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: Non-distored (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. Distorted

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.

  • \$\begingroup\$ 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. \$\endgroup\$ Feb 11, 2016 at 1:42
  • \$\begingroup\$ 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). \$\endgroup\$
    – brhans
    Feb 11, 2016 at 2:12

1 Answer 1


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.

  • \$\begingroup\$ 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. \$\endgroup\$
    – manu
    Feb 11, 2016 at 21:35
  • \$\begingroup\$ 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. \$\endgroup\$ Feb 11, 2016 at 21:42

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