simulating opamp virtual ground

yefj Hello, I have build the opamp as shown bellow, the right input is 180 phase to the other AC input.
negative feedback is supposed to do virtual ground,which means that if i connect the output to the inverting input as shown bellow i get totay different DC values on both inputs of the opamp.
Where did i go wrong?
Thanks.  Dominik Przyborowski It is not opamp, nor pseudo differential Amplifier. Both branches are completely independent, so there is no reason to have gate voltage of one mosfet following gate voltage of other one.

yefj Thanks.
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UPDATE:
I have updated the schemtics as shown bellow,but stiil the DC input voltages differ.
Where did i go wrong inplementing virtual ground?
Thanks. Last edited:

Super Moderator
Staff member A differential detector commonly is a long-tail pair. The tail can be a resistor, or transistor, or current source. Attach it to a negative supply voltage, so that your input signals can be zero volts (ground) and still cause the amplifier to operate. Then your output voltage is referenced to 0V ground.

Adjust operation so that the left half of the detector influences the right half (and vice versa). When current increases in the left half, it causes current decrease in the right half. Etc.
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Basic example of differential detector. 2 NPN, with tail resistor. Last edited:

FvM

Super Moderator
Staff member To make a virtual ground, the amplifier must have high differential gain and sufficient common mode rejection. Your simple amplifier has neither.

crutschow As noted, you can't have a differential amp if the transistor sources are connected to ground.
The sources need to be connected to a common, high impedance current source.

guntherleet

Newbie level 6 You connected two common source amplifiers in parallel.
introduce a current tail providing both branches to create differentially Basic virtual ground Above we have a Aol of 100K, a feedback of 10%, so overall G = 10. But notice the Vir Gnd V,
its 99 uV. In other words at the diff input the two inputs are essentiually same V due to the
feedback.

Now lets reduce the Aol (Slider Gain) to 100 - Here we see large error in Vout, and we are losing the Vir Gnd, its jumped to 90 mV, almost
100X what it was when we had large Aol.

The importance of Vir Gound, one example, is the classic summer - Basically when Virtual Ground is solid then the current thru each input is unaffected by
the other input currents, eg I = V1/R = V2/R = V3/R in this case.

So lots of Slider G (forward path gain) here makes the simple summer work well. Or higher
Feedback Factor (which applies more error correction when closed loop G
is lowered).

Regards, Dana. Commands Quick-Menu: 