potentiometer burnout

MaxHeadRoom

Joined Jul 18, 2013
28,619
I don't understand what you mean by "one outer connection".

Not sure I do. It doesn't make sense to me that it would matter which way the current flows through the pot as long as it's flowing through the right connections. I tried it anyway like I said and same result.
There are three connections to a pot, the outer ends are the two ends of the pot connected to power +v and Common, the third is the slider which taps off this 0 to +ve.

There is only one correct way to connect it, there is usually a low voltage supply connected to the outer ends, the slider connection usually goes to a high impedance input of some kind, if you connect the slider to the +v say, and turn the pot to 0v, you are effectively causing high current to flow and end up shorting the supply.
Max.
 

bwilliams60

Joined Nov 18, 2012
1,442
I'm a newb to electronics but have been playing with DC all my life so I will take a stab at this. The pot you are replacing is rated for 10 to 2M ohms and you are replacing it with a 50K pot. So here's my logic and somebody tell me if I'm in left field. A pot burns up because too much current is passing through it, regardless of how he is hooking it up. Too much current = too little resistance. Am I off base saying that his pots are just too low for the circuit? They could have been set on the board at way over 50 K to begin with. Just a thought. I think you need higher resistance pots.
 

MrChips

Joined Oct 2, 2009
30,720
I'd say you're in left field.
How do you know the pot is 10 to 2M ohms?

According to the photos shown, component side and solder side, it appears that the pot is wired as a variable resistor with two leads shorted.

What is the value of the original pot?

What is the function of the pot?
Why does the pot blow? We need to see a circuit diagram.
 

Thread Starter

Synaps3

Joined Jun 5, 2013
99
I tried tshuck's suggestion, but I was only able to find one 33k and one 100k ohm resistor. Nothing else came close. I tried it and it made no difference, so I did some more investigating. I measured the voltage with the pot fully open and found that one terminal is ground, the second is 8.2v and the third is always the same as the second. When I begin to turn the pot, the RF power output increases and the voltage drops to about 8v on both wires at which point it begins to burn out.

I'll get a better picture of the back. There is one on the site I linked earlier, however the back of mine is a little different (I think the circuit went through some revisions since). The front is the same though.
 

MrChips

Joined Oct 2, 2009
30,720
If the pot is rated at 1/2W the resistance can be as low as 130Ω.

Put a 220Ω resistor in series with the pot.

But before you do that I want to see the top side and bottom side of the circuit board.

What is the resistance of the original pot?
 

bwilliams60

Joined Nov 18, 2012
1,442
Hey Mr Chips,
Before you go throwing me out into left field, perhaps you should look at the facts. Here, let me get them for you.
http://www.bourns.com/pdfs/3296.pdf
I know I'm new but you can treat me with a little respect. Just makes sense to me. If I'm wrong, just explain why without being ignorant. Have a good day and happy reading.
 

Thread Starter

Synaps3

Joined Jun 5, 2013
99
I thought I found the problem on the back of the board there was a really tiny amount of solder that was shorting one of the pins to ground. It was so small I figured it couldn't be intentional, so I removed it and connected another new pot. This time it didn't burn out, but it would have these surges where when set to full the output power would slowly drop and then when it hit zero, it would spike to full and then do the whole thing over again. The other pots would do something similar when they were burning out, but this one wasn't so I'm not sure if it was the change I made or if this was just a better quality pot. Also the full output power doesn't seem to be 1W like it should anymore.

I don't have that 220 ohm or anything. I'd have to buy it and I still don't even have a clue if that's going to fix the problem. If the original pot didn't have any resistors on it WHY SHOULD THIS NEED IT? I'll upload the pictures of the back tomorrow because Im pissed right now.

What is the resistance of the original pot?
The creator of the circuit said I should use a 50K. Maybe he doesn't know what he's talking about. He can barely speak english at all.
 

MrChips

Joined Oct 2, 2009
30,720
Hey Mr Chips,
Before you go throwing me out into left field, perhaps you should look at the facts. Here, let me get them for you.
http://www.bourns.com/pdfs/3296.pdf
I know I'm new but you can treat me with a little respect. Just makes sense to me. If I'm wrong, just explain why without being ignorant. Have a good day and happy reading.
Hey, don't get your knickers in a twist. You said you were a newb to electronics and you did ask for somebody to tell you if you're in left field. Well I did tell you and I am not ignorant.

