# Opamp input resistance simulate this circuit – Schematic created using CircuitLab

In analysing an ideal op-amp circuit I'm asked to state the input resistance seen by an input voltage.

Some of this may be irrelevant but a quick summary of the circuit:

Two unknown voltages, VinA and VinB are connected to the inverting and non-inverting inputs, respectively. Both have a 10k resistor between Vin and the input. Vout feeds back to the inverting input, through another 10k resistor. VinB, after the first 10k resistor, is connected in parallel to the non-inverting input (as stated) and another 10k resistor which continues to ground.

So, the question is what is the input resistance seen by VinB.

Is it

a) the two resistors in series (since no current flows to the input), i.e. 20k?

b) the resistance used in a voltage divider formula, i.e. 10k/10k+10k = 0.5?

c) something else I've overlooked?

• There is a "schematic" button on the question editor. Yes, the schematic is required. May 15 '18 at 16:11
• plz upvote if you like my answer May 15 '18 at 16:35
• I did, but it won't show publicly since my reputation is <15! May 15 '18 at 16:36

It is (a) i.e. 10k + 10k = 20k.

No current flows into the Op amp terminal. So the only path in which the current can flow from VinB is through R1 and R4 to ground.

Since the opamp has very little current flowing into it, it can in most cases be ignored. For an ideal opamp (you need to erase the TL081 the input bias current is zero. For a real op amp it is uA to fA and for most cases can be ignored. This is why we use op amps, to buffer impedance.

The supply would then see the load as 20k for the VinB.

For VinA there are to cases to analyze, but the supply will see 20k. If you want to know how much max current will flow, then consider when vout is 0V

Side note:However for the VinA the impedance will not be buffered and the current will change with Vout, which is what happens with an inverting configuration. A non inverting configuration would not see the effects of VinA

• I understand that it might not have any practical value but I'm asked to "State the input resistance seen by VinB". May 15 '18 at 16:27
• Ah so my first instinct, it just sees the two resistors in series? May 15 '18 at 16:29
• For an ideal opamp (you need to erase the TL081 the input bias current is zero. For a real op amp it is uA to fA and for most cases can be ignored. This is why we use op amps, to buffer impedance May 15 '18 at 16:30
• Okay; but assuming an ideal opamp, the answer to my question is that VinB sees an input resistance of 20k? May 15 '18 at 16:31
• Yes, if you look at the currents through the node, there is no current flowing into or out of V- or V+ May 15 '18 at 16:33