I'm At A Loss: Variable Resistor

I have a breadboard mount 1 kOhm variable resistor.

By itself (not plugged into the breadboard) it works fine.

If I plug it into the breadboard, with nothing attached, and nothing in the same column, and nothing powered up, adjusting it does absolutely nothing.

I have 4 more of these, and nothing else similar. It's for the voltage adjustment for an LM317T, which it is not connected to at this point.

Have I been abducted by aliens and don't know it yet?

Most potentiometers have 3 pins--The two outside ones are the ends of a "fixed" resistor that is the same as the maximum value for the pot. For example, a 1k potentiometer would measure a constant 1k between the outer pins regardless of the position of the knob. The middle pin is the "wiper", and that's what you want to measure--measure between the middle pin and one of the outer pins (it doesn't matter which).

If you're already doing that then I expect you inserted it into the breadboard incorrectly. Each pin needs to be in its own row of 5 holes.

I guess a clearer description of how you're trying to measure it would be really helpful here

Regards,
Matt

DerStrom8 said:

Most potentiometers have 3 pins--The two outside ones are the ends of a "fixed" resistor that is the same as the maximum value for the pot. For example, a 1k potentiometer would measure a constant 1k between the outer pins regardless of the position of the knob. The middle pin is the "wiper", and that's what you want to measure--measure between the middle pin and one of the outer pins (it doesn't matter which).

If you're already doing that then I expect you inserted it into the breadboard incorrectly. Each pin needs to be in its own row of 5 holes.

I guess a clearer description of how you're trying to measure it would be really helpful here

Regards,
Matt

Click to expand...

TYVM. That may well have solved the problem. Grats for the three terminal explanation. I've never used these before.

I'm just using a VOM to measure the resistances.

HOWEVER, the three terminals are not in line That's where I messed up. When I plugged the VR into the breadboard, only one lead was connected. Hmmm. No variation in resistance? I need to move jumper.

BTW, I DXed the third terminal, hoping to avoid complications. Is that going to cause any?

I have no idea what "DXed" means, but it depends on your application. If you want a simple variable resistor than you only need to connect the wires to the center and to one of the outer pins. You don't need the second outer pin. However, some circuits require all three pins--as you turn the wiper, the resistance decreases with respect to one pin, but increases with respect to the other. This is especially useful when you're creating a simple voltage divider--Connect one outer pin to +5v and the other outer pin to ground. Then as you turn the knob on the potentiometer, the voltage on the wiper (center) pin varies between 5 and 0 volts.

Regards,
Matt

DerStrom8 said:

I have no idea what "DXed" means, but it depends on your application. If you want a simple variable resistor than you only need to connect the wires to the center and to one of the outer pins. You don't need the second outer pin. However, some circuits require all three pins--as you turn the wiper, the resistance decreases with respect to one pin, but increases with respect to the other. This is especially useful when you're creating a simple voltage divider--Connect one outer pin to +5v and the other outer pin to ground. Then as you turn the knob on the potentiometer, the voltage on the wiper (center) pin varies between 5 and 0 volts.

Regards,
Matt

Click to expand...

DX is Marine Corps slang for Direct Exchange. It means you turn in a troublesome piece of gear to the supply section and never get it back. Kind of like 'binning' it.

I never saw any signs of life when I connected the VOM to the terminal opposite the screw adjustment, so I cut that one off. If I measure the resistance between the first and second terminal, I can adjust it. I tried connecting the third terminal to ground, as the schematic suggests, but I was stuck at 18 volts. No adjustment

The only markings I can find are 2101Y, 1K, CW, 1NA

It still doesn't work. The breadboard is too small. I need to turn it, but it is too long for a 5 hole breadboard. If I put it parallel to the center divider, I need to remove all components on one end of the board. If I have it perpendicular to the centerline, the leads cover up the adjustment screw. If I slant it . . . well, it still doesn't work.

The problem is getting the screwdriver into the adjustment screw.

WTH, I'm wasting my time and yours. I need something different. Sorry.

I've learned a lot, and I appreciate all of the input I have received. Building power supplies on breadboards is starting to look more and more like a waste of time. I built 10-20 on Wal-Mart wooden plaques with no problems. Trying to bread board them has cost me months without a single power supply being built.

