About this calculator
This fixed bias PNP BJT common collector calculator analyzes a common-collector amplifier built with a PNP bipolar junction transistor and fixed base-resistor bias. Enter your component values and it immediately finds the DC operating point (IC, VEC and the operating region), the small-signal parameters, voltage, current and power gain, input and output impedance, the output waveform with its clipping limits, the frequency response and the power dissipated in every component. Each result can be expanded to show the equation used and the numbers substituted into it, so you can follow or check the working by hand.
How to use it
- Enter the supply rails (VCC and VEE; use 0 V for a grounded lower rail).
- Type the resistor values and pick Ω, kΩ or MΩ, or type suffixes directly, such as
4.7kor1M. Enterinffor an open-circuit load. - Set the transistor parameters β, VBE(on), VCE(sat), temperature and Early voltage.
- Enter the signal amplitude, frequency and source resistance Rsig.
- Switch on Frequency Response to add the coupling and internal capacitances and see the Bode plot, fL and fH.
- Use the link button to copy a URL that reopens the calculator with your values.
How the common collector amplifier works
In the common-collector (CC) amplifier, better known as the emitter follower, the signal enters at the base and the output is taken from the emitter. The collector is at AC ground, connected to the supply directly or through a small resistor. The emitter voltage follows the base voltage one base–emitter drop away, so the voltage gain is slightly less than one and the output is in phase with the input.
Its value is impedance transformation. The resistance in the emitter is multiplied by (β + 1) when seen from the base, which gives a high input impedance. The output impedance is the source resistance divided by (β + 1) plus re, often only a few ohms to a few tens of ohms. That makes the emitter follower the standard buffer between a high-impedance stage and a low-impedance load such as a speaker, a cable or the next stage.
The gain approaches unity when gm·RL′ is much larger than one, where RL′ = RE ∥ RL ∥ ro. With heavy loads the output swing can become limited on the side where the transistor would cut off, because the emitter current cannot fall below zero. Check the “maximum undistorted swing” result when RL is small.
Fixed (base-resistor) bias
Fixed bias sets the base current with a single resistor RB from a bias source VBB. In many textbook circuits that source is simply the supply VCC, so enter the same value in both fields to model that case. Including the emitter resistor, the base loop gives IB = (VCC − VEB − VBB)/(RB + (β + 1)·RE), and then IC = β·IB.
Without RE the collector current is directly proportional to β. Because β varies by a factor of two or more between transistors of the same type, and also changes with temperature, a plain fixed-bias design can move far from its intended Q-point when the transistor is replaced. An emitter resistor helps: when (β + 1)·RE is large compared with RB, the current is set mostly by RE instead of β.
PNP polarity and conventions
A PNP transistor is the mirror image of an NPN: it conducts when the base is about 0.6–0.7 V below the emitter, and current flows into the emitter and out of the base and collector. The emitter therefore connects toward the positive supply and the collector toward the negative rail or ground. To keep the numbers positive, the calculator reports VEB, VEC and VBC for PNP stages, and every bias equation is the NPN equation with all polarities reversed. The small-signal model (gm, rπ, ro) is identical, so a PNP stage biased at the same current has the same gain and impedances as its NPN counterpart.
For the complementary circuit, see the Fixed Bias NPN BJT Common Collector calculator. Comparing the two is a quick way to check your PNP/NPN sign conventions.
Equations used
The calculator follows the standard two-step method taught in electronics courses. It first solves the DC bias circuit with all capacitors open, then replaces the transistor with its small-signal model and the capacitors with short circuits to find the mid-band AC behavior. These are the equations for this common-collector circuit with the default component values.
DC analysis
- Base Current (IB)
IB = (VCC − VEB − VBB) / (RB + (β + 1)·RE) - Collector Current (IC)
IC = β × IB - Emitter Current (IE)
IE = (β + 1) × IB - VEC (assuming active)
VEC = (VCC − VEE) − IC·RC − IE·RE
Small-signal analysis
- Thermal Voltage (VT)
VT = k·T / q = T / 11604.5 - Transconductance (gm)
gm = IC / VT - Base–Emitter Resistance (rπ)
rπ = β / gm - Output Resistance (ro)
ro = VA / IC - Emitter AC Load (RL′)
RL′ = RE ∥ RL ∥ ro - Resistance Looking into Base (Rib)
Rib = rπ + (β + 1)·RL′ - Voltage Gain (Av = vo / vb)
Av = (β + 1)·RL′ / Rib - Overall Gain (Gv = vo / vs)
Gv = Av × Zin / (Zin + Rsig) - Current Gain (Ai = io / ii)
Ai = Av × Zin / RL - Power Gain (Ap)
Ap = |Av × Ai| - Input Impedance (Zin)
Zin = RB ∥ Rib - Output Impedance (Zout)
Zout = RE ∥ ro ∥ (rπ + Rsig ∥ RB)/(β + 1)
The frequency-response section adds one low-frequency pole for each coupling or bypass capacitor, using the resistance that capacitor sees (short-circuit time constants), and estimates the upper cut-off frequency from the transistor’s internal capacitances using open-circuit time constants. The exact −3 dB points are then found numerically from the combined response.
