Simple Light/dark sensitive LDR to LED Activator Circuit

[alexs415]

success!!

hallelujah!

Thank you so much for all of your help!

I might be picking you brains for some hysteresis but for the time being it works!

speak soon!

Alex

 

[Reloadron]

Glad it is working so far. Every now and then when something works for me I just attribute it to the fact that ocasionally even blind squirrels find an acorn.

Ron

 

[alexs415]

Hey guys,

Regarding Ground. There is no ground implemented to any of the cables joining the circuit so do i just make a common ground from scratch? Therefore just joining up all unused terminals and things like the third leg of the pot together?

Regards

Alex

 

[Reloadron]

Everything marked (-) in the original drawing is ground or common. Everything on the VCC line marked (+) is VCC. Also, what I was getting at with the pot is attached to give you an idea. Hope that helps.

Ron

 

[alexs415]

Arr Thanks for that.

So when using batteries you will always connect earth to the ' - ' terminal?

Regards

Alex

 

[Reloadron]

Yes, battery negative (-) terminal can be common, ground or earth depending on the naming conventions used.

Ron

 

[alexs415]

Hey guys, different question this time.....what if I want to include a second led? Can I just connect it in parrell with the current one?

Regards

Alex

 

[MikeMl]

Hey guys, different question this time.....what if I want to include a second led? Can I just connect it in parrell with the current one?

Regards

Alex

Duplicate the LED current-limiting resistor, too.

 

[alexs415]

Hey guys, regarding the LED's can I use virtually any standard type?

Thanks again

Alex

 

[Reloadron]

Hey guys, regarding the LED's can I use virtually any standard type?

Thanks again

Alex

As long as you know the LED forward voltage and current to calculate the series resistors yes.

Ron

 

[alexs415]

and that would be done using ohms law?

Therefore if using an LED with a forward voltage of 2v and a forward current of 30ma that would mean i need around a 100ohm resistor in-place?

thanks

Alex

 

[ericgibbs]

and that would be done using ohms law?

Therefore if using an LED with a forward voltage of 2v and a forward current of 30ma that would mean i need around a 100ohm resistor in-place?

thanks

Alex

hi alex,

For any LED resistor calculation.

Let
Vled = the forward drop of the LED
Aled= the rated current for the LED
Vs= the supply voltage to the resistor/LED string.
Rs = the required resistor

So:

Rs= [Vs-Vled]/ Aled
eg: Say Vs=12v and Vled= 3V and Aled=15mA

Rs= [12-3]/0.015 = 600R

 

[alexs415]

so therefore with this LED **broken link removed**

I would be looking at a 330R resistor?

Regards

 

[ericgibbs]

so therefore with this LED **broken link removed**

I would be looking at a 330R resistor?

hi,
Its rated at 2.5Vfwd at 30mA, I would run it at 25mA.

So thats [Vs-2.5v]/0.025

I don't now your Vs.?? But you should be able to work out Rs..

Regards

hi,
Its rated at 2.5Vfwd at 30mA, I would run it at 25mA.

So thats [Vs-2.5v]/0.025

I don't now your Vs.?? But you should be able to work out Rs..

 

[alexs415]

Thanks for that.

It's 12v DC although when measured its 13 therefore....

[13-2.5v]/0.025 = 420R

i assume in this scenario its better to have a little more resistance than less as we dont want to over load so a 470R should be sufficient?

Many Thanks

Alex

 

[ericgibbs]

Thanks for that.

It's 12v DC although when measured its 13 therefore....

[13-2.5v]/0.025 = 420R

i assume in this scenario its better to have a little more resistance than less as we dont want to over load so a 470R should be sufficient?

Many Thanks

Alex

hi,
A 470R would be OK, that will ~22mA.

 

[MarkSchlicher]

I have a similar need, for a remote cue light, powered by 9V battery, activated by a photocell taped to a camcorder's onboard record/tally light.

A friend sketched me a wiring diagram (I don't read schematics). A very simple circuit; good, because I have very limited electronics knowledge. It's basically a battery, wired in series with a photoresistor and a red LED. I am pretty good with soldering, it's just the circuit details have me confused. Specifically, it appears that the photoresistor will basically substitute for the current-limiting resistor, true? I bought the RS grab-bag and in bright light the most sensitive one has a minimum resistance of 1K ohm.

This is much higher than the calculated R for the current-limiting resistor.
Given my 9v supply voltage and 1.7V forward voltage, and 20mA current rating on the LED, it's calling for 390R, 1/2watt

Questions:
do I also need a current-limiting resistor, since it will be in series with the photo-resistor?
am I wasting a bunch of current and possibly heating up the photoresistor unecessarily?
am i wasting significant potential brightness?
is there a better, but not too complex, way to use the photocell to trigger the LED? Is there a wiring diagram available?