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Transistor Biasing

Calculate base current, collector current, and collector voltage for a fixed-bias BJT transistor circuit.

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Fixed-Bias BJT Circuit

In a fixed-bias bipolar junction transistor (BJT) amplifier, a single base resistor sets the base current from the supply voltage. The collector current follows the transistor current gain, and the collector voltage drops across the collector resistor.

Key Formulas

$$I_b = \frac{V_{cc} - V_{be}}{R_b}, \quad I_c = \beta I_b, \quad V_c = V_{cc} - I_c R_c$$

Base current \(I_b\) depends on supply voltage \(V_{cc}\), base-emitter drop \(V_{be}\) (typically 0.7 V for silicon), and base resistor \(R_b\). Collector current \(I_c\) equals \(\beta\) times \(I_b\). Collector voltage \(V_c\) is the remaining voltage after the collector resistor drop.

Design Notes

Fixed bias is simple but sensitive to temperature and \(\beta\) variation. For stable operation, emitter bias or voltage-divider bias circuits are often preferred. This calculator helps you quickly estimate operating point values for classroom or prototype work.

Related tools: Op-Amp Gain Calculator and Logic Gate Calculator.

Frequently Asked Questions

What is Vbe for a silicon BJT?

For a forward-biased silicon transistor, \(V_{be}\) is typically about 0.7 V at room temperature. Germanium devices use roughly 0.3 V.

What does beta (β) represent?

Beta is the DC current gain: \(I_c = \beta I_b\). It varies with temperature, collector current, and device lot. Typical small-signal values range from 50 to 300.

Why is collector voltage important?

\(V_c\) sets the transistor operating region. For amplification, the collector should stay between saturation and cutoff, usually mid-supply for maximum swing.

What units should I use for resistors?

Enter base and collector resistors in kilohms (kΩ). The calculator converts to ohms internally for the current and voltage equations.

Can Ic exceed the supply-limited maximum?

Yes. If \(I_c R_c\) exceeds \(V_{cc}\), the transistor saturates and the simple fixed-bias model no longer holds. Check that \(V_c\) stays positive for active-region operation.