Worked example
The calculator opens with a ready-to-use example circuit: VCC = 12 V, VEE = 0 V, VBB = 0 V, RB = 100 kΩ, RC = 0 Ω, RE = 1 kΩ, RL = 1 kΩ, β = 100, VEB(on) = 0.7 V, VEC(sat) = 0.2 V, T = 300 K, VA = 100 V, signal amplitude = 10 mV peak, signal frequency = 1000 Hz, Rsig = 50 Ω. The analysis of that circuit, with frequency response switched on and the default capacitor values, gives:
| Operating region | Forward-active |
|---|---|
| Collector Current (IC) | 5.622 mA |
| VEC | 6.322 V |
| gm | 217.5 mS |
| rπ | 459.8 Ω |
| ro | 17.79 kΩ |
| Voltage gain Av | 0.9907 V/V |
| Overall gain Gv (from Rsig) | 0.9892 V/V (−0.094 dB) |
| Input impedance Zin | 33.15 kΩ |
| Output impedance Zout | 5.021 Ω |
| Output amplitude at 1 kHz | 9.891 mV peak |
| Lower cut-off fL | 15.85 Hz |
| Upper cut-off fH | 1.064 GHz |
| Transistor power PT | 35.58 mW |
| Total supply power | 68.14 mW |
The transistor is in the forward-active region with IC = 5.622 mA and VEC = 6.322 V, which leaves 2.839 V of undistorted output swing. The stage has a voltage gain of 0.9907 V/V, so a 10 mV peak input signal produces an output of about 9.891 mV peak, in phase with the input. The mid-band region extends from 15.85 Hz to 1.064 GHz. Change any value above to see how the results move.
Key characteristics of the common-collector stage
| Property | Typical behavior | Default example |
|---|---|---|
| Voltage gain | ≈ 1, non-inverting | 0.9907 V/V |
| Current gain | High (≈ β + 1) | 32.84 A/A |
| Power gain | ≈ current gain | 32.53 |
| Input impedance | High (≈ (β + 1)·RL′ ∥ bias) | 33.15 kΩ |
| Output impedance | Very low | 5.021 Ω |
| Phase shift | 0° | 0° |
| Typical uses | Buffers, output stages, impedance matching | |
Design tips
- Bias the emitter near the middle of the available voltage range so the output can swing symmetrically.
- For a low output impedance the source driving the base should itself have a low resistance, since Zout ≈ (Rs′ + rπ)/(β + 1).
- With a small load, raise the bias current so that IE·(RE ∥ RL) exceeds the required peak output. Otherwise the peaks on the cut-off side clip.
Frequently asked questions
Why is the voltage gain of an emitter follower less than one?
The output is the emitter voltage, which is the base voltage minus the base–emitter voltage. For signals, a small part of the input appears across re instead of the load, so the gain is gm·RL′/(1 + gm·RL′), always slightly below one.
Why use a common-collector stage if it has no voltage gain?
It provides large current gain and turns a high-impedance source into a low-impedance one. A weak signal source that could not drive a low-resistance load directly can drive it through an emitter follower with almost no loss of amplitude.
Does the emitter follower invert the signal?
No. The emitter moves in the same direction as the base, so the phase shift is 0° in mid-band.
Why is fixed bias considered unstable?
Its collector current is roughly β times a base current that the circuit sets independently of the transistor. Any change in β, from part-to-part spread or temperature, moves the collector current and the Q-point by the same proportion, and can push the transistor into saturation.
Why would I use a PNP transistor instead of an NPN?
PNP stages let a circuit work “upside down”: the emitter ties to the positive rail, which is convenient for high-side switching, current sources that push current into a load, level shifting between stages, and complementary push-pull output stages. Silicon PNP devices have somewhat lower β and speed than matched NPN parts, but the analysis is the same.
What voltage gain does this CC amplifier give with the default values?
With the default components the calculator gives Av = 0.9907 V/V (−0.0809 dB) from the base to the load, and an overall gain of 0.9892 V/V from the 50 Ω signal source, with Zin = 33.15 kΩ and Zout = 5.021 Ω.