BJT Amplifier & Bias Calculator
Work out the DC operating point of a voltage-divider biased transistor, then the gain, input and output impedance and bandwidth of a common-emitter stage or an emitter follower.
How to Use
- Pick a mode: Bias for the DC operating point, Common emitter for a voltage amplifier, or Emitter follower for a buffer.
- Enter the supply and the resistors, each with its own unit (kΩ, Ω or MΩ). You can also type 4.7k or 56k straight into a field.
- Enter the current gain β (hFE). The 2N3904 data sheet guarantees 100 to 300 at 10 mA, so 100 is a safe design value.
- In the amplifier modes, choose whether the bias comes from the divider or from a collector current you already know, then add the load, the source resistance and how much of RE is bypassed.
- Read the operating point or the gain, Zin, Zout and bandwidth beside the schematic, and follow every step in Show Work.
Worked Example
Bias. A 12 V supply, R1 = 56 kΩ, R2 = 10 kΩ, RC = 3.3 kΩ, RE = 1 kΩ, β = 100 and VBE = 0.7 V. The divider is a 1.818 V source behind RTH = 56 kΩ ∥ 10 kΩ = 8.485 kΩ. The base loop gives IB = (1.818 − 0.7) ÷ (8.485 kΩ + 101 × 1 kΩ) = 10.21 µA, so IC = 100 × IB = 1.021 mA and IE = 1.032 mA. Then VE = 1.032 V, VB = 1.732 V, VC = 12 − 1.021 mA × 3.3 kΩ = 8.63 V and VCE = 7.598 V, comfortably in the active region.
Common-emitter gain. At 25 °C, VT = kT ÷ q = 25.69 mV, so re = 25.69 mV ÷ 1.032 mA = 24.91 Ω and rπ = 101 × re = 2.516 kΩ. With RE fully bypassed and a 10 kΩ load, RC ∥ RL = 2.481 kΩ and Av = −100 × 2.481 kΩ ÷ 2.516 kΩ = −98.63. The input impedance is 8.485 kΩ ∥ 2.516 kΩ = 1.94 kΩ, so a 600 Ω source sees an overall gain of −75.34. Leaving 100 Ω of RE unbypassed drops the gain to −19.67 but raises Zin to 5.073 kΩ.
The common mistake: the stiff-divider shortcut. Treating the divider as an ideal 1.818 V source gives IE = (1.818 − 0.7) ÷ 1 kΩ = 1.118 mA, which is 8.4 % high. The 10.21 µA base current flows through RTH = 8.485 kΩ and drops 87 mV on the way. The shortcut only works when R2 carries much more than the base current; here it carries 17 times IB, which is not enough to ignore it.
Show Work
Formulas
Background: Biasing and the Small-Signal Model
The point-contact transistor was demonstrated at Bell Labs by John Bardeen and Walter Brattain in December 1947, and William Shockley worked out the junction transistor in early 1948. The bipolar junction transistor that followed is a current-controlled device whose current gain β varies from part to part: the 2N3904’s data sheet allows anything from 100 to 300 at 10 mA. Voltage-divider bias with an emitter resistor answers that spread with negative feedback, because any extra collector current raises the emitter voltage and throttles the base.
For signals, the transistor is replaced by its small-signal model. Its key value is the transconductance gm = IC ÷ VT, where VT = kT ÷ q is the thermal voltage (25.69 mV at 25 °C, from the exact SI values of Boltzmann’s constant and the elementary charge). From it follow the emitter resistance re = VT ÷ IE and the base resistance rπ = β ÷ gm = (β + 1)re. The formulas here are those in W. Marshall Leach’s Georgia Tech notes The BJT (2010) and MIT course 6.071’s Bipolar Junction Transistor Circuits (Chaniotakis and Cory, 2006), with the transistor’s output resistance ro taken as very large.
The bandwidth estimate uses the Miller effect, first described by John Milton Miller in 1920 for vacuum tubes: a capacitance from input to output of an inverting amplifier looks (1 + |Av|) times larger from the input. The transistor’s capacitances come from its data sheet: fT = gm ÷ 2π(Cπ + Cμ), as in MIT course 6.012’s lectures on amplifier frequency response, which also give the open-circuit time-constant sum used here. The defaults are the Nexperia PMBT3904 (the 2N3904 in SOT-23) limits: Cμ at most 4 pF and fT at least 300 MHz. Because they are worst-case limits and fT falls at low current, the result is an estimate.
