[SOLVED] Differential Amplifier Input Voltage

[calsonic2009]

I have a problem dealing with differential amplifier input

According to this circuit we have to applied the same input(+-) voltage at 2 side of the circuit (vd/2) and (-vd/2) in order to get voltage gain.

**broken link removed**

- My question is why the input voltage have to be the same and (+-).

- Second question is if comparing the first circuit to the second circuit below, Is it the same circuit?

- If it's the same why can we applied any different voltage, but the on the first circuit we have to applied only the same voltage(+-).

Thank you

**broken link removed**

**broken link removed**

 

[FvM]

There's only 1 valid attachement in your post.

To analyze differential behaviour, you distinguish between common mode and differential input signals to simplify the calculation. The circuit shows a differential signal. Real operation usually involves both kinds of signals superimposed.

 

[calsonic2009]

Picture of the second circuit

**broken link removed**

**broken link removed**

 

[FvM]

Differential and common mode signal, as said. I guess, that the diagram title already clarifies about their specific purpose.

 

[LvW]

There's only 1 valid attachement in your post.

To analyze differential behaviour, you distinguish between common mode and differential input signals to simplify the calculation. The circuit shows a differential signal. Real operation usually involves both kinds of signals superimposed.

Hi calsonic,

perhaps the following explanation can help to understand FvM`s reply:

*Assume any two different input voltages Vx and Vy (same sign or not) ,
* You always can split these voltages in two new parts Vcm=(Vx+Vy)/2 and Vd/2=(Vx-Vy)/2 ,
* By superimposing both parts you will get again Vx and Vy: Vx=Vcm+Vd/2 and Vy=Vcm-Vd/2 ,
* Now you can amplify (separat for common mode and diff. signals) with the respective gain values Gcm and Gd:
Common mode Vcm*Gcm=Vout,cm
Diff. mode Vd*Gd=Vout,d
* Then, both output voltages are added (normally, the common mode signal is very small).
_______________
This procedure shows that there is a common mode voltage also for the case Vx=0 (or Vy=0).

 

[calsonic2009]

Crystal clear now, thanks to everyone!