Koford (Slot Car) Bench Power Supply Project

[TheChad]

If you are talking about the switch mode module, then don't change the value of the pot - it is part of the design. That particular module I mentioned needs a 50K linear. It is worth remembering that it will give a max output voltage of supply voltage minus1V, and is up to 98% efficient so only 2% of the input power is dissipated in heat. Switch mode modules' outputs are very stable with changing loads and power out = power in x efficiency. So with a 12.6V 8A transformer you will get up to 11.6v at 8A. At 6V output or lower, you can get the max 15A.

Is it for sure a 50k potentiometer? Do you know that for sure?

I have the module on order, Now I need to order the potentiometer..


What should I do about the smoothing capacitor? What value is the correct size? 6800uF? Bigger? Smaller?

I am going to order one either way, so the price is basically the same, I might as well get the correct size..

Thanks!

-TheChad

 

[davenn]

anything from 6800 to ~ 15000uF will do


am looking forward to seeing pics of your final construction

 

[TheChad]

am looking forward to seeing pics of your final construction

Will definitely!

 

[pebe]

Is it for sure a 50k potentiometer? Do you know that for sure?

I have the module on order, Now I need to order the potentiometer..


What should I do about the smoothing capacitor? What value is the correct size? 6800uF? Bigger? Smaller?

I am going to order one either way, so the price is basically the same, I might as well get the correct size..

Thanks!

-TheChad

I am going on the 503 marked on the pot shown in the picture. I have no reason to think they will have changed it, but if you want to be sure wait till your module arrives.

 

[TheChad]

anything from 6800 to ~ 15000uF will do

Okay, Going with a 12000uF 25v Capacitor... Any objections, Thoughts, Opinions?

Thanks!


-TheChad

 

[TheChad]

I am going on the 503 marked on the pot shown in the picture. I have no reason to think they will have changed it, but if you want to be sure wait till your module arrives.

Okay, yeah that's what I was going by also... I didn't know if maybe you had this module or something...

I will wait till it shows up to order the potentiometer just to make sure it's right..

Thanks,

-TheChad

 

[TheChad]

Okay, while I am awaiting the DC/DC Converter module.. I was using the power supply to break in a motor drawing 1.3A @ 4v. Of course the regulators got hot, as we already know..

But the Bridge Rectifier also got hot. I know Bridge Rectifier's don't run cool. Would changing the Rectifier to one that handles say 15A do anything? Will it run any cooler?

Otherwise, I assume there is no solution to the heat that the Bridge Rectifier produces?

Thanks!

-TheChad

 

[Gryd3]

There is a slight voltage drop with the bridge rectifier, so it will get warm.
How 'hot' did it get? It should only drop about a volt and a half, and at 1.3Amps, that should be just shy of 2Watts.
I'm under the impression it was mounted to the bottom of the case for the power supply, so that should help carry away the heat.

 

[TheChad]

There is a slight voltage drop with the bridge rectifier, so it will get warm.
How 'hot' did it get? It should only drop about a volt and a half, and at 1.3Amps, that should be just shy of 2Watts.
I'm under the impression it was mounted to the bottom of the case for the power supply, so that should help carry away the heat.

It is mounted (well caulked) to the bottom of the power supply case..

I'm still looking for that darn infrared temp gun, but it was hot enough that after 5-10 seconds It wasn't comfortable to leave my finger on it still.

-TheChad

 

[pebe]

Okay, while I am awaiting the DC/DC Converter module.. I was using the power supply to break in a motor drawing 1.3A @ 4v. Of course the regulators got hot, as we already know..

But the Bridge Rectifier also got hot. I know Bridge Rectifier's don't run cool. Would changing the Rectifier to one that handles say 15A do anything? Will it run any cooler?

Otherwise, I assume there is no solution to the heat that the Bridge Rectifier produces?

Thanks!

-TheChad

Bear in mind that a switch mode converter does for DC the same that a transformer does for AC, ie. it changes the ratio of input to output voltages without changing the power.

So supplying 4V 1.3A (5.2W) for your motor will take only 0.41A from your 12.6V supply, and the bridge rectifier will run cooler.

