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Koford (Slot Car) Bench Power Supply Project

Discussion in 'Power Electronics' started by TheChad, Sep 24, 2014.

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  1. TheChad

    TheChad

    114
    3
    Sep 23, 2014
    Hi,

    Let me start off by introducing myself! My name is Chad. I had several years of electronics classes 10+ years ago but I went threw stage 4 cancer/chemo which wiped out most of the knowledge I had of electronics! Though I typically begin picking up on the things that were wiped pretty quickly.. So please be patient with me!

    Anyway, I have a variable power supply, used to test/break in motors and run accessories (motor & tire lathe's) for Slot Car's.

    The power supply as it came from the manufacture had a 6.3v 6a transformer. I needed a bit more power, so I installed a 12.6v 8a transformer in the case, as well as increased the capacitor to a 4700uF (I don't remember what size the original capacitor was).

    The Linear Voltage Regulators that came with the power supply are LM350T's, 3 of them wired in parallel giving me 9a capacity. But the power supply is designed to use the case as a heat sink and it runs REALLY hot, there is no powered fan to move air over the case. I wanted to see if there was something I could do to cool the power supply down, so I started looking at the components and found that there is a Linear Voltage Regulator LM388T available that has basically the same specs as the LM350T's, except they are rated at 5A. So I thought I would switch out the LM350T's for the LM388T's there by increasing the amount of amp capacity and hopefully they will run cooler?

    Is that a valid thought? Also I don't remember why I upgraded the capacitor, but is the 4700uF capacitor good? or should that be different?

    As you see in the picture, they had a resistor soldered between the 1st and center pin of the linear Voltage Regulator, what is the purpose of that resistor?

    Again I apologize if these questions are dumb, but I'm kind of re-learning everything..

    Thanks in advance for the help!

    -TheChad
     

    Attached Files:

  2. Gryd3

    Gryd3

    4,098
    875
    Jun 25, 2014
    Welcome to the forums!

    I'm going to jump the gun a little and say that no, it will most likely run just as hot. Here's why I think that.
    You are using a linear voltage regulator, these function almost like a variable resistor. So if you feed them 12V, and they provide 5V, they are eating the remaining 7V as heat. It does not matter if they are rated for 1A or 100A, they will still dissipate the same amount of heat.
    Here are some ways that I can think of to fix your problem:
    -Forced air cooling. Add some fans.
    -Do some research and see if you can swap out the linear regulators for switch-mode regulators. (This will most likely require that a single switch-mode regulator be used... I have heard that using them in parallel is tricky, but I have not tried)
    -Swap out the transformer to supply a lower voltage to the regulators you have. This will result in less wasted power to drop the transformer voltage to what you want to use.

    I mentioned them in the order I would try.
    Fans are easy to add, and easy to upgrade.
    Switch mode regulators run a lot cooler, but require some extra parts to run. They switch on/off quickly to 'average' out the voltage to be what you desire. Because they are mostly all the way on or all the way off, they are not 'eating' the extra voltage like the linear ones so they run cooler. They are also capable of providing more amperage because of this. They are not as easy as a swap in/out though.
    This is tricky to get right... and may not work as intended.. This would most likely be the most expensive solution as well... mentioned merely for completeness. This will also limit the upper limit that you can set your voltage regulators too if they are meant to be adjusted.
     
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,401
    2,777
    Jan 21, 2010
    Your idea to use a higher powered transformer was on the right track, but you should have gone for one with the same voltage (i.e. 6.3V) and a higher current.

    Doubling the input voltage in this case will probably triple the heat produced for a given current. Go back to a 6.3V transformer.

    Adding extra regulators in parallel is not really a great idea, there are other ways to get higher current. However, given that you've done it this way, you'll need three times the heat sink size (for the higher voltage) and an additional 25% on top of that due to the additional current. Actually you'd need more than that, but it gives you an idea.

