PDA

View Full Version : Calling all electronics experts



davidwdyer
07-30-2011, 01:07 PM
I bought three "2HP" variable speed motors from some surplus place in the U.S. some years ago. One I mounted on a small, table top drill press. (I also lengthened the round stand.) It really works well.

The other in on my son's grinder I made for him. The third is still waiting a use, maybe a medium size Chinese lathe I bought.

Anyway, the potentiometer on the grinder burned out. I have no idea of the specs. on it. The (probably useless) numbers on the back are:
00281921 and R1378602. I can find all sorts of potentiometers here in Brazil, but... how many K should I buy?

The pot hooks up to the three terminals near it in the pic, in the upper left hand corner, labeled P1, P2, and P3.

I'm hoping by looking at the circuit board, general knowledge, shooting in the dark etc. someone may be able to help me.

I have used the pot from the third motor for the time being, but I can see I will need another one soon.

http://www.johndyerco.com/Dad/Circuit1.JPG

http://www.johndyerco.com/Dad/Circuit2.JPG


Another problem is that the green resistors seem to be betting too hot and will probably go south on me soon. Are some some similar ones with some kind of heat sink?

These boards were never very good, with lots of components needing re-soldering, but they do work. Also, the motors, while rated at 2HP don't seem to really give that kind of power. I think these were surplus treadmill motors.

b2u44
07-30-2011, 01:18 PM
Remove one of the remaining good potentiometers from its circuit board. Measure the resistance across the two outermost terminals on the good potentiometer. Round that number to the nearest standard potentiometer value. For example, if you measure a resistance of 9987, that would be rounded up to 10,000 = 10k.

As far as the other two resistors go: You could always upgrade them to ones with a higher watt rating. Typically, heatsinked resistors exceed a 25-watt rating, which is quite large.

davidwdyer
07-30-2011, 01:25 PM
Please excuse my ignorance here.

When you measure the resistance, do you do it with the "knob" turned all the way to the left (counter clockwise) or to the right (clockwise)?

mechanicalmagic
07-30-2011, 01:30 PM
Please excuse my ignorance here.

When you measure the resistance, do you do it with the "knob" turned all the way to the left (counter clockwise) or to the right (clockwise)?
Makes no difference.
(That's the center terminal.)

DJ

b2u44
07-30-2011, 01:35 PM
Yeah, what DJ said.

The center terminal is what varies in resistance.

Within the pot, there's basically a resistive wire running in a circle from one of the side terminals to the other side terminal. The center terminal connects to a 'wiper' which slides along this wire. As the wiper is moved closer to one terminal, the resistance between the wiper and that terminal decreases, while it increases between the wiper and the other terminal. Potentiometers are spec'd by the overall resistance of that wire (i.e. from one end to the other) which is why you measure from the two outer terminals and ignore the middle one.

MaxHeadRoom
07-30-2011, 01:37 PM
Speed control pots are usually from 5k to 10k, make sure you get a linear version, not logarithmic.
If you ever have to replace these drives, the KB and Baldor versions have been around for a long time and are relatively cheap.
Max.

JoeFin
07-30-2011, 01:39 PM
I got a "nothin" on that part number from google which is pretty rare

davidwdyer
07-30-2011, 01:39 PM
Aha!!!!!:)

That explains it. I think the one on the drill press will be easier to get to.

How about the one which I took out? Could I just measure that? The defect was that it would start slowly as normal, but when you moved it a little bit, the motor took off uncontrollably. Could I still use it to measure?

davidwdyer
07-30-2011, 01:42 PM
Still another dumb question. How do you know it the pot is "linear" or logarithmic?

Here, the people who "help" you at the counter don't know much and are not very helpful.

Black_Moons
07-30-2011, 01:43 PM
The remaining pot on the circuit board has absolutely 0 relation to the one thats fryed and hence has absolutely NO REASON to be the same value.