You don't have enough info to work on.

Suppose the voltage across the pot is 12V.
If the resistor is rated at 0.5W

Power = VxV/R
R = 12 x 12 /0.5 = 288Ω

which means that the resistance can be as low as 288Ω and still survive.

If the 50kΩ pot was set to halfway, then 25kΩ would dissipate 12 x 12/25000 = 5.76mW, i.e. it would even warm a flea.

The op still has not told us the resistance of the original pot.
 
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Thread Starter

Synaps3

Joined Jun 5, 2013
99
The op still have not told us the resistance of the original pot.
@MrChips - keep reading

The creator of the circuit said I should use a 50K
As I said, the creator said 50K. That should be the resistance of the original.

The voltage across the pot is 8.2V and drops to 8V before it starts to burn out.
 

MrChips

Joined Oct 2, 2009
30,720
We still don't know what the circuit does and what is the purpose of the pot.
According to the photos, the pot is connected to a small TO92 transistor which I presume controls a larger TO220 transistor.

If you look closely at the bottom side of the board there is an SMD resistor in series with the pot to ground. This will prevent the pot from burning out.

You can help by drawing the circuit for what is there around the two transistors and labeling the transistor markings and the value of the resistances.
 

MrChips

Joined Oct 2, 2009
30,720
@MrChips - keep reading



As I said, the creator said 50K. That should be the resistance of the original.

The voltage across the pot is 8.2V and drops to 8V before it starts to burn out.
What is the resistance marked on the original pot?
What is the resistance as measured with an ohmmeter?
 

bwilliams60

Joined Nov 18, 2012
1,442
The OP did provide a picture of the original POT and I provided a datasheet for that POT. Am I missing something here? Thank you for clarification Mr Chips. I just still think that there is too much current passing through it and it is a direct result of not enough resistance. Maybe I am in left field but components die from too much current. That much I know. I am interested to hear the end result.
 

MrChips

Joined Oct 2, 2009
30,720
You are correct. Too much current will destroy a device.

Did you calculate the current? I did.
There is not enough current to destroy a 50kΩ 1/2W resistor.

To give you some perspective, 120VAC is not enough to destroy a 50kΩ 1/2W resistor.

There is missing information and that is why the pots are being destroyed.

The marking on that Bourns trim pot is the model number, not the resistance.

If the op looks more closely he will find the resistance written on the pot.
 
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Thread Starter

Synaps3

Joined Jun 5, 2013
99
There is missing information and that is why the pots are being destroyed.

The marking on that Bourns trim pot is the model number, not the resistance.

If the op looks more closely he will find the resistance written on the pot.
The resistance is not written anywhere on the pot, only the model number and another number that isn't the resistance. I measured it with a meter like you said though. 49.56K. Whats weird is no matter how much I turn it down, it doesn't go below 49.2K. I think it's broken because my 50K had to get down to at least around 30 for me to notice the power output.
 

BobTPH

Joined Jun 5, 2013
8,813
Hey Mr Chips,
Before you go throwing me out into left field, perhaps you should look at the facts. Here, let me get them for you.
http://www.bourns.com/pdfs/3296.pdf
I know I'm new but you can treat me with a little respect. Just makes sense to me. If I'm wrong, just explain why without being ignorant. Have a good day and happy reading.
The datasheet is saying that that model of potentiometer comes in a range of resistances from 10 Ohms to 2 Meg. There is table on the same pages listing all of the resistances it comes in. The particular one in use is 50K Ohms. Somewhere on the pot, it will be marked with 503, meaning 50 with 3 more zeros.

Bob
 

tracecom

Joined Apr 16, 2010
3,944
The datasheet is saying that that model of potentiometer comes in a range of resistances from 10 Ohms to 2 Meg. There is table on the same pages listing all of the resistances it comes in.
Correct! What a shame that we are 38 (now 39) posts into this thread, and the OP still hasn't told us what is written on the pot he took out, so we could be certain what its value is.
 
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Thread Starter

Synaps3

Joined Jun 5, 2013
99
Correct! What a shame that we are 38 (now 39) posts into this thread, and the OP still hasn't told us what is written on the pot he took out, so we could be certain what its value is.
I TOLD YOU I measured it with a meter 49.5K. I don't see what more info you need. Here is everything on it: 3296 W503 302B
 
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