Potentiometer as a Variable Resistor

[MrClamperSir]

I do use a linear supply with a LM317 adj reg however I also use a 12VDC battery. both for the same purpose, to drive my tools. The reason I don't use buck, switched..... is because I've been using the same supply's for 16 years. I like them but they don't make them anymore. I'm trying to understand how they work/function etc. so I can continue to repair/rebuild and build them.

 

[Harald Kapp]

Which model? Have you tried to find a schematic for the one you use?

 

[MrClamperSir]

There is no model with these. They were handbuilt in the early 90's. The schematic for the one I am using currently is posted here.

I paid a guy awhile ago to create the schematic for me based off of pictures and what little info I could gather. That's why it's off slightly and I've been struggling to get it up and running properly.

 

[Harald Kapp]

So there is already a third thread of yours on basically the same topic .
Plese don't open several threads on one topic. It easily leads to confusion and double answers. unnecessarily so.
As he merge of the two above threads already created a jumble, I will refrain from merging this 3rd thread.

 

[MrClamperSir]

So there is already a third thread of yours on basically the same topic ........ I will refrain from merging this 3rd thread.

I apologize for any confusion. The 2 posts eventually turned into 1 discussion but my intent was to get 2 questions answered and didn't want them blending in the same conversation.

I'm not sure how there is a third post on basically the same topic. post #1 is about a variable bench top power supply. #2 and #3 are about which rheostat to use and using a pot instead.

 

[MrClamperSir]

Using Ohm's law, I've calculated that I use on the low end of 3V and 1A I need 3Ω at a 3 watt rating and at the upper end of 8V and 1.5A I need 5.3Ω at a rating of 12 watts. I found two rheostats that seem like a good fit, 12.5watt 6ohm and 12.5watt 10ohm.

It's been suggested that I leave some room in the values so the components don't have to work at their max values. I'm not sure if this applies to this setup or not.

 

[mgrass]

The simplest thing you can do (and it's pretty cheap too)...
http://www.aliexpress.com/item/Chea...52&btsid=c2008ae5-8f14-4930-8fe1-cad4f87e2be4
This is a stepdown dc-dc converter that can handle 3amps. DC in 5-35v and output is 0-something slightly less than the input voltage. It has an LED readout on it too. $1.90usd. free shipping. If you really don't car about the voltage readout, buy http://www.ebay.com/itm/DC-DC-LM259...376552?hash=item1c5de50d68:g:0EsAAOSw585WUETY for $0.84 It has the same specs. You could replace the 10 turn trimmer with a standard size pot so it would be easier to adjust...like with a knob instead of a screwdriver.

 

[Harald Kapp]

Using a resistor to lower the power to your tools is equivalent to using the regulation of your linear power supply. Only you shift some of the power dissipation from the power supply to the resistor. As suggested in your other thread, simply use a bigger heatsink or a fan to cool the existing one.

 

[MrClamperSir]

Using a resistor to lower the power to your tools is equivalent to using the regulation of your linear power supply. Only you shift some of the power dissipation from the power supply to the resistor. As suggested in your other thread, simply use a bigger heatsink or a fan to cool the existing one.

The current rheostat is not heating up. My tools are simply not engaging until the knob is turned 3/4 of the way up. I'm guessing it's because I've got 12VDC coming in, 25Ω 1A rheostat to control the current and only drawing between 3V-8V at 1A.

That's why I'm asking is replacing the 25Ω rheostat with a 6Ω or 10Ω (12.5watt) rheostat would help my tools draw current sooner AND the rheostat would remain cool and operational?

 

[Harald Kapp]

That's why I'm asking is replacing the 25Ω rheostat with a 6Ω or 10Ω (12.5watt) rheostat would help my tools draw current sooner AND the rheostat would remain cool and operational?

The tools should start earlier, but the rheostat will become hot anyway as the power dissipated is the same (voltage drop*current). The rheostat may in total become not as hot as the 25Ω type because the power is distributed among a larger part of the rheostat.


You do not even need a new rheostat, which is probably expensive at this level of power. Add a fixed power resistor in parallel to the rheostat. Using e.g. a 15Ω resistor (10W or more) will result in an adjustable range of 9.4Ω...0Ω (the latter being limited by the smallest ressitance of the rheostat).
The equation used is Rtotal=(R1*R2)/(R1+R2) where R1, R2 are the resistances of the two resistors in parallel (fixed and variable).

 

[MrClamperSir]

The tools should start earlier, but the rheostat will become hot anyway as the power dissipated is the same (voltage drop*current). The rheostat may in total become not as hot as the 25Ω type because the power is distributed among a larger part of the rheostat.


You do not even need a new rheostat, which is probably expensive at this level of power. Add a fixed power resistor in parallel to the rheostat. Using e.g. a 15Ω resistor (10W or more) will result in an adjustable range of 9.4Ω...0Ω (the latter being limited by the smallest ressitance of the rheostat).
The equation used is Rtotal=(R1*R2)/(R1+R2) where R1, R2 are the resistances of the two resistors in parallel (fixed and variable).

Ahhh yes, of course!!! Thank you very much. I will try this tomorrow and test it at work. And you're right about the rheostats being pricey. They're around $50 a piece and these ratings.

Thanks again!!!

 

[MrClamperSir]

The tools should start earlier, but the rheostat will become hot anyway as the power dissipated is the same (voltage drop*current). The rheostat may in total become not as hot as the 25Ω type because the power is distributed among a larger part of the rheostat.


You do not even need a new rheostat, which is probably expensive at this level of power. Add a fixed power resistor in parallel to the rheostat. Using e.g. a 15Ω resistor (10W or more) will result in an adjustable range of 9.4Ω...0Ω (the latter being limited by the smallest ressitance of the rheostat).
The equation used is Rtotal=(R1*R2)/(R1+R2) where R1, R2 are the resistances of the two resistors in parallel (fixed and variable).

So I put in a fixed 10watt 10Ω resistor. The results gave me a better range earlier on in the rheostat however the ceramic resistor was HOT (scary hot)!!! Is this to be expected?

 

[Alec_t]

Yes, if it's dissipating several Watts without a heatsink or forced air cooling.

 

[Harald Kapp]

Something's got to get hot. The resistor, the rheostat or the transistor in the linear power supply.
That's the problem with linearly regulating power.
Use a suitable cooling mechanism as suggested by Alec.

 

[dorke]

What a mess this thread has turned into...
I don't recall posting in the middle of it.
How did that happen?
Plus-don't get any alerts...

 

[davenn]

What a mess this thread has turned into...
I don't recall posting in the middle of it.
How did that happen?
Plus-don't get any alerts...

Harald did a merge that would have been better off staying separate
as it would have saved much confusion
not to worry


Dave

 

[cjdelphi]

Without going to page 2...

This what pwm was designed for, to eliminate an expensive coil of wire..

Changing the period in which it stays on for during it's cycle determines how hard your power tools work, even a simple 555 could be used..

Avoid rehostats, use little cheap potentiometers to vary the pwm to drive your power tools (110v + tools and a triac and diac and a 1/2w Pot)

 

[Harald Kapp]

Sorry for the confusion this merge has cause. At the time I thought it to be a good idea as the contents are closely related. In hindsight it wasn't as the two threads already had developed a good measure of content.

Harald