Question

if $i_b = 100 \mu a$ and $\beta = 125$, what is $i_c$? (the bjt is connected and not saturated). be sure to include the units!

collector
base
emitter
npn bjt

answer:

Explanation:

Step1: Recall BJT current relations

For a BJT, the current gain \(\beta=\frac{I_C}{I_B}\), so \(I_C = \beta\times I_B\). Also, \(I_E=I_B + I_C\) (KCL at emitter - base junction).
Given \(I_B = 100\ \mu\text{A}\) and \(\beta = 125\), first find \(I_C\).
\(I_C=\beta\times I_B=125\times100\ \mu\text{A}=12500\ \mu\text{A}=12.5\ \text{mA}\)

Step2: Calculate \(I_E\)

Using \(I_E = I_B+I_C\), substitute \(I_B = 100\ \mu\text{A}\) and \(I_C = 12500\ \mu\text{A}\)
\(I_E=100\ \mu\text{A}+ 12500\ \mu\text{A}=12600\ \mu\text{A}=12.6\ \text{mA}\) (or we can use \(I_E=I_B(1 + \beta)\) directly: \(I_E=100\ \mu\text{A}\times(1 + 125)=100\ \mu\text{A}\times126 = 12600\ \mu\text{A}=12.6\ \text{mA}\))

Answer:

\(12.6\ \text{mA}\) (or \(12600\ \mu\text{A}\))