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How OpenLoop Gain

Ideal Gain  Kcl' = 1/B = R1+R2/R1 = 10.000  
Actual Gain  Kcl = A/(1+AB) = 9.999  
Ideal Output  vo' = vin ·
Kcl' = vin · 1/B = vin · (R1+R2)/R1 = 10.000 

Actual Output  vo = vin · A/(1+AB) = 9.999V  
Output Error  vo_err = vo'  vo = 10.000V  9.999V = 0.001V 
Is there an easier way to predict this error? Simply use the Kerr equation and the ideal gain Kcl'..
vo_err = vin · Kcl' · Kcl_err
= vin · Kcl' · 1/(AB)
= 1V · 10 · 10/100000
= 0.001V
The 1mV error has spilled over our error budget of 0.305uV. However, suppose in the best case, your system calibrates out this 1mV initial error at the factory. But we're still not out of the weeds yet  the spec needs to hold even a 25 deg temperature change!
GAIN DRIFT ERROR
The bugger here is that the op amp you've chosen has an openloop gain A
that changes from 100k to 200k over 25C! What effect does that have on the
output error? Lets find the output error with A=200,000.
vo_err = vin · Kcl' · Kcl_err
= vin · (R1+R2)/R1 · 1/(AB)
= 1V · 10 · 10/200000
= 0.0005V
Bad news! The upshot is that even though we calibrated the 1mV error out when A=100k, the error changed by 0.5mV as the gain increased to 200k with a 25C temperature change. The bottom line is that the system will not hold the spec of 0.305mV!
So what can you do? The solution lies in choosing an op amp with a larger openloop gain A. This reduces the initial error and, most likely, the change in error as A drifts with temperature.
SIMULATION
Let's simulate the noninverting amp with
ideal resistor values R1 = 1000 and R2 = 9000 for a gain of +10.. With the
input signal at VS=1V, we expect to see an output voltage close to
V(3) = VS * (R1+R2)/R1 ≈ 10V.
Inside the simplified op amp model, the openloop gain is set by the gain of EGAIN. Initially, set EGAIN to 100k. The ideal output at V(5) is generated a simple voltage controlled source EAMP1 set to an exact gain of 10.0
CIRCUIT INSIGHT Run a Transient Simulation and plot the output V(3). Place a cursor on the trace to get a precise reading of the output voltage. Is should be very close to, but not quite +10.0V. To see the ideal output, plot V(5). Finally, to find the output error, open a new window and plot V(3)V(5). The error should be as calculated as above.
Now change EGAIN to 200k to simulate the change in openloop gain over temperature. What is the new error? Is it what you expected?
HANDSON DESIGN How can we save this design? To lower the output error, you need to choose an op amp with a larger openloop gain. Choose a new value for A that is 5x or 10x the original. Set EGAIN to this value. Rerun the simulation with your new value. Then double it to simulate a change in A versus a temperature change. Does the change in output fall below your design target of 0.305mV? If not, choose another op amp with an even higher openloop gain.
INVERTING AMPLIFIER
The gain equation for the inverting amplifier takes a little more
effort to calculate.
But it's very similar to the noninverting amplifier.
Kcl = vo / vin = R2/R1· AB /(1+AB)
Again, the power of feedback control systems occur when A*B >> 1, making the ideal closed loop gain
Kcl' = R2 / R1
You can predict the output error as follows.
vo_err = vin · Kcl' · Kcl_err
= vin · R2/R1 · 1/(AB)
Copy the SPICE file for the noninverting amp to a new file. Then create a simulation for the inverting amplifier.
NEXT UP
What about the effect of openloop gain
on the AC accuracy of your amplifier? The next topic builds on this topic to
see the accuracy of your circuit when passing a 20 kHz sine wave, for
example.
SPICE FILE
Download the file or copy this netlist into a text file with the *.cir extension.
OP_AOL_DC1.CIR * INPUT VS 1 0 ac 1 PWL(0US 0V 0.01US 1V) * * NONINVERTING AMP R1 2 0 1K R2 3 2 9k XOP1 1 2 3 op_001 * * IDEAL NONINVERTING GAIN EAMP1 5 0 1 0 10 * * BASIC OP AMP MODEL * In+ In Vout .SUBCKT op_001 1 2 3 RIN 1 2 1e12 * OPEN LOOP GAIN EGAIN 3 0 1 2 100000 * .ENDS * * ANALYSIS ************************************* .TRAN 1us 100us .PROBE .END
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