11 min readConcept sheetSection A MCQ (1 mark) and objective (2 marks, two parts); Section B very short answer (2 marks) on cut-off and saturation; Section B short answer (3 marks) on the oscillator and its feedback

Transistor as a switch and as an oscillator

As a switch, a transistor jumps between cut-off (off) and saturation (on). As an oscillator, an amplifier with positive feedback keeps an LC circuit oscillating.

Cut-off (switch off)IB = 0, so IC ≈ 0 and the output Vo = VCC

Saturation (switch on)Large IB, VCE ≈ 0, so Vo ≈ 0 and IC = VCC/RL

Sustained oscillationLoop gain Aβ = 1, with a total phase shift of 360° (positive feedback)

Frequency of an LC circuitf = 1/(2π√(LC))

The two states of a transistor switch

Off: cut-offThe input VBB = 0, so VBE is below the barrier and IB = 0. No collector current flows. Nothing drops across RL, so the whole supply appears across the transistor: Vo = VCC (high).

On: saturationA high input sends a large base current. The transistor conducts fully, VCE falls to about 0, and IC = VCC/RL. The output is low, near 0 V.

Between the two lies the active region used for amplifiers; a switch jumps across it quickly. A high input gives a low output, so this circuit is also a NOT gate.

Worked example: is the switch on?

Problem

In a transistor switch, VCC = 9 V, RL = 1 kΩ, RB = 20 kΩ and β = 100. Take VBE = 0.7 V when it conducts. Find the output (a) when VBB = 0 and (b) when VBB = 5 V.

Working

(a) VBB = 0: no base current flows, so the transistor is cut off. Vo = VCC = 9 V.
(b) Input loop: IB RB + VBE − VBB = 0, so IB = (5 − 0.7) V ÷ 20 000 Ω = 2.15 × 10⁻⁴ A = 215 μA.
The most current RL allows is IC = VCC/RL = 9 V ÷ 1000 Ω = 9 mA.
β × IB = 100 × 0.215 mA = 21.5 mA. This is more than 9 mA, so the transistor is saturated.
So VCE ≈ 0 and IC = 9 mA.

Answer

(a) Vo = 9 V (off); (b) Vo ≈ 0 V and IC = 9 mA (on)

Using the switch: an LED indicator

Put an LED in series with the collector resistor. A high input (+5 V) saturates the transistor, collector current flows and the LED lights. A zero input cuts the transistor off, and the LED goes dark. The same on-off action handles the 0 and 1 signals inside computers.

How an oscillator keeps going

An **oscillator** makes a continuous AC signal of a chosen frequency, with no input signal. Its heart is an **LC tank circuit**.

1
A charged capacitor joined to an inductor makes the charge surge back and forth. Energy swaps between the electric field of C and the magnetic field of L.
2
Resistance and radiation waste energy, so these oscillations die away (they are damped).
3
So a transistor amplifier with gain A is added. A feedback circuit sends a fraction β of the output back to the input. Here β is the feedback factor, not the current gain.
4
If Aβ < 1 the oscillations die out; if Aβ > 1 they grow; if Aβ = 1 they stay steady.
5
A CE amplifier shifts the phase by 180°. The feedback circuit adds another 180°, so the fed-back signal is in phase with the input: positive feedback.
6★
Each cycle, the amplifier makes up the energy lost, so the tank oscillates continuously at f = 1/(2π√(LC)).

★ key step

How to draw the oscillator (Colpitts)

Block diagram: a box labelled amplifier, gain A, with input Vi and output Vo. From Vo, draw a loop back through a box labelled feedback, β to the input. Write Aβ = 1.
Colpitts circuit: an npn transistor in CE mode with VCC. Across its output, draw a tank circuit: an inductor L in parallel with two capacitors C1 and C2 in series.
Join the point between C1 and C2 to the common (earth) line.
Take the output across C1, and feed the voltage across C2 back to the base.
Label the tank circuit, the feedback path, the output and VCC.

Worked example: frequency of an LC oscillator

Problem

The tank circuit of an oscillator has L = 1 mH and C = 10 nF. (In a Colpitts circuit, C is C1 and C2 combined in series.) Find the frequency of oscillation, taking π = 3.14.

Working

LC = 1 × 10⁻³ H × 10 × 10⁻⁹ F = 1 × 10⁻¹¹ s².
√(LC) = √(10 × 10⁻¹²) ≈ 3.16 × 10⁻⁶ s.
f = 1/(2π√(LC)) = 1 ÷ (2 × 3.14 × 3.16 × 10⁻⁶ s) = 1 ÷ (1.98 × 10⁻⁵ s).
f ≈ 5.0 × 10⁴ Hz.

Answer

About 50 kHz

In real life

The quartz watch on your wrist, the clock inside a phone and an FM radio transmitter all depend on oscillators. A transistor switch flashes the small LED that shows a phone is charging.

Exam angle

2 marks: “Explain how a transistor works as a switch. In which two regions is it operated?” Cut-off: IB = 0, IC ≈ 0, Vo = VCC, so the switch is off. Saturation: large IB, VCE ≈ 0, IC = VCC/RL, so the switch is on. It jumps between the two; a high input gives a low output.

3 marks: “Why do the oscillations in an LC circuit die away, and how does a transistor amplifier with feedback keep them going? State the condition for steady oscillations, with a block diagram.” Energy is lost in resistance and by radiation, so LC oscillations die away. An amplifier (gain A) with feedback (factor β) makes up the energy lost in the LC tank. Condition: Aβ = 1, with 360° total phase (180° from the amplifier and 180° from the feedback): positive feedback. Block diagram.

1 marks: “What type of feedback is needed in an oscillator?” Positive feedback: the fed-back signal is in phase with the input.

Remember it

Switch: cut-off is off with a high output; saturation is on with a low output. Oscillator: Aβ = 1 and 180° + 180° = 360°.