Beginner LDR question - I just don’t get it

Hi I’ve decided this week I’d like to start learning electronics and I have a few components from a raspberry pi project that never got off the ground so decided to build my first circuit, exciting! Naturally I pulled out a resistor, LED and 9v battery, classic. Revelling in my “current” success (get it?) I thought I’d be clever and increase the resistance so this light was off then add an LDR to the circuit in parallel to the resistor expecting the light to get brighter when I cover the LDR, disappointment ensued.

My understanding is; if I add more resistors in series the total resistance increases but if I add more resistors in parallel the total resistance decreases.

My theory was; adding the LDR in parallel then covering it up would increase the resistance value of the LDR in turn decreasing the total resistance in the circuit and lighting up the LED

What happened was; the exact opposite

My question; where did I got wrong? Does the LDR behave differently from other resistors (aside from the light dependant part)? I have seen another circuit with a transistor involved why does this need to be added?

any help would be much appreciated, thanks,
Marc

Resistor in series add. Your two resistors (A resistor and an LDR.) in parallel so the formula is 1/R = 1/R1 +1/R2 +1/R3 and so on.. R is the combined resistance and R1, R2, R3 and any other resisters are the resistors connected in parallel.. So in your case 1/R = 1/R1 + 1/R2 so (Multiplying both sides by R1 x R2)
R1 x R2/R = R2 + R1 So( multiplying both sides by R) R1 x R2 = R(R2 +R1) so R = R1 x R2/(R1 +R2). The LDR behaves the same way as other resistors and you are correct that it's resistance increases when it is in the dark, So the resistance of the parallel resistors also increases.

Les.

Your mistake is that the LDR is probably far too high a resistance to feed an LED, you would normally use an opamp (as a comparator) to connect the LDR to the LED.

TheMarcKnight said:

My understanding is; if I add more resistors in series the total resistance increases but if I add more resistors in parallel the total resistance decreases.

Click to expand...

Exactly correct; but the proportions of change depend on the exact values and ratios of resistors in the actual circuit.

The LDR is effectively a different value resistor depending on how much light it is exposed to.

Totally dark, it's very high, probably enough to consider it's not doing anything significant - so you just have the fixed resistor providing current for the LED.

At the other extreme, with bright light, it will be quite a low resistance, probably a lot lower than the fixed resistance so providing a lot more current than that & the LED is brighter.

At in-between light levels, it will allow some extra current to the LED.

When thinking of resistors in parallel, always keep in mind that the equivalent resistance would be smaller than the smallest one.

TheMarcKnight said:

My theory was; adding the LDR in parallel then covering it up would increase the resistance value of the LDR in turn decreasing the total resistance in the circuit and lighting up the LED

What happened was; the exact opposite

Click to expand...

Yes, adding a resistance in parallel will lower the total resistance, but if you increase the value of that resistance, then the total parallel resistance will also increase, which is what you observed.

Think about it.
If you add a resistance in parallel, the total resistance will drop, but the higher the resistance added, the less the total will drop.

TheMarcKnight said:

My theory was; adding the LDR in parallel then covering it up would increase the resistance value of the LDR in turn decreasing the total resistance in the circuit and lighting up the LED

Click to expand...

let's say you have a LDR in parallel with a 1k resistor, and that with light on it it, the LDR is 1k, and in the dark, the LDR is 10k. with light on it, the total resistance is 500 ohms, with the LDR in the dark, the total resistance is 909 ohms. remember the formula for resistances in parallel is 1/(1/R1+1/R2)