# Transistor Amplifier & Switch - BJT Curves, Q-point, CE Amp, MOSFET

> NPN BJT simulator: output curves, load line, draggable Q-point and operating regions, CE amplifier bias, gain and clipping, plus BJT and MOSFET switch design.

- URL: https://qilingtools.com/en/physics/transistor
- Category: Physics Tools (https://qilingtools.com/en/physics)
- Tags: [Physics](https://qilingtools.com/en/tags/physics), [Education](https://qilingtools.com/en/tags/education)
- Last updated: 2026-10-29
- Price: Free, no sign-up
- Privacy: Runs locally in the browser; files are never uploaded

This tool models an NPN transistor with the Ebers–Moll equations. The first tab draws the family of I_C–V_CE output curves with the DC load line, and you can drag the Q-point, the supply and the collector resistor right on the chart to watch the transistor move between cutoff, active and saturation. The second tab is a voltage-divider-biased common-emitter amplifier with emitter resistor and bypass capacitor: it gives the DC node voltages, small-signal parameters, gain, input and output resistance, and draws the output waveform from the large-signal model, so a badly placed Q-point visibly clips. Two more tabs size the base resistor for an LED or relay driver and check a MOSFET switch for conduction loss, switching loss and junction temperature.

## How to use

1. On Output curves & Q-point, set V_CC, R_B, R_C and β, then drag the Q-point and watch the region change
2. Switch to Common-emitter amp and adjust R1, R2, R_C, R_E and the load to see the bias point and gain
3. Raise the input or push the Q-point toward saturation or cutoff to see clipping and the AC load line
4. Use the Switch tab for a standard base resistor and saturation margin, and the MOSFET tab to check gate drive and heating

## Features

- Output curve family from the Ebers–Moll model with optional Early effect, with saturation and cutoff regions shaded
- Drag the Q-point, V_CC and R_C on the chart; I_B, I_C, V_CE, dissipation and region update live
- Divider-biased CE amplifier: V_B, V_E, V_C, g_m, r_π, A_v (exact and approximate), R_in, R_out and maximum undistorted swing
- Input and output waveforms solved point by point, with warnings for saturation clipping, cutoff clipping and distortion, plus DC and AC load lines
- BJT switch: LED series resistor, E12 base resistor, forced β, IO pin current check and flyback diode guidance
- MOSFET switch: square-law output curves, load-line operating point, conduction, switching and gate-drive losses, and junction temperature

## FAQ

### How can I tell whether a BJT is in cutoff, active or saturation?

Look at the two junctions. If the base-emitter junction is not forward biased (V_BE below about 0.5 V) the transistor is in cutoff, with I_B ≈ 0 and I_C ≈ 0. With base-emitter forward biased and base-collector reverse biased it is active, I_C = βI_B, nearly independent of V_CE. With both junctions forward biased it is saturated: V_CE is only 0.1–0.3 V and the circuit limits I_C to less than βI_B. In practice, assume active, compute I_C = βI_B, then check V_CE = V_CC − I_C R_C. If it is above about 0.3 V the assumption holds; otherwise the transistor is saturated.

### Why use voltage-divider bias with an emitter resistor?

With fixed bias I_C = βI_B is directly proportional to β, which varies several-fold between parts of the same type and rises with temperature, so the Q-point easily drifts into saturation. Divider bias pins the base voltage with R1 and R2, and the emitter resistor R_E adds DC negative feedback: if I_C rises, V_E rises, V_BE falls and I_C is pulled back. The result is I_C ≈ (V_B − 0.7)/R_E, almost independent of β. A bypass capacitor across R_E keeps the AC signal off R_E, so the feedback does not cost gain.

### How is the voltage gain calculated?

At midband, compute g_m = I_C/V_T (V_T ≈ 26 mV) and r_π = β/g_m from the bias current. With the emitter resistor fully bypassed, A_v = −β(R_C‖R_L)/r_π = −g_m(R_C‖R_L). With an unbypassed part R_E1, A_v = −β(R_C‖R_L)/[r_π + (1+β)R_E1] ≈ −(R_C‖R_L)/(r_e + R_E1). The minus sign means the output is inverted. Adding a load R_L lowers the gain; keeping some R_E1 lowers it too but makes it more predictable, reduces distortion and raises the input resistance.

### Why is the output clipped at the top or bottom?

A common-emitter stage inverts. If the negative half of the output is flattened, the collector has fallen to near the emitter voltage and the transistor is saturating, usually because the bias current is too high. If the positive half is flattened, I_C has reached zero and the transistor is cut off, usually because the bias current is too low. The ideal Q-point sits in the middle of the AC load line so both swings are equal; if both halves clip, the input is simply too large. Even without clipping, the exponential base-emitter law makes the two halves slightly unequal, and unbypassed emitter resistance reduces that distortion.

### Why does a transistor relay driver need a diode?

A relay coil is an inductor. When the transistor switches off, the current cannot stop instantly, so the coil generates L·di/dt that drives the collector far above the supply, often tens to hundreds of volts, enough to destroy the transistor. A flyback diode across the coil (cathode to the supply) lets the current circulate at turn-off and clamps the collector to V_CC + 0.7 V. Also drive the transistor into hard saturation: make the base current 2–10 times I_C/h_FE(min), so V_CE stays low and the transistor stays cool even with a low-gain part.

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