my future fellow Electrical engineers.

I can't figure out how one gets -500(b) as the Open-loop gain.

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Using node analysis:

\$i_1 = \frac{v_- - v_{in}}{100 Ohms}\$

\$i_2 = \frac{v_- - v_{out}}{100k Ohms}\$

and then substituting in \$v_- = v_+ - \frac{v_{out}}{A}\$. As the open loop gain of an op amp is not infinite. However, using this method, I got a wildly different answer (assuming I solved for \$\frac{v_{out}}{v_{in}}\$ correctly).

  • \$\begingroup\$ \$ v_+=0\$, \$ i_1+i_2=0\$, and \$ V_{out}>>\frac{V_{out}}{A}\$ gives -500. \$\endgroup\$ – Chu Oct 6 '19 at 21:20
  • \$\begingroup\$ Sorry, but can you please elaborate further? \$\endgroup\$ – Throw Away Oct 6 '19 at 21:42

The question is not asking for the open loop gain. The question is telling you the open loop gain is 1000. You are supposed to calculate the closed loop gain, given that the open loop gain is 1000.

Let's assume Vout is 1V. Then V- must be -0.001V (because of open-loop gain). Then the current through the 100k will be 1.001V/100k = 10.01uA. The same current flows through the 100 Ohm resistor. So Vin is -0.001V - 10.01uA * 100 Ohms = -0.002001V.

So the closed loop gain is 1 /(-0.002001), which is about -500.

In an ideal op-amp, the gain for this inverting configuration would be Gideal = -R2/R1 = -100k/100 = -1000.

There is also a general formula for op-amps when open-loop gain is not infinite. The formula is:

Gain, G = Gideal * ( A / (A + 1 + R2/R1))

Where R2 is the feedback resistor, R1 is the other resistor, A is the open-loop gain. This also holds true for non-inverting op-amps.

If that formula looks familiar, maybe you were supposed to memorize it for this class.

  • \$\begingroup\$ The last formula - given by mkeith - is very helpful ......it would be a good exercise for the questioner to verify it - starting with H. Blacks general feedback formula. \$\endgroup\$ – LvW Oct 7 '19 at 8:32

You have all the necessary equations:

\$i_1+i_2 = \frac{v_--v_{in}}{100}+\frac{v_--V_{out}}{10^5}=0\$

Multiply by \$10^5\$

\$ 10^3( v_--v_{in})+(v_--v_{out})=0\$

substitute: \$ v_-=-\frac{v_{out}}{A}=-\frac{v_{out}}{10^3}\$

\$ 10^3( -\frac{v_{out}}{10^3}-v_{in})+(-\frac{v_{out}}{10^3}-v_{out})=0\$

but \$ v_{out}>>\frac{v_{out}}{10^3} \$, hence:

\$ (-v_{out}-10^3\:v_{in})-v_{out}=0\$

\$\therefore\: 2\:v_{out}=-10^3\: v_{in}\$

giving: \$ \frac{v_{out}}{v_{in}} =-500\$


IMHO this is not the type of question where a "long" mathematical analysis is intended.

You can use the following approach to calculate the closed loop gain:

  • This is an inverting amplifier, with ideal opamp inputs (no current).
  • We do not care about voltage limits, we do a linear analysis.
  • The output voltage is A times (1000 times) the difference on the inputs.

  • We imagine that the minus input is -1V (this is an educated choice which I made because the + input is at GND and I want a simple positive input differential).

  • The +/- difference is 1 and the output is at A*1 or 1000V.
  • So the voltage across the 100k resistor is 1001V (1000V - (-1V) ).
  • The voltage across the 100 Ohm resistor is one thousand of that, so 1.001V (because the current flowing in both resistors is the same).
  • The voltage at \$V_{in}\$ is therefore (-1.001V-1V)=-2.001V
  • \$V_{out}/V_{in} = 1000V/-2.001V = {almost} -500\$
  • Conclusion: the gain is close to -500, answer (b).

The trick is that I "force" the voltage on the opamp input, and I calculate the other values from there.

To be safe, you should verify the result by checking that -2.001V on the input and 1000V on the output are correct by finding the voltage on the negative input and make sure that it is -1V. If not, you made a mistake in the analysis.


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