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Koford (Slot Car) Bench Power Supply Project

TheChad

Sep 23, 2014
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Good find @pebe, this would be helpful and should run a lot cooler than the OPs current linear regulators.

That looks simple enough to install..... What else would I need to make that work?

How would that be adapted/installed in my power supply?

-TheChad
 

davenn

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That looks simple enough to install..... What else would I need to make that work?

How would that be adapted/installed in my power supply?

-TheChad

just remove the reg circuit you currently have keep the bridge rectifier and large smoothing cap
feed its output into the switchmode cct, adjust the output voltage to what you want and you are ready to go

ohhh and its a buck converter, so the input probably needs to be higher that the required output voltage

Dave
 

TheChad

Sep 23, 2014
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so are you using resistors on their outputs as shown ?

Not currently, but I will get some resistors and install them. I just wanted to make sure the 1 regulator wasn't getting hot because it was faulty or something.. I would *think* typically the FIRST regulator in the group would get the hottest not the last if current sharing was a problem... That's another reason I'm kind of confused..

So I need to put the resistor between the regulator and the output wire? (Right now there is 1 wire soldered to all 3 regulators (see pictures in previous post).. I would unsolder that wire from all 3 regulators install a resistor to each of the regulators then solder that wire back to the other side of the resistors? (Just checking..)

were they really using 33000uF or was it really 3300 uF ? 33,000uF is a BIG cap !
for an 8 A supply somewhere from 10,000 to 20,000 uF would be ample ( keeping in mind the minimum of 1000uF / amp)
and DONT FORGET the 0.1 uF caps on the input and output of each of the reg chips ( mounted as close to the chip pins as possible)

cheers
Dave

Yep, 33000uF. (See Picture)
 

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davenn

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So I need to put the resistor between the regulator and the output wire? (Right now there is 1 wire soldered to all 3 regulators (see pictures in previous post).. I would unsolder that wire from all 3 regulators install a resistor to each of the regulators then solder that wire back to the other side of the resistors? (Just checking..)
yup ... just as with the diagram :)

Yep, 33000uF. (See Picture)

wow OK probably a bit of an overkill ;) ..... note ... having more doesn't hurt ( in this situation)
 

TheChad

Sep 23, 2014
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just remove the reg circuit you currently have keep the bridge rectifier and large smoothing cap
feed its output into the switchmode cct, adjust the output voltage to what you want and you are ready to go

ohhh and its a buck converter, so the input probably needs to be higher that the required output voltage

Dave

So I could keep the 12.6v 8A transformer..?

But it only works at a set voltage? whatever the potentiometer on the board is set to?

I need to be able to adjust it with the potentiometer on the front of the power supply.. (I could unsolder the potentiometer on the board and solder in jumper wires from the potentiometer on the power supply case?)

Thanks!

-TheChad
 

davenn

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So I could keep the 12.6v 8A transformer..?

yup

But it only works at a set voltage? whatever the potentiometer on the board is set to?

yes ... do as suggested below if you need to be continuously adjusting

I need to be able to adjust it with the potentiometer on the front of the power supply.. (I could unsolder the potentiometer on the board and solder in jumper wires from the potentiometer on the power supply case?)

Yes you could do that :)
 

TheChad

Sep 23, 2014
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Yes you could do that :)

Sounds simple enough, actually seems easier than messing with the regulators at all..... Really I'm only switching out 1 component, the regulators for the Switch mode...

So this seems like the new plan... So, does the potentiometer rating matter? The potentiometer in the power supply case is a B1K (http://www.mouser.com/ProductDetail/Alpha-Taiwan/RV170SF-10-15R1-B1K/?qs=/ha2pyFaduh8J5PgS9Jj1hkVQUW3cV56rsfYPuuXJluABbwvL72FoA==), I was going to Switch it out to a B5K (http://www.mouser.com/Search/Produc...15R1-B5Kvirtualkey14860000virtualkey31CQ305-F) for more precise adjustment.. Will this work?

From what I can come up with on Google, I *think* the potentiometer that they are using on the board is this one: http://www.farnell.com/datasheets/1815120.pdf which is a 10ohm to 1MegOhm.. Will using a 5k Ohm potentiometer make it so I loose adjust ability? Or just make the adjustments more sensitive?

