Amplifier Inputs
Use quiescent-point small-signal values. The calculation assumes the intended middle frequency region.
Formula Used
The tool first finds the effective output resistance. It then calculates the device stage gain, input divider loss, total gain, decibels, and output voltage.
BJT, unbypassed: Av,stage ≈ −βRout / [rπ + (β + 1)RE]
rπ = β / gm
FET, unbypassed: Av,stage ≈ −gmRout / (1 + gmRS)
Gain(dB) = 20 log10|Av,total|
Vout = Av,totalVin
The negative sign represents phase inversion. These are midband small-signal estimates, not large-signal limits.
How to Use This Calculator
- Choose the transistor stage that matches your circuit.
- Select whether the emitter or source resistor is bypassed.
- Enter small-signal transconductance at the intended operating point.
- Enter all resistances as AC values in kilo-ohms.
- For a BJT, enter the estimated current gain beta.
- Enter the available source signal in millivolts.
- Submit the form and review total gain, loading loss, and output voltage.
- Compare the loaded result with the unloaded result before finalizing parts.
Example Data
| Parameter | Example value | Why it matters |
|---|---|---|
| BJT transconductance | 40 mS | Sets the basic current-to-voltage conversion strength. |
| Collector resistor | 4.7 kΩ | Provides the main AC voltage developing resistance. |
| Load resistor | 10 kΩ | Reduces the collector or drain resistance in parallel. |
| Signal source resistance | 0.6 kΩ | Determines how much input signal reaches the amplifier. |
| Emitter resistor | 0.47 kΩ | Controls local feedback when not bypassed. |
Mid-Frequency Gain Basics
Mid-frequency gain describes an amplifier where coupling capacitors act almost like shorts. Device capacitances are still too small to dominate the signal. The circuit therefore behaves mainly through resistances and small-signal parameters. This range is useful because it shows the stable voltage gain. It also helps engineers compare designs without low-frequency rolloff or high-frequency losses.
Device Transconductance
A common-emitter stage inverts the waveform. A common-source stage also inverts the waveform. Their gain magnitude depends strongly on transconductance. Higher transconductance creates greater output current for a given input voltage. That current flows through the effective output resistance. The resulting voltage is the amplified output. Collector or drain resistance matters greatly. A connected load reduces the available resistance. This effect lowers voltage gain.
Input Divider Effects
Input attenuation is equally important. A real source has resistance. The amplifier input resistance forms a divider with that source resistance. The transistor may provide a low input resistance. Bias resistors can lower it further. The voltage reaching the active device can become much smaller than the source voltage. The calculator includes this divider. It reports the stage gain and total gain separately. This distinction prevents overly optimistic results.
Degeneration and Stability
Emitter or source degeneration improves stability. It uses an unbypassed resistor beneath the device. The resistor creates negative feedback. Gain decreases, but linearity often improves. Input resistance can rise in a bipolar stage. The gain becomes less dependent on device variations. A bypass capacitor can remove much of this feedback at mid frequencies. Then gain becomes higher. The selected bypass condition therefore changes the calculation.
Loading and Real Circuits
Output resistance also affects practical designs. The collector or drain resistor appears in parallel with the external load. Transistor output resistance can add another parallel path. A smaller effective resistance gives lower gain. A very large output resistance has little effect. The calculator lets you include it when known. Enter zero when it is ignored. Compare loaded and unloaded figures to see the loading penalty.
Interpreting the Results
Use consistent units while entering values. Transconductance is entered in millisiemens. Resistances are entered in kilo-ohms. The input signal is entered in millivolts. The output result uses millivolts. A negative voltage gain indicates phase reversal. Gain in decibels uses the magnitude only. Confirm that the chosen parameters represent the quiescent operating point. Small-signal gain is valid only near that point. Large signals can cause distortion, clipping, or device nonlinearity.
Design Checks
This estimator uses standard small-signal approximations. It assumes the circuit operates in its intended midband region. It does not model capacitor tolerances, temperature drift, or complex frequency response. Use measured values for final design checks. Simulate the complete circuit before production. Then compare predicted and measured gain. Differences can reveal loading, bias, or layout problems. This process leads to stronger amplifier designs. Check supply voltage, thermal limits, and allowable signal swing as well. These limits affect useful gain. Consider every expected operating condition before finalizing the circuit. Reliable designs always need adequate headroom.
Frequently Asked Questions
1. What is mid-frequency voltage gain?
It is the amplifier voltage gain in the central operating band. In this region, coupling capacitors act nearly as short circuits and high-frequency capacitances have limited influence.
2. Why is the calculated gain negative?
A common-emitter or common-source stage reverses signal polarity. The negative sign shows a 180-degree phase shift. Gain magnitude remains the absolute value.
3. Why does the load resistor reduce gain?
The load sits in parallel with the collector or drain resistance. The effective output resistance decreases, so the same small-signal current develops less output voltage.
4. What does transconductance mean?
Transconductance measures output current change per input voltage change. A higher value usually gives more voltage gain when the effective output resistance stays unchanged.
5. When should I use a finite output resistance?
Use a finite value when a device model, measurement, or datasheet estimate is available. Enter zero when you intentionally want to ignore that parallel path.
6. Why does an unbypassed emitter resistor lower gain?
It introduces local negative feedback. Current changes produce voltage across the resistor, reducing the effective controlling voltage. This trades gain for stability and linearity.
7. Does the input source resistance matter?
Yes. It forms a divider with amplifier input resistance. Higher source resistance can greatly reduce the voltage reaching the transistor or FET gate.
8. Why does the BJT option need beta?
Beta helps calculate rπ and the effect of emitter degeneration. It also determines the small-signal input resistance for the simplified BJT model.
9. Can I use this for large signal design?
No. The result is a small-signal estimate around the bias point. Large signals can encounter clipping, distortion, changing transconductance, and supply limitations.
10. How is gain in decibels calculated?
The calculator uses 20 times the base-ten logarithm of the absolute voltage gain. The sign is excluded because decibels represent magnitude.
11. What should I check after calculating gain?
Verify bias voltages, signal swing, power limits, capacitor behavior, source loading, load resistance, and measured frequency response before accepting the design.