Id measure the two most outer terminals of the fryed pot, they should retain the pots value if its only the wiper that went (Most common failure mode)

If the resistive material itself is fryed in a spot, Use the outside terminal and wiper and put it about midrange, measure the resistance, and double it. (One outside terminal will likey read open to the wiper, thats the damaged side)

As for the smoken hot resistors, they look to be.. 1/2W? Maybe 1W.

You can replace em with higher wattage unheatsinked resistors, but you'll likey need to stand them up verticaly, or maybe at an angle, to make room.

JoeFin
07-30-2011, 01:45 PM
Still another dumb question. How do you know it the pot is "linear" or logarithmic?

The application

MaxHeadRoom
07-30-2011, 01:49 PM
Still another dumb question. How do you know it the pot is "linear" or logarithmic?

Here, the people who "help" you at the counter don't know much and are not very helpful.

If you have someone who serves you over the counter and are not that electronic savvy they may not know what you are talking about, it should be listed in the spec if they have it.
I don't think 5k/10k it is going to be very critical in your application, I have used 5k or 10k with no problem, the input impedance of the circuit is probably fairly high.
Max,

davidwdyer
07-30-2011, 01:50 PM
Uhhh, maybe I didn't ask the right question. I meant, when I am going to buy one, is there some number, symbol or distinguishing characteristic for me to know which is which?

I'm not doubting what you recommend I should use, but how am I, a rank beginner, to recognize what I am looking for. As mentioned, the store clerks probably don't even know the difference.

Black_Moons
07-30-2011, 01:52 PM
PS you can check if its linear or log by moving the wiper while measuring resistance. Or if the wipers dead, open it up and measure one of the outside contacts and put your probe right onto the resistive material at verious spots.

You also might be able to just SEE if its linear/log after opening it up. Linear typicaly has a consistant resistive material, Log will have one that gets wider all of a sudden, or thicker. (Usally log isent really log, but more like a log aproximated by 2 or 3 lines on a graph)

When you go to buy one, Its what you ask for. Linear will be defacto normal.
Log might be called 'audio taper' or 'audio pot' as that is there most common use.

MaxHeadRoom
07-30-2011, 01:53 PM
Well I guess if the store does not have any details, you could take a meter with you and just measure the resistance in the centre and half way either side, the reading should be fairly linear, which is what you want.
In a log or antilog type, the resistance will increase or decrease exponentially.
You need the linear type for your application.
Max.

danlb
07-30-2011, 01:54 PM
Just in case the picture represents how you use this controller...

It's vital that you rotate the board 90 degrees and reattach the two black things to the aluminum frame. The frame is the heat sink that keeps the black chips from overheating.

Dan

davidwdyer
07-30-2011, 01:56 PM
Sorry about that. I do attach the chips to the heat sink. I just slipped it in there for the picture. Not to worry.

davidwdyer
07-30-2011, 01:59 PM
Taking a closer look at the resistors, the say 4.7K Ohms. What should I jump up to on the replacement. As long as this board is not being used, it seems like a good time to replace them.

Will a larger resistor affect the performance of the motor?

tyrone shewlaces
07-30-2011, 02:09 PM
Checking out this link may help:
http://www.geofex.com/Article_Folders/potsecrets/potscret.htm
notably this picture illuminates the innards and help show what to measure:
http://www.geofex.com/Article_Folders/potsecrets/pot5.JPG

online component suppliers are the best places to get stuff in my opinion. I almost never buy stuff locally anymore because the local stores either don't exist anymore or the "help" doesn't know a pot from a toilet. Online suppliers will have way more options than you can ever need sou you could get exactly what you want, whether what you want is what will work or otherwise.

Black_Moons
07-30-2011, 02:16 PM
Taking a closer look at the resistors, the say 4.7K Ohms. What should I jump up to on the replacement. As long as this board is not being used, it seems like a good time to replace them.

Will a larger resistor affect the performance of the motor?

Incress the *wattage*, the wattage changes the size of the resistor, you'd keep the same resistance, and hence the circuit will never know the diffrence.