Will the response on the potentiometer adjustment be instant as it is now? With out any delay/sluggishness?

Thanks!

-TheChad
 
Last edited:

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Sometimes what you say has been contradictive to me... You said the regulators "DON'T" run away, but then you said not to run them in parallel unless I make sure they are balancing the load.. So that to me seems like a contradiction.

It's not.

They don't run away, but they don't share current either.

As I've said, one will get hotter than the rest and start to go into shutdown eventually. It is not recommended to operate the devices regularly in shutdown mode.

I also am not sure you are fully reading my posts... I don't know how many times I can express.. *I* did not build this power supply. It was built by an company which happens to be an Electrical Engineering company that also has a Slot Car division..

Yeah, but you're completely re-engineering it for more current and with a higher input voltage.

The fact that you don't know what compromises and limitations the original design has doesn't mean it has none.

You are basically saying that even UNMODIFIED, the power supply was not built right and shouldn't have been built that way... Well you'll have to take that up with the manufacture...

All I am relying on is your information.

The apparent fact that the existing regulator was simply glued to the rear panel is appallingly bad design. Would I suggest you repeat this?

Was the original power supply variable? This on its own makes the design far more demanding because the amount of heat dissipated at low output voltages and high output currents increases rapidly over what you get at higher output voltages.

All I want to do is modify it to work better for my needs. Is it going to be perfect? No, not unless I completely start from scratch. I don't need or want to do that at this point, just make it work BETTER for my needs...

However you have pretty much ignored everything I've suggested.

At this point, I have identified that 1 of the regulators is hotter then the other 2, which means I need to fix that so they are equally sharing the load..

Great, you've shown that what I predicted would happen would happen.

I can't replace the 3 regulators with 1, It's not possible for the amount of current I have. The best I could do is replace the 3 regulators with 2, which may or may not help with the load sharing? I am going to order a lower voltage transformer, which from the advice given, will remove some of the heat issue.

I've covered this in detail in various earlier posts.

But I'm trying to figure out if I should order a 10v 8A transformer, or a 7.5v 15A Transformer. Either of them will have enough voltage and enough amperage. But will the extra Amperage of the 15A transformer cause more heat? or does it matter as long as I'm not actually drawing that much amperage?

I'm sorry. I've tried to answer but all that's happened is that I've been criticised for doing calculations.

I presume the power supply needs to be able to tolerate a shorted output. I assume you want reasonably good voltage regulation (i.e. you don't want it to sag significantly under load). I assume you want it to be reliable. I'm NOT assuming you want a super accurate bench power supply (although having said that, you could probably purchase a bench power supply capable of the sort of power you require for under $100). And given all that, I assume you need to keep it affordable.

Honestly, I've read and appreciate every bit of advice you have given. I have no doubt you are telling me the BEST and PROPER way to do things, but what I know is what I already have works, I just want to change a couple of things to make it a little better, with out rebuilding or redesigning the entire power supply...

You appear to be going beyond what is easily possible for a linear design. You WILL be thermally limited. Power supplies with these power ratings using linear regulators employ various techniques including switching input voltages and forced air cooling to achieve it. But all you want to use is passive cooling.

A switchmode regulator will solve many of the thermal problems by virtue of the fact that it is simply far more efficient. But it is more complex.

I recently made a 6V 5A regulator running from a 12V lead acid battery which used a very tiny heatsink. In providing 30W of energy, it wasted only a handful of watts, compared to a linear supply that would have wasted over 30W.

With the 6.6v 6A transformer they used a 33000uF capacitor, So I *think* a 56000uF capacitor would be good for the 10v 8A transformer I am looking at ordering. But that is a guess and probably wrong?

That's a pretty large capacitor, and it will have a side effect of drawing HUGE currents from the transformer near the peaks of the AC waveform, so you may find that the transformer, rectifiers and even the capacitor may get quite warm.

Incidentally a switchmode regulator allows you to have a higher voltage input that is filtered far less well, whilst not suffering from the problem that this excess voltage is converted directly into heat.

