DC Potentiometer in AC circuit to control DC output.

[abador1]

I finally finished a power supply last night after finally finding a working step down transformer. I put the 5K audio DC potentiometer in one of the AC legs from the transformer (series) and was amazed to find that I could get anywhere from about 18.6V to 3.6V. More recently I tried to charge a battery with the rectified and filtered DC output but I couldn't seem to get it above 3V and it was jumping around a lot. Could this unpredictable nature be because I have the potentiometer outside of the DC output? Could the AC ruin the DC potentiometer? I also have the fuse on the other leg of the transformer, would this protect from over current in the load that the power supply is connected to?

 

[Hero999]

Please post a schematic.

You can't use a potentiometer to vary the voltage to a large load because the output impedance of the potentiometer is too high so the output voltage will drop. Furthermore, the potentiometer is probably only rated to something like 0.25W and will burn out if the current is too high.

 

[abador1]

I'll see what I can do about the schematic, I'll have to draw it up. Could you please explain the impedance of the potentiometer a little bit? I know about the impedance from a coil of wire but never from a potentiometer.

 

[abador1]

Ok, I got it. Sorry it's a little messy. I tried cleaning it up but the simulation software I was using wouldn't let me take off some labels, the blued box at the bottom was only a command for the program. Basically the fuse is on one side of the circuit and the potentiometer is on the other side. The load obviously is not part of the power supply. I hope this helps give a better view of it.

View attachment 41442

 

[audioguru2]

You drew the connections of the diodes upside-down which makes the bottom wire of the 1000uF capacitor and the bottom wire of the 1k resistor positive. Then the LED has backwards polarity.

The 5k pot will probably get hot and burn out.

 

[Hero999]

The resistor in series with the LED is 1k and the load is 1k, giving a total load impedance of 500R, ignoring the LED's forward voltage drop.

The output voltage can be calculated using the potential divider formula which can be found on Wikipedia.
http://en.wikipedia.org/wiki/Potential_divider

 

[abador1]

I actually have a rectifier that is in one unit so I guess I drew the rectifier wrong, sorry about that. I ended up removing the potentiometer from the AC part of the circuit and put it in the DC in series with the output. I have still had trouble with the power supply though. I changed out the Capacitor to a 3300 uf just because that was the only other high capacitance cap that was within the voltage range. (more capacitance than I really need) I then checked the resistance on ether side of the DC circuit (+ and -) to check for shorts and the resistance was pretty high so I doubt that is the problem. I even checked the potentiometer which came out clean. I get a voltage reading every so often where it is supposed to be but this part is unpredictable and when ever I hooked it up to a 6V battery to charge, it would go back down to less than 1 volt. I am hoping if something is bad that it is anything other than the transformer just because it is so rare to find a working transformer when salvaging them. Could it maybe the 1K resistor and LED?

 

[Hero999]

Here's another way to control the brightness of an LED. It's slightly more complicated but the current through the LED is regulated (it doesn't change much as the voltage varies) and the filter capacitor is much smaller.

View attachment 41447

 

[abador1]

I probably should have labeled the out terminals in the schematic instead of labeling the resistor as load. Basically I'm just trying to use the LED as an indicator that it is working right/is on. The "load" is actually just supposed to simulate a load that could have been used on the output. I  think that made it a little more confusing.

 

[Hero999]

The load can be connected in parallel with C1 which can be increased in size, if needs be.

The circuit you posted is a bad idea because the potentiometer also increases the impedance of the power supply so the current and voltage to the load will also fall. The brightness of the LED will vary depending on the load connected to it.

 

[abador1]

Thanks for all the replies everyone. I ended up putting the potentiometer in the DC part of the circuit and in series with the load. I still don't know why the readings I kept getting were so unpredictable, all I can think of is that maybe it's the fact that my multimeter only cost me $12. I put a 4.7K resistor in series with the potentiometer to bring the voltage to a more workable level but I will probably increase that to a 10K. I was able to charge a battery up to a certain point but then found out it was bad after the voltage kept dropping fast when a lead was removed. I kept trying to measure the voltage across the output and was surprised that the potentiometer was not changing the voltage at all, then I remembered that the voltage will always measure the the total across an open and put a 10K resistor across the output to get a better reading. I am now getting a stable output. Thanks for all the input.

 

[Hero999]

I doubt there's a problem with the meter. The voltages were unpredictable because they depended on the current drawn by the load which is unpredictable.