Extra wattage just means the resistors stay cooler at X power because they are bigger and hence have more surface area to disipate heat.

Pertty sure those are 1/2W resistors. Might be 1W.
2W resistors come in square packages typical btw.

dp
07-30-2011, 02:22 PM
Non-linear pots are often referred to as audio taper. Volume controls are designed to allow for our logarithmic hearing and to prevent a little bit of adjustment from knocking down the walls of your condo.

With a linear pot any repeated fraction of rotation of the dial will result in exactly the same change in resistance. The first 1/8th turn is identical to the last. Not so for others. This is measured between either end terminal, and the center terminal.

As an aside, pots are nice voltage dividers but pi$$ poor current controllers unless care is taken because the area of the device in use changes with the rotation of the knob. At either end, very little of the physical device is being used and so a little current goes a long way toward smoking it. The workaround is to add series resistance either to the wiper where it will affect the full range of the pot, or at either or both end terminals of the pot.

This site seems to have wrapped it all up:

http://pdf.directindustry.com/pdf/bourns/the-potentiometer-handbook/11910-40708.html

MaxHeadRoom
07-30-2011, 02:36 PM
Also, the motors, while rated at 2HP don't seem to really give that kind of power. I think these were surplus treadmill motors.

One. Be very suspicious of the HP quoted for ex-treadmill motors, you just have to compare the size with a normal industrial rated one.
Two. These kind of drives often have current limit built in, in the case of the KB or Baldor, they come with plug in current limit resistor that is selected according to the size of the motor.
Max.

davidwdyer
07-30-2011, 03:04 PM
Sorry I'm still a little confused about the resistor question.

I seems that they are marked in Ohms with a sort of "Omega" symbol but everyone is talking about watts (as in James). How will I know what sort of wattage the present ones are and how will I know how to chose another one?

MrSleepy
07-30-2011, 03:12 PM
Hi There..

It looks like your board uses the TDA1085 chip in a Triac based circuit.

In the early 90s I used to use the TDA1085 quite often..

I still have the original Plessey design data with example circuits (before Motorola revived it from Plesseys ashes).

Judging by the components you have on board..It looks most like a derivation of the Fixed Field Motor application circuit.

Here are a few photos of the application manual that may help you (they are around quite large 3-4Mb,so I've shown a link instead of showing them direct).

Fixed Field Motor application
http://i897.photobucket.com/albums/ac180/MrSleepy123/IMG_1508.jpg
again in two parts
http://i897.photobucket.com/albums/ac180/MrSleepy123/IMG_1509.jpg
http://i897.photobucket.com/albums/ac180/MrSleepy123/IMG_1510.jpg

Pin Out..
http://i897.photobucket.com/albums/ac180/MrSleepy123/IMG_1511.jpg
Universal motor circuit..
http://i897.photobucket.com/albums/ac180/MrSleepy123/IMG_1507.jpg


HTH Rob

gary350
07-30-2011, 03:14 PM
Use your Volt Ohm Meter to check resistance across terminals 1 and 3 of the variable resistor your using now. Buy the same watt rating it will work.

The value is sometimes stamped on the outside.

Do a Google search for the part number stamped on the back it might tell you the Resistance and watt rating.

MaxHeadRoom
07-30-2011, 03:15 PM
I would guess those are 2watt?
Look close, it may be marked on them.
Anything physically bigger usually means higher wattage.
Size is their ability to dissipate heat (wattage).
Also it would pay the mount them higher on the board, not only does it take the heat away from the P.circuit, longer leads tend to act as heat radiators.
Max.

davidwdyer
07-30-2011, 03:33 PM
Now I get it. I buy resistors with the same Ohm value, but capable of taking higher wattage.

Thanks so much everyone for the info. This forum is great. I imagine I could probably get a diagnosis for anything here, including irritable bowl syndrome.

I think I can take it from here.

Black_Moons
07-30-2011, 03:35 PM
I would guess those are 2watt?


