EE330 Final Exam Fall 04

EE330 Final Exam Spring 15

Given: = 30 = 0.97VBE = 0.7 V

Note: is NOT = 1, and the base current IB is NOT negligible !!!!

the is VERY low to clearly see the difference between IC and IE .

R1 = 20KR2 = 30KR3 = 2.5KI1 = 1 mA

R4 = 120KR5 = 3.8K

R6 = 2KR7 = 100KI2 = 2 mA

1) For the figures shown above, determine the voltages V1 thru V8. Assume all BJTs are in the ACTIVE (linear) region.

2) For the BJT current source shown below, determine the following:

a) Value of Rref to set the current in the slave transistor to 2 mA.

b) Maximum value of Rload while keeping the slave in the active region.

Given: = 25Vt = 25 mVVBE = 0.7 V

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3) A Common-Base amplifier is biased with a current source, IE , resulting in REext = , RLoad = 15 K, RC = 10 K, and RSig = 350 .

Assume VT = 25 mV = 100VBE = 0.7V

a) At what current IE should the transistor be biased so the amplifier Rin = Rsig ? b) What is the resulting overall voltage gain Gv ?

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4) For the figure shown below, determine the following:

a) Rinb) Routc) Avod) Gv

Given: = 200ro = 1/gm re = 25 NPN

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5) WITHOUT modifying any of the values in (4) above, add a resistor to the emitter in such a way that the Rin is changed to 10 K.

6) For the figure shown above, determine the following:

a) Rinb) Routc) Avod) Gv

Given: = 200VT = 25 mVro = PNP

RB1 = 40 KRB2 = 80 KRC = 8 K

RLoad = 15 K Ro = Rsig = 100

Ibias = 0.5 mA(assume BJT is operating in the active region)