Hi

I have got a 230V power supply with a 10k pot controlling the output voltage. I want to replace the pot with a LDR with the limits of when its completely dark, the power supply outputs 6V, and in sunlight, the output is 12V.
The current potentiometer outputs a 1.3V signal to the internal circuitry to give a 6V output, and 7.3V to give the p/s output of 12V. What circuitry would I need to add in this LDR to give the same values.
The LDR i have is has a resistance range of 5.5 - 1000kOhms.
Any other information needed, let me know, a quick response will be greatly appreciated.
Ollie

 

We would need to know the exact resistance of the LDR when it is at the illumination level you want 12V out of the supply. And the resistance of the LDR when it is in what you call "dark".

The is one other test you will have to do. Set up the supply to produce 6V. Temporarily connect a second 50K or 100K pot (wired as a two terminal rehostat, start with maximum resistance) between the wiper of the existing pot an the end of the existing pot that has the highest voltage on it. Slowly reduce the resistance of the add-on pot until the supply reaches 12V. Disconnect the add-on pot, and measure the resistance that it took to trick the supply into putting out 12V instead of 6V.

Knowing the resistance that it takes to increase the output voltage will allow using the LDR to create a corresponding resistance change...

 

Right, the resistance of the LDR at full illumination is 0.5K, and at no illumination its 60K Ohms. Ill do the other test tomorrow and get back to you. Thank you.

 

I think ive done it correctly, the pot has a resistance change of 2.9kOhms. From 0.526 @ 6V to 3.391 @ 12V. What circuitry will i need to allow the LDR to do the same job?

 

Can you post the power supply schematics?

 

Im afraid I cant post the scematic as i cant find one online. It is an old power supply, and the following link has the specifications if that helps at all.
http://www.lee-dickens.biz/process/sigcon_pdf/DIN710-720.pdf

 

I would perhaps use a setup with a comparator that trigger a relay. Would that be something you can do

 

I want it as basic as possible, all i need really is the circuitry needed to convert the LDR resistance into the resistance expected from the current pot to give the 6 - 12V output values.

 

If you want the system to switch from 12 to 6 volt you have to include some logic. This do not have to be very complicated.

 

Ok, change of plan, if I have a constant 15V supply, and using a LDR to vary the outputbetween 6-12V, so the output is 12V at full illumination and 6V at no illumination. How can I achieve this?

 

How do you want it? Shall the voltage gradual change from 12 to 6 volt, or do you want only two voltage levels 12 and 6 volt

 

I believe that you can do this with just the LDR, possibly a resistor or two, wired directly to the existing power supply. Go ahead and make the measurements I asked you to make.

 

just from the info you gave in post #3, #4

The LDR has 500ohms iluminated
the pot has 500ohms @ 6v.

The LDR will increase to 60K NON iluminated.
the pot only needs 3.4K @ 12v.

So if it is just a change in step voltage outputs 6v. or 12v.
then the LDR if it has the proper power handling should work, just in itself, as a replacement.

If I understand your question corrdctly.

 

I think ive done it correctly, the pot has a resistance change of 2.9kOhms. From 0.526 @ 6V to 3.391 @ 12V. What circuitry will i need to allow the LDR to do the same job?

I didn't see this post till just now. Is the pot wired as a rehostat (only one end and the wiper hooked up), or are all three terminals hooked up?