# Pots don't zero?

Hi! Are potentiometers supposed to go zero? I’ve tested a few and all can’t go below .5 Ohms. I’m going to use this on LM338 Constant current regulator, I won’t be able to set it higher than 2.5A since pots don’t go low.

0.5 ohms out of a few thousand Ohms is pretty darn good.

Keep in mind that every contact between two conductors has resistance.

What potentiometers are you using? What resistance and power? The LM338 is not a constant current regulator, it is a voltage regulator. It can be set up as a constant current source, but it is not ideal for it as the entire current must pass through the current set resistor.

2.5A x 1.25V = 3.9W, so you'd need at least a 10W potentiometer. But wait, it isn't that simple! The power rating for a potentiometer is only good at maximum resistance. So if it is set at 1/10th the resistance, it is only good for about 1/10th the power rating.

Let's go back to what you are trying to do. What is it? Are you trying to drive an LED, or a heater, or a battery charger?

Here is a switching regulator with both a voltage -and- current adjust. So you can use the current adjust as a constant current regulator.

http://www.ebay.com/itm/like/121108371961

Here is one with digital readouts for voltage and current:

http://www.ebay.com/itm/5A-Constant-Current-Voltage-LED-Driver-Battery-Charging-Module-Voltmeter-Ammeter/231238256589

Both are rated up to 5A, and will run a -lot- cooler than a linear voltage regulator being used as a constant current source.

polymorph: 0.5 ohms out of a few thousand Ohms is pretty darn good.

Keep in mind that every contact between two conductors has resistance.

What potentiometers are you using? What resistance and power? The LM338 is not a constant current regulator, it is a voltage regulator. It can be set up as a constant current source, but it is not ideal for it as the entire current must pass through the current set resistor.

2.5A x 1.25V = 3.9W, so you'd need at least a 10W potentiometer. But wait, it isn't that simple! The power rating for a potentiometer is only good at maximum resistance. So if it is set at 1/10th the resistance, it is only good for about 1/10th the power rating.

Let's go back to what you are trying to do. What is it? Are you trying to drive an LED, or a heater, or a battery charger?

Thanks for the reply! Im gonna use it for bench power supply and to test 2-2.5A laser diode. 10W pot :( thats big lol. Gonna search how to build the one you linked, I can buy, but I just wanna learn how : )

Multimeters can’t reliably read small resistances as they have internal contact resistance
in the multi-way switch. Any reading < 5 ohms is suspect, anything under 1 ohm is bogus.
The best you can do is measure the value for shorted-out test leads before and after the
test measurement and subtract this error from the measurement.

For measuring a low resistance you should be considering a Kelvin 4-terminal arrangement
where a known test current is fed to the DUT via two wires and you connect the multimeter
on voltage setting directly across the component (two more wires). Useful to have a bench
supply with variable current limit.

DUT = Device Under Test.

However with a potentiometer its easy to take a measurement of the wiper voltage, with
the track supplied from a know voltage. This provides a constant (calculable) current
with which to do the measurement, then place the voltmeter probes directly on the
relevant two terminals of the pot.

Tamulmol: Hi! Are potentiometers supposed to go zero? I've tested a few and all can't go below .5 Ohms. I'm going to use this on LM338 Constant current regulator, I won't be able to set it higher than 2.5A since pots don't go low.

In that sort of circuit all the current has to go through that pot. That means the pot has to have a very high wattage rating. These are expensive.

I thought I covered that.

Hi, when measuring low resistance, below 10R, first short the probes of DMM together and note the reading, say 0R4. This 0R4 is the resistance of your leads and switches in the measuring circuit. Then measure the resistor, say it reads 0R9. The real resistance will be about 0R9 - 0R4 = 0R5 (R is used as Decimal Point)

Tom... :)