I am wiring in a digital amp meter in the same way I would an analog meter and when I adjust the wiper arm towards High end (50 ohms, 12 volt supply) I smoke the pots. They get real hot Draw excesive current. Whats going on?

 

If you have an answer for me, please e-mail me at <SNIP> A schematic would be greatly appreciated.

 

It would be helpful if you provided us a schematic of your set up. For istance, telling us that you are wiring in your digital ammeter the same way you would an analog meter tells us nothing because we have no idea you you would wire in an analog meter.

I am guessing that you have the fixed ends of the pot connected between the +12V and ground (other side of the 12V supply). Is the ammeter then connected between the wiper and ground?

And, no, I'm not going to send you an e-mail. This is not a free private consulting business (if you want to pay me my hourly chargeout rate, that is a different story and we can most certainly discuss it). The whole point of the forum is for lots of people to have the opportunity to learn from the discussion about other peoples' issues.

 

The way I'm hooked up is: 12 volts to one side of pot, the power side of the digital meter is attached there also, wiper arm going in to meter red input side, output side hooked to load then load to gnd, also neg. side of meter hooked to gnd. When I turn the power on it will not allow current to flow thru meter, it seems to read the pot setting, its like its blocking and adding power back onto the wiper arm, because as soon as I adjust the wiper arm towards the high side (50 ohms) the pot starts smoking and the meter shows for a second that I'm drawing around 5 Amps and then the pot burns up. I dont know how to attach a diaghram to show you my set up, but it's the simple way, just have the meter hooked between the pot and load.

 

Again, it would be helpful if you provided us a schematic of your set up.

Even a sketch in MSPaint is better then your word description.

 

I don't know how to draw on this sight. So let me make it really simple. I hooked up one side of my wiper arm to the input of the amp meter and the other side to the load.

 

I don't know how to draw on this sight. So let me make it really simple. I hooked up one side of my wiper arm to the input of the amp meter and the other side to the load.

Your descriptions don't make sense. What you described this time makes it sound like the wiper arm has two sides.
Here is how you attach an image to a post.
Just above the Manage Attachments button is a list of all file extensions that can be attached.

 

OK, Where's the "manage attachments button?

 

Under the normal Reply Box is a button labeled "Go Advanced". Click it and then go to the section below the Reply to Thread section lebeled "Additional Options". Then click "Manage Attachments".

 

I tried attaching, but not sure it got thru. So let me try to explain what I have and what I'm trying to do. I have a 50 ohm pot and a digital amp meter and a load. The digital amp mtr. has 4 wires, two 30 Ga. and 2 18 Ga. I'm hooking the small Ga. wires (one red and one black) to my 12 volt power source to provide power for the LED's to light. That leaves me two 18 Ga. wires left(one red and one blk.) Now to my Pot. I have 12 volts hooked to one end of the pot. and the other end is just open. I take the red 18 Ga. wire and hook it to my wiper arm and the black wire to the load, and the other end of the load to B- or gnd. or return, what ever you want to call it. Does this make anymore sense? Its the simplest way to measure current in a circuit, just cut it open between the pot and load and stick an amp meter in between.

 

should this not be measuring voltage drop across a shunt resistor ?

 

Smoking pot is illegal!

Seriously though, your circuit is just a power source connected to a pot and a load. Forget about the current meter; it should have negligable effect. Now, I don't know what your load is, but if the pot is adjusted to it's minimum level, then you only have a voltage source and load, as the pot is now at a minimum, almost zero, ohms. So the current is Vs/RL. If the current is too high for the pot, then it will smoke. What's the purpose of the pot in the first place? If you want to vary the current, either use a higher power pot or a series fixed resistor to limit the current to a safe level.

 

I tried attaching, but not sure it got thru. So let me try to explain what I have and what I'm trying to do. I have a 50 ohm pot and a digital amp meter and a load. The digital amp mtr. has 4 wires, two 30 Ga. and 2 18 Ga. I'm hooking the small Ga. wires (one red and one black) to my 12 volt power source to provide power for the LED's to light. That leaves me two 18 Ga. wires left(one red and one blk.) Now to my Pot. I have 12 volts hooked to one end of the pot. and the other end is just open. I take the red 18 Ga. wire and hook it to my wiper arm and the black wire to the load, and the other end of the load to B- or gnd. or return, what ever you want to call it. Does this make anymore sense? Its the simplest way to measure current in a circuit, just cut it open between the pot and load and stick an amp meter in between.

What Brownout said.
I think most of the confusion resulted from the fact that you have a 4-wire ammeter, which I had never heard of until now.

 

How close is this? If load is 5Ω & pot is set to 5Ω, then there's 1.2A flowing, so if pot is 1/2 W, it will get hot.

 

How close is this? If load is 5Ω & pot is set to 5Ω, then there's 1.2A flowing, so if pot is 1/2 W, it will get hot.

Pot power dissipation will be I^2*R = 1.44*5 = 7.2 Watts = smoke.

 

Okay, I think I have a pretty good picture of how you have it hooked up.

What is the load?

What is the make/model of the meter? I would like to check to be sure that it is what you think it is and not something intended for making 4-wire measurements.

What is the power rating on your 50Ω pot?

When you say that you move the wiper toward 50Ω, I suspect you are actually moving it so that the resistance between the 12V terminal and the wiper is approaching 0Ω.

Turn the wiper fully the other way (or disconnect the wiper and hook the other fixed end of the pot to the meter). What is the current that is indicated?

 

Why do you have a pot in series with the meter? Are you trying to limit the amount of current being supplied to the load? What is the load? (What current would you expect the load to draw at the full supply voltage?) More detail results in better answers.

 

Pots are lower wattages than resistors. They have a rating, which it is obvious you are exceeding.

I have to admit the title caught my attention, which is the only reason I'm posting. Everyone seems to have it under control here, better than I do.

 

How close is this? If load is 5Ω & pot is set to 5Ω, then there's 1.2A flowing, so if pot is 1/2 W, it will get hot.

Your exactly dead on except I'm using a 50 ohm 5 watt pot to heat a loads that vary from 16-72 ohms. to dissipate dew from my telescope and different eyepieces. So why is this amp. meter not passing current, I'm bewildered that such a simple circuit is drawing up to 4+ amps in the pot before it smokes? Thanks for seeing the simplicity of the circuit.

 

The problem is,you are talking the total wattage. When the wiper is mostly on one end you have focused the power to a small area, which it can not handle.

Pot are not meant to handle power, rheostats are much higher capacity.

You need to use the pot to control something else. If we had a schematic we could help.