Why Does the Early Effect Give a BJT a Finite Output Resistance?

Studyloaf Team•October 10, 2026•Last verified October 10, 2026•7 min read

Because a larger collector-emitter voltage widens the collector-base depletion region, which narrows the effective base and lets a little more collector current through at the same base-emitter voltage. The collector current therefore rises slowly with VCE instead of staying flat, and the inverse of that slope is the output resistance, ro = (VA + VCE) ÷ IC ≈ VA ÷ IC. With an Early voltage of 50 V and 1 mA at VCE = 5 V, ro is 55 kΩ, or about 50 kΩ by the usual approximation.

What you'll learn

  • What base-width modulation is and why it makes IC depend on VCE
  • How the Early voltage VA is defined from the output characteristics
  • How to get ro from the model, exactly and approximately
  • A worked example: ro, and how much it lowers a common-emitter stage's gain
  • Why gm × ro is a fixed ceiling for a single transistor's gain

What physically changes when VCE rises?

In the active region the collector-base junction is reverse biased, and its depletion region extends partly into the base. Raising VCE (with VBE held fixed) raises that reverse bias, so the depletion region pushes further into the base and the neutral base gets narrower. That is base-width modulation, first described by James Early. With a narrower base, the electrons injected from the emitter (for an npn transistor) cross a shorter distance to the collector, so their concentration gradient across the base is steeper and the diffusion current, which is the collector current, is larger. Fewer of them recombine on the way too, so β rises slightly as well.

How is the Early voltage defined?

Plot IC against VCE for several fixed base drives. In the active region each curve is a nearly straight line with a small upward slope. Extend those lines backwards past VCE = 0 and, to a good approximation, they all meet the voltage axis at the same point, VCE = −VA. VA is the Early voltage. The usual model that captures this is IC = IS e^(VBE/VT) (1 + VCE/VA), where VBE is fixed for each curve and VT is the thermal voltage.

BJT output characteristics extended back to the Early voltageCollector current against collector-emitter voltage for three base drives, computed from IC equals IS times e to the VBE over VT times 1 plus VCE over VA with VA equal to 50 volts. In the active region each line rises slowly with VCE. Extended backwards as dashed lines, all three meet the voltage axis at minus 50 volts, minus VA. The middle line carries 1 milliamp at VCE equal to 5 volts.-50-40-30-20-1001000.511.5VCE (V)IC (mA)dashed lines meetat −VA = −50 V1 mAat 5 V
Output characteristics computed from IC = IS e^(VBE/VT)(1 + VCE/VA) with VA = 50 V, for three base drives. The dashed extensions of the active-region lines meet at VCE = −50 V. The middle line carries 1 mA at VCE = 5 V.

How do you get ro from the model?

Differentiate the model with respect to VCE at fixed VBE: dIC/dVCE = IS e^(VBE/VT) ÷ VA. Since IC = IS e^(VBE/VT)(1 + VCE/VA), this equals IC ÷ (VA + VCE). So ro = (VA + VCE) ÷ IC exactly, within the model. When VA is much larger than VCE the VCE term is dropped to give the familiar ro ≈ VA ÷ IC, which is the form most textbooks and GATE solutions use.

Worked example: VA = 50 V, IC = 1 mA at VCE = 5 V

  • Exact output resistance: ro = (50 + 5) V ÷ 1 mA = 55 kΩ. The approximation VA ÷ IC gives 50 kΩ, about 9% lower.
  • Check against the model: holding VBE fixed, IC is 0.9145 mA at VCE = 0.3 V, 1 mA at 5 V and 1.0909 mA at 10 V. The rise from 5 V to 10 V is 0.0909 mA for 5 V, a slope of 1 ÷ 55 kΩ.
  • Now use the transistor in a common-emitter stage with RC = 10 kΩ. At 1 mA and VT = 25.85 mV (300 K), gm = IC ÷ VT = 38.68 mA/V.
  • Ignoring ro, the gain is −gmRC = −386.8.
  • Including ro, the collector sees RC in parallel with ro: 10 kΩ ∥ 55 kΩ = 8.46 kΩ, so the gain is −38.68 mA/V × 8.46 kΩ = −327.3, about 15% less.
Output side of the CE small-signal model with r o includedAt the collector, the dependent current source gm vbe with gm equal to 38.7 milliamps per volt sits in parallel with the output resistance r o of 55 kilohms and the collector resistor RC of 10 kilohms. Together the two resistances are 8.46 kilohms, so the gain magnitude falls from 386.8 to 327.3.gm·vberoRCvogm = 38.7 mA/V, ro = 55 kΩ, RC = 10 kΩAv = −gm(RC ∥ ro) = −327
The output side of the small-signal model with ro included: the dependent source gm·vbe, ro = 55 kΩ and RC = 10 kΩ all share the collector node, so the gain becomes −gm(RC ∥ ro) = −327.

Why is gm × ro a ceiling on gain?

Even with an ideal current-source load (RC infinite), the collector still sees ro, so the largest voltage gain one transistor can give is gm × ro. With the approximations gm = IC/VT and ro = VA/IC, the current cancels: gm × ro = VA ÷ VT, about 1934 for VA = 50 V at 300 K. This intrinsic gain does not depend on the bias current. Raising IC raises gm but lowers ro in proportion: at 0.5 mA ro is about 100 kΩ, at 2 mA about 25 kΩ.

Where else does ro matter?

Anywhere a transistor is meant to act as a current source. A current mirror's output current changes with its output voltage because of ro, so ro is the mirror's output resistance. Active loads in amplifiers rely on it for their gain, and cascode stages exist largely to multiply it. In GATE questions, ro is usually stated as 'neglect the Early effect' (ro infinite) or given through VA; read the question for which.

Common mistakes

  • Writing ro = VA × IC or VCE ÷ IC. It is the slope of IC against VCE: (VA + VCE) ÷ IC, or VA ÷ IC.
  • Forgetting ro in parallel with RC when the question gives VA, which overstates the gain.
  • Taking VA as a positive point on the plot. The lines meet at VCE = −VA; VA itself is quoted as a positive voltage.
  • Assuming ro is a fixed device constant. It scales as 1/IC, so it changes with the bias point.
  • Applying the Early-effect slope in saturation, where the steep rise near VCE = 0 is a different region entirely.

Check your understanding

  • Q1. A BJT with VA = 100 V is biased at IC = 2 mA. Roughly what is ro? Answer: ro ≈ VA ÷ IC = 100 V ÷ 2 mA = 50 kΩ.
  • Q2. At fixed VBE, IC rises from 1.1 mA at VCE = 4 V to 1.2 mA at VCE = 9 V. What is ro? Answer: 5 V ÷ 0.1 mA = 50 kΩ.
  • Q3. Using the same two points, what is the Early voltage? Answer: the line falls to IC = 0 at VCE = 4 − 1.1 mA ÷ 0.02 mA/V = −51 V, so VA = 51 V.

Related

Official source

Last verified

The GATE 2027 EC syllabus lists the BJT under Section 3, Electronic Devices, and small signal analysis of BJT amplifiers under Section 4, Analog Circuits (checked against the official PDF from IIT Madras on 10 October 2026). The output resistance, the currents on the characteristic and both gains were computed in code, with ro also checked by numerical differentiation.

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