    Forced air cooling can make heat sinks far more efficient, so a fan would be of great help.

    However, your new transformer (rated at 8A) compared to the old one (rated at 6A) doesn't really add much in terms of current capability.

    You would probably be far better off getting an old computer power supply which can give you perhaps 15A to 25A at 5V and produce far less heat. They have the advantage of being easily available (often $free).
     
  4. TheChad

    TheChad

    114
    3
    Sep 23, 2014

    Thanks soo much for the information!

    I didn't think of it like that, I was thinking on the line that I was pushing them too hard causing more heat!

    A fan is not possible as the power supply case is small.

    I am afraid the switch mode option may be a bit too much for me to take on!

    I will look look to see if I can find a transformer that puts out less. I don't need 12+ volts, only about 8v max.

    Any idea what the resistor is doing?

    Thanks again!

    -TheChad
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,401
    2,777
    Jan 21, 2010
    How are you connecting those devices to the case? they should be bolted on (with a *small* amount of heatsink compound), and you should probably add a heatsink to the outside (again, with heatsink compound between it and the case), so there is more surface area to allow better convective cooling. The fan could be placed on the heatsink outside the case.

    It would also be far better to place the devices as far apart from each other as practical.

    I can't see exactly where that resistor is connected, looks like between pins 1 and 2. This is an adjustable regulator and relies on 2 resistors to set the output voltage. This is almost certainly one of them. there may be another one elsewhere (or have you made the output variable)
     
  6. TheChad

    TheChad

    114
    3
    Sep 23, 2014

    They are attached to the bottom of the case with either caulk or thermo sealant, it's a silver caulk like material... This was done from the manufacturer..

    If I separated them, I could install a heat sink on each of them. But it would still be on the inside of the case with no fan..

    It is a variable power supply, so there is a potentiometer on the front of the case that dials the power...

    So what function does that resistor do? And what determines the size of the resistor used in that location?

    I will post some more/better pictures of the whole power supply in a bit when I get back home....

    Thanks

    -TheChad
     
  7. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,401
    2,777
    Jan 21, 2010
    That is a very poor way of connecting the device to the case. drill a hole and use a bolt. Using proper heatsink compound is also advisable.

    what is the range of voltages?

    It is part of the voltage regulation. Look up a datasheet on the LNm350 for more information. Essentially there is 1.25V across that resistor, thus a certain current flows. That same current also flows through the potentiometer used to set the output voltage. Because the current is known, the potentiometer setting is converted reasonably accurately into a voltage, and this allows the output voltage to be varied.

    Is there no room on the back of the case for a proper heatsink?
     
  8. TheChad

    TheChad

    114
    3
    Sep 23, 2014
    It goes from about ~3v to ~12v. It's a 1k Potentiometer


    I don't know if the caulk like material they used is thermo or not.. I went to radio shack (It's the only place anywhere in Central Illinois that sells anything electronics related!) and bought some Silicon Thermo Compound (http://www.radioshack.com/product/index.jsp?productId=12729884&retainProdsInSession=1), which may or may not be what they already used, but if it's not, then it should help with heat transfer..

    -TheChad
     

    Attached Files:

  9. Colin Mitchell

    Colin Mitchell

    1,417
    312
    Aug 31, 2014
    The only way to spread the heat is to add a 10 amp bridge between the transformer and the first bridge.
    I have hundreds of these as scrap and its a cheap way to solve the problem.
    You can add two or more until you get the maximum voltage-drop allowable for the output voltage you are requiring.
     
  10. Gryd3

    Gryd3

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    875
    Jun 25, 2014
    Are you talking about Daisy chaining bridge rectifiers together so each one drops about 1.4V each to get closer to the desired output?
     
