3. The figure below shows a differential amplifier, biased using a 1 mA current

Question

3. The figure below shows a differential amplifier, biased using a 1 mA current source (the current source has a design similar to that in problem 1). The current source has an output resistance, REE, VA 100 V, and B S 100. 10V 10 KS2 10 KS2 Rc Rc VCI OA O VC2 Vbl Vb2 1 mA. EE 10V (a) Find REE. (b) What is the differential gain, Ad? (c) What is the common mode gain, Acm (d) Suppose two sinusoidal voltages of the same frequency (10 KHz) and in phase are applied at the bases of Q1 and Q2. The peak amplitude of vbl is 11 mV and that of vb2 is 5 mV. Write vbl and vb2 as the sum of a common-mode signal (viem and a difference mode signal.(vid). (e) Find the voltages (AC DC) at the collectors of Q1 and Q2. Plot these voltages (you may ignore the outputs due to the common mode signals, vicm). (f) Find voD Vc2 Vc Plot voD

Explanation / Answer

As Problem 1 is not given so I am not able to calculate external resistance of current source so I will use value of Ree as variable in my answers please read the answer till end

A) As I dont know the design of the current mirror I can't calculate value of Ree but going by the natural way of doing it maybe the circuit used as a current source is a basic BJT current mirror in that case Ree= (Va+Vce)/Ic in this case Ic would be 1ma and Va = 100 so basically Vce + Va = 100 V approx

so Ree = 100V/1ma =100 kilo-ohm counter check it and use this value for later parts however as I am not sure about problem 1 i will use Ree as a variable for later parts

B) differential gain = Ad = IcRc/Vt where Ic = current in Rc = half of 1ma =0.5 ma and value of Vt (Thermal Voltage)= 25mv (generally assumed value)

so Ad= 0.5ma *10K-ohm/25mv = 200

C) Common mode gain = -(IcRc/Vt) / [1 +(1+?)IcRee/2Vt?] = -0.198 (assuming Ree= 100kilo-ohms if Ree is different use that value)

D) vicm = (vb1 +vb2 )/2

vid = (vb1 - vb2 )

E) For this part u will have to do small signal analysis which is quite lengthy so please add it as another question and i will answer it

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