Could you post a photo of your potentiometer? I just want to make sure you have the right pins hooked up the right way.

Matt

DerStrom8 said:

Could you post a photo of your potentiometer? I just want to make sure you have the right pins hooked up the right way.

Matt

Click to expand...

Sore subject. I've been perma-banned from three electronics forums because I don't have photographic capability. I can scan schematics. That's it.

- understanding how to measure continuity and verifying measurements is key to communicating questions.,
- if Pins are numbered 123 on POTs then- 2 is the tap and if marked with CW, it means 2 goes from 1 to 3 in the CW direction.
- Looking up similar shaped part specs ( if you can't find the Murata or Bourns Spec or whatever)
- it will help you to help yourself then us when you check the obvious. ( faults )
- beware some trimmers are thin film metalized plastic and burn out easily with modest currents and very low power , such as 1/10W.
- others are robust Cermet conductors with rugged wipers. and others Wirewound.
- verify the resistance between 2 and 1 and 3 and then the voltage in circuit on each pin to see it meets expectations.
- mechanical construction is as important as electrical.

Do you have a cell phone that can take pics?

Your terminals 1K, 1CW 1NA probably means.

between 1NA and 1K there is ALWAYS 1000 ohms.
With the pot fully CW, there is 1K between 1 NA and 1K.
With the pot fully CCW there is 1K between CCW and 1NA

(1CW-1K)+(1CW=1NA)=1000 ohms

i.e. mid position: between 1NA and 1CW there is ~500 ohms and between 1K and 1C there is aprox 500 ohms.

Usually the terminals are marked 1, 2 and 3 or CW, wiper and CCW.
1K=CCW
1CW = CW
1NA - Wiper

Now, for the what to do with the orphaned terminal when you only want a 2 terminal resistor? For RELIABILITY reasons, you connect the CCW and wiper positions together if you want a 0-1000 ohm resistor. What this does, is it prevents burning out the pot at the end points. It also reduces noise at the endpoints.

I can't find a datasheet quickly, BUT this is what a potentiometer datasheet look like: https://www.bourns.com/data/global/pdfs/3310.pdf

A breadboard power supply: https://www.robotshop.com/en/sfe-breadboard-power-supply-kit.html?gclid=CKeZiLbe4MMCFUkV7AodkREAYQ

Solderless breadboarding of power supplies isn't a good idea.

KeepItSimpleStupid said:

Do you have a cell phone that can take pics?

Your terminals 1K, 1CW 1NA probably means.

between 1NA and 1K there is ALWAYS 1000 ohms.
With the pot fully CW, there is 1K between 1 NA and 1K.
With the pot fully CCW there is 1K between CCW and 1NA

(1CW-1K)+(1CW=1NA)=1000 ohms

i.e. mid position: between 1NA and 1CW there is ~500 ohms and between 1K and 1C there is aprox 500 ohms.

Usually the terminals are marked 1, 2 and 3 or CW, wiper and CCW.
1K=CCW
1CW = CW
1NA - Wiper

Now, for the what to do with the orphaned terminal when you only want a 2 terminal resistor? For RELIABILITY reasons, you connect the CCW and wiper positions together if you want a 0-1000 ohm resistor. What this does, is it prevents burning out the pot at the end points. It also reduces noise at the endpoints.

I can't find a datasheet quickly, BUT this is what a potentiometer datasheet look like: https://www.bourns.com/data/global/pdfs/3310.pdf

A breadboard power supply: https://www.robotshop.com/en/sfe-breadboard-power-supply-kit.html?gclid=CKeZiLbe4MMCFUkV7AodkREAYQ

Click to expand...

TYVM.

I do not have phone (of any kind)

The Bourne looks nice, but it isn't what I have, and I can't afford anything else this month. The three terminals are connected to the VR at front, middle and rear. The rear one is the one I cut off, but, as I said, I have 4 more. They are also staggered. (front and middle do not plug into the same column) I think the third terminal lines up with the first.

I need a variable resistance. One side will go to ground, the other connected to the junction between the 240 Ohm resistor and the ADJ pin of the regulator. I wanted to use the two terminal type, but don't have any.

