# Voltage through resistor connected to positive terminal of op amp

I have done quite a bit of research on this but I cannot seem to find an example or an answer at all. If I connect a resistor to the positive terminal of an op amp, since no current flows from the negative to the positive terminal, how does this affect the voltage of the two inputs of the op amp?

See the below figure:

Is the input voltage of the op amp just 0 volts, or does the 10 ohm resistor have an effect?

since no current flows from the negative to the positive terminal, how does this affect the voltage of the two inputs of the op amp?

In reality, all op-amps do have a small current flowing into or out from their input terminals. This parameter is listed in their datasheet and is called input bias current ($I_{B}$) .

If the 10 ohm resistor wasn't there, and V1 = 0, your output voltage wouldn't be exactly zero, and one of the main reasons for this is the effect of $I_{B}$.

Assuming the rest of the op amp is ideal (infinite open loop gain, zero input voltage offset, etc), we can calculate this effect:

Lets call the 500 ohm resistor $R_{F}$, the 50 ohm resistor $R_{IN-}$ and the 10 ohm resistor $R_{IN+}$.

$I_{B}$ would have to come through $R_{F}$, so $V_{o}=I_{B}R_{F}$. The larger your feedback resistor, the bigger the deviation from the expected output voltage.

But if you add $R_{IN-}$, it will drop a voltage of $V_{IN+} = -I_{B}R_{IN-}$.

The op amp will adjust Vo so that $V_{IN-} = V_{IN+}$, so now $R_{IN-}$ can also contribute with some current. After solving this you'll find that if you choose $R_{IN+} = R_{IN-}||R_{F}$, it will properly compensate the effects of the input bias currents, and your output Vo will be as expected.

As a last remark, I'd like to mention that with this new knowledge you may want to always add input bias current compensation, but this is not always a good thing, because of the effects of thermal noise of resistors. A classic example is when amplifying a small ac signal. It is better to live with a DC offset (that you can later remove or compensate for) than to introduce the noise of the additional resistor right at the input of the op-amp, which will be amplified by the gain of the configuration.

1. The resistors values are little bit small for typical OpAmp. You have to work in the kOhm and MOhm range.

2. The resistor on the non-inverting input is generally not needed if the opamp is close to ideal. Using this resistor compensates the input current of the real OpAmp. Its value should be equal to the resistors connected to the inverting input in parallel - 50||500 = 45ohm. So, in this schematic, the resistor has too small value.

Anyway, for the modern JFET or MOSFET input OpAmp, this resistor can be safely omitted (set to 0).

As long as this resistor works with second order effects, you simply can ignore it on simplified analysis of the circuit.