Finding Quiescent Collector Current IC in BJT Differential Amplifier for gm Calculation

I want to perform the small signal AC analysis of the circuit. To calculate gm. I need IC ,ie the DC quiescent collector current. But in the differential amplifier stage, the AC input voltages are not connected via capacitors....So while performing DC analysis ,we cannot open circuit the AC voltage sources. Then how we are to find the quiescent collector currents?...Also even if the AC voltage sources are connected via capacitor, then while DC analysis, the base would be floating....!Please help me solve this problem.

 

 

The circuit is wrong. Q3 should be PNP and its emitter should go to Vcc. Its collector will either be loaded by a constant current source or a resistor (which you have labelled as RE, which in turn should return to -VEE instead of ground). RC on Q1 collector is not needed. Then there should be a feedback network (or a simple resistor which will give unity gain) between Q3 collector and Q1 base. Q2 base should be returned to ground via a resistor that has the same value as the feedback resistor.

With that done, assuming the Q3 collector to be at zero volts because of the diff pair and feedback, Q3 collector current = |Vee| / RE. Q3 base current (Ib3) = Iq3/Hfeq3

Q2 collector current (Icq2) = 10*IB3 = 0.7/RC (calculate RC)

Total diff pair current is approx. 2 * Icq2 - that is the current mirror's output.. Calculate R1 as (|Vcc| + |Vee| - 0.7)/(2 * Icq2)

Sir, This is a problem on differential amplifiers from Neamen Electronic Circuits Third edition.

My bad, though I still feel that its good only as a theoretical exercise. To find the DC conditions, work back from Q3 - since its emitter is returned to ground, and o/p is being taken from the collector, assume the collector to be mid-way of Vcc and ground with no signal applied. Also assume that at peak swing Q3 has at least 2V across it to avoid saturation.

VceQ3 = 2V
Vcc = 12V
Vo3 = 6V (mid-way between Vcc and ground)
VRE = Vo3 - 2 (Vce at peak swing) = 4V
IRE = IcQ3 = 4/3.3k = 1.21 mA
VRC2 = IcQ3 * RC2 = 4.85V
Vo2 = VRE + 0.7 = 4.7V
IcQ2 = (Vcc - Vo2)/RC = (12-4.7)/8k = 912uA (approx. 1mA)
Assuming the diff pair is balanced, total Q5 current
IcQ5 = IcQ2 * 2 = 1.824 mA (almost 2 mA - [A])
IR1 = (|Vcc| + |Vee| - 0.7) / 12k = (12 + 12 - 0.7)/12k = 1.942 mA, almostt 2 mA [B] (a bit more than IcQ5 as expected, since this is a simple current mirror and there will be an imbalance). Compare [A] and [B] we can see that they are the same - 2 mA which checks out if things are ideal.