Solving 3 Electronics Engineering Questions

Q1

SHAPE \ MERGEFORMAT

Assumptions

IB2 0 , VBE2 0 .5V

a . VRB VB2 VZ VBE2 4 0 .5 4 .5V

IRB IRA VRB /RB 4 .5V /20 k 0 .225 mA

VRA RA IRA 20 k 0 .225 mA 4 .5V

VOUT VRA VRB 4 .5V 4 .5V 9V

b . IOUT VOUT / RL 9V /10 900 mA

IE IOUT IRA 900mA 0 .225mA 900 .225mA

IB IE 1 900 .225mA /101 8 .913mA

c . IZENER IE2 ( IC2 20mA

Using KCL

IR IC2 IB 20mA 8 .913mA 28

.913mA

Since VOUT VE 9V , assuming VBE 0 .5V

VB VE VBE 9 0 .5 9 .5V USING OHM 'S LAW

R (VIN - VB /IR (30-9 .5 /28 .913mA 7109 .0236

d . VRB 4 .5V all the time , and IB2 0

Starting with VOUT 15V

VRA VOUT - VRB 15 - 4 .5 10 .5V

IRA IRB VRA /RA 10 .5V /12 k 0 .875 mA

RB VRB /IRB 4 .5V /0 .875 mA 5 .1429 k ( this will be RC

VOUT 6V

VRA VOUT - VRB 6 - 4 .5 1 .5V

IRA IRB VRA /RA 1 .5V /12 k 0 .125 mA

RB VRB /IRB 4 .5V /0 .125 mA 36 k

Rv 36 k ( - RC 36 k ( - 5 .1429 k 30 .8571 k . Q2

a . VE RW IO 2 k 1mA 2V

VB VRB VBE VE 0 .5 2 2 .5V

IRB1 IRB2 VB /RB2 2 .5V /5 k 0 .5mA

RB1 (6 - VB / IRB1 (6 - 2 .5 / 0 .5 mA 7 k

b . Small Signal Model re VT /Io 25mV /1mA 25 Applying Voltage Vx at Zin and injecting current Ix will result to the circuit below

Applying KVL at the loop shown

Vx (Ix - Bib )ro (Ix Ib )Rw

Vx (Ix - 100Ib )20k (Ix Ib )2k

Vx (Ix 20k ) - (20k 100Ib (2k Ix (2k Ib

Vx /Ix Zin 20k ( - (20k 100Ib /Ix 2k (2k Ib /Ix

Zin 22k Ib /Ix (2k ( - 2000k , but since Ix Ic BIb

Zin 22k ( - (1998k /B 22k ( - (1998k /100

Zin 22k ( - 19 .98k

Zin 2 .02k ( which is approximately RW

Q3

SHAPE \ MERGEFORMAT

a . Since IG 0A , at DC analysis the equivalent circuit is Also since VS 0V

VGS VDS VD VG VO VDD - IDRD 6 - (3 k ID (1

ID k (VGS - VT )2 0 .00025 (VGS - 1 )2 (2 Solving equation (1 ) and (2 ) simultaneously yields

ID 1mA and VGS 3V

IG 0 , VD VDS VGS 3V

IS ID 1 mA

VRD 6V - 3V 3V b Since no value of ro is given , assume ro is open circuit

gm 2k (VGS - VT 2 .00025 (3-1 1mS Av -gm (RF RD -1mS (8G 3k -1mS 3k -3

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Electronics Engineering

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