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What is the value of Vout if we know the value of Vin in this case?

It is a buffer amplifier, but it has two extra resistors so I think that the gain is smaller than 1, but how small?

(Vin-V3)/R1=0 <=> Vin-V3=0 => Vin=V2=V3

(Vout-V2)/R2=0 <=> Vout-V2=0 => Vout=Vin=V2=V3

  • \$\begingroup\$ Sounds like homework: What have you figured out so far? \$\endgroup\$ Commented Jan 5, 2013 at 12:28
  • \$\begingroup\$ Looks to me like a unity gain buffer, so Vout = Vin \$\endgroup\$
    – m.Alin
    Commented Jan 5, 2013 at 12:28
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    \$\begingroup\$ There are better ways to explain this circuit, but since you accepted a answer already and after just 16 minutes, doing so is not worth the trouble. \$\endgroup\$ Commented Jan 5, 2013 at 14:04

2 Answers 2


For an ideal op-amp, the following analysis holds. For a practical answer you will need to simulate your circuit using something like LTSpice.

(Input - V3) / R1 = 0, (op-amps have infinite input impedance)

(Output - V2) / R2 = 0, (op-amps have infinite input impedance)

V2 = V3 (op-amps work to mantain Zero potential difference between their inputs when configured for negative feedback is in this circuit)

I have explained the important aspects of an op-amp for solving this problem. I leave the substitution and result for you to find.

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    \$\begingroup\$ Bingo. Notice that would not be the case if R1 != R2. \$\endgroup\$
    – NickHalden
    Commented Jan 5, 2013 at 12:42
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    \$\begingroup\$ Ugh, right you are. I can't think straight right now. \$\endgroup\$
    – NickHalden
    Commented Jan 5, 2013 at 12:59
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    \$\begingroup\$ No, opamps do NOT "work to maintain 0 potential difference between their inputs". All they do is multiply the difference between the two inputs times a large value at put that on the output. In some circuits, feedback may be employed for the result you state, but getting that behavior is not inherent to the opamp itself. There are plenty of opamp circuits where your statement is not true. \$\endgroup\$ Commented Jan 5, 2013 at 14:03
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    \$\begingroup\$ @OlinLathrop This is a pretty standard way to teach op-amps with the requirement that it is in a negative feedback configuration. \$\endgroup\$
    – Kortuk
    Commented Jan 7, 2013 at 16:13
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    \$\begingroup\$ @Kortuk: There a lot of bad ways to teach electronics, and some of them are common. The problem is that too many will just follow the "rule", not think about what the circuit is really doing, and forget whatever constraints may have been mentioned along with the rule to make it valid. I have seen people get this wrong often. \$\endgroup\$ Commented Jan 7, 2013 at 17:43

With the ideal opamp, we cam assume:

  • infinite gain
  • infinite input impedance
  • zero offset voltage

So, whatever we input to the positive (or negative) input is multiplied by infinite gain. If we leave the opamp open loop, the output will just hit one of the rails (positive or negative depending on the polarity of the input signal)

However, if we feed some of the output back to the inverting input, we can use this to control the gain (and negative feedback has some other useful effects also)

With your example and an ideal opamp, it doesn't matter what value the resistors are, or whether they are equal. Since there is no current flowing through either of them, the result is always the same (a gain of 1).

With a real opamp, you have an input bias current (we will ignore the myriad of other non-ideal parameters and just focus on this one), so matching the impedance both inputs see is a good idea (unless there is already internal compensation, which some opamps have - in some cases matching the impedances can make things worse due to the input bias currents being unequal)

So for the example in your question, say we have an input impedance of 1MΩ (a very low value, but some opamps can have very low input impedances, make sure to check the datasheet), we use 10kΩ for the input resistor, but no resistor in the feedback loop. We will choose an input voltage of 1V.

We now get an input current of 1V / 1MΩ = 1uA.

So we now have a voltage drop across the input resistor of 1uA * 10kΩ = 10mV, which is present at the output (which will be 990mV) instead of 1V

If we want to prevent this, we need to match the voltage drop in the feedback loop to cancel out the offset caused by the input bias currents. So we use 10kΩ for the feedback resistor, it drops 10mV too so output is now 1V again.

Here's an example of matching the parallel combination of the feedback resistors when you have some gain:

Opamp Bias Current Compensation

This app note from Analog Devices is worth a read for more in depth discussion.

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    \$\begingroup\$ +1 for explaining why R2=R1 matters in the presence of input bias currents. @Cristi : read that bit carefully! \$\endgroup\$
    – user16324
    Commented Jan 5, 2013 at 22:59

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