Unequivalent Input Impedance in Differential OpAmp Circuit

[Clive Arthur]

On 10/03/2020 12:54, Rick C wrote:
<snip>

My simulations have shown no situation where the input impedances are not equal and in fact the value appears to be around 40 kohms on each input leg. I just can't explain that for a differential input signal.

Using your sim and setting V5 and V6 amplitudes to 2V so they're the
same amplitude opposite phase and the op-amp doesn't clip, and using
your .tran simulation.

For your 1kHz burst INA and INB are pretty much the same amplitude
because C101 and C100 are only about 160R at 1kHz.

The current through R41 is 300uA p-p and through R43 it's 100uA - the
input impedance seen at INA is 13.3kR and at INB it's 40kR.

But for your common mode signal, the currents through R41 and R43 are
the same. When INA and INB are equal, so are the input impedances at
40kR. If you increase the CM amplitude you can see in the 1kHz burst
that the in-phase common mode current is the same in both resistors,
while the opposite phase differential currents are 3:1.

--
Cheers
Clive

 

[Rick C]

On Tuesday, March 10, 2020 at 2:45:21 PM UTC-4, Clive Arthur wrote:

On 10/03/2020 12:54, Rick C wrote:
snip


My simulations have shown no situation where the input impedances are not equal and in fact the value appears to be around 40 kohms on each input leg. I just can't explain that for a differential input signal.


Using your sim and setting V5 and V6 amplitudes to 2V so they're the
same amplitude opposite phase and the op-amp doesn't clip, and using
your .tran simulation.

For your 1kHz burst INA and INB are pretty much the same amplitude
because C101 and C100 are only about 160R at 1kHz.

The current through R41 is 300uA p-p and through R43 it's 100uA - the
input impedance seen at INA is 13.3kR and at INB it's 40kR.

But for your common mode signal, the currents through R41 and R43 are
the same. When INA and INB are equal, so are the input impedances at
40kR. If you increase the CM amplitude you can see in the 1kHz burst
that the in-phase common mode current is the same in both resistors,
while the opposite phase differential currents are 3:1.

--
Cheers
Clive

Thanks for your reply. My signal sources are at 1 kHz (now) but I have also used AC analysis over a range of frequencies, typically 20, 1k and 20k as well as locating the -6 dB frequency.

I have tested the circuit in common mode, differential mode and each of the two inputs separately. In every case the frequency response is the same within experimental error (3 decimal places). Since the impedance of the two input capacitors must be the same over frequency regardless of mode of test, I have to conclude the input impedance of the circuit does not actually change with test mode.

Or... in another thread someone pointed out that the gain of the two inputs is not equal and so the result is the appearance of equal frequency response at the two inputs. He did not provide a quantitative description so I am not about to verify this. In fact the gain of the two inputs should be identical. That is the purpose of the resistors in the non-inverting input arranged as a voltage divider, to equalize the gain of the two inputs.

Another poster did a hand waving analysis using Z1 and Z2 without explaining what he was using them for, so I assume they are part of some standard way of analyzing amplifier circuits. He wrote an equation for gain G= and one for T= which I don't recall what it stood for. He then massaged the equations and produced a result, none of which made sense to me since there was inadequate info.

--

Rick C.

+-+ Get 1,000 miles of free Supercharging
+-+ Tesla referral code - https://ts.la/richard11209