Now design a BJT circuit which turns a red LED on when the room lights are off.
Conversely, the LED should extinguish when the room is lit again. (Cup your handsaround the circuit to simulate darkness.)

You will need a light-sensitive resistor for this design.

A cadmium sulfide (CdS) photocell has a low resistance when exposed to light (typically 1-4 k) and a high resistance in darkness (typ. 50k-100 k).

Materials -> R1 = 100 Ω, NPN BJT 2N2222, R3 = light-sensitive resistor, R4 = 120 KΩ, 5 V battery, red led

Design -> please see the image attached

Explanation:

We use LTspice to simulate the circuit that consists of a voltage divider in the gate of the transistor to change the operation region of the BJT accordingly to the resistance of the light-sensitive resistor; when R3 (light-sensitive resistor) has low resistance the voltage of the gate and emitter is approximately the same and the transistor is in a Cut-off state, when R3 increases its resistance the voltage of the gate start to rise and as a result, there is in a Forward-active state; a diode is placed on the collector with a small resistance R1 and whenever the transistor is active the led is going to turn on.

Note: If you need more current for the led you could reduce the value of R4, this will produce that lower values of R3 activate the transistor, you can adjust the circuit just by changing the value of R4, therefore, I highly recommend to use a potentiometer for R4 in the range of 10-300 K.

no he didnt use it

explanation:

he said no

answer: make you irritable