Transistor Amplifier

One NPN transistor in a common-emitter stage. Choose how the base is biased, set β and the emitter resistor, and the page solves the resting current, the gain, and one period of the output for a sine wave at the input.

Worked example — Transistor Amplifier

Collector current at rest	1.0612 mA
Base–emitter voltage that current needs	0.6563 V
Voltage left across the transistor	5.9405 V
Transconductance	41.05 mS
Small-signal voltage gain	−4.54
The two-resistor estimate	−4.70
Output rises by	0.454 V
Output falls by	0.455 V

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Transistor Amplifier

A bipolar transistor passes a collector current set by the voltage across its base–emitter junction, I_C = I_Se^V_BE/V_T, and draws a base current I_C/β to do it. In a common-emitter stage the signal goes in at the base and comes out at the collector, where R_C turns a changing current into a changing voltage. The operating point is the current and voltage the stage rests at with no signal. The voltage gain is how far a small change at the base moves the collector: A = −g_mR_C / (1 + g_mR_E(β+1)/β), with the transconductance g_m = I_C/V_T.

What each symbol means

I_C: collector current, in amperes. The current through R_C, and the quantity the bias is designed to pin down
V_BE: base–emitter voltage, in volts. Near 0.65 V for a small transistor at a milliampere, and 60 mV more for every tenfold increase in current
V_T: the thermal voltage kT/q, 25.85 mV at 300 K
β: current gain, I_C/I_B. A property of the individual transistor, and the number a designer cannot rely on
g_m: transconductance, I_C/V_T, in siemens. How much the collector current changes per volt across the base–emitter junction
R_E: the emitter resistor, in ohms. It carries the emitter current and feeds part of the input back against itself

Assumes: The forward-active Ebers–Moll model with the −1 dropped. β is a constant for the part, and there is no Early effect: the collector current does not depend on V_CE, which on a real transistor it does by a few percent per volt. I_S = 10⁻¹⁴ A and V_T = 25.85 mV are held fixed, so nothing on the page warms up, and real β also drifts with current and temperature. Below V_CE = 0.2 V the page calls the transistor saturated and stops treating it as an amplifier; within a period it holds the output on that floor rather than modelling what saturation does. The signal arrives through a coupling capacitor that blocks DC and passes the signal untouched, from a source with no resistance of its own. There is no load on the collector and no bypass capacitor across R_E.

A threefold spread in β, a 6% change in current 🖖

One part number can carry a β of 100 or of 300: the MMBT3904 datasheet guarantees nothing narrower at 10 mA. Press the two Divider buttons in turn. The resting current goes from 1.0612 to 1.1248 mA and the gain from −4.54 to −4.58. Then the two Base resistor buttons: 0.5165 to 1.5455 mA, a factor of 2.99, and the gain follows from −93.90 to −280.99. A single resistor fixes the base current, and β multiplies whatever it is handed. The divider fixes a base voltage instead, the emitter resistor turns that voltage into a current, and β is left with one small job: the base current it draws through the divider's 8.48 kΩ.

The emitter resistor steadies the gain; the current needs the divider 🖖

With R_E = 1 kΩ the gain is −4.54, and −R_C/R_E is −4.70. The exact gain is −g_mR_C / (1 + g_mR_E(β+1)/β), and with g_mR_E at 41 the 1 in the denominator hardly counts, so the transistor's own numbers nearly cancel out. The two jobs come apart if you let them. Choose Single base resistor and leave R_E at 1 kΩ: from β = 100 to 300 the current still moves by a factor of 2.75, from 0.4938 to 1.3599 mA, while the gain moves only from −4.42 to −4.60. The price is in the table. Without R_E the stage amplifies 21 to 61 times more, and that extra gain is different in every transistor you fit.

Without an emitter resistor, 10 mV already comes out bent 🖖

Press 10 mV, no emitter resistor. The output falls 1.146 V and rises only 0.779 V. The collector current is I_Se^V_BE/V_T with V_T = 25.85 mV, so 10 mV up multiplies the current by e^10/25.85 = 1.47 and 10 mV down divides it by the same 1.47, and an exponential gains more than it loses. The small-signal gain treats that curve as the straight line touching it at the resting point, which is why the dashed prediction agrees with the output only near the middle. Put an emitter resistor back and most of the input lands across the resistor, which is linear, rather than across the junction: the divider stage given 0.2 V falls 0.911 V and rises 0.907 V.

