How do Differential amplifiers work?

Thread Starter

wouter368

Joined Sep 22, 2022
20
Hello,

For a school project I need to know how a differential amplifier works. I watched several videos on it and tried to understand them, but there's a few things that are just very unclear to me. For reference, I'm going to talk about the circuit below:

1667607942388.png

The problems that I encounter are the following:

1. The differential amplification formula
1667608160878.png
The formula above, I don't understand what they mean by Vout. Do they refer to Vout in the reference image I showed above? Because in another nearly identical formula, it's stated another way:
1667608218052.png
These are basically the exact same formulas, except Vout has been replaced for (Vout-Vout'). How does this work? What is the actual output voltage on a differential amplifier? When I analyze circuits with a differential amplifier embedded into it, I only see one output. However, on the reference image there's two outputs. How are these two outputs connected so that they form one output? And how are they connected so that the output voltage is equal to the difference between Vout and V'out?

2. The 'normal mode'

So, for the normal mode the formula is as following:
1667608542429.png
or
1667608552546.png
First of all, how are these two formulas connected? How is Vout in the first formula equal to Vout+Vout' in the second formula?
Then the second question I have is; how do you make Acm a small number? Because I read that to make a differential amplifier, Acm should be very small, and ideally 0. How could you make Vout very small, while keeping Vin and Vin' high?

3. Inverting and non-inverting input.

I read that differential amplifiers have inverting and non-inverting inputs, but I don't really understand how one input can be inverted in the reference image. It would make a lot of sense if one input was inverted, since it would probably solve a lot of the troubles I'm currently having but I can't seem to find any explanation that uses formulas/math to explain how one input (for example Vin2) can be inverted.

Sorry for all this rambling, but I have been struggling to find any explanation to make differential amplifiers more clear to me. Thanks a lot in advance for helping me out!
 

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Papabravo

Joined Feb 24, 2006
21,258
A differential amplifier definitely takes the difference of the two inputs. The output can be either differential or single ended. The single ended output can be taken from either collector. the following simulation may clear things up.
1667610043145.png

It is important to recognize that both the input and the output have a common mode component and a differential mode component. They also have separate gains. There is a very small common mode gain and a rather large differential mode gain. The ratio, often expressed in dB, between the two gains is the Common Mode Rejection Ratio.
 

WBahn

Joined Mar 31, 2012
30,235
So let's think about the circuit you are starting with.

1667615674031.png

Would you agree that if Vin and Vin' are equal, that Vout and Vout' will be equal? (We are going to assume that everything is perfectly matched).

Now, what happens if Vin increases by a small amount (while Vin' is held fixed)? Does Vout go up or down? Does Vout' go up or down?

So what is the output, Vo, of our amplifier?

We actually have four obvious choices:

Vo = Vout
Vo = Vout'
Vo = Vout -Vout'
Vo = Vout' - Vout

The first two are single-ended outputs. We just pick one of them to be the output and forget that the other exists. Out output will have a significant DC offset (a non-zero voltage even when there is zero input signal), but we can deal with that.

The second two are differential outputs.

In both cases, our choice of polarity is arbitrary. We simply pick one.

Now, what about the input, Vi?

We have essentially the same four choices. We can fix one of the inputs (i.e., either Vin or Vin') at a constant voltage and use the other for our input signal. That would make it a single-ended input. Or we can apply a differential signal to both Vin and Vin'.

Given our decision as to how we define Vo, whichever input causes Vo to increase (become more positive or less negative), that is our non-inverting input. The other one is the inverting input. That's really all these labels mean -- increasing the signal at the non-inverting input causes an increase in whatever we have chosen to be our output voltage, while the reverse is true for the inverting input.

Another thing that might help you is to realize that ANY two voltages can be written in terms of a differential mode and common mode component:

Given Va and Vb

Vcm = (Va + Vb)/2 // The common mode component is just the average.

Vdm = (Va - Vb) // The differential mode component is just the difference (but we need to decide which signal gets subtracted from the other and then be consistent with that choice).

You should be able to verify that

Va = Vcm + (Vdm/2)
Vb = Vcm - (Vdm/2)
 

Thread Starter

wouter368

Joined Sep 22, 2022
20
A differential amplifier definitely takes the difference of the two inputs. The output can be either differential or single ended. The single ended output can be taken from either collector. the following simulation may clear things up.
View attachment 279957

It is important to recognize that both the input and the output have a common mode component and a differential mode component. They also have separate gains. There is a very small common mode gain and a rather large differential mode gain. The ratio, often expressed in dB, between the two gains is the Common Mode Rejection Ratio.
Hey, thanks for helping me. I was wondering how it works that on the third terminal (with Vout1 en Vout2 in green and blue) Vout2 is shifted in phase relative to Vout1. It seems as if either Vin1 or Vin2 has been inverted and then amplified, but how can you calculate this?
 

Papabravo

Joined Feb 24, 2006
21,258
vout1 and vout2 are shifted in phase and have different offsets. This leads to different results, depending on the order of subtraction, as shown in the red and gray traces. Q3 has a constant current going through it and the sum of the emitter currents must equal the current in Q3.

Given the two input sources you should be able to compute the common mode input voltage and the differential input voltage components. That makes it easier to see what is going on
 
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