Question

2. (20 points) In the circuit shown below the silicon transistor has B=70. (a) Determine Ib, Ic and Vce (the operating point). YOUR ANSWERS: 18 = Ic = VCE = 5.6ko 100k2 3 5.6k2 - 12 V Circuit for problem 2.

Answer 1

Solution:

Given that

—4-2 tockh SS-KL Се), с B: 0 Bio) - VER+

Given:

• 
  Ve=0V
  
• $V_{B}=-12\,V$
• 
  VEB= 12V >0.7V
  
• Base-Emitter Junction is forward biased.
• Therefore, transistor operate either in saturation (or) in active region.
• Assume, it is in active region.
  • $I_{C}=\beta I_{B}$
  • $I_{E}=I_{C}+I_{B}=(1+\beta)I_{B}$

KVL: $V_{EB}+100I_{B}-12=0$

$I_{B}=\frac{12-0.7}{100}\,mA$

1B = 0.113 mA

$I_{c}=7.91\,mA$

$I_{E}=8.023\,mA$

$V_{EB}=0.7\,V$

$V_{C}=5.6\,I_{c}-12$

$=5.6(7.91)-12$

$V_{C}=32.296\,V$

$V_{E}=-12\,V$

$V_{CB}=32.296+12>0$

$\therefore$ Collector junction is forward biased.

$\therefore$ Transistor is in saturation region.

$\underset{(sat)}{V_{EC}}=0.2\,V$

KVL : $+12\,V-5.6\,I_{C}-0.2=0$

$I_{C}=\frac{12-0.2}{5.6}\,mA$

$I_{C}=2.1\,mA$

KVL:- $+12-100\,I_{B}-0.7=0$

$I_{B}=\frac{12-0.7}{100}\,mA$

1B = 0.113 mA