11 min readConcept sheetSection A MCQ (1 mark) and objective (2 marks, two parts); Section B very short answer (2 marks) numerical on β; Section B short answer (3 marks) on input and output characteristics

Common emitter characteristics and current gain

In CE mode, the input characteristic is IB against VBE and the output characteristic is IC against VCE. From them we find input resistance and β = ΔIC/ΔIB.

Input characteristicIB against VBE at constant VCE; like a forward diode curve, rising steeply after about 0.6 V for Si

Input resistanceri = ΔVBE/ΔIB at constant VCE

Output characteristicIC against VCE at constant IB; rises quickly, then is nearly flat

AC current gainβ = ΔIC/ΔIB at constant VCE

The circuit

To study CE characteristics, an npn transistor has its emitter common to both circuits. The base circuit has a variable supply VBB (about 0 to 3 V), a microammeter for IB and a voltmeter for VBE. The collector circuit has a variable supply VCC (about 0 to 15 V), a milliammeter for IC and a voltmeter for VCE. Potentiometers set each voltage. For a pnp transistor, every supply is turned round.

Input and output characteristics

Input characteristicKeep VCE fixed. Raise VBE in steps and read IB. Below about 0.5 V, IB is almost zero. Above about 0.6 V it rises steeply, like a forward-biased diode. The input resistance ri = ΔVBE/ΔIB is the reciprocal of the slope. In CE mode it is typically of the order of 1 kΩ. (Your lesson also quotes 20 to 100 Ω in one place; about 1 kΩ is the usual CE value.)

Output characteristicKeep IB fixed (say 20 μA). Raise VCE from 0 in steps and read IC. IC rises quickly at first, then becomes nearly flat. Repeat for larger IB: each curve lies higher. The slope ΔIC/ΔVCE is the output admittance; its reciprocal is the output resistance.

The flat part of the output curves is the active region used in amplifiers. The bottom curve (IB = 0) marks cut-off, and the steep part near VCE = 0 is saturation.

How to draw the CE circuit and the curves

Circuit: an npn symbol with E joined to the common line. Base side: VBB, a potentiometer, a μA meter in series with B, and a voltmeter VBE across B and E.
Collector side: VCC, a potentiometer, an mA meter in series with C, and a voltmeter VCE across C and E.
Input curves: x-axis VBE (V), y-axis IB (μA). Flat near zero, then rising steeply after about 0.6 V. Label each curve with its VCE.
Output curves: x-axis VCE (V), y-axis IC (mA). Draw four or five curves, each rising quickly and then levelling off. Label them IB = 20, 40, 60, 80 μA from the bottom up.

Activity: npn transistor in CE mode: current gain and voltage gain

Aim: draw the output characteristics of an npn transistor in CE mode, then find its current gain β and its voltage gain AV with a 1 kΩ load. Use a medium-power npn transistor, 9 V and 1.5 V supplies, a 0 to 300 μA meter, a 0 to 30 mA meter, a 0 to 10 V voltmeter, rheostats, a 20 kΩ base resistor, and 1 kΩ and 4 kΩ resistors.

1
Identify the leads (the emitter is next to the tab) and connect the CE circuit. Set both rheostats to zero; every meter should read zero.
2
Set IB = 100 μA. Raise VCE in small steps up to about 9 V and record IC each time. Keep IC below 30 mA.
3
Repeat with IB = 150 μA and 200 μA. Plot IC against VCE for all three.
4
Current gain: draw a vertical line at VCE = 5 V. Read IC where it cuts each curve. Find β = ΔIC/ΔIB for each pair of curves and take the mean.
5
Voltage gain: put Ro = 1 kΩ in the collector and Ri = 4 kΩ in the base. Set VCE near 5 V. Change the input voltage by a small step δVi (0.05 to 0.2 V) and note the change δVo in VCE.
6★
Record AV = δVo/δVi for five or six readings. Observe that it is nearly constant, and check that AV ≈ β × Ro/Ri.

★ key step

Worked example: β and voltage gain from readings

Problem

At VCE = 5 V, a transistor gives IC = 15 mA when IB = 100 μA, and IC = 22 mA when IB = 150 μA. (a) Find β. (b) With Ro = 1 kΩ and Ri = 4 kΩ, find the expected voltage gain, and the output change for a 0.1 V input change.

Working

(a) ΔIC = 22 − 15 = 7 mA = 7 × 10⁻³ A.
ΔIB = 150 − 100 = 50 μA = 5 × 10⁻⁵ A.
β = ΔIC/ΔIB = 7 × 10⁻³ ÷ 5 × 10⁻⁵ = 140.
(b) AV = β × Ro/Ri = 140 × 1000 Ω ÷ 4000 Ω = 35.
δVo = AV × δVi = 35 × 0.1 V = 3.5 V.

Answer

(a) β = 140; (b) AV = 35, so the output changes by about 3.5 V

Worked example: input resistance

Problem

At VCE = 2 V, raising VBE from 0.70 V to 0.72 V raises IB from 20 μA to 40 μA. Find the input resistance.

Working

ΔVBE = 0.72 − 0.70 = 0.02 V.
ΔIB = 40 − 20 = 20 μA = 2 × 10⁻⁵ A.
ri = ΔVBE/ΔIB = 0.02 V ÷ 2 × 10⁻⁵ A = 1000 Ω.

Answer

ri = 1000 Ω, that is 1 kΩ

In real life

Engineers who design mobile phones and music systems read these curves on a transistor's data sheet. They pick a base current that keeps the transistor in its flat, active region, so voice and music are amplified cleanly.

Exam angle

3 marks: “Draw the circuit to study the characteristics of an npn transistor in CE configuration. Draw its typical input and output characteristics.” Circuit: VBB, μA meter and VBE voltmeter in the base circuit; VCC, mA meter and VCE voltmeter in the collector circuit. Input: IB against VBE at constant VCE, near zero to about 0.5 V, then steep. Output: IC against VCE at constant IB, rising then flat, higher for larger IB.

2 marks: “From the output characteristics, IC changes from 4 mA to 6 mA when IB changes from 20 μA to 30 μA at constant VCE. Find β.” β = ΔIC/ΔIB = 2 × 10⁻³ A ÷ 10 × 10⁻⁶ A = 200.

1 marks: “Name the quantity kept constant while drawing an output characteristic of a transistor in CE mode.” The base current IB.

Remember it

In: IB against VBE (VCE fixed). Out: IC against VCE (IB fixed). Gain: β = ΔIC/ΔIB.