1960's Allen Bradley Potentiometer

Thread Starter

jrisebo

Joined Nov 29, 2011
8
I asked this on the electronics part of the forum, but only got a phone number to call. They on the other end of the phone were no help. I see much more activity here on this part of the forum than teh electronics resource part.

So, I am trying to figure out what was used,it is a AB Pot with the only marking or number 8466 Type J. Im trying to figure out the resistance of it.
 

SgtWookie

Joined Jul 17, 2007
22,230
Is it in a circuit that is not working, or what?
If it is in a circuit, try removing the wire from one end terminal or the other, then measure the resistance across the two end terminals using an ohmmeter.

If it is not in a circuit, just measure across the two end terminals.
 

Adjuster

Joined Dec 26, 2010
2,148
On another forum you mentioned that a schematic for the equipment shows the value of the resistor as 8k, but that this Allen-Bradley device is a replacement. You also said that there is a resistor in parallel with the pot, and that you plan to take a picture of it.

It is helpful to provide as much information as possible in making such enquiries. Perhaps it would also be helpful to post the circuit schematic?

8kohms is not a very common value for a potentiometer. Possibly a somewhat higher value such as 10kohms has been used, with a fixed resistor connected in parallel as a (partial) correction. With fuller information, somebody may be able to advise further.
 

Thread Starter

jrisebo

Joined Nov 29, 2011
8
On another forum you mentioned that a schematic for the equipment shows the value of the resistor as 8k, but that this Allen-Bradley device is a replacement. You also said that there is a resistor in parallel with the pot, and that you plan to take a picture of it.

It is helpful to provide as much information as possible in making such enquiries. Perhaps it would also be helpful to post the circuit schematic?

8kohms is not a very common value for a potentiometer. Possibly a somewhat higher value such as 10kohms has been used, with a fixed resistor connected in parallel as a (partial) correction. With fuller information, somebody may be able to advise further.
Here is the pictures. Its a 6K resistor across the terminals. Not sure what the POT is. How do I measure it? I have bought a 10K POT that will fit the chassis, and wonder do I just put a 2K resistor across as they have?

Thanks
 

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Thread Starter

jrisebo

Joined Nov 29, 2011
8
Ok, Maybe this is all Moot. I have ordered a 10K POT to go where the schematic calls for a 8K. I should be ok right? its a voltage control on a DC power supply, so will it just mean I get a little less top end voltage?
 

MrChips

Joined Oct 2, 2009
30,720
You cannot measure the resistance of the pot while it is still connected unless you are certain that whatever it is connected to has a much higher resistance. You will have to unsolder the pot in order to measure the resistance.
 

Adjuster

Joined Dec 26, 2010
2,148
I would agree with that: if you do the arithmetic, 10kΩ in parallel with 39kΩ comes to very nearly 8kΩ.

Rtotal = R1*R2/(R1+R2) = 7.96kΩ.

If the old fixed resistor looks too mangled up after you have removed it, or measures incorrectly, fit a new one. You might like to do that anyhow, as it may be an old carbon composition type and they are not too stable especially when messed about with.
 

SgtWookie

Joined Jul 17, 2007
22,230
Even if that 39k resistor does measure within 10% of 39k, I still would not solder it on the new pot.

Those cylindrical bodied resistors are the old plain carbon resistors, and as Adjuster mentioned, they can change their value quite radically. I had a bunch of that type from back in the 60's and 70's, many of them were 1% tolerance - some of them were over 2x their marked value.

The newer carbon film resistors are very reliable in comparison. If you want even better stability and lower noise, use a metal film resistor.

Avoid getting the resistor body hot when soldering. Use a hemostat on the leads of the part to absorb the heat.
 

Adjuster

Joined Dec 26, 2010
2,148
The arrangement shown in that picture will behave as a variable resistor with a maximum value of approximately 8kΩ. The variation of resistance with shaft position will be subtly different than if the track resistance were really 8kΩ, but I doubt that the change would be very obvious to the user. Since a real 8kΩ device probably comes under the heading of "unobtainium", best to use 10k and be content with it.

Strictly speaking, the connection from 1 (end) to 2 (wiper) turns this from a potentiometer to a simple variable resistor. On first sight the connection of one end of the track to the wiper may seem to be unnecessary, but in practice it can reduce noise and it certainly avoids the resistance going above the maximum value if the wiper contact is poor. This arrangement is also convenient for wiring up. Note that the sense of variation of the control with rotation can be reversed by linking the slider (middle) contact to the other end instead.
 

MrChips

Joined Oct 2, 2009
30,720
It is always good engineering practice to connect the unused terminal to the center wiper for the reasons you described.
 
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