  11. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,401
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    Jan 21, 2010
    The amount you should use should be sufficient to fill up the air gaps between the smooth metal surface of the transistor and the smooth metal surface of the heatsink. Generally speaking it's a smear. Any more than that and it will seriously reduce the heat transfer. What you're trying to do is to get metal to metal contact, and to fill any small air spaces (microscopic ones) with the heatsink compound. You do this because the heatsink compound is way better than air, but way worse than metal at transferring heat. It is NOT a glue. You MUST bolt the device to the heatsink -- even if that heatsink is just the metal case.

    As you reduce the voltage, for a given current, the amount of heat generated increases.

    As I've said before, it's a poor design (because the regulators will not share the current equally), and if you don't attach a larger heatsink, or a fan (or both) then you WILL run into problems, especially with the combination of low output voltages and high currents.
     
  12. Colin Mitchell

    Colin Mitchell

    1,417
    312
    Aug 31, 2014
    Daisy chain the bridges.

    Feel each regulator with your finger after a minute of high current and if one is hotter than the rest, they are not sharing.
    It's pointless doing anything unless you KNOW the problem.
     
  13. Gryd3

    Gryd3

    4,098
    875
    Jun 25, 2014
    Daisy chaining multiple bridges together will merely result in only half of the other bridges actually being used... You could do away with that idea, and if reducing the voltage to the regulators are the proposed solution, wouldn't the better options be a new transformer, reducing the coils on the secondary, or cascading diodes after the bridge?

    I do agree with the temperature test, but I don't think the op has experienced any failures. So even if all 3 dissipated equal heat, would the sum of all 3 on the same heat-sink not be the close enough to the same that it makes little difference?
    I've seen that poor design in amplifiers, and the shared heat-sink is required to prevent thermal runaway of only a single piece of the parallel network. If they were on separate heat-sinks, they would require additional modifications (ie, current mirror)
    As long as the temperature on all three is maintained close to the same (ie, by the same heat-sink) it should work well enough for most applications.
    I better mounting method though is a must have. Computer geek wannabes plaster lots of heatsink paste on the processor and end up over-heating it... Metal to Metal is ideal, a nice firm connection with very little paste only filling the tiniest of dimples and scratches in the metals that wouldn't touch each other anyway. This will help draw heat away from the regulators and into the metal housing. If the metal housing runs hot, you may need to modify the housing to allow for the use of a fan, or to increase the surface area by creatively modifying the shape or adding a finned heatsink.
     
  14. Gryd3

    Gryd3

    4,098
    875
    Jun 25, 2014
    I looked at your response more of a 'There are 4 diodes in a bridge and only 2 will be used' .. you meant it more for the heat-sinking capability. My point is still valid in that the bridge will only be half utilized, but I overlooked the heat-sink ability of the bridge.
    As far as your website is concerned, I'm sure you've got lots of your quotes on there. Calm down, grow up, and if your going to be on a community forum like this, help people learn and understand. You can't just snap, call something wrong and stick a quote on your blog complaining about the knowledge of others when you have had no hand in trying to help anyone understand any of it.
     
  15. TheChad

    TheChad

    114
    3
    Sep 23, 2014
    Thanks for all the information guys!

    So it sounds like the best solution is really to reduce the transformer, because I am creating unnecessary heat by having 12.6v, especially when 90% of the time I'm using it at less than 6v..

    The model of the transformer I installed is this: http://www.mouser.com/ProductDetail/Hammond/166R12/?qs=/ha2pyFadugM2ZdsbPYCrHHTiduN43vPqBcM32c3ioQ= ( 100.8(VA), 115 V 60 Hz (Primary). 12.6 C.T. 8A (Secondary), CH16H Dimension ).

    I am thinking of changing that one out for this: http://www.mouser.com/ProductDetail/Hammond/166R10/?qs=/ha2pyFaduheCjKp2H8aHCZ37F4ytt9TNSJgihGUnxI= ( 80(VA), 115 V 60 Hz (Primary), 10 C.T. 8A (Secondary), C15H Dimension ).

    Will reducing the voltage by 2.6v make a significant different in heat from the Voltage Regulators?

