performance limits.
op-amp circuit to find its gain).
3
12
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Operational Amplifier Basics
Non-ideal Operational Amplifiers Calculation of circuit input resistance, RIN P U T Let’s think back to the days of Thévenin equivalent circuits.. If you look at the figure below, what is the first thing that you notice that differs from other circuits that we have Thévenized before? The DEpendent voltage source. Traditional Thévenization rules explicitly state that we can’t simply “kill” dependent sources, so there must be another way of solving for the equivalent resistance of the circuit below. vd
vIN
IIN
vOUT
Rf
R1
rOUT µvd
vIN
+ +
rIN
–
vOUT
−µv −µ d
--
Figure 9: RIN P U T calculation for a non-ideal op-amp One way to solve for the equivalent resistance involves first expressing the equivalent resistance in terms of the total current through and the voltage across the circuit. In other words, by Ohm’s Law, we can express resistance as RIN P U T =
vIN IIN .
In the circuit above, we
have all it takes to find the equivalent resistance: IIN , which is simply the current through the input resistor (and the entire circuit), R1 ; and vIN , which is the voltage across the entire circuit. Based on this information, we can simply solve KCL equations at node vd and we can express total circuit resistance as a function of the input voltage and current:
KCL @ vd vd − vIN vd − (−µvd ) vd + + =0 R1 Rf + rOU T rIN
13
(16)
Operational Amplifier Basics
Non-ideal Op-Amps : Calculation of RIN P U T (cont’d) By looking at KCL equation (16), above, what is the only “lone” term that we see? The only thing that breaks a long chain of vd terms, is the vIN term. Since it’s a symbolic variable, we don’t know anything about it and we would rather eliminate it now than have it stick around and cause problems later on. It would be even better, if we can express vIN in terms of vd : vIN − vd = IIN R1 The equation above happens to be your first current in KCL equation (16). We rearrange the terms to get vIN by itself on one side of the equation (for easier substitution): (17)
vIN = vd + IIN R1 Now, we simply substitute equation (17) into (16): vd − (vd + IIN R1 ) vd + µvd vd + + =0 R1 Rf + rOU T rIN
From the above equation, we see that the vd terms go to zero, cancelling out the R1 term on both the top and the bottom of the first part of the KCL. We’re left with the equation below:
IIN =
vd vd (1 + µ) + Rf + rOU T rIN
Now, the vd term seems suspiciously repetitive.. as if it’s trying to tell us something (it is). To get ourselves a little closer to our goal of finding the equivalent resistance, we divide all of the terms of the equation by vd . We’ll also bring the (1 + µ) term on the top into the denominator: IIN 1+µ 1 = + vd (Rf + rOU T ) rIN 1 1 1 + = R +r OU T f RIN rIN
(18)
1+µ
14
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Operational Amplifier Basics
Non-ideal Op-Amps : Calculation of RIN P U T (cont’d, 2) What does equation (18) look like? Parallel resistor combinations (i.e.
1 Req
=
1 R1
+
1 R2
+ ... +
1 RN ).
To get the appropriate value of the equivalent resistance, we simply convert the above equation (18) into a more recognizable form: RIN =
Rf + rOU T 1+µ
k (RIN )
(19)
[!]
Remember that equation (19) above is not your final resistance value! Analyzing the circuit from node vd completely neglects R1 , which (as we see from the diagram) is one of the biasing resistors in this inverting configuration. We have to manually add the value of R1 to the equivalent resistance that we found from analyzing our KCL equations, as the last step: RIN P U T = R1 +
Rf + rOU T 1+µ
k (RIN )
(20)
A similar process can be applied to the non-inverting op-amp configuration. The value of µ will always be given to you in the non-ideal op-amp model (it will usually be somewhere in the tens of thousands and will be a unitless quantity).
