Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

simulating opamp virtual ground

yefj

Advanced Member level 2
Joined
Sep 12, 2019
Messages
559
Helped
0
Reputation
0
Reaction score
2
Trophy points
18
Activity points
3,007
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.
1624216501890.png


1624217638971.png
 

Attachments

  • 1624216416610.png
    1624216416610.png
    106.2 KB · Views: 36

Dominik Przyborowski

Advanced Member level 4
Joined
Jun 6, 2013
Messages
1,075
Helped
481
Reputation
964
Reaction score
456
Trophy points
1,363
Location
Norway
Activity points
8,054
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

Advanced Member level 2
Joined
Sep 12, 2019
Messages
559
Helped
0
Reputation
0
Reaction score
2
Trophy points
18
Activity points
3,007
Yes i will add PMOS load connected by gates together.
Thanks.
--- Updated ---

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.

1624219648992.png
 
Last edited:

BradtheRad

Super Moderator
Staff member
Joined
Apr 1, 2011
Messages
14,303
Helped
2,823
Reputation
5,654
Reaction score
2,774
Trophy points
1,393
Location
Minneapolis, Minnesota, USA
Activity points
106,688
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.
--- Updated ---

Basic example of differential detector. 2 NPN, with tail resistor.
long-tail pair 2 NPN (tail resistor) 5V supply sine-n-sawtooth create SPWM.png
 
Last edited:

FvM

Super Moderator
Staff member
Joined
Jan 22, 2008
Messages
49,412
Helped
14,439
Reputation
29,142
Reaction score
13,236
Trophy points
1,393
Location
Bochum, Germany
Activity points
284,363
To make a virtual ground, the amplifier must have high differential gain and sufficient common mode rejection. Your simple amplifier has neither.
 

crutschow

Advanced Member level 5
Joined
Feb 22, 2012
Messages
3,981
Helped
939
Reputation
1,876
Reaction score
945
Trophy points
1,393
Location
Colorado USA Zulu -7
Activity points
22,524
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
Joined
May 12, 2019
Messages
11
Helped
4
Reputation
8
Reaction score
1
Trophy points
3
Activity points
140
You connected two common source amplifiers in parallel.
introduce a current tail providing both branches to create differentially
 

danadakk

Advanced Member level 4
Joined
Mar 26, 2018
Messages
1,458
Helped
257
Reputation
530
Reaction score
314
Trophy points
83
Activity points
6,681
Basic virtual ground

1624316428571.png


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 -

1624316655613.png


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 -

1624316930705.png


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.
 

LaTeX Commands Quick-Menu:

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Top