Definately NOT 2W! 2W are much bigger.

Most likey 1/2W or maybe 1W.

davidwdyer
07-30-2011, 03:39 PM
Does the "J" written on them give a clue?

It says 4.7K Ω J.

MaxHeadRoom
07-30-2011, 03:47 PM
Definately NOT 2W! 2W are much bigger.

Most likey 1/2W or maybe 1W.

I just bought some 2w that appear exactly the same size.
Digikey P220W-2BK-ND. 0.157" x .42"
Max.

psomero
07-30-2011, 04:44 PM
Sorry I'm still a little confused about the resistor question.

I seems that they are marked in Ohms with a sort of "Omega" symbol but everyone is talking about watts (as in James). How will I know what sort of wattage the present ones are and how will I know how to chose another one?

the omega is the symbol for ohms.

the power dissipation capability of a resistor isn't always denoted on the case, but you can usually get a good feel for what it'll be based just upon size and construction. stuff below a watt is usually carbon film and looks like a round lump with axial leads (i.e. http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=P1.0KBATB-ND)

the four colored bands on the package denote the resistance and precision of the resistor.

larger resistors around 10 watts can often be had in a ceramic package that's white and square shaped. they'll still have axial leads and are about an inch or two in length usually. in my experience, they'll print the resistance and power of the resistor right on the side of the case.

larger stuff than that gets more complicated. there's wirewound resistors that come in a little extruded aluminum case with heatsink fins on it and various other designs, all of which pretty much need some sort of feature for eliminating the bunches of heat a higher power resistor is designed to handle.

Evan
07-30-2011, 05:36 PM
Those resistors can run that hot normally without failure. The problem is that they may melt the solder and do damage the circuit board. You can replace them with the same type and wattage but leave them standing above the circuit board by a centimetre or so. Going to a higher wattage is a good idea as they will run cooler.

A 5K or 10K pot will work fine. It's just a voltage divider. The operational difference between a linear and log pot is that the log pot will cram all the resistance change at the counter clockwise end of the rotation. That means that most of the speed control will occur in the first 20% of rotation. It won't hurt anything to use a log pot but the control won't be easy to use. A linear pot will distribute the resistance change evenly over the entire range of rotation.

Pots may or may not be marked as to the type. There is no standard for potentiometer markings.

davidwdyer
07-30-2011, 05:47 PM
I'm sure I can get either a 5K or 10K potentiometer. Any thoughts about which would be preferable? Or perhaps it just doesn't make any difference at all. I actually bought a 100 K one, just shooting in the dark, but I'm sure I can exchange it. It'll be quite easy. Just about an hour and a half to two hour trip through the big city, fighting for a parking place, walking about a half a mile (which is good for the old waistline) spending about 30 minutes trying to exchange something which in the U.S. would take 3 minutes. But... it's got to be done. I'll look for some resistors while I'm there.

darryl
07-30-2011, 08:47 PM
Couple things- first the resistors. Those look like 3 watters to me, and the J means something, don't recall what, but maybe flame-proof. Just replace them with 5 watt rated of the same resistance value. The five watters will likely be in a white ceramic rectangle about a half inch square and between one and two inches long. There's a couple ways to do that- one is to just solder them in standing off the board a short distance, and the other is to put a kink in each lead before inserting them in the holes. Use the kink method as that gives each lead a positive stop where it goes through the holes, and that helps to support the component and minimize connection problems in the future. Of course you have to be able to get a good solder connection to the leads in the first place. Many boards that look as heat-damaged as that one will be difficult to resolder nicely, so you might have to add some strands of copper on the trace side, going from points on the same traces where it's easier to solder.

