Continue to Site

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.

Register Log in

Differential Amplifier with a BJT input stage gain

Status
Not open for further replies.
A

aryajur

Advanced Member level 3
Joined
Oct 23, 2004
Messages
793
Helped
124
Reputation
248
Reaction score
37
Trophy points
1,308
Location
San Jose, USA
Activity points
7,788
Differential Amp gain

I have a simple Bipolar Differential Amplifier with a BJT input stage. The equivalent circuit I have drawn and attached. When I do an AC simulation of the diff amp, giving a 500m and -500m signals at the inputs, I get the output which looks like the one I have attached.
The differential gain of the output looks around 0.4 which was about right for my circuit. But it shows a common mode gain of 5, why I am I getting such a huge common mode gain? The tail current is implemented using a MOS current source.
 

Re: Differential Amp gain

My be you have to insert common mode feedback when simulate the amplifier!!!
 

Re: Differential Amp gain

tyanata said:
My be you have to insert common mode feedback when simulate the amplifier!!!
Click to expand...
The load of the diff amp is made by a resistor so the output common mode is precisely defined by the current times the resistance. So this would not need a common mode feedback.
 

Re: Differential Amp gain

Hi

My reply is a bit late (Maybe you have already figured out the answer by now)

The outputs you are seeing are not corresponding to a common mode gain of 5.

It's just that you get larger swings at the output when the frequency of the input is lower as compared to a higher frequency.

As for the differential output it is wrong to predict the value just by looking at the amplitude plots. You need to plot (voutp-voutn)/(vinp-vinn). This will take care of phase as well.

For common mode change of output you need to see (voutp+voutn). And it must be zero provided there is no un-symmetry in your ckt.

P.S. In my post, all the voutp, voutn, vinn, and vinp are not just the mag. response. They include total info. (i.e. the phase also).
 

Re: Differential Amp gain

did u check the input common mode which bias the bjt in active mode
also 7µ amp is too low this may cause the gm of the transistor to be very small
 

Re: Differential Amp gain

Hello Advaita,
yes I did figure out the problem. By common mode gain I meant that from the transfer function you see a common mode signal which will be the average of the 2 outputs Which is 5. The input differential signal is 1 so that means the differential to common mode gain is 5. And the differential gain is as you said (Voutp-Voutn)/(Vinp-Vinn) which is 0.4, because Vinp-Vinn =1 . So yes the observations I gave in my 1st post are correct.
The common mode gain was actually coming because of a common mode feedback circuit in the next stage that was moving the common mode voltage of the nodes. This was the 1st stage of a folded cascode amp with common mode feedback.
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top