Making a power supply with L200 to regulate both voltage and current.

[cthulhudarren]

I've been working with L200CV Pentawatts to get a DC power supply working. It would be adjustable from just under 3V to 36V. The current is set by a small value resistor, around 1 ohm or less, but if you had the right rheostat in parallel with that resistor you could make that adjustable too.

So this should work great.


http://www.zen22142.zen.co.uk/Circuits/Power/l200.html

I've been having an issue where I am using 2 alkaline 9V batteries in series and have just over 18v at the L200 input, so theoretically I should be able to get an output of ~17v max. So far I haven't been able to get the thing working, my adjustable range is only up to 2.9volts. I don't know if it is because the batteries can't supply the IC with enough current or what (can't find any minimum input requirements for current in the spec-sheet which is linked to in the above link), but I'm going to have to breadboard this to troubleshoot why it doesn't work.

But looking at the website it says the Quiescent current is 4.2mA Two batteries should be able to supply that.

Has anyone ever successfully used these for a power supply?

[iccaros]

you RSO is a .45 ohm 1% resistor?
This is strait from the datasheet so I do not see why it would not work

[JRB]

I just happened to have soldered my power supply I made 3 years ago at uni so it was usable for pedals today and it happened to use the L200.

Comparing your schematic to mine the difference I see are:
I don't have Rsc its just a connection between pin 5 and 2.
I connected the open pin of R2 to pin 2.
And both my caps are polarized.
C1 = 470u
C2 = 220u

I haven't tested my new design yet, but the only change I made is that the old design had a common ground and I split that up so I had isolated outputs.

[cthulhudarren]

I'm actually using a 1 ohm 1% resistor, which should limit the current to <500mA. I don't think 2  9V batteries can supply that much current anyway. But I am not getting the .45 v from pin 2 to pin 5. I get essentially 0v. I also don't feel the IC heating up. So I don't think it is in operating mode.

I have tried making R1 smaller with not much success either. And this is with 3 different ICs. I even tried hooking up the IC backwards in hope that somehow my pinout was wrong. I also tried taking out C1 to no avail.

The only thing I have not done is put the capacitor from pin 2 to ground. I figured that was just for smoothing anyway and shouldn't effect the voltage I read on the meter.

I was going to veroboard this but then I realized I should be breadboarding it first. I need to buy a new breadboard. They are pretty cheap online.

At this point I am practically praying that the problem is that my batteries aren't providing enough power to get the IC to function. But the IC Quiescent value tends to point to otherwise. I am also assuming that the + side of my battery terminal goes to +Vcc and he negative terminal goes to the Ov node which connects to pin 3.

It has got to be something simple!

EDIT: I found the proper line on the datasheet:

Id Quiescent drain Current (pin 3)   
Vi = 20V
typical
4.2 mA
max
9.2 mA

But I'm not sure what the drain is at Vi=18v, and it now occurs to me that I should measure this current in my circuit to see if anything is flowing.

[cthulhudarren]

I just happened to have soldered my power supply I made 3 years ago at uni so it was usable for pedals today and it happened to use the L200.


Comparing your schematic to mine the difference I see are:
I don't have Rsc its just a connection between pin 5 and 2.
I connected the open pin of R2 to pin 2.

I think this just means that you are not limiting the output current.

I haven't tested my new design yet, but the only change I made is that the old design had a common ground and I split that up so I had isolated outputs.

I don't understand this.

[JRB]

Basicly 3 years ago we had to make a power supply as a assignment.
The power supply needed to have two outputs.
One output was +15V the other one -15V.
This power supply worked.

Since you can dial in what ever voltage you want with the L200 I figured why not change that to two times isolated +9V.
What I had to do was split up the common ground between the two L200's so I could get two +9V outputs instead of two.
This power supply hasn't been tested yet.

If you want I can try and upload the schematic I used it doesn't have the values on it but at least you get a view of what components go where.

[cthulhudarren]

Basicly 3 years ago we had to make a power supply as a assignment.
The power supply needed to have two outputs.
One output was +15V the other one -15V.
This power supply worked.

Since you can dial in what ever voltage you want with the L200 I figured why not change that to two times isolated +9V.
What I had to do was split up the common ground between the two L200's so I could get two +9V outputs instead of two.
This power supply hasn't been tested yet.

If you want I can try and upload the schematic I used it doesn't have the values on it but at least you get a view of what components go where.

Sure, thanks. The circuit I was integrating this supply into had a DPDT (labeled "NPN/PNP" ) switch wired to change the supply voltage from "+" to "-". That part worked, thankfully.

[iccaros]

I'll try and bread board this, I have 1 L200c I got in a regulator Pack.. see If I can repeat what you get or what I think you should get..

[cthulhudarren]

I'll try and bread board this, I have 1 L200c I got in a regulator Pack.. see If I can repeat what you get or what I think you should get..

Thank you very much in advance!

[PRR]

> the Quiescent current is 4.2mA

This means the LOAD must be at least 4+mA. (The internal circuitry "steals" the power to run itself from the load current.)

Your voltage-set network pulls about 2mA.

Hang a 1K resistor across the Output and see if it works better.

[JRB]

Here it is.



X1-1 and X1-2 are the 230VAC connection.
X2-1, X2-2, X2-3 and X2-4 are the 9VDC outputs (or what ever you set the L200's to).

[cthulhudarren]

Okay, so as Fred Sanford used to say, I'm a big dummy.

I started from scratch and bread-boarded the power supply circuit. It worked great, even without the output cap since I'm not rectifying AC. I used a 12VDC 400mA power supply so it was actually outputting over 16 volts, which is great. Then I went back in to re-verify everything yet again. And then I found that I had made a mistake in the wiring color scheme I used to ensure that I connected everything up correctly. I had essentially swapped the wires from pin 4 and pin 5 in the IC. I can't believe that I hadn't seen it before. I had "verified" it a dozen times.

It took me this long to be able to look at my own circuit objectively and see my problem. The human psyche is a wonderful thing.

So I swapped those wires and *boom*, I'm golden.

Thanks for reading along.