Brief summary of how potential dividers work?

1. Feb 24, 2006

Chazzer3

Hi guys,

Does anyone have a good summary of how potential dividers work, (no flames!) Particuarly if you could, in a circuit which has two resistors in series, (with an LDR second), and coming off of the LDR (Potential divider?) is an Invertor...

The experiment is supposed to prove how the HIGH and LOW can let you turn an LED on and off...

Is that clear?

Thanks,

Charles -

2. Feb 24, 2006

Hootenanny

Staff Emeritus
http://www.doctronics.co.uk/voltage.htm [Broken]

I suggest doing a google search.

Last edited by a moderator: May 2, 2017
3. Feb 24, 2006

Chazzer3

Thanks for that, but I've already looked at that, (searched!) and it doesn't help with my specific request, please can you help? (1st post^^!)

Thanks,

Charles -

4. Feb 24, 2006

chroot

Staff Emeritus
What's an LDR?

A voltage divider is simply two resistors in series. Since the voltage drop across each resistor is proportional to its resistance (Ohm's law), the ratio of voltage drops between the resistors is equal to the ratio of the their resistances.

In other words, suppose V is the voltage applied to the divider. V1 is the voltage drop across resistor 1, which has resistance R1. V2 is the voltage drop across resistor 2, which has resistance R2.

The ratio of the voltage drops (V1/V2) is equal to the ratio of the resistances (R1/R2).

- Warren

5. Feb 24, 2006

Hootenanny

Staff Emeritus
It is easier to use a variable resistor and an LDR. You also need to use an OP-AMP to compare the voltages or a transitor. In truth your probably better asking the engineering guys.

6. Feb 24, 2006

Chazzer3

Thanks a lot, I only just grasp that, but thanks!

Charles -

mr burke...?

7. Feb 24, 2006

Hootenanny

Staff Emeritus
An LDR is a Light Dependant Resistor. Usually has a low resistance in bright light and a high in dark light, if I remember correctly it makes use of the photoelectric effect.

8. Feb 24, 2006

chroot

Staff Emeritus
I moved this to the Electrical Engineering Forum. I'm one of the so-called engeering guys, or at least that's what my business cards say.

- Warren

9. Feb 24, 2006

Chazzer3

Thanks, nice forum!

Anyway, basically, I have the question: explain how or why changing light level give a different voltage. (Explain how the potential divider works).

It is in this circuit:

So, the question is based on that circuit, I hope that's clear, I drew it up quickly, because I needed to!

Thanks,

Charles -

10. Feb 24, 2006

Tom Mattson

Staff Emeritus
Are these photoresistors? If so then you should mention that. When you say "resistor" most people think "chunk of carbon with little colored bands painted on it".

11. Feb 24, 2006

chroot

Staff Emeritus
Well, I don't know what the black box is, or why it's connected to your inverter, so I probably can't answer anything decisively about that part of the circuit. It looks like a power supply, however.

However, the voltage divider part is simple: as the photoresistor ("LDR") changes resistance, the ratio between it and the fixed resistor changes. Thus, the ratio between their voltage drops changes, thus the voltage presented to the inverter changes.

To help cement this in your mind, think about what the voltage divider does in extreme cases. (I'm going to call the voltage presented to the inverter the "output voltage" from the divider.)

If the photoresistor were zero resistance, there would be no voltage drop across it. (Ohm's law). If there were no voltage drop across the photoresistor, then the divider's output voltage must be equal to the voltage at its bottom.

On the other hand, if the photoresistor were infinite resistance, there would be no current at all through the divider. If there's no current, there's no voltage drop (Ohm's law), and the output of the voltage divider would be equal to the voltage at its top.

To take a more realistic case, let's imagine that the photoresistor had twice the resistance of the resistor up top. The photoresistor is thus 2/3 of the total series resistance, and drops 2/3 of the voltage. The fixed resistor up top is 1/3 of the total series resistances, and drops 1/3 of the voltage.

- Warren

12. Feb 24, 2006

chroot

Staff Emeritus
He's using slightly different terminology ("LDR" in place of "photoresistor"), but, according to his schematic, the lower resistor is a photoresistor, while the top resistor is a fixed resistance.

- Warren

13. Feb 24, 2006

Chazzer3

Thanks for the lengthy replies, they are really nice, but I just can't seem to get this, (I mainly just don't get the complete answer, sorry, I mean, I don't get exactly what the question asks for in an answer, which you can't really help (?)) Thanks again,

Charlie -

14. Feb 24, 2006

chroot

Staff Emeritus
Well, the question is a relatively open-ended question, after all Chazzer3.

How about we approach it this way: let's have you do an example for us. Maybe you just need to put pencil to paper to make it clear.

Let's imagine the top resistor is 100 ohms, while the bottom photoresistor (at some particularly illumination) happens to be 200 ohms. Further, let's imagine the potential at the top of the divider is 5V, while the potential at the bottom is 0V.

Can you calculate the ratio of the resistances? Can you calculate the ratio of the voltage drops? Can you therefore tell me how much of the voltage is dropped by the top resistor, and how much is dropped by the bottom photoresistor?

- Warren

15. Feb 24, 2006

Chazzer3

Oh, well, that does seem to help now, thanks, I can probably figure it out now to be honest, thanks so much for your help, this is a great forum!

Charlie -

16. Feb 24, 2006

chroot

Staff Emeritus
Can you answer the questions I just posed you? I think you'd understand this a lot better as soon as you start playing with it with some real numbers.

- Warren