Electrical – Light dependent resistor (LDR) switched LED-circuit

First basic error: you are driving the OPAMP inputs with single resistors connected to the supply lines: this simply ties the inputs to the same voltages, as there is virtually no current flowing in them.

So, first thing to do is understand what is a voltage divider.

Then, you have to use the principle of the voltage divider to create voltages to the inputs of the OPAMP. Of these voltages, one will be fixed, and will be used as a reference, while the other will be variable in function of the value of the photoresistor.

The photoresistor

Let's talk about the photoresistor: this is a resistor that decrease its value when exposed to light; let's consider it in a binary way (for your purpose): low when there is light, high when it's dark. Now you have to specify what do you want as 'light', and what do you want as 'dark'. Then measure the value of the resistor in the two cases. If you are simulating, I don't know how to create these situation, try to take a look or use a variable resistance to simulate the effect.

Depending on the values chosen for the photoresistor, you will have to size the other 3 resistors such as they will generate a positive differential voltage in one case, and a negative in the other.

The AND gate

There is a "small" problem also with the AND inputs: the switch input, when this is open, will be floating and with an unpredictable value; so you should put a (quite big - 10K should work, for sure in the simulation) resistor, between that input of the AND gate and ground. This will pull down the voltage when the switch is open; but you really need the 680 Ohm? For sure not in the simulation, you can think about it in a real implementation.

(Sorry but I'm busy, I'm adding the answer a piece at a time...but try to understand those parts)

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.