Basic LED circuit with LDR

Hi folks,

my first built. I was trying to get the sensitivity of the LDR (light dependent resistor, marked R4) down, but could not manage it. I would like it to switch the led "on" at dark (around 15kOhm LDR resistance). Ideas?

by olafesq
October 05, 2012

I assume you want to vary R4. To do so, you need to connect the center point of it to one of the end-points.

by carlos1w
October 05, 2012

Hi, sry for the confusing notation. I changed the name R4 to LDR now, I use the potentiometer symbol to show that the LDR resistance varies with light. When sunny then led off, when dark the led is switched on.

by olafesq
October 06, 2012

You still need to connect the slider (the unconnected end of the pointer) either to ground or the junction of the three resistors otherwise the resistance between the end points of the potentiometer will not change as you vary the slider position parameter 'k'.

That said, in your DC Sweep simulation, you have just swept the total resistance value of the pot you have called LDR:

LDR_R

That works OK.

You could just replace the pot symbol with a 2 terminal resistor and achieve the same thing.

Note however that you will not get a fast switching action between light on and light off. You can see this in the led current plot from your sim.

You will need to use some sort of comparator with hysteresis to do that.

:)

by signality
October 07, 2012

The sort of thing illustrated here ....

For more info:

http://en.wikipedia.org/wiki/Schmitt_trigger

by signality
October 08, 2012

Thanks signality, you are answers are exactly on point. Schmitt trigger is a good keyword here (I was wondering how something like this is called).

by olafesq
October 08, 2012

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