 

[TheChad]

Bear in mind that a switch mode converter does for DC the same that a transformer does for AC, ie. it changes the ratio of input to output voltages without changing the power.

So supplying 4V 1.3A (5.2W) for your motor will take only 0.41A from your 12.6V supply, and the bridge rectifier will run cooler.

But the switch mode converter is AFTER the rectifier in the circuit... So wouldn't that make the rectifier still be doing the same amount of work?

-TheChad

 

[Gryd3]

But the switch mode converter is AFTER the rectifier in the circuit... So wouldn't that make the rectifier still be doing the same amount of work?

-TheChad

Nope
If we ignore the laws of physics and think of an ideal switch-mode converter, the same power that goes in, comes out.

So.... if you are drawing 5V, and 2A from the supply, that is 10W.
With the current setup, the 2A will need to go through the bridge, then the linear regulators. This is what is causing the heat.

With a switch-mode supply, the 5V 2A being drawn out, will only require 12V 0.83mA. This will reduce the current through the bridge, resulting in a cooler running part.
[12V @ 0.83mA = 10W ]
Of course, it not really going to be as low as 0.83mA due to losses in the circuit, and the switch-mode efficiency.

 

[TheChad]

Okay! So I got the parts in...

I got the DC/DC Converter board installed, I got the built in rheostat removed and wire leads soldered in...

I temporarily connected the new 50k ohm rheostat for testing..

There are 3 leads from the board, 3 pin's on the rheostat I removed from the board and 3-pins on the external rheostat.... So I connecting all 3 pins, what happens is that when the rheostat knob gets to about 50% the voltage begins to go back down instead of climing.... So I disconnected the 3rd pin on the rheostat, using only 2 pins as it was connected before with the old linear regulatos. That fixes that problem! (Weird but Okay)

Next, the rheostat get's to about 90%-95% and it's already at max voltage, which shows @ 19v. So there is still 5%-10% rotation left in the rheostat.... I'm guessing that's because it's capable of up to 35v? but Would be nice to have the extra control...

Okay, so I don't want 19v, that is MUCH too high... Is there any way of limiting that? Adding resistor(s)? Worse case a different rheostat? I'd like to have the 100% rotation of the rheostat @ 12v-13v...

Also, Would it be normal to not have anything connected to the negative terminal on the DC out side? The negative test lead is connected to the AMP draw pin on the digital meter...

It works with nothing connected to the negative terminal on the DC out side, but is that normal? Where should that negative terminal be connected to?

Thanks!

-TheChad

 

[Gryd3]

Glad to see the progress you have made so far!

First thing to address is the rheostat. Depending on the design, it may or may not have required 3 pins. You may just be getting lucky with using two... but it all depends on which ones are used.
If all it does is compare a resistance, than two is fine. (As long as it's the middle pin, and a side pin)... otherwise, if the rheostat is supposed to operate as a voltage divider, all 3 pins will be required. From the sounds of things though... I'm pretty sure 2 pins is fine. (The board may have two of the three pins shorted together anyway... which is pretty much like using two pins. You may have made an error when you originally connected all 3 pins.)

The range of the potentiometer... It's hard to say for sure what will allow you to fine tune the voltage you want for when the dial is at 100%... easy answer is to adjust the dial until the power supply puts out the highest voltage you 'want', then measure the resistance it is set to. A little bit of math can be done to determine the best solution to solve this... the solution could be as easy as adding a resistor, or it could require a new rheostat, but we will find out when you post your numbers. (Also please post the resistance when you have it turned all the way down... it could be 50KΩ or 0Ω...)
** One of the two used wires may need to be removed to take a proper resistance measurement... or you could use a trick... There are 3 pins... Left, Center, Right... The Left or Right pin sounds like it is currently unused. If this is the case, measure the resistance from the Left and Right pins. It should read 50KΩ, if this is the case, you can measure the unused pin, and the center pin 'while it is connected to your circuit!' (but powered off to be safer) The number you get from the center pin to the unused pin will be the opposite of what the used pin will be... ( 50KΩ - reading = CurrentSetting ) So if you read 10K, that means the power supply is seeing and responding to the 40K on the opposite side of the rheostat.

