(a).
The small-signal transistor parameters gm,rπ,ro .
The values of small-signal parameters are,
gm=7.73 mAVrπ=15.5 kΩ, ro=∞
Given Information:
The circuit diagram is shown below.
β=120, VBE(on)=0.7 V, VA=∞VCC=VEE=3.3 V, RS=500 Ω,RL=6 kΩ, RB=100 kΩ, RE=12 kΩ, RC=12 kΩ
Calculation:
The coupling and bypass capacitor act like open circuit for DC analysis. The AC voltage source ( vs ) is short circuited for DC analysis.
The modified figure is,
Applying Kirchhoff’s voltage law in base-emitter loop,
IBQRB+VBE(on)+IEQRE+VEE=0IBQRB+VBE(on)+(1+β)IBQRE+VEE=0IBQ=−VEE−VBE(on)RB+(1+β)REIBQ=−(−3.3)−0.7100 k+(1+120)12 kIBQ=3.3−0.71552 kIBQ=1.675 μA
Determining the collector current,
IBQ=1.675 μAICQ=βIBQICQ=(120)(1.675 μA)ICQ=0.201 mA
Determining the Trans-conductance,
gm=ICQVTgm=0.2010.026gm=7.73 mAV
Determining the diffusion resistance rπ ,
gmrπ=βrπ=βgmrπ=1207.73 mAVrπ=15.5 kΩ
Determining the small-signal transistor output resistance ro ,
ro=VAICQro=∞0.201×10−3ro=∞
(b).
The small signal current gain (Ai=ioii) and small signal voltage gain (Av=vovs) .
The values of current and voltage gain,
Ai=0.654Av=6.25
The coupling and bypass capacitors are short circuited for small-signal analysis. The DC voltage source is short circuited for small−signal analysis.
Applying current division rule in output,
Io=−(RCRL+RC)(gmVπ) ...(1)
Considering input circuit,
Vπ=Ib(rπ1+β)
Applying current division rule in input,
Vπ=−(rπ1+β)(RERE+rπ1+β)IiVπ=−(rπ1+β||RE)Ii
From equation (1),
Io=(RCRL+RC)(gm)((rπ1+β||RE)Ii)Ai=IoIi=gm(RCRL+RC)(rπ1+β||RE)=(7.73 mAV)(12 k12 k+6 k)(15.5 k1+120||12 k)=(5.153)(0.127)=0.654
Determining the small-signal voltage gain,
Vo=−(Rc||RL)(gmVπ)
Applying voltage division rule in input,
Vπ=−((RE||rπ1+β)Rs+(RE||rπ1+β))Vs
Plugging the value of Vπ ,
Vo=(Rc||RL)(gm)(((RE||rπ1+β)Rs+(RE||rπ1+β))Vs)Av=VoVs=gm(Rc||RL)((RE||rπ1+β)Rs+(RE||rπ1+β))=(7.73 mAV)(12 k||6 k)((12 k||15.5 k1+120)0.5 k+(12 k||15.5 k1+120))=7.73×4×0.202=6.25
(c).
The values of input and output resistance.
The values of input and output resistance are,
Ri=127 ΩRo=12 kΩ
The circuit diagram is,
Considering the small signal equivalent circuit,
The value of input resistance is,
Ri=RE||rπ1+β =12 k || 15.5 k1+120=12 k||0.128 k=12×0.12812+0.128 k=127 Ω
Determining the output resistance,
Ro=Rc=12 kΩ
Microelectronics: Circuit Analysis and Design