Problem solved in full

Twice the gain, without giving up the stability 6 steps
Same 12 V supply and the same 4.7 kΩ collector resistor, but you need a gain of about −10, and the resting current must stay near 1 mA for any transistor whose β lies between 100 and 300. Choose R_E and a new divider.

Open this problem in the tool
1. The gain is close to −R_C/R_E, so R_E = 4.7 kΩ / 10 = 470 Ω.
  
  $A \approx - \frac{R _{C}}{R _{E}} \Rightarrow R_{E} = \frac{4 . 7 kΩ}{10} = 470 Ω$
2. Try that on this page first. With R_E at 470 Ω and β at 300 the stage saturates: the resistors ask for 2.2847 mA, which leaves 0.184 V across the transistor. The divider was sized for 1 kΩ under the emitter, and it cannot simply be kept.
  
  $I_{C} = 2.2847 mA, V_{C E} = 0.184 V < 0.2 V$
3. Work backwards from 1 mA. The emitter current is 1.01 mA, so the emitter sits at 1.01 mA × 470 Ω = 0.475 V, and 1 mA needs V_BE = 25.85 mV × ln(10⁻³/10⁻¹⁴) = 0.655 V. The base has to be at 1.13 V.
  
  $V_{E} = 1.01 mA \cdot 470 Ω = 0.475 V, V_{B E} = 25.85 mV \cdot ln \frac{1 0 ^{- 3}}{1 0 ^{- 14}} = 0.655 V, V_{B} = 1.13 V$
4. For β to stop mattering, the divider's own resistance must be small against β·R_E, which is 100 × 470 Ω = 47 kΩ at the bottom of the range. Aim for a tenth of that, R_TH ≈ 4.7 kΩ. The base current, 10 µA at β = 100, drops 0.047 V across it, so the divider must deliver V_TH ≈ 1.18 V.
  
  $R_{T H} \approx \frac{β _{m i n} R _{E}}{10} = \frac{100 \cdot 470}{10} = 4.7 kΩ, V_{T H} = 1.13 + 0.01 mA \cdot 4.7 kΩ \approx 1.18 V$
5. R₁ = R_TH·V_CC/V_TH = 4.7 kΩ × 12 / 1.18 = 47.8 kΩ and R₂ = R_TH·V_CC/(V_CC − V_TH) = 5.21 kΩ. The nearest standard values are 47 kΩ and 5.1 kΩ.
  
  $R_{1} = \frac{R _{T H} V _{C C}}{V _{T H}} = \frac{4 . 7 \cdot 12}{1 . 18} = 47.8 kΩ, R_{2} = \frac{R _{T H} V _{C C}}{V _{C C} - V _{T H}} = 5.21 kΩ$
6. Now solve the finished circuit with the page's model. At β = 100 it rests at 0.9985 mA with a gain of −9.39; at β = 300, 1.0644 mA and −9.48.
  
  $β 100300 I_{C} 0.9985 mA 1.0644 mA A - 9.39 - 9.48$
Answer

R_E = 470 Ω, R₁ = 47 kΩ, R₂ = 5.1 kΩ. Across β from 100 to 300 the current moves 6.6% and the gain 1.0%, close to what the page's own stage does at half the gain. The page cannot build this circuit, because its divider is fixed, and that is why the problem exists: keeping the page's divider with R_E = 470 Ω saturates the stage at β = 300, which is the mistake this problem is set to catch.

$R_{E} = 470 Ω, R_{1} = 47 kΩ, R_{2} = 5.1 kΩ$

Learning path

What is not a resistor

References (1)

The β spread insight 1 rests on (DC current gain 100 minimum, 300 maximum at 10 mA for one part number), and the 0.2 V saturation floor the page uses: Diodes Incorporated, "MMBT3904: 40V NPN Small-Signal Transistor in SOT23," datasheet DS30036 Rev. 27-2, March 2025. Electrical Characteristics, p. 4.

By the 42Math team · Reviewed against published sources. Corrections welcome.