But in terms of value, it's probably OK. You'll get about 12V under load and up to about 5.5A. At 5.5A the ripple would be just under 1V, and assuming 2.5V overhead for the regulator, you could reliably adjust the voltage from 1.25V to 8.5V. The maximum power dissipation would be in the region of 70W (assuming short circuit output and current limiting not too much higher than 6A. With an output of 5A at 4V, the power dissipation would be around 40W. Given that the TO220 device has a thermal resistance between the junction and case of 4 degC/W, the maximum dissipation of the device (assuming ambient temp of 25C and die temperature of 115C would be well under 20W per device. The TO-3 devices have a lower thermal resistance so they could have a larger dissipation. Also, increasing the temperature you allow the junctions to reach will increase this. The operating junction temp maxes out at 125C, so there's only so far you can go.

If you assume the load is not continuous (and you really haven't said much at all about this) then you can raise the dissipation limit a little because you assume it has time to cool down. However this is somewhat risky as you really need to limit the duration of these high loads such that the device doesn't overheat. It doesn't matter how much time it has too cool down again if exceeds the max temp when under load.

I don't completely understand the schematics and what not, so while you may look at them and it makes perfect since, a lot of it might as well be in another language to me. So I apologize if my questions or project are upsetting you,

I am trying to keep it as simple as possible (for example I showed you a fairly complex design from the datasheet and then suggested something a lot simpler). If you don't understand what the schematics mean it's far better to ask than to remain silent and have me assume you're just ignoring me.

What you need to do is to fully understand and specify the nature of the load so we can recommend the best way to drive that load.
 

(*steve*)

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If you want 5A, I'd look at the LM2587.

It's only slightly more complex than a three terminal regulator, but you will need an inductor and a special diode. This will work fine with your 12V transformer and existing capacitor(s), although you will need some additional low value capacitors and possibly a low ESR capacitor for the output.

You will also probably get away with connecting it to the back panel for cooling.

I believe these also have current limiting, so you won't find yourself amidst smoke if you accidentally short it (but it probably won't do it any good either).

edit: I'll take that back. The 2587 isn't the simple device I thought it was. You might have to wind your own transformer... I was thinking about another similarly named device.
 

TheChad

Sep 23, 2014
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It's not.

They don't run away, but they don't share current either.

As I've said, one will get hotter than the rest and start to go into shutdown eventually. It is not recommended to operate the devices regularly in shutdown mode.



Yeah, but you're completely re-engineering it for more current and with a higher input voltage.

The fact that you don't know what compromises and limitations the original design has doesn't mean it has none.



All I am relying on is your information.

The apparent fact that the existing regulator was simply glued to the rear panel is appallingly bad design. Would I suggest you repeat this?

Was the original power supply variable? This on its own makes the design far more demanding because the amount of heat dissipated at low output voltages and high output currents increases rapidly over what you get at higher output voltages.



However you have pretty much ignored everything I've suggested.



Great, you've shown that what I predicted would happen would happen.



I've covered this in detail in various earlier posts.



I'm sorry. I've tried to answer but all that's happened is that I've been criticised for doing calculations.

I presume the power supply needs to be able to tolerate a shorted output. I assume you want reasonably good voltage regulation (i.e. you don't want it to sag significantly under load). I assume you want it to be reliable. I'm NOT assuming you want a super accurate bench power supply (although having said that, you could probably purchase a bench power supply capable of the sort of power you require for under $100). And given all that, I assume you need to keep it affordable.



You appear to be going beyond what is easily possible for a linear design. You WILL be thermally limited. Power supplies with these power ratings using linear regulators employ various techniques including switching input voltages and forced air cooling to achieve it. But all you want to use is passive cooling.

A switchmode regulator will solve many of the thermal problems by virtue of the fact that it is simply far more efficient. But it is more complex.

I recently made a 6V 5A regulator running from a 12V lead acid battery which used a very tiny heatsink. In providing 30W of energy, it wasted only a handful of watts, compared to a linear supply that would have wasted over 30W.



That's a pretty large capacitor, and it will have a side effect of drawing HUGE currents from the transformer near the peaks of the AC waveform, so you may find that the transformer, rectifiers and even the capacitor may get quite warm.

Incidentally a switchmode regulator allows you to have a higher voltage input that is filtered far less well, whilst not suffering from the problem that this excess voltage is converted directly into heat.

