About this calculator
This PNP BJT common base calculator analyzes a common-base amplifier built with a PNP bipolar junction transistor and a split-supply emitter 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 base amplifier works
In the common-base (CB) amplifier the base is held at AC ground, the signal drives the emitter, and the output is taken from the collector. The input source has to supply the entire emitter current, so the input impedance is very low, about re = VT/IE. That is often only a few tens of ohms. The collector, by contrast, behaves like a high-impedance current source.
The emitter current passes almost unchanged to the collector (α = β/(β + 1) ≈ 0.99), so the current gain is slightly less than one. The voltage gain, however, is gm·(RC ∥ RL), which is the same magnitude as a bypassed CE stage. It is also non-inverting: raising the emitter voltage reduces the base–emitter voltage, reduces the collector current and raises the collector voltage.
With the base grounded there is no Miller multiplication of Cµ, so the CB stage has excellent high-frequency performance. It is used in RF and wideband amplifiers, as a current buffer, and as the upper transistor of the cascode, where it combines with a common-emitter stage to give high gain and wide bandwidth together.
Common-base bias with a split supply
Here the base is grounded and the emitter resistor RE returns to the positive rail VCC, while the collector resistor RC returns to the negative rail VEE. The emitter therefore sits one VEB above ground and the emitter current is IE = (VCC − VEB)/RE.
This bias is very stable. The emitter current depends only on the supply, RE and the nearly constant junction voltage, not on β. The collector voltage then follows from VC = VEE + IC·RC, and the calculator checks that the collector–base junction stays reverse-biased so the transistor remains in the forward-active region.
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 NPN BJT Common Base 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-base circuit with the default component values.
DC analysis
- Emitter Current (IE)
IE = (VCC − VEB) / RE (base grounded) - Base Current (IB)
IB = IE / (β + 1) - Collector Current (IC)
IC = α·IE = β·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 Resistance (re)
re = VT / IE - AC Load (RL′)
RL′ = RC ∥ RL - Voltage Gain (Av = vo / ve)
Av = gm·RL′ (non-inverting) - 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 = RE ∥ re - Output Impedance (Zout)
Zout = RC ∥ ro·[1 + gm·(RE ∥ Rsig ∥ rπ)]
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 = 5 V, VEE = −12 V, RC = 4.7 kΩ, RE = 4.7 kΩ, RL = 10 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) | 905.8 µA |
| VEC | 8.443 V |
| gm | 35.04 mS |
| rπ | 2.854 kΩ |
| ro | 110.4 kΩ |
| Voltage gain Av | 112 V/V |
| Overall gain Gv (from Rsig) | 40.3 V/V (32.1 dB) |
| Input impedance Zin | 28.09 Ω |
| Output impedance Zout | 4.627 kΩ |
| Output amplitude at 1 kHz | 394.9 mV peak |
| Lower cut-off fL | 203.8 Hz |
| Upper cut-off fH | 24.21 MHz |
| Transistor power PT | 7.654 mW |
| Total supply power | 15.44 mW |
The transistor is in the forward-active region with IC = 905.8 µA and VEC = 8.443 V, which leaves 2.896 V of undistorted output swing. The stage has a voltage gain of 112 V/V, so a 10 mV peak input signal produces an output of about 394.9 mV peak, in phase with the input. The mid-band region extends from 203.8 Hz to 24.21 MHz. Change any value above to see how the results move.
Key characteristics of the common-base stage
| Property | Typical behavior | Default example |
|---|---|---|
| Voltage gain | High, non-inverting | 112 V/V |
| Current gain | ≈ α (slightly below 1) | 0.3147 A/A |
| Power gain | Medium | 35.25 |
| Input impedance | Very low (≈ re) | 28.09 Ω |
| Output impedance | High (≈ RC) | 4.627 kΩ |
| Phase shift | 0° | 0° |
| Typical uses | RF and wideband amplifiers, current buffers, cascodes | |
Design tips
- Drive the CB stage from a low-impedance source or a current. With a 50 Ω generator and an input impedance of 25 Ω, most of the signal is lost across Rsig. The “Overall gain” result shows the effect.
- The emitter current is set almost entirely by the supply and RE, so the bias point is nearly independent of β.
- For the widest bandwidth keep the collector load small. The dominant high-frequency pole is Cµ·(RC ∥ RL).
Frequently asked questions
Why is the input impedance of a common-base amplifier so low?
The signal enters at the emitter, and the emitter must carry the full transistor current. A small change in emitter voltage causes a large change in emitter current, so the stage looks like a small resistor of about re = VT/IE. At 1 mA that is about 26 Ω at room temperature.
Is the common-base amplifier inverting?
No. The output is in phase with the input. A rise in emitter voltage reduces the base–emitter voltage, which lowers the collector current, which raises the collector voltage.
Where is the common-base configuration used?
In RF and high-frequency amplifiers because it avoids the Miller effect, as a current buffer that accepts a signal current at low impedance and delivers it at high impedance, and as the top transistor of a cascode amplifier.
Why does the common-base calculator use two supplies?
Grounding the base and returning RE to the opposite rail is the simplest way to forward-bias the emitter junction while keeping the base at AC and DC ground. A single-supply version needs a base divider and a bypass capacitor, which gives the same small-signal results.
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 CB amplifier give with the default values?
With the default components the calculator gives Av = 112 V/V (41 dB) from the emitter to the load, and an overall gain of 40.3 V/V from the 50 Ω signal source, with Zin = 28.09 Ω and Zout = 4.627 kΩ.