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Design voltage divider biased circuit to give Icq=5mA, VCEq=5V and $\beta$=100.
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written 8.0 years ago by
teamques10
★ 70k
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Step 1: Circuit Diagram
Step 2: Finding Vcc
$Vcc = 2V_{CEQ}$
Vcc = 10V
Step 3: Finding RE
$V_{RE} = 10% of Vcc$
$V_{RE} = 1V$
We know that,
$V_{RE} = I_ER_E$
Finding $I_E,$
$I_C= \beta I_B$
$I_B = I_C/\beta = 5mA/100 ………………..Ic=5mA(Given)$
$I_B= 50\mu A$
$I_E = I_B+I_C$
$I_E …
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