But in terms of value, it's probably OK. You'll get about 12V under load and up to about 5.5A. At 5.5A the ripple would be just under 1V, and assuming 2.5V overhead for the regulator, you could reliably adjust the voltage from 1.25V to 8.5V. The maximum power dissipation would be in the region of 70W (assuming short circuit output and current limiting not too much higher than 6A. With an output of 5A at 4V, the power dissipation would be around 40W. Given that the TO220 device has a thermal resistance between the junction and case of 4 degC/W, the maximum dissipation of the device (assuming ambient temp of 25C and die temperature of 115C would be well under 20W per device. The TO-3 devices have a lower thermal resistance so they could have a larger dissipation. Also, increasing the temperature you allow the junctions to reach will increase this. The operating junction temp maxes out at 125C, so there's only so far you can go.

If you assume the load is not continuous (and you really haven't said much at all about this) then you can raise the dissipation limit a little because you assume it has time to cool down. However this is somewhat risky as you really need to limit the duration of these high loads such that the device doesn't overheat. It doesn't matter how much time it has too cool down again if exceeds the max temp when under load.



I am trying to keep it as simple as possible (for example I showed you a fairly complex design from the datasheet and then suggested something a lot simpler). If you don't understand what the schematics mean it's far better to ask than to remain silent and have me assume you're just ignoring me.

What you need to do is to fully understand and specify the nature of the load so we can recommend the best way to drive that load.


Thank you again for your information...

I don't know what other information to provide about the load.... All this power supply is used for is running electric motors... The MAX Amps I will likely ever pull is 5A and that would be at 4+ volts... These motors pull their max amps at 4v, past 4v I notice very little change in the amps that are drawn.. The motor gets faster obviously with higher volts, but doesn't draw any more amps..

The switch mode does seem like the best option, I thought it was going to be much more involved than it's seeming to be.. So I think that will be the best option here.. It will get rid of the regulators problem all together.

-TheChad
 

Gryd3

Jun 25, 2014
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Thank you again for your information...

I don't know what other information to provide about the load.... All this power supply is used for is running electric motors... The MAX Amps I will likely ever pull is 5A and that would be at 4+ volts... These motors pull their max amps at 4v, past 4v I notice very little change in the amps that are drawn.. The motor gets faster obviously with higher volts, but doesn't draw any more amps..

The switch mode does seem like the best option, I thought it was going to be much more involved than it's seeming to be.. So I think that will be the best option here.. It will get rid of the regulators problem all together.

-TheChad
You can tackle this anyway you want.

If you want to keep the parts you have currently, you can make the design more robust by putting a low value resistor in-between the regulator and the output line. This needs to be done for each regulator you have in parallel (so each one has it's own resistor on the output) This will help keep the output current on each regulator the same, and prevent one from heating up more than the others.

There are of course, two options for a switch mode supply.
You can build one yourself, which at this point I am sure you are capable. It requires a component much like the regulators you are using but with a couple extra pins. You need to connect an inductor, capacitor, and a couple miscellaneous components and your set.
You can buy a pre-made module as well, and they should be designed in such a way as there is no severe delay or lag between adjusting the potentiometer and seeing the output change. Note I say 'should' here.
If the components such as capacitors and inductors being used are too large, there will be a slight delay. This delay will be reduced the more current you are drawing. The whole point of these power supplies though, is that it compares the output voltage to a reference, and tries to hold it there. If the reference changes, it will compensate, even with large value inductors and capacitors. I cannot give you numbers, but I am very certain that you will not notice any delays with the switch-mode module you posted a few posts back. (If you buy a pre-made module, you will most likely need to keep the bridge rectifier in your current supply unless the switch mode is designed to work with an AC input.)
 

TheChad

Sep 23, 2014
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You can tackle this anyway you want.

If you want to keep the parts you have currently, you can make the design more robust by putting a low value resistor in-between the regulator and the output line. This needs to be done for each regulator you have in parallel (so each one has it's own resistor on the output) This will help keep the output current on each regulator the same, and prevent one from heating up more than the others.

There are of course, two options for a switch mode supply.
You can build one yourself, which at this point I am sure you are capable. It requires a component much like the regulators you are using but with a couple extra pins. You need to connect an inductor, capacitor, and a couple miscellaneous components and your set.
You can buy a pre-made module as well, and they should be designed in such a way as there is no severe delay or lag between adjusting the potentiometer and seeing the output change. Note I say 'should' here.
If the components such as capacitors and inductors being used are too large, there will be a slight delay. This delay will be reduced the more current you are drawing. The whole point of these power supplies though, is that it compares the output voltage to a reference, and tries to hold it there. If the reference changes, it will compensate, even with large value inductors and capacitors. I cannot give you numbers, but I am very certain that you will not notice any delays with the switch-mode module you posted a few posts back. (If you buy a pre-made module, you will most likely need to keep the bridge rectifier in your current supply unless the switch mode is designed to work with an AC input.)


