Vi1 = Vi2 = 3V dc
It = 2mA

For the diff amp above I have to perform a DC analysis and check if Q3 and Q4 are in active mode, so I have two equations

KVL at input
3 = Ib*R3 + Vbe + Ve
Ve = 3 - Ib*R3 - Vbe

KVL at supply
12 = Ic*R1 + Vce + Ve

Ie = It/2 = 2/2 = 1mA

If Ic approx = Ie then Ic = 1mA

The Beta value wasn't given so how do I find Ib ?

 

The tail is a 2ma ccs. So each leg gets 1ma. Base is at 3v, emitters at 3 minus 0.7v, and collector at 12v minus 4.7k x 1ma.

Pretty clear and shut.

As to beta, you typically assume that it is sufficiently large here.

 

The tail is a 2ma ccs. So each leg gets 1ma. Base is at 3v, emitters at 3 minus 0.7v, and collector at 12v minus 4.7k x 1ma.

Pretty clear and shut.

As to beta, you typically assume that it is sufficiently large here.

Yes thanks, that's what I thought initially. The resistance R6 (200 ohm), is that the emitter resistor for both the transistors ?

 

Hi,

Just a quick note...

R6 is a pot and with a completely symmetrical circuit that would be adjusted exactly half way, which puts 100 ohms in series with each emitter, but they would be considered separate resistors too.

Find the Beta from the data sheet for whatever transistors you are using.

Watch out that the collector voltage calculation is reasonable.

 

R6 allows you to counter the imbalance between the two legs, and also to. Provide some degenerative power (DC feedback).

 

R6 allows you to counter the imbalance between the two legs, and also to. Provide some degenerative power (DC feedback).

R6 also provides AC negative feedback.

 

Ok, for the second part I have to determine the differential and common mode voltage gains. So is that R6 resistor just a separate 100 ohm resistor for each emitter of the bjt's ?

 

Ok, for the second part I have to determine the differential and common mode voltage gains. So is that R6 resistor just a separate 100 ohm resistor for each emitter of the bjt's ?

That's the equivalent circuit, yes.

 

Will Rid (input differential resistance) be

Rid = Rib(input resistance from base) + R3 ?

Where Rib = ie(re + R6/2)/ib

 

What's C2 for please

 

What's C2 for please

In this case (AC signal amplifier) you can treat the capacitor as a frequency-dependent resistors. Called the capacitive reactance Xc to distinguish it from the resistor resistance because the resistance of resistors does not change with frequency. But the capacitive reactance Xc will is change with frequency.
Xc = 1/(2Π*F*C)
The large the frecuancy the smaller the capacitive reactance Xc. The reactance is large at low frequencies and small at high frequencies.

So, now you can "mentally" replace the C2 capacitor with his "equivalent" Xc reactance. And examine its influence on the diff-amp gain at the frequency when the Xc is in the range of 4.7k. Because at low-frequency Xc is very high so we can neglect it.