Now to address the negative lead. I can't give you a yes/no answer, but I can tell you that I'm not surprised that it works without explicitly using the negative side of the output... Take a look at your board if you can, the negative input is most likely connected directly to the negative output. If this is not the case, take a closer look to be absolutely sure... If it is not directly connected even with a jumper or something else then we are missing a detail, or you have gotten very lucky and should not risk running a load without connecting the negative side of the DC output. (It sounds like the negative is common though, so you should be fine... always doublecheck someone elses intel though. I don't loose a thing if you pop anything, but you do)

Keep us posted!

 

[TheChad]

Glad to see the progress you have made so far!

First thing to address is the rheostat. Depending on the design, it may or may not have required 3 pins. You may just be getting lucky with using two... but it all depends on which ones are used.
If all it does is compare a resistance, than two is fine. (As long as it's the middle pin, and a side pin)... otherwise, if the rheostat is supposed to operate as a voltage divider, all 3 pins will be required. From the sounds of things though... I'm pretty sure 2 pins is fine. (The board may have two of the three pins shorted together anyway... which is pretty much like using two pins. You may have made an error when you originally connected all 3 pins.)

Should I try shorting the 2 outside pin's? I'm pretty sure it is only using the potentiometer to compare resistance.. Connecting all 3 pins obviously doesn't work.. so either it's only using 2 pins, or the 3rd pin is shorted to the 1st pin?

The range of the potentiometer... It's hard to say for sure what will allow you to fine tune the voltage you want for when the dial is at 100%... easy answer is to adjust the dial until the power supply puts out the highest voltage you 'want', then measure the resistance it is set to. A little bit of math can be done to determine the best solution to solve this... the solution could be as easy as adding a resistor, or it could require a new rheostat, but we will find out when you post your numbers. (Also please post the resistance when you have it turned all the way down... it could be 50KΩ or 0Ω...)

At 12v the potentiometer measures 16,300Ω. Turned all the way down it measures 0Ω and outputs 1.2v as the spec's say it should.

Now to address the negative lead. I can't give you a yes/no answer, but I can tell you that I'm not surprised that it works without explicitly using the negative side of the output... Take a look at your board if you can, the negative input is most likely connected directly to the negative output. If this is not the case, take a closer look to be absolutely sure... If it is not directly connected even with a jumper or something else then we are missing a detail, or you have gotten very lucky and should not risk running a load without connecting the negative side of the DC output. (It sounds like the negative is common though, so you should be fine... always doublecheck someone elses intel though. I don't loose a thing if you pop anything, but you do)

Keep us posted!

I'm pretty sure the negative is common... The negative test lead is connected directly to the amp meter lead. Should I connect the negative output lead to the same wire? If it's common, it wouldn't make any difference connecting the negative output terminal to the amp meter lead/negative test lead right? I guess would there be any benefit? or completely pointless/useless?

Thanks!

-TheChad

 

[Gryd3]

potentiometer to compare resistance.. Connecting all 3 pins obviously doesn't work.. so either it's only using 2 pins, or the 3rd pin is shorted to the 1st pin?
If it's working, don't short the unused pin to anything. No need to mess with it if it's working
(You could always take a closer look at the DC/DC converter to confirm if two pins are shorted together if you really want... but the fact that it didnt work when you had 3 pins connected has me a little worried that something was overlooked when it was connected.

At 12v the potentiometer measures 16,300Ω. Turned all the way down it measures 0Ω and outputs 1.2v as the spec's say it should.
So... I have a make-shift solution you can use if you want at the expense of linearity. Just means that the dial won't be 100% consistant as you turn it, but you may not even notice.
If you have a 24KΩ resistor, you can connect that in parallel with the rheostat you have connected.
The 24KΩ resistor in parallel (leave the rheostat connected, but add the resistor to the center pin and used pin.) will result in roughly a 16.3KΩ max resistance. If you don't have a 24K, you can also use the original rheostat. Just connect each pin to the other rheo. You can then turn the main rheo all the way up, and adjust the second one (the original) until the voltage is what you want the max to be. You can then ignore the second rheo unless you want to later increase the voltage to something higher than 12V.
Did I make sense? I'm a little tired XD

