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BJT Emitter Current

In a bipolar junction transistor, the three terminal currents are related by Kirchhoff's current law: the emitter current equals the collector current plus the base current. The base current is very small, typically less than 1% of the collector current. This tiny base current controls a much larger collector current, which is what makes the transistor amplify signals.

Written by the Vorzep Team

Semiconductor Devices BJT

BJT Emitter Current Calculator

Kirchhoff’s current law for a bipolar transistor: IE = IC + IB.

Inputs

Drag the handle (or focus it and press ↑/↓) to reorder. Scientific notation such as 2e-5 is accepted. Pick any unit; conversion is automatic.

Presets

Results

Enter values and press Calculate. Your answer will appear here.

Collector and emitter current vs. base current

For your current gain β, IC = β·IB and IE = (β + 1)·IB. The two lines are almost on top of each other when β is large, because IB is small. The dot is your input.

What is being calculated

NPN transistor showing base, collector and emitter currents An NPN bipolar transistor. Base current IB and collector current IC flow in, and the emitter current IE flows out. By Kirchhoff's current law, IE equals IC plus IB. IB IC IE BCE IE = IC + IB IC = β · IB IE = (β + 1) · IB α = IC / IE

Common examples

CaseIEβ

Illustrative inputs, not manufacturer data.

Relation between β, α and IE/IB

βα = β/(β + 1)IE/IB = β + 1

Higher β means α is closer to 1, so IE is only slightly larger than IC.

Unit conversions

QuantityConversion
Current1 A = 1,000 mA · 1 mA = 1,000 µA · 1 µA = 1,000 nA · 1 nA = 1,000 pA
Gain relationsα = β/(β + 1) · β = α/(1 − α)
Thermal voltageVT ≈ 25.85 mV at 300 K · 1 V = 1,000 mV

Formula

IE = IC + IB
  • IE: emitter current (A), leaving the emitter of an NPN
  • IC: collector current (A)
  • IB: base current (A)
  • β = IC/IB: common-emitter current gain
  • α = IC/IE: common-base current gain

Example: IC = 1 mA and IB = 10 µA give IE = 1 mA + 0.01 mA = 1.01 mA, with β = 100 and α = 1 mA ÷ 1.01 mA ≈ 0.9901.

This is Kirchhoff’s current law applied to the transistor, so it holds in every operating region. When you know β instead, IC = β·IB and IE = (β + 1)·IB. When you know α, IE = IC/α. For a PNP transistor all currents reverse direction; enter magnitudes. With an optional VT, Advanced mode gives the small-signal emitter resistance re = VT/IE and gm = IC/VT.

References

How to Use This Tool

  1. Pick a preset, or enter the collector current IC and base current IB with their units.
  2. In Advanced, choose what you know instead: IB and β, IC and β, or IC and α.
  3. Optionally add VT for the emitter resistance and gm.
  4. Click Calculate.
  5. Read IE in the highlighted result and check β, α and the chart.

What This Tool Offers

  • Basic and Advanced modes on one calculation engine
  • Current unit conversion from A down to pA
  • Four ways to enter the known values (Advanced)
  • β, α, current shares, re and gm
  • IC and IE versus IB chart and labelled transistor diagram
  • Reorderable inputs, presets and a mobile-friendly layout

Variables

Symbol Quantity Unit
I_E Emitter current (sum of collector and base currents) A
I_C Collector current (majority carrier current) A
I_B Base current (recombination current) A

What it means

In a Bipolar Junction Transistor (BJT), the emitter current I_E is the sum of the collector current I_C and the base current I_B: I_E = I_C + I_B. This follows from KCL at the emitter node. Typically, I_C = β·I_B, where β is the current gain (typically 50‑300). Since I_B is much smaller than I_C, I_E ≈ I_C. The emitter current is the total current entering the transistor, and it splits into base and collector. This relationship is essential for biasing BJT circuits and for calculating power dissipation. Example: If I_C = 10mA and I_B = 0.1mA, then I_E = 10 + 0.1 = 10.1mA. The β = 10/0.1 = 100, which is a typical value for a small‑signal transistor.

Worked example

BJT Emitter Current – Practical Example

Real‑World
Scenario: A BJT has collector current IC = 2 mA and base current IB = 20 µA. Find the emitter current.
ParameterValue
IC2 mA
IB20 µA = 0.02 mA
FormulaIE = IC + IB
1IE = 2 + 0.02 = 2.02 mA
Final Design IE = 2.02 mA ✓ Emitter current
Why: The emitter current is the sum of collector and base currents – a fundamental Kirchhoff's law applied to the transistor.

Common mistakes

  • Currents: I_E = I_C + I_B – the emitter current is the sum of collector and base.
  • Approximation: For high β, I_E ≈ I_C.
  • β: I_C/I_B – typical values 50‑300.
  • Sign: For NPN, currents are positive into the device; for PNP, opposite.
  • KCL: This is simply a Kirchhoff current law at the transistor node.

Applications

BJT emitter current I_E = I_C + I_B describes the total current entering the emitter, which splits into collector and base currents. Since β is large, I_E ≈ I_C. This relation is used to bias BJT amplifiers, to design current sources, and to analyse circuit performance. Engineers use it to compute voltage drops and to ensure proper operation. The base current is also used to determine input impedance. By understanding this equation, professionals can design reliable BJT circuits and predict their behaviour. This is a core equation in transistor circuit analysis.

  • BJT amplifier biasing and DC operating point design
  • Current source and mirror circuit design
  • Input impedance and loading effect analysis
  • Temperature compensation in analog circuits
  • Educational understanding of BJT currents

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  • Published July 10, 2026
  • Last updated October 9, 2026