Answer: Rc = 400 Ω and Rb = 57.2 kΩ
Explanation:
Given that;
VCE = 5V
VCC = 15 V
iC = 25 mA
β = 100
VD₀ = 0.7 V
taking a look at the image; at loop 1
-VCC + (i × Rc) + VCE = 0
we substitute
-15 + ( 25 × Rc) + 5 = 0
25Rc = 10
Rc = 10 / 25
Rc = 0.4 k
Rc = 0.4 × 1000
Rc = 400 Ω
iC = βib
25mA = 100(ib)
ib = 25 mA / 100
ib = 0.25 mA
ib = 0.25 × 1000
ib = 250 μAmp
Now at Loop 2
-Vcc + (ib×Rb) + VD₀ = 0
-15 (250 × Rb) + 0.7 = 0
250Rb = 15 - 0.7
250Rb = 14.3
Rb = 14.3 / 250
Rb = 0.0572 μ
Rb = 0.0572 × 1000
Rb = 57.2 kΩ
Therefore Rc = 400 Ω and Rb = 57.2 kΩ