I'm pretty sure the negative is common... The negative test lead is connected directly to the amp meter lead. Should I connect the negative output lead to the same wire? If it's common, it wouldn't make any difference connecting the negative output terminal to the amp meter lead/negative test lead right? I guess would there be any benefit? or completely pointless/useless?
If negative is common, then it makes no difference if the output is connected or not. Always best to confirm though

 

[TheChad]

(You could always take a closer look at the DC/DC converter to confirm if two pins are shorted together if you really want... but the fact that it didnt work when you had 3 pins connected has me a little worried that something was overlooked when it was connected.

What could be connected wrong? I tried moving all 3 pins around and the same results.... When the potentiometer get's to 50%, the voltage stops climbing and begins to go back down.... Which makes since in that it's pulling the resistance off the other side at 50%.... You can't really tell by looking at the DC/DC board, obviously the 3rd pin connects to something on the board otherwise it would have no effect at all... but I'm not sure why it functions the way it does... Maybe because the little potentiometer that came on the board is suppose to be "infinitely" adjustable, so maybe after you adjust it all the way in one direction, if you keep turning it, it starts goes back the other way instead of bottoming out..

So... I have a make-shift solution you can use if you want at the expense of linearity. Just means that the dial won't be 100% consistent as you turn it, but you may not even notice.
If you have a 24KΩ resistor, you can connect that in parallel with the rheostat you have connected.
The 24KΩ resistor in parallel (leave the rheostat connected, but add the resistor to the center pin and used pin.) will result in roughly a 16.3KΩ max resistance. If you don't have a 24K, you can also use the original rheostat. Just connect each pin to the other rheo. You can then turn the main rheo all the way up, and adjust the second one (the original) until the voltage is what you want the max to be. You can then ignore the second rheo unless you want to later increase the voltage to something higher than 12V.
Did I make sense? I'm a little tired XD

Make's perfect sense! As a matter of fact, I just did this as you suggested with the original rheo, and it works fantastic... I don't notice any inconsistency in the main potentiometer... it seems consistent as the voltage climbs or falls. If I turn it quick, the voltage jumps/spikes, which I guess is normal and can't really be avoided? Maybe the small capacitor (that I have not yet installed) will help that?

I'm pretty sure the negative is common... The negative test lead is connected directly to the amp meter lead. Should I connect the negative output lead to the same wire? If it's common, it wouldn't make any difference connecting the negative output terminal to the amp meter lead/negative test lead right? I guess would there be any benefit? or completely pointless/useless?
If negative is common, then it makes no difference if the output is connected or not. Always best to confirm though

Negative is definitely common.... Just for giggles, I went ahead and connected the negative from the DC/DC board to the same wire (That connects from the AMP meter to the negative test lead) and absolutely nothing happens. Which pretty much just confirms that they are the same common.... So I guess there really is no reason to leave it connected.

It occurred to me now, that perhaps I should have just left the original rheostat connected to the board and just soldered to the back the external rheostat, which would give me the same as I have connected now, but would basically give me a permanent ability to open the case and adjust the max output voltage on the DC/DC board itself..... Maybe I will just solder the rheostat back into the board... It seems like a perfect solution..

I haven't installed the new 15000uF capacitor yet or the output capacitor yet, Those will get done last... I am thinking I may also still install the resistor you suggested earlier to drain off the capacitor, it does self drain in the current configuration, but it takes a couple minutes.. The new main capacitor is much larger and I'm guessing it is going to take quite awhile to drain off...

Thanks!

-TheChad

 

[Gryd3]

What could be connected wrong? I tried moving all 3 pins around and the same results.... When the potentiometer get's to 50%, the voltage stops climbing and begins to go back down.... Which makes since in that it's pulling the resistance off the other side at 50%.... You can't really tell by looking at the DC/DC board, obviously the 3rd pin connects to something on the board otherwise it would have no effect at all... but I'm not sure why it functions the way it does... Maybe because the little potentiometer that came on the board is suppose to be "infinitely" adjustable, so maybe after you adjust it all the way in one direction, if you keep turning it, it starts goes back the other way instead of bottoming out..

