# Why does an diff-amp being fed by two separate op-amps with low input resistances distort the signal on the inverting input?

I'm using the following circuit. In1 and In2 come from two separate amplifiers (Max9643), both of which are fed by a square pulse but have their inverting and non-inverting terminals criss-crossed with each other. The first stage amplifiers work well separately. Now coming to the second stage (i.e. the circuit shown below). simulate this circuit – Schematic created using CircuitLab

When I use R1=R2=20k and R3=R4=100k, they work properly and In1 and In2 are shown below: (This's only a spice simulation and practical output varies, especially with respect to limiting high frequency).

Now if I use R1=R2=20 and R3=R4=100, Vin2 somehow gets affected. I realize I'm doing something wrong with the resistor values but what exactly is happening? On a practical circuit, more than voltage reduction, the edges seem to become contaminated and are jagged. I am using an op-amp with 10s of uA bias currents so the high value resistors might not be suitable for me. But if it's an input current/voltage issue why does this occur only at the inverting terminal? I've been beating myself up over this since I think this must be a basic issue, but I'm not able to put my finger on the exact cause. I'm planning to check this with a fully differential amplifier IC instead of the op-amp but I would really like to know what I'm doing wrong.

• Even at these low signal levels, your op amps do not have the current capacity to handle your low resistances. Try multiplying your resistors by 10, and see what happens. – WhatRoughBeast Feb 11 '16 at 1:42
• The output of your opamp in this schematic and the output of the opamp driving In2 are connected together through R2 & R3 in series. With those low resistor values they'll be fighting each other quite hard and you'll certainly start seeing effects like that (possibly due to the opamps' internal output impedances). – brhans Feb 11 '16 at 2:12

## 1 Answer

Realize that the input impedances are quite different on the two inputs. On the non-inverting input it will be 120 ohms to ground, and on the inverting input only 1/6 of that.

The outputs of your max9643 chips can only supply a few mA so the inverting input will load down the source excessively with <100mV input.

• Thank you, that helps. Can you tell a bit more about how you arrived at 1/6 of the resistance on the inverting input? I'm trying to calculate the minimum resistance I'll need to put without loading the Max9643 beyond 1mA. This's for high-frequency applications so I can't put a big feedback resistor without it significantly affecting the frequency response, so I am trying to keep it at minimum+margin. – manu Feb 11 '16 at 21:35
• Well, I divided 120 by 20.. and got 6. The input Z to the inverting input is only 20 ohms because there is a virtual ground at the inverting terminal of the amplifier. Except it's not really a ground, it's at 20/120 of the voltage on the non-inverting input to the amplifier. So the difference of the two MAX9643 outputs cannot exceed some tens of mV (maybe 50-60mV). Increase the resistor values considerably and this problem should go away. – Spehro Pefhany Feb 11 '16 at 21:42