About This Calculator
This calculator brings three transistor stages into one page. Bias mode solves the voltage-divider circuit exactly, including the base current through the divider, and reports the region: active, saturated (with the real saturated currents) or cut off. Common-emitter mode adds the gain, input and output impedance and an upper −3 dB frequency, with any part of RE bypassed. Emitter-follower mode gives the near-unity gain, the high input and low output impedance, and how far the output can swing before the transistor cuts off. The amplifier modes can take their current from the divider or from a collector current you already know.
Every resistor has its own unit, and you can type 4.7k or 56k directly. VT follows the temperature you enter; VBE does not, so enter the VBE you expect at that temperature yourself. Everything runs in your browser; nothing you enter is sent anywhere.
Related calculators: BJT Power Dissipation Calculator, JFET Bias Calculator, and Op-Amp Gain Calculator.
Frequently Asked Questions
How do I find the Q-point of a voltage-divider biased transistor?
Replace the divider with its Thevenin equivalent and write the base loop. With 12 V, R1 = 56 kΩ and R2 = 10 kΩ, VTH = 1.818 V and RTH = 8.485 kΩ. With RE = 1 kΩ, β = 100 and VBE = 0.7 V, IB = 1.118 V ÷ (8.485 kΩ + 101 × 1 kΩ) = 10.21 µA, so IC = 1.021 mA and, with RC = 3.3 kΩ, VCE = 7.598 V.
Why does β hardly change the collector current?
The emitter resistor feeds back: more current raises VE and takes voltage away from the base-emitter junction. In the 12 V example, tripling β from 100 to 300 (the 2N3904’s whole data-sheet range at 10 mA) only moves IC from 1.021 mA to 1.084 mA, about 6 %.
What is re, and why does the gain change with temperature?
re = VT ÷ IE, where VT = kT ÷ q is the thermal voltage: 25.69 mV at 25 °C and 30.00 mV at 75 °C. At IE = 1.032 mA, re is 24.91 Ω at 25 °C. A fully bypassed stage has gain ≈ −(RC ∥ RL) ÷ re, so the example falls from −98.63 at 25 °C to −84.47 at 75 °C.
What does the emitter bypass capacitor do?
It shorts RE at signal frequencies but not at DC, so the bias stays stable while the gain rises. In the example (RC 3.3 kΩ, RL 10 kΩ, RE 1 kΩ) the gain is −2.397 with RE unbypassed, −19.67 with 100 Ω of RE left in, and −98.63 fully bypassed. The input impedance falls from 7.842 kΩ to 5.073 kΩ to 1.94 kΩ as it does.
Why is an emitter follower’s gain just below 1?
The output is the base voltage minus VBE, divided between re and the emitter load: Av = (RE ∥ RL) ÷ (re + RE ∥ RL). At 10 mA, re = 2.556 Ω, so into 1 kΩ ∥ 50 Ω the gain is 0.949. The follower’s value is its impedance change: about 10.09 kΩ in and 49.71 Ω out from a 10 kΩ source.
How do I use the BJT Amplifier & Bias Calculator?
Simply type your numbers and read the result, which refreshes the instant you change something. There is nothing to submit and nothing to wait for.
Does it cost anything or need an account?
No. The tool is completely free, there is no account to create, and it keeps working offline after the page first loads.
Is anything I type uploaded?
No. The tool works entirely on your device, so the values you enter never leave your browser.
Common Use Cases
Audio and sensor preamps
A bypassed common-emitter stage at 1.02 mA gives a gain of −98.63 into a 10 kΩ load, or −75.34 from a 600 Ω source once the 1.94 kΩ input impedance loads it.
Checking a bias design
See whether the Q-point leaves room to swing: the 12 V example has 3.37 V of headroom upward and 7.4 V downward, and RC = 12 kΩ would drive it into saturation.
Setting gain with emitter degeneration
Leaving 100 Ω of the 1 kΩ emitter resistor unbypassed sets the gain near −19.7 and lifts the bandwidth estimate from 794 kHz to 2.84 MHz.
Buffers and line drivers
An emitter follower biased at 10 mA drives a 50 Ω load with gain 0.949 and an output impedance of 49.71 Ω from a 10 kΩ source.
Headphone drivers
At 20 mA into 220 Ω ∥ 32 Ω the follower gives gain 0.9565 and 10.63 Ω out, but its negative peak is limited to 0.564 V before the transistor cuts off.
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