    Why would the regulators not be sharing the load? Assuming all 3 are good?

    Changing the transformer is the most expensive cost of any component in the power supply. It's about $40. If that is the solution, then I will do it. But I hate to waste $40 if there are other option(s) or if the 2.6v isn't going to make a noticeable difference...

    Thanks for all the help guys!

    -TheChad
     
  16. Gryd3

    Gryd3

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    875
    Jun 25, 2014
    That $40 investment will cut down on heat... but I don't want to tell you to go for it, as I personally don't feel $40 justifies running your power supply slightly cooler... This is your call though. Mounting the regulators properly to the case will help draw the heat from the parts, but will cause the metal housing to warm up. If this is acceptable, it should be left as is... Adding additional components inside to spread it out will most likely result in diminished returns where a portion of the case runs slightly cooler, the the rest of the case that the other components are mounted to will be warmer.

    Electronics components have manufacturing tolerances, and these tolerances result in the devices behaving very slightly differently... This could cause one regulator to be working a little harder than the others. The solution to this is adding more components to individually control and adjust each regulator to so that they are all operating almost exactly alike. I would not go through the trouble with linear regulators to build a current mirror though. Especially since building a switch mode will give you way better results.
    Because they are all mounted to the same heat-sink though they should operate well enough for your purpose... just make sure that they are all fastened firmly to the metal case to draw heat away. If one is not firmly mounted, it will heat up faster than the others and 'runaway' causing itself to fail.

    Depending on how you use your power supply.. have you looking into re-purposing an old desktop power supply? They are not adjustable, but they offer a few fixed voltages at an impressive current. They can usually be picked up or scavenged for free ;)
     
  17. TheChad

    TheChad

    114
    3
    Sep 23, 2014

    The power supply must be variable. I have to be able to set the voltage depending on what I am doing, Breaking in a motor, reving a motor to check for amp spikes, running an armature or tire lathe..

    I really don't use the power supply for long periods of time, but when I do, typically it's sitting at say 5v breaking in a motor. That's why I was hoping I could just change or add regulators to reduce the heat enough to make it "good enough". When I first changed the transformer I was thinking in my typical fashion of "Just because you have more power, doesn't mean you have to use it!". Because I knew I didn't need 12.6v, but It fit in the case and I said "What could it hurt?", But I never thought of the fact that the whole time I'm NOT using the additional voltage, I was burning it off!

    I am going to run some tests and see how hot the regulators are actually getting, Maybe they are HOT, but OKAY under normal operating conditions.

    -TheChad
     
  18. Gryd3

    Gryd3

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    Jun 25, 2014
  19. TheChad

    TheChad

    114
    3
    Sep 23, 2014
    LOL, I have a little better way..

    Digital Laser Thermometer!

    Right now I am installing a new Digital Volt/Amp meter so I have wires everywhere.. Tomorrow I hope to get it all put back together and I can run some real tests.

    The case has never been soo hot I felt like I burned my hand touching it, but I could feel heat radiating off of it!

    Right now I'm trying to work out the problem of, when I turn the potentiometer all the down, so effectively "off", it's still @ almost 4v. I'd like to get that down to 0 volts, but I don't know if that will happen?

    I just checked and I am getting 14v output at my connection leads... That is much higher than I need. I think the transformer I posted earlier thats 10v 8a, should give me about ~11.5v output after rectification. which is really where I wanted to be when I did this whole change of transformers in the first place..

    Thanks again for all the help!

    -TheChad
     
  20. Gryd3

    Gryd3

    4,098
    875
    Jun 25, 2014
    Most likely not... The potentiometer and the resistor form a voltage divider.
    The regulator will adjust it's output based on a formula that essentially boils down to a a minimum voltage always being present... Old formula was 1.25 x ( 1 + Potentiometer / Resistor ) .. I could not find the data sheet for the other, but it's impossible to set the output to 0 unless you start adding more electronics.
     
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