D 9104 Consider the BiCMOS amplifier shown in Fig P9104 The

D 9.104 Consider the BiCMOS amplifier shown in Fig. P9.104. The BJT has BE 0.7 V, B 200, C 0.8 pF, and fr e600MHz. The NMOS transistor has V. 31V, k\' WIT 2 mA/V and C 1 pF Consider the dc bias circuit. Neglect the base current o Q2 in determining the current in Q1. Find the dc bias currents in Q1 and 02, and show that they are approximately 100 A and 1 mA, respectively. (b) Evaluate the small-signal parameters of Q1 and O2 at their bias points. Consider the circuit at midband frequencies. First, deter- mine the voltage gain Vo/V- (Note that RG can be neglected in this process.) Then use Miller\'s theorem on RG to determine the amplifier input resistance Rin. Finally, determine the overall voltage gain V/V sig (d) Consider the circuit at low frequencies. Determine the frequency of the poles due to C1 and Ca, and hence estimate the lower 3-dB frequency, ft. (e) Consider the circuit at higher frequencies. Use Miller\'s theorem to replace RG with a resistance at the input. (The one at the output will be too large to matter) Use open-circuit time constants to estimate fr (f To considerably reduce the effect of RG on Rin and hence on amplifier performance, consider the effect of adding another 10-M2 resistor in series with the existing one and placing a large bypass capacitor between their joint node and ground. What will Rin, AMD and f become? +5 V 3 km 10 Mn 1 HF 100 kn vi 0.1 HF sig 6.8 km 1 km

Solution

As there I\'d no current through RG. Hence VG=VC

Where VG= gate voltage and VC= collector voltage.

Given VBE= 0.7 v , as VE=0, hence VB=0.7 v

Now Vs= source voltage= VB=0.7 v

Id=drain current= Vs/6.8=102 micro A= 100 micro A.

Now Id=(kn\'*w/L)/2*(VGS-Vt)^2

After solving. VGS=1.316 V

VG=1.316+Vs=2 v

Ic= collector current= (5-VC)/3=(5-VG)/3= 1 MA

(b) re= 26 MV/IE=26/IC= 26 ohm

Where IE= emitter current

gm=[2*kn\'*w/L*Id]^0.5= 0.63 mA/v

(C) AC analysis.

Neglect RG.

Vo= -(Rc|| RL)*beta*gm*via/(beta*re/6.8+1+gm*beta*re)

Vo/vi=-23.25

Ri= RG/(1-vo/vi)= 412 k ohm

Ro=RG

Henc . Vo/Vs= Ri/(Ri+100)*Ro/(Ro+Rc||RL)*Vo/Via

Vo/Vs= 412/(512)*10/(10 )*-23.25

= -18.7

(d) fc1= cutoff frequency due to c1= 1/(2**(Ri+100)*c1)