Light sensor switch using LDR

Question

What is the reason for using the transistor (2N3904) is this circuit? What good is this transistor doing in this circuit.What happens if we remove the transistor from this circuit? And what will happen if we use a PNP transistor instead of a NPN transistor.

so my explaination about the original circuit which I implemented because I don't have 50 OHm resistor.When there is light in the room, the circuit inside LDR is complete and hence all the current flows through 1K and LDR branch, and very low current is at base so the amplified current by 2N3904 is insufficient to lightup the LED. When there is dark in the room, the circuit of LDR is open and hence the base current of 2N3904 is maximum (2.3mA) which after amplification is 232.2mA which is sufficient for the LED to work. The LED has voltage drop of 1.8V.

Answer 1

The transistor amplifies the current. You have a small current from base to emitter, and the transistor creates a larger current from collector to emitter. The amplification factor can be found as \$H_{FE}\$ in the datasheet, and for small signal transistors is often around 100. So 1 mA base current will result in 101 mA emitter current (that's 100 mA collector current + the 1 mA base current).

I'd like to repeat that this is not the best circuit. There should at least be a small resistor in series with the LED for regulation. If you replace the transistor with another one of the same type you suddenly may have two or three times the LED current. That's because the collector current in your circuit is only determined by base current and \$H_{FE}\$, there's not else limiting it. But for a BC337 \$H_{FE}\$ can vary between 100 and 600! So you can have a 1:6 variation in LED current. That's not good. Do it this way:

(By the way, drawn in 2 minutes with CircuitLab)

If you leave out Q1 you only have the current through R2 and Q2, and that increases with the light level. So you can't use that directly for the LED, for that you want the current to decrease, and also the current will be too low.

The voltage across R2 is constant: 3 V - 0.7 V = 2.3 V, so it's current will be constant too. The increase/decrease inversion is done by phototransistor Q2: if its current increases the base current to Q1 has to decrease, since the total is constant.

A PNP transistor works like an NPN, but with the currents reversed: a low current from emitter to base will cause a larger current from emitter to collector. If we would replace Q1 with a PNP then the circuit turns upside down:

This circuit does exactly what the other does: if it's dark there won't be any current through Q2, and R2 will cause base current in Q1. The current flows from the emitter of Q1 through its base to R2 and ground. That base current will cause a higher collector current which will light the LED. R1 will limit the current to a safe value. If there falls light on Q2 it will cause a higher current through R2, but that current was constant at 2.3 mA ((3 V - 0.7 V) / 1 kΩ), so the base current will decrease, and so will the LED current.