variable voltage output

[scsn]

Hi,
What would the best way to produce 0-12v with maximum of 1A? I am able to produce 0-12v using a digital potentiometer connected to a processor, or D/A then using opamp to scale it up.

But I am not sure the best way to be able to then supply 1A maximum from the 12v output , as I know I cant do it with opamp and D/A etc.

Many Thanks
Scott

 

[alec_t]

0-12v

AC or DC (I'm guessing DC)? Derived from a mains supply (presumably), or what? Mains isolated (for safety)?

 

[DerStrom8]

Hi Scott. Welcome to ETO!

If you're using AC, you'll need a rectified supply that can produce between, say, 12 and 18 volts, and an LM317 regulator. You'll need a large heatsink on it, though. I believe the 317 (adjustable voltage regulator) can supply up to 1 amp, but I can't be sure. If you're using DC, then it's easier, since you don't need any rectification. Simply follow the diagram below. Changing the value of R2 (the potentiometer), you can change the voltage output.

View attachment 61126

Be careful with the wiring, though. The diagram there does not show a very good symbol for the pot. You'll need to connect one side to the 'Adj' pin, and the wiper (middle of the pot) needs to connect to ground. Do not connect only the outer pins, or else it will not be adjustable.

I hope this helps!
Regards,
Der Strom

 

[crutschow]

Note that the LM317 circuit, as shown by DerStrom8, has a minimum output voltage of 1.2V. If you really need to go completely to 0V you will need a different circuit.

 

[scsn]

Thanks for all your help. It will be running from a 24v DC power supply (I should have mentioned..).

1.2v is actually low enough. Will it be this when the pot is adjusted to 0R?

Scott

 

[ronv]

Here is the data sheet for the 317. Lots of good info there. Yes the 1.25 volts is with the pot at 0. But, with such a high input voltage (24 volts) you will have trouble getting 1 amp out of it because of the high power loss in the regulator. If you have a 15 volts supply that would be better.

https://www.electro-tech-online.com/custompdfs/2012/02/lm317-n.pdf

 

[audioguru]

It will be running from a 24v DC power supply (I should have mentioned..).

Then the input voltage is too high since you must throw away (24V - 12V - 1.2V)/1A= 10.8W of heat. How will you cool it?

1.2v is actually low enough. Will it be this when the pot is adjusted to 0R?

Almost. It will be when the pot is turned down to zero ohms, not 0R. What the heck is "OR"? An Operating Room in a hospital?

 

[DerStrom8]

Then the input voltage is too high since you must throw away (24V - 12V - 1.2V)/1A= 10.8W of heat. How will you cool it?

I agree. That's WAY too much power for the 317 to dissipate. Even with a huge heat sink it will still be very risky. That's a lot of heat! Do you need the full 12 volts? If not, you could take a 12 volt supply, but when the voltage is all the way up, it will still be 12v - dropout voltage of the regulator. I think anywhere around 14 or 15 volts will probably be enough, though you will definitely need a heat sink, preferably with a cooling fan blowing air across it.

Almost. It will be when the pot is turned down to zero ohms, not 0R. What the heck is "OR"? An Operating Room in a hospital?

'R' is a common notation for "ohms", just as 'k' is for "kilohms", and 'M' for "megohms". I don't know why they use R, but it is a popular way of signifying the value is in ohms.

 

[audioguru]

'R' is a common notation for "ohms"

No it is not. Some other weirdos say E or any other letter instead of saying OHMS.

 

[DerStrom8]

No it is not. Some other weirdos say E or any other letter instead of saying OHMS.

It could be national conventions, or just personal preference, but it is very common. I have seen many other notations for 'ohms', but the important thing is that the person reading the schematic, parts list, etc know it's not K or M.