NPN Common Base Calculator

Q: 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.

Q: What values of β and VBE should I use?

Use the datasheet’s typical hFE at your collector current, often 100 to 300 for small-signal parts such as the 2N3904 or BC547. VBE(on) of 0.7 V is the standard hand-analysis value for silicon. Try the minimum and maximum β from the datasheet to see how sensitive your design is.

Q: 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.

About this calculator

This NPN BJT common base calculator analyzes a common-base amplifier built with an NPN bipolar junction transistor and a split-supply emitter bias. Enter your component values and it immediately finds the DC operating point (I_C, V_CE 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 (V_CC and V_EE; use 0 V for a grounded lower rail).
Type the resistor values and pick Ω, kΩ or MΩ, or type suffixes directly, such as 4.7k or 1M. Enter inf for an open-circuit load.
Set the transistor parameters β, V_BE(on), V_CE(sat), temperature and Early voltage.
Enter the signal amplitude, frequency and source resistance R_sig.
Switch on Frequency Response to add the coupling and internal capacitances and see the Bode plot, f_L and f_H.
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 r_e = V_T/I_E. 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 g_m·(R_C ∥ R_L), 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 C_µ, 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 R_E returns to the negative rail V_EE, while the collector resistor R_C returns to the positive rail V_CC. The emitter therefore sits one V_BE below ground and the emitter current is I_E = (0 − V_BE − V_EE)/R_E.

This bias is very stable. The emitter current depends only on the supply, R_E and the nearly constant junction voltage, not on β. The collector voltage then follows from V_C = V_CC − I_C·R_C, and the calculator checks that the collector–base junction stays reverse-biased so the transistor remains in the forward-active region.

NPN polarity and conventions

An NPN transistor conducts when its base is about 0.6–0.7 V above its emitter. Conventional current flows into the collector and the base and out of the emitter, so in a single-supply circuit the emitter is connected toward ground and the collector load toward the positive rail. All voltages in this calculator follow that convention: V_BE, V_CE and V_CB are positive in the forward-active region. Saturation is detected when V_CE falls to V_CE(sat), and the calculator then re-solves both loop equations instead of simply flagging the problem.

For the complementary circuit, see the PNP 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 = (0 − VBE − VEE) / RE (base grounded)
Base Current (IB)IB = IE / (β + 1)
Collector Current (IC)IC = α·IE = β·IB
VCE (assuming active)VCE = (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: V_CC = 12 V, V_EE = −5 V, R_C = 4.7 kΩ, R_E = 4.7 kΩ, R_L = 10 kΩ, β = 100, V_BE(on) = 0.7 V, V_{C_E(sat)} = 0.2 V, T = 300 K, VA = 100 V, signal amplitude = 10 mV peak, signal frequency = 1000 Hz, R_sig = 50 Ω. The analysis of that circuit, with frequency response switched on and the default capacitor values, gives:

Operating region	Forward-active
Collector Current (I_C)	905.8 µA
V_CE	8.443 V
g_m	35.04 mS
r_π	2.854 kΩ
r_o	110.4 kΩ
Voltage gain A_v	112 V/V
Overall gain G_v (from R_sig)	40.3 V/V (32.1 dB)
Input impedance Z_in	28.09 Ω
Output impedance Z_out	4.627 kΩ
Output amplitude at 1 kHz	394.9 mV peak
Lower cut-off f_L	203.8 Hz
Upper cut-off f_H	24.21 MHz
Transistor power P_T	7.654 mW
Total supply power	15.44 mW

The transistor is in the forward-active region with I_C = 905.8 µA and V_CE = 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 (≈ r_e)	28.09 Ω
Output impedance	High (≈ R_C)	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 R_sig. The “Overall gain” result shows the effect.
The emitter current is set almost entirely by the supply and R_E, so the bias point is nearly independent of β.
For the widest bandwidth keep the collector load small. The dominant high-frequency pole is C_µ·(R_C ∥ R_L).

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 r_e = V_T/I_E. 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 R_E 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.

What values of β and V_BE should I use?

Use the datasheet’s typical h_FE at your collector current, often 100 to 300 for small-signal parts such as the 2N3904 or BC547. V_BE(on) of 0.7 V is the standard hand-analysis value for silicon. Try the minimum and maximum β from the datasheet to see how sensitive your design is.

What voltage gain does this CB amplifier give with the default values?

With the default components the calculator gives A_v = 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 Z_in = 28.09 Ω and Z_out = 4.627 kΩ.

NPN BJT Common Base Calculator

Circuit Schematic

DC Operating Point

Gain Analysis

Impedance

AC Signal Analysis

Power Dissipation

Model & Assumptions