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Potentiometer Identification

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Musicmanager

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Morning Guys

I hope one of you experts would just confirm I'm thinking straight .. .. ..

I've recently built myself a PSU which includes a chinese 0 - 30 vdc kit I got from Banggood .. .. very pleased with the results except that the variable volts and amps controls are a bit approximate and very difficult to adjust accurately. I've been reading and think that the answer is to change the pots for multiturns - quite straightforward.
However, the existing pots are not marked at all and the chinglish parts lists describes them as 10k W .. Am I right in thinking they are 10k 1W ??

Thanks

S
 
The Chinese power supply kit is a copy of the horrible old Greek kit power supply we joked about on Electronics-Lab forums:
1) Many of its parts are overloaded, especially its driver and output transistors.
2) With the recommended 24VAC transformer that is also overloaded it cannot produce 30V at 3A and the TL081 opamps operate at a total supply voltage higher than their maximum allowed voltage.
3) I can't remember if the Chinese kit has a main filter capacitor value that is way too low like the original Greek kit does.

In the forums I fixed it by using TLE2141 opamps with a much higher voltage rating, a BD139 better driver transistor with heatsink and two output transistors to share the heat.
 
Hi AG

Good to hear from you.

Yes, I'm aware of your modifications to this kit, you sent me details including a revised parts list and schematic some time ago. I decided that since I needed something to get me up and running I would build the kit 'as is' and then .. .. I've ordered another kit and some of the revised parts, still got some to find, and then I'll build your revised kit and if I'm successful I can then swap them over.

My immediate concern is that the 2 pots used for adjusting volts and amps are single turn and difficult to use so I wanted to fit some multiturns instead. The pots in the kit are not marked, but the parts list says 10k W which I interpreted as 10k 1 watt. I was just trying to check out my theory.

I hope the weathers better in Canada than it is here :(

S
 
Why not just measure the resistance and measure the maximum voltage across them when in the circuit. You can then work out the actual power dissipation in the pots.
 
Hi Les

I'm very much a novice - that all sounds a bit testicle for me ! :)

No need anyway :D

The 30V adjustment pot won't have more than 30V across it, and it could be less (the opamps might amplify the setting) - applying VxV/R gives 900/10000 - so less than 1/10th watt.

The current pot is likely to be even less voltage.
 
Nigel, I like your quick and dirty reasoning.

Taking the sophisticated approach for a change, I calculated that the voltage across potentiometer P1 is 10.2 volts, which gives a dissipation of 10.4mW.

As for the potentiometer being marked up as "10k W", I vaguely remember seeing the usual omega symbol being written as a W in one or two places.

My thoughts for what they are worth.

JimB
 
I think I had a hat when I came in :) hopefully, it's a tin one !!

If one of you guys would explain the calculation process that gives you these figures to work from, I wouldn't have to trouble you, I'd be able to work it out for myself :D

S
 
My calculation went like this...

Refering to the schematic of the Bangood PSU, which Music sent to me in a conversation a few days ago:
Bangood PSU.JPG

We are concerned with the dissipation in potentiometer P1.
The voltage across P1 is the putput voltage of op-amp U1.
R5 = R6 = 10k
D8 is a 5.1v zener.

To calculate the output voltage (Vout) of the op-amp:
First find an experssion for the voltage at pin2 of the op-amp V2 = Vout - 5.1
And find an expression for the voltage at pin3 of the op-amp V3 = Vout x R6/(R5+R6)
If the op-amp is working correctly V2 = V3 (or very near enough, I am not going to argue about it).
So, that gives:
Vout - 5.1 = Vout x R6/(R5 + R6)
using basic algebra, this can be kicked around to give Vout = 10.2 volts.

The power dissipated in P1 can be calculated using W = V^2 /R.

The voltage across P2 will be less than that across P1, and can be calculated by considering the effects of R17 and R18.

And that is how I did it.


Nigels quick and dirty method also has merit, using that method he found that the dissipation in the pot was reasonably low and could be tolerated by most 10 turn panel mounted potentiometers, end of problem.
If quick and dirty gave a high power disipation which would entail buying an expensive pot, then it is time to go sophisticated and get the true value of dissipation.

JimB
 
I would simply use big knobs on the pots. A big knob gives better resolution than a skinny knob. Then the knobs can have arrows and the panel can be marked with the output voltage and current settings.
 
Morning

True, but I don't have room on the front panel for big knobs ! I've ordered two 10k multiturns with 4mm shafts and a pair of knobs to suit :angelic:

S
 

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Not Sure How Accurate you want your output voltage to be, Or at what Current?
But Wire Size and Lengths between the Supply and the Load will create voltage drops.
So a Volt Meter at the Load is Better than the Volt Meter on the Power Supply.
 
I realize that you are panel space limited, but my preference is for two pots, one course and one fine. It gives you the resolution for setting the voltage you want, but you can get there faster.

The way to implement that is to add a 1K pot, wired as a rheostat, is between the top of the 10K pot and it's supply voltage. You don't get the same trim range across the full operating range, but it is better than a single, one turn pot.
 
MusicM;
I also built a variable power supply kit from Elenco.

Like yourself, I immediately found that the resolution of a single turn pot left much to be desired.
So I also replaced it with a 10-turn pot, and have been very satisfied with the result for over 5 years. I also installed larger diameter knobs.
I had also initially considered ChrisP's solution, but like yourself did not have enough front panel space.
 
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