Another thing- almost every pot will have a resistance rating stamped on the side of it. It will seldom be printed on. If there's no stamped number indicating the value or type, then you should test to see what the actual value is. The test is easy, so why not do it- you need to remove at least one lead to one end terminal. If you rotate the pot to put the slider right at the end that's still connected, you can get an accurate reading without any of the circuitry interfering with it. If you want to find out if linear or log, also disconnect the center terminal then turn the pot exactly halfway. Now measure the resistance from the center terminal to each end- if it's a linear pot, the readings will be close to the same. If it's a logarithmic pot, the readings won't be close.

Replacing it with some nominal value without knowing the actual resistance may work, but there's a good chance it won't work properly. That's because the pot may have some other resistor in series with it, and using a pot of some other value would put the voltage values out of design range. The pot is not just a voltage divider when it's in a circuit. There's going to be some finite resistance in the circuit where the controls tap wires to, and that will in part determine what the voltage range will be when the control is rotated from end to end. You could end up with it working but having limited range, or it might end up being too sensitive over a small part of the rotation- either way it won't be right.

One other thing- it's very likely that some control cleaner will fix it right up and you won't need to replace it at all. Nutrol is good for that, and there are other control cleaners that work also. There's one that has some lube in it specially made for linear slider type controls, and that will help to keep the pot rotating smoothly. Using a non-lube cleaner on a linear slider will probably make it hard to operate- this is lesser of a problem with a rotating control, but the option to use a cleaner with or without lube in it is yours to decide. Don't use WD or you will have to replace the control soon.

Nutrol can be used for many things around the shop, so it's not like you buy a can for 8 or 10 bucks and only use it once. You wouldn't normally use it for linear slider controls, but control cleaner with lube can be used for either type. Cramolin makes one, but MG Chemicals probably does as well.

If you do decide to replace the control, try to get one made in USA. CTS is one brand. Apparently (just heard this from a friend today) the china made pots are crap.

MaxHeadRoom
07-30-2011, 09:13 PM
I'm sure I can get either a 5K or 10K potentiometer. Any thoughts about which would be preferable?
In that application I am sure either will work, if you do have a choice, go with the 10k.
Max.

davidwdyer
07-31-2011, 05:44 AM
I've tried the crappy little tester I have here at home, a "Sunwa" brand, and it doesn't seem to be sensitive enough to measure anything.

The 100K pot I erroneously bought just wiggles the needle enough to imagine that it is registering correctly. The old one does nothing. I have another tester at the shop but haven't gotten there to test it yet.

Darryl,

Thanks for all the information. I plan to get some new resistors too. I'm not sure I have the cleaner. Will spray brake cleaner work?

The Artful Bodger
07-31-2011, 06:23 AM
At the risk of confusing the issue, did anyone mention that if you cannot get higher wattage resistors just double up two in parallel that are twice the resistance? That will be the same combined resistance but twice the heat dissipation capacity.

davidwdyer
07-31-2011, 08:01 AM
At the risk of confusing the issue, did anyone mention that if you cannot get higher wattage resistors just double up two in parallel that are twice the resistance? That will be the same combined resistance but twice the heat dissipation capacity.


Thanks for that tip, but I's quite sure I can find those resistors.

darryl
07-31-2011, 03:30 PM
You might get away with a quick spritz of brake cleaner, but if anything's going to remove the lube, that will. A quick spritz followed by blowing out with air might be the ticket. Just don't give it time to dissolve the lube or soften the resistance element. Control cleaner would be better and safer.

A tip with the meter- if it's not sensitive enough to read high k ohms, you can still use it to make comparisons, if not exact readings. A 9v battery will be needed, and all you do is read on a voltage scale. First reading is the battery- it should read roughly 9v. Then find a resistor of about 100k and put that in series with the battery and read the voltage again. Find another resistor of about 500k and do the same test. You might have to switch the meter to a lower full-scale voltage range to get a significant deflection on the needle.

The point of this is to do comparisons. Change between the resistor and the control, and when the needle moves the same amount for each, the resistances are matched.

If the 100k pot barely moves the needle- well, is there a battery in that meter that's dead? They need a battery to supply current for resistance readings. You might find that replacing it makes the meter work properly on ohms. First thing to check out-