I will try connecting the terminals in different orders, with a 1.5 volt supply

Scan your potentiometer.

You might have something like this: **broken link removed**

Use your ohmmeter to figure out which is CW CCW and wiper. Two terminal potentiometers for the most part don't exist.

Your probably using an LM317 regulator. The pins might be too big for the breadbord AND the regulator MIGHT require a heat sink. The heat sink is required for the thermal protection to work properly.

KeepItSimpleStupid said:

Scan your potentiometer.

You might have something like this: **broken link removed**

Use your ohmmeter to figure out which is CW CCW and wiper. Two terminal potentiometers for the most part don't exist.

Your probably using an LM317 regulator. The pins might be too big for the breadbord AND the regulator MIGHT require a heat sink. The heat sink is required for the thermal protection to work properly.

Click to expand...

Your link shows exactly what I have,

The middle and front terminals are the only ones that change resistance with the adjustment screw.

The VR is not connected to anything except the VOM. The circuit is not powered up, but works perfectly except for the VR.

Basically, I can't get the output voltage down to less than +18. I need a 0-3 vdc voltage source. Then I can try to get the current down to less than 0.5 Amps.

The middle and the back should change too but that all depends on how you have it plugged into the breadboard. I think you want the longest dimension of the POT parallel to the longest dimension of the breadboard and the pot on the divided side if you have the standard breadboard.

Here is a breadboard pic. http://wiring.org.co/learning/tutorials/breadboard/ Note the break in the parallel busses. A lot of people think that there are 4 bussed parallel to the longest dimension. Note that there are eight busses.

Let's ask what are your resistor values? Something about 220-270 and what else?

The regulator sometimes needs a minimum load to work, otherwise it tends to stick close to Vin.

Honduras said:

TYVM.

I will try connecting the terminals in different orders, with a 1.5 volt supply

Click to expand...

a LM317 need about 2 volts more than the desired output to work and can not go below 1.24 volts, so it's probably not what you want to be using.
what is you power supply, what what is your goal?

there are several ways to connect a LM317 here's 4

Bear in mind that the screw of the trimmer will continuously turn forever. You will feel a clicking at the stops.

Kinarfi's right, 1.2 to about 30V and not 0 at the low end.

You MAY actually want a variable REFERENCE rather than a power supply? A reference is stable, but usually cannot source or sink much current.

If the input is 18V and the output is 2V at 0.5A then the poor little LM317 will be dissipating (18V - 2V) x 0.5A= 8W which is a lot of heat so it needs a pretty big heatsink.
When the trimpot is set to 1k then the 240 ohm resistor has 1.25V across it so its current is 1.25V/240 ohms= 5.2mA and the 1k trimpot will have a voltage across it of 5.2mA x 1k ohms= 5.2V. Then the maximum output voltage from the LM317 is 1.25V + 5.2V= 6.45V.

Why is the input voltage so much higher than 3V? All it does is make lots of heat. For a maximum output of 3V the input should be about 5V then the heating when the LM317 output is at its 1.25V minimum is (5V - 1.25V) x 0.5A= 1.875W and only a small heatsink will be needed.

If the input is 18V and the output is 2V at 0.5A then the poor little LM317 will be dissipating (18V - 2V) x 0.5A= 8W which is a lot of heat so it needs a pretty big heatsink.
When the trimpot is set to 1k then the 240 ohm resistor has 1.25V across it so its current is 1.25V/240 ohms= 5.2mA and the 1k trimpot will have a voltage across it of 5.2mA x 1k ohms= 5.2V. Then the maximum output voltage from the LM317 is 1.25V + 5.2V= 6.45V.

Why is the input voltage so much higher than 3V? All it does is make lots of heat. For a maximum output of 3V the input should be about 5V then the heating when the LM317 output is at its 1.25V minimum is (5V - 1.25V) x 0.5A= 1.875W and only a small heatsink will be needed.

How come I click "post" and nothing happens? So of course I click it again and a warning says "Wait at least 3 seconds" but it does not say why and does not show that the first post worked?

audioguru said:

How come I click "post" and nothing happens? So of course I click it again and a warning says "Wait at least 3 seconds" but it does not say why and does not show that the first post worked?

Click to expand...

I've seen this as well.