Differential Amplifier circuit

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,786
LOL, are you kidding me ?? Please read post 30 again. Or find out what "differential output" means.
The one where you explain how to get AoE equations? Yes, but how that explains LTSpice results? Sorry, I'm not getting the point!

Differential output voltage is the difference between Q1 and Q2 collectors voltage!
 

Jony130

Joined Feb 17, 2009
5,598
Differential output voltage is the difference between Q1 and Q2 collectors voltage!
Exactly. And this is the answer to your question. Adm is a gain for Differential output. And the gain for single-ended output is two times smaller than Adm. We could "recover" this lost in gain if we use current mirror as a load.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,786
Exactly. And this is the answer to your question. Adm is a gain for Differential output. And the gain for single-ended output is two times smaller than Adm. We could "recover" this lost in gain if we use current mirror as a load.

Ok, I got it... The problem is that for the previous problem I used LTSpice RMS values to check Adm and Acm calcs... That's why I didn't noticed this detail before!

And how would that could be made? Make the current source as Load? Connect it between both Q1 and Q2 collectors?

edited;
Checking AoE...
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,786
I just built the Diff Amp and I think I got it working... Of course things gets a little deviated because of the BJT's built-in differences and also because of the inaccuracy of the resistors, but I think it's working...

As I don't have a proper power supply capable of delivering such low voltages as 100 mV or 200mV, I used an rpot to get a voltage divider as low as I could... I ended up getting an output voltage between 0V and the saturation of around 10 V. And yes, this was with DC voltages. I don't have a function generator!

http://i.imgur.com/m7ndevO.jpg
 
Last edited:

Bordodynov

Joined May 20, 2015
3,431
Look signal at "1". You will see an amount equal to half of the input signal. For the calculation can not be in the equivalent circuit grounding point "1".
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,786
Look signal at "1". You will see an amount equal to half of the input signal. For the calculation can not be in the equivalent circuit grounding point "1".
Sorry, I didn't understood what you mean. You mean to check the voltage in the circuit or you mean in the book or in my schematics?
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,786
Morning @Bordodynov

I'm not sure what those commands in blue means and if both those circuits are working "together" so that one of them is affecting somehow the other one because I'm not such an expert in LTSpice... But I would like to know what is happening there!

Thanks


Edited;

Are those commands forcing the voltages at those nodes?
 

Bordodynov

Joined May 20, 2015
3,431
Blue I have marked with comments.
Comments do not affect the calculation.
In these the comments I recorded what happened as a result.
See results on the graphics side.
Virtually grounding point "1" only for the differential signal. The magnitude of the differential ssignala is the voltage difference at the bases of transistors (Vsignal=(input+--input-)=0.5-(-0.5)=1.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,786
Ah ok, I got it!

Ok, one practical question now!

Where are these Diff amplifiers used? And how are they used? I mean, I have calculated Adm and Acm where the transistor bases where connected accordingly. But in a real/practical situation how are those bases connected? And where is the circuit that is needing this amplifier, connected?

And if, for a real/practical situation, the bases are connected just as we have done for the calcs, what means the CMRR? I ask this because we cannot evaluate those 2 gains simultaneously, right? Because the transistor's bases are connected differently for each gain!
 

Bordodynov

Joined May 20, 2015
3,431
All your calculations you do for unbalanced signal paths. Those. output signal from the collector of the left transistor. "Acm" you counted correctly, and "Adm" is not correct. I wanted to intervene, but ahead of me and said that "Acm" should be halved. In reality, the input signal is divided between the transistors of the differential stage equally. Ie each transistor gets half of the input signal.
If remove the output signal symmetrical (the differential signal with two collectors, that output signal doubles). But gain factor "Acm"=0. This occurs therefore that transistors identical (alike spice models is used). Real "Acm" is not a zero. The Scatter parameter transistor and resistor of collectors does this ("Acm" is not a zero).
 
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