A switch worked by darkness

Using that principle, activity 11.6 builds a switch that turns a lamp on automatically when it gets dark. Its light sensor is a light-dependent resistor (LDR): in light its resistance is very low (ohms), in the dark very high (around 100 kΩ).

Activity 11.6

You will need: a D400 or D313 transistor, an LDR, a 10 kΩ volume control V_R, a 2.5 V bulb, a 3 V battery and a project board

Cover the LDR with a fingertip to darken it and adjust V_R until the bulb lights.
Take your finger away and let light fall on the LDR.

Figure to be added

Activity 11.6 — V_R from + to the base, the LDR from the base to −, the bulb in the collector lead.

The observation

The bulb goes out in the light. V_R can be set so the bulb comes on at the darkness wanted.

The potential divider

V_R and the LDR form a potential divider across the 3 V supply. With R₁ on top and R₂ below, the current is I = 3 ÷ (R₁ + R₂), and the base voltage is V_B = R₂I = 3R₂ ÷ (R₁ + R₂). Keeping R₁ fixed and changing R₂ gives any voltage from 0 to 3 V at the base.

With R₁ = 10 kΩ, find R₂ for V_B = 0.7 V:

0.7 = 3R₂ ÷ (10 000 + R₂)

7000 + 0.7R₂ = 3R₂

7000 = 2.3R₂

R₂ = 7000 ÷ 2.3 ≈ 3043 Ω

As the light on the LDR fades and its resistance climbs past 3043 Ω, the bulb just starts to glow; darker still, the base passes 0.7 V and I_C rises to its maximum — the switch is ON.