15
Operational Amplifier Basics
Non-ideal Operational Amplifiers Calculation of circuit output resistance, ROU T P U T vd
vOUT
Rf
R1
rOUT
µvd +
rIN
–
−µv −µ d
Figure 10: ROU T P U T circuit with killed independent source The calculation of output resistance is the same nodal analysis problem with a twist.. we have to kill the independent voltage source (by shorting it), but since we cannot Thévenize circuits with dependent current sources, we have to leave the −µvd dependent source in the circuit (just like we had done in the previous problem). What should we do? vd
vOUT
Rf
R1
rOUT
µvd
rIN
+ –
0
ITEST −µv −µ d
vTEST
Figure 11: ROU T P U T calculation for a non-ideal op-amp 16
10/15/2008 v.2 (JJHS) Copyright © Akhan Almagambetov May be reprinted without permission.
Operational Amplifier Basics
Non-ideal Op-Amps : Calculation of ROU T P U T (cont’d) Once again, our trusty little ohmmeter comes into play. As we see from Figure 11, we simply add a test voltage source vT EST and a 0 or 1Ω resistor in series with the test source. This way, we can find the total current in the circuit. We immediately see that since we’ve added a negligible resistance to the source branch, the voltage at the vOU T node will be the same as vT EST . We therefore can analyze the KCL currents leaving the node vOU T , which will completely eliminate vT EST from our equations (when writing our KCL equations for the node, we can state that the current leaving is equal to IT EST , as shown below:
KCL @ vOU T vOU T + µvd vOU T + = IT EST rOU T R1 + Rf
(21)
Note that almost no current goes through the rIN resistor (if you remember from our discussions during lecture, the input resistance of the op-amp is VERY large), so we can safely assume that node vT EST is connected directly to ground through resistors Rf and R1 .. hence the second term in equation (21). We immediately see that our “lone” term here is vd so we’ll try to express it in terms of what we need to make this equation play nice and give us a value of the output resistance. In particular, we need to express vd in terms of vT EST . Once again, we rely upon the fact that the input resistance of the op-amp is HUGE , therefore having the R1 resistor in parallel with rIN would roughly give you the value of R1 .. this fact turns thevd node into R1 a simple voltage divider calculation (for simplicity’s sake we express R1 +Rf as α, which is approximately equal to α ≈ 0.1 for op-amps): vd =
R1 R1 + Rf
vT EST = αvT EST
(22)
Since we have vd expressed in terms of vT EST , we can now substitute equation (22) into our KCL equation (21) and get the following relationship: vT EST
1 + µα rOU T
+ vT EST
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1 R1 + Rf
= IT EST
Operational Amplifier Basics
Non-ideal Op-Amps : Calculation of ROU T P U T (cont’d, 2) We now divide through by vT EST and place the (1 + µα) in the denominator of the first term to get an equation that resembles parallel resistor combinations: 1 + µα 1 IT EST + = rOU T R1 + Rf vd 1
rOU T 1+µα
+
1 1 = R1 + Rf ROU T P U T
All we have left is to express the equation above in a more concise form and we’re done! ROU T P U T = (R1 + RF ) k
rOU T 1 + µα
(23)
Now that we have calculated both the input and output resistance of this inverting opamp configuration, we’ll move on to gain calculation!
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10/15/2008 v.2 (JJHS) Copyright © Akhan Almagambetov May be reprinted without permission.
Operational Amplifier Basics
Non-ideal Operational Amplifiers Calculation of circuit gain, A =
vIN
IIN
vOU T vIN
vd
R1
vOUT
Rf
+ rOUT
vIN
vOUT
µvd
rIN
+ –
µvd --
Figure 12: Gain calculation for a non-ideal op-amp Compared to the other characteristics (whew!), gain calculation is relatively simple to do.. you will have two nodal equations, which you would need to solve to get vOU T expressed in terms of vIN (I would suggest first expressing vOU T in terms of vd , and then simply substituting an expression for vIN .. otherwise you might get tripped up by the math).
KCL @ vd vd − vIN vd − 0 vd − vOU T + + =0 R1 rIN Rf
(24)
vOU T − vd vOU T − µvd + =0 Rf ROU T
(25)
KCL @ vOU T
We solve the above equations to get vOU T in terms of vd and vOU T . We then get rid of the remaining vd terms by substitution. 19