Thanks soo much for the reply/information!

At this point with the use of that module, changing to a switch mode is a no brainier! Now my only concern is the built in potentiometer...

-TheChad
 

Gryd3

Jun 25, 2014
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Thanks soo much for the reply/information!

At this point with the use of that module, changing to a switch mode is a no brainier! Now my only concern is the built in potentiometer...

-TheChad
Well, step one is identifying the value.
There is usually a 3 digit number printed on the case. The first two numbers are taken literally, and the third number is usually how many 0s to add. So a 502 becomes a 50(00) which is very nice 5kΩ value.

Once you know the value, you can swap it out.
 

TheChad

Sep 23, 2014
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Well, step one is identifying the value.
There is usually a 3 digit number printed on the case. The first two numbers are taken literally, and the third number is usually how many 0s to add. So a 502 becomes a 50(00) which is very nice 5kΩ value.

Once you know the value, you can swap it out.

I have a module on order... So I will know for sure once I get it... But from the pictures, it looks like 503 (50,000) ohm.. That seems like a whole lot... But I checked and my potentiometer is available in a 50k ohm...

I have to order another potentiometer no matter what... I may just order both the 5k ohm and the 50k ohm, they are only a few $ each.. Then I can test them both!

That potentiometer that comes on the module turns 25 full turns to cover the full range... My potentiometer turns only 260 degrees..

Am I thinking wrong to think that it should still adjust the same as my current setup with the same potentiometer, as the LM350 regulators are 1.2-33v and and the switching board is 1.2-32v..

So my thought is with the same potentiometer the adjustment should be the same? Am I wrong?

Thanks!

-TheChad
 

davenn

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the 25 turn pot just gives finer setting (adjustment) of the voltage

ohhhh BTW ... you know there are 2 different types of pot ?
a logarithmic one and a linear one ... you need a linear one
 

Gryd3

Jun 25, 2014
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the 25 turn pot just gives finer setting (adjustment) of the voltage

ohhhh BTW ... you know there are 2 different types of pot ?
a logarithmic one and a linear one ... you need a linear one
Thanks for that Dave. I've only ever seen 'log' on audio applications, but Ive been surprized before.
 

davenn

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yeah, on audio gear, generally they use log pots for volume controls and linear ones for tone/balance controls
 

TheChad

Sep 23, 2014
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the 25 turn pot just gives finer setting (adjustment) of the voltage

ohhhh BTW ... you know there are 2 different types of pot ?
a logarithmic one and a linear one ... you need a linear one


The potentiometer that is on the power supply right now is http://www.mouser.com/Search/m_Prod...15R1-B1K/&qs=sGAEpiMZZMsb7Mk3aR4n05bmgoB3qUuV

I was going to install the 5k version of the same potentiometer to give me finer control of the voltage...

They do make a 50k ohm version of the same potentiometer...

They are linear potentiometers...

-TheChad
 

davenn

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The potentiometer that is on the power supply right now is http://www.mouser.com/Search/m_ProductDetail.aspx?Alpha-Taiwan/RV170SF-10-15R1-B1K/&qs=sGAEpiMZZMsb7Mk3aR4n05bmgoB3qUuV

I was going to install the 5k version of the same potentiometer to give me finer control of the voltage...

They do make a 50k ohm version of the same potentiometer...

They are linear potentiometers...

-TheChad


yes, that is a linear pot
a 5k wont give you finer adjustment, pretty sure it will just give you less range
 

pebe

Sep 3, 2013
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The potentiometer that is on the power supply right now is http://www.mouser.com/Search/m_ProductDetail.aspx?Alpha-Taiwan/RV170SF-10-15R1-B1K/&qs=sGAEpiMZZMsb7Mk3aR4n05bmgoB3qUuV

I was going to install the 5k version of the same potentiometer to give me finer control of the voltage...

They do make a 50k ohm version of the same potentiometer...

They are linear potentiometers...

-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.
 
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