1)Problem 12-2: Find AC collector resistance and AC saturation current of fig 12-38.
Soln: In the AC equivalent circuit we have Rc || R(Load).As such the Ac resistance = ((680*2700)/(680+2700)) ohm=543 ohm.
The Voltage at base of transistor=(470*15)/(470+2000)=2.85V.Drop across Base emitter=0.7V.
Hence Ve=2.85-0.7=2.15 V. hence IE= (2.15/220)A=9.7 ma.Since Ie appx equal to Ic hence AC saturation current = 9.7 ma.

2)Problem 12-6 DC collector resistance and DC saturation current of Fig 12-39

Soln: The DC collector resistance is same as Rc i.e 100 ohm.
For DC saturation current assuming Vce negligible.Idc(sat)=30/(Rc + Re)=30/(100+68)=178.5 ma.

3)Problem 12-10: All resistances tripled in Fip 12-39.
The Ac collector resistance = Rc=(300*300)/600=150ohm.
Base Voltage={30/(300+600)}*300=10V.
Hence Ie=Ic=(10-0.7)/(Re).Here Re is 68*3=204 ohm.Hence Ic=Ie=(9.3/204)A=45.5 ma.
Due to Current Ic=45.5ma The Dc collector voltage Decreases to Vc=(30-(45.5ma*300)) V=16.35V
The Vceq for the transistor=Vc-Ve=(16.35-9.3)=7.05V.
Also The IcRc value is (150 ohm * 45.5 ma)=6.8V.
The mp is the lower value from Vceq and IcRc ie 6.8V.
As such the MPP=2*6.8V=13.6V.

4)The DC power supplied to circuit in 12-38.
Soln: The bias current thru voltage divider=15/(2+0.47) ma=6.07 ma.
The Voltage at base=6.07 ma * 0.47 k=2.85V.
Voltage at emitter=2.85V-0.7V=2.15V.
The Quiescient Ie=Ic= (2.15/220)=9.7ma.
As such total current from the 15V source=Current for Voltage divider + Collector Current=(6.07+9.70)ma =15.77 ma. Hence Power supplied=(15*15.77)mW=236.55 mW.

5) Dc power supplied to amplifier in 12-39.
Voltage Divider Current=(30/300)=100 ma.
Base Voltage=Vb=(100*0.1)V=10V.
Ie=Ic=(10-0.7)/68 =136.7 ma.

Hence Total current from 30V supply=(136.7+100)ma =236.7 ma.
Hence Power Supplied= (30*236.7)ma=7.1W.

6)efficiency is Pout/Pin.The Pin is the DC input Power and Pout is the Ac output power.

The AC Peak to Peak is 5V.As such AC RMS=5V(2*sqrt(2)).The AC op power=(Vrms*Vrms)/Rload.
Hence OP AC=25/(8*3.2) =0.976 W.
Voltage Divider Current=0.82 A.
Base voltage=(0.82* 2.2 )V=1.8V.
Ie=(1.8-0.7)/1 =1.1A.
Hence Total Current from 10V supply= (1.1 + 0.82)=1.92A.
Hence Pdc input=10v*1.92A=19.2W.

Hence Efficiency=(0.976/19.2)*100=5.08 %.

7)For Fig 12-42.
Soln: The diode Drop across each diode is 0.7V.
As such the total Bias current=(30-1.4)/200=143 ma.
This is the value of Quiescient Collector current assuming diode drop matches emitter diodes in diode curves.