I have a voltage divider feeding into an op-amp voltage follower buffer circuit. I'm trying to calculate the voltage level at the non-inverting input of the op-amp vs. the source voltage. The possible supply voltage is roughly in the range [0V, 5V].

Amplifier circuit

First I tried using a simple voltage divider, so:

$$ V_A = R_2 / (R_1 + R_2) * V_S $$

However, when I tried simulating this circuit in TINA-TI, the nodal voltages don't match those calculated.

I thought that was because the notion that there is no current flowing through the op-amp input wasn't truly non-zero, so I tried compensating for the op-amp input resistance and capacitance.

No amplifier

This time, the output voltage should be:

$$ V_A = (R_2 || R_3) / (R_1 + R_2||R_3) * V_S $$

While this equation does match TINA-TI's simulation of the no op-amp circuit, it still doesn't match the op-amp circuit.

What else am I missing?


further information: I'm using the LMH6612 op-amp.

Suppose I want to target an output voltage of 2.5V. According to my first equation the required source voltage is ~2.82V. By recursively updating my simulation I calculated that the actual required source voltage is ~3.47V. Using the second circuit the required source voltage is closer, but only marginally at ~2.87V.

  • \$\begingroup\$ How far off is the simulator from the analytical solution? Did you account for the op-amp offset current. At 0.01 uA, this would cause about 1 mV difference. \$\endgroup\$
    – The Photon
    Commented Aug 18, 2012 at 17:00
  • \$\begingroup\$ I updated my post with a test scenario. The difference is on the order of ~0.5V. \$\endgroup\$ Commented Aug 18, 2012 at 17:08
  • \$\begingroup\$ Is this happening at DC? \$\endgroup\$
    – s3c
    Commented Aug 18, 2012 at 18:06

1 Answer 1


After further reading into non-ideal op-amp behavior it looks like it's actually the op-amp input bias current that's the culprit.

R1||R2 ~= 100kohm, which created about a 0.6V drop (close enough to what the difference was). Adding a resistor to the inverting input equivalent to R1||R2 fixed the problem.

Reference: Practical Considerations: Bias current

The fixed circuit:

enter image description here

  • \$\begingroup\$ A schematic of where you added that resistor might be useful to other users in the future whom find your question with similar problems. \$\endgroup\$
    – Kortuk
    Commented Aug 18, 2012 at 18:42
  • \$\begingroup\$ good suggestion, done. \$\endgroup\$ Commented Aug 18, 2012 at 20:06

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