This could happen if it was hooked up incorrectly. The 3 pins I mentioned earlier is almost always Left/Center/Right.
The Center pin adjusts with the dial, and will slowly slide from the Left pin to the Right pin to change the resistance. So if the Left/Right pins were accidentally shorted together when it was wired wrong... the mid point of the dial would be the furthest away from the Left and Right.
It is quite common to short the unused side of the potentiometer to the Center pin on the circuit board... so if you connected the Center pin of your rheo to the wrong spot, this would explain things.

Make's perfect sense! As a matter of fact, I just did this as you suggested with the original rheo, and it works fantastic... I don't notice any inconsistency in the main potentiometer... it seems consistent as the voltage climbs or falls. If I turn it quick, the voltage jumps/spikes, which I guess is normal and can't really be avoided? Maybe the small capacitor (that I have not yet installed) will help that?

Yes, a small output capacitor will help.. but remember is 'smooths' it out. The spike will be less, but it will also limit how fast you can change the voltage. The bigger the output capacitor, the more it will smooth things out. The recomended value above would be a good start.

Negative is definitely common.... Just for giggles, I went ahead and connected the negative from the DC/DC board to the same wire (That connects from the AMP meter to the negative test lead) and absolutely nothing happens. Which pretty much just confirms that they are the same common.... So I guess there really is no reason to leave it connected.

Your call. On the flip side, there is no reason to disconnect it either. For completeness sake, it may be better to leave it connected if it can be done cleanly.

It occurred to me now, that perhaps I should have just left the original rheostat connected to the board and just soldered to the back the external rheostat, which would give me the same as I have connected now, but would basically give me a permanent ability to open the case and adjust the max output voltage on the DC/DC board itself..... Maybe I will just solder the rheostat back into the board... It seems like a perfect solution..

Agreed, take care when soldering and resoldering to a board. The pads and traces have a tendancy to lift off of they are overheated/heated too much.

I haven't installed the new 15000uF capacitor yet or the output capacitor yet, Those will get done last... I am thinking I may also still install the resistor you suggested earlier to drain off the capacitor, it does self drain in the current configuration, but it takes a couple minutes.. The new main capacitor is much larger and I'm guessing it is going to take quite awhile to drain off...

You can take care of this two ways.
Connecting a resistor across a capacitor will help drain it quicker, but it is always connected to the circuit and will still try to drain it while it's running.
Using 10k should not affect the output, and is still high enough that it should only loose about 1/10th of a milliamp when your running at 12V.
Alternatively, you could leave your dc test motor connected when you turn it off, and the residual power in the capacitor should make it's way through the motor.

 

[TheChad]

This could happen if it was hooked up incorrectly. The 3 pins I mentioned earlier is almost always Left/Center/Right.
The Center pin adjusts with the dial, and will slowly slide from the Left pin to the Right pin to change the resistance. So if the Left/Right pins were accidentally shorted together when it was wired wrong... the mid point of the dial would be the furthest away from the Left and Right.
It is quite common to short the unused side of the potentiometer to the Center pin on the circuit board... so if you connected the Center pin of your rheo to the wrong spot, this would explain things.

Nothing was connected wrong, that I know. I even tried moving all the pins around. Either way, I guess at this point it doesn't matter as I've re-soldered the original potentiometer back into the board and connected just the 2 wires from the external potentiometer. (See Picture).


Here is the picture of the 3-wires soldered in place of the potentiometer, which is what I origionally did.... and a picture of how it is now, Re-soldered the origional potentiometer back to the board and soldered the 2 terminals from the external rheostat.. Do you think there could be any possibility of shorting where the 2 wires are soldered on the bottom side? they are solid, they don't move, but they are close together, then again, the pins from the PCB rheostat are/were just as close... I guess arcing isn't a concern?

Thanks,

-TheChad

 

[Gryd3]

It takes a very high voltage to arc.
http://en.wikipedia.org/wiki/Dielectric_strength
I would not worry about it, just make sure it does not break off and short anything it touches.