Make an Old PC PSU Your Bench Variable Power Supply

Discussion in 'The Projects Forum' started by hazim, Nov 30, 2010.

  1. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    Hi all.

    That can be done easily by using the 5V, -5V, 10V, 12V, -12V rails. the 12V rail can supply up to 10A current and the -12V can supply about 1A or less. These varies according to the manufacturer quality.
    It can be used as a "constant voltages supply" but a voltage regulator circuit can be built to make it a variable voltage supply.
    With considering the needed output current, something like this can be done:
    [​IMG]

    But I didn't build anything yet. I'm looking around the PWM IC (TL494/KIA494 or other) and specifically at the feedback circuit. I think a small change in components, like replacing a resistor with a potentiometer will make us able to vary the output voltage up and down. Using the 12V and -12V rails gives us a variable power supply with max voltage up to 24 at 0.8 or 1A. Using a switch between -12V rail and 0V makes the max. voltage 12V with current up to 10A. This will be very easy and helpful and cheap, and it should work properly :)

    To make this thread a completed project, I hope we find what component can be replaced or what changes should be done. The feedback pin is pin#3

    Regards,
    Hazim

    [​IMG]
     
    Last edited: Nov 30, 2010
  2. Wendy

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  3. tom66

    Senior Member

    May 9, 2009
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    hazim likes this.
  4. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    Ok, I decided to use both +12V and -12V outputs for high voltage (12 to 24V), and use the 12V output with ground for lower voltages and higher current. A switch can simply change between them... I will build an "external" voltage regulator circuit that should hold current up to 10A. For the voltage regulator circuit above in my first post, it my hold around 5A not more right? to increase it I can make some changes and add another (parallel) 2N3055 with adding series resistors... But I want to ask if I could use MOSFET instead of the BJTs.
    Can I use IRFZ44 instead of the 2N3055 resistor? It can easily hold 10A. I think it will not require a heatsink since its ON Rds is 24mOhm which makes P about 2.4W only at 10A. Also, if it works with a MOSFET, what will be the use of 2N2905? there is no more need for current amplification. And what changes should be made if I used a MOSFET.

    Even if I'm wrong in my idea, I hope you don't give me another circuit to use or another idea before explaining the problems of my idea. I want to know more about replacing BJTs with MOSFETs in switching because I want to learn and because I see the dissipation power will decrease remarkably when using MOSFETs and as a result the efficiency increases.

    Best regards,
    Hazim
     
  5. tom66

    Senior Member

    May 9, 2009
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    You have to take into account power dissipation as well. The MOSFET or transistor is used as an electronically stabilised resistor, and not in its saturation band, so power dissipation is usually quite high. Power dissipation in a linear regulator can be estimated by (Vin - Vout) x Iout. So if you want 5V out at 10A with 12V input, you are producing 50W but dissipating well over 70W in the pass transistor, which is pretty poor efficiency, about 41%. You will need a MASSIVE heatsink and at least a fan to do this safely; ideally, you'd also have two transistors, to distribute the load. The circuit will probably not work with a MOSFET, but you won't gain any efficiency either way.
     
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  6. wayneh

    Expert

    Sep 9, 2010
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    A MOSFET is indeed superior when driven to be FULLY on, or off, as a switch. But in your application, you're intentionally using the transistor to be at an intermediate level between full on or full off. That's linear regulation and, as you've heard, will result in a LOT of power being turned to heat in the transistor. Exactly the same amount in the BJT or the MOSFET. So you need to choose the device best suited to dealing with all that heat.

    Just FYI, I have a 2N3055 mounted on a nice slab of aluminum with some "fins" for dissipating heat. I think I could pass 1-2A thru it, dropping a few volts, without much concern but I'm certain it would overheat at currents not much above that. I'd need a true heat sink and maybe a fan to get to 5A or more.
     
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  7. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    Thank you for the information. Yes its saturation.. I'll use two 2N3055s or two TIP35Cs. I have to find transistor to use instead of the 2N2905 with higher collector current, around 1A.

    Regards,
    Hazim
     
  8. tom66

    Senior Member

    May 9, 2009
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    No, the transistor is not in saturation, it is acting in its linear mode, and so wastes a lot of power. There is no way to get away from it - that 70W has to go somewhere. If it were switch mode then less would be wasted, but that's not always an option.
     
  9. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    Yes. For Q1 (2N2905), I didn't find a good replacement with higher Ic. I'm thinking about using the 2N2905 with an "on hand" suitable npn transistor. I found 2SC1568 (see datasheet). I'm not sure if this part of the circuit I designed need any modifications or not.

    [​IMG]

    The power supply is in its way to be finished and I'll upload photos and post the final circuitry.

    Regards,
    Hazim
     
  10. tom66

    Senior Member

    May 9, 2009
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    How much current do you actually need? 10A with 12V in and 5V out is a LOT of power, no matter how many transistors you have.
     
  11. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    Yes 10A with 12V in and 5V out means about 70W... Actually I'll not use such a high current especially on low voltages. I don't need definite amount of current to use at a definite voltage, what I need is a variable power supply for my workbench, but I prefer to have it with higher current capability, 10A for the power supply I will build.
    So what do you see about the circuit?
     
  12. tom66

    Senior Member

    May 9, 2009
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    The circuit should work as is, but you'd be pushing it for 10A.
     
  13. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    Pushing? why? I will use it mostly at low currents, from few mA's to say 3A. But as I said, sometimes I may need to draw higher current. I hope It should work fine.
    Thanks for your help tom.

    Regards,
    Hazim
     
  14. tom66

    Senior Member

    May 9, 2009
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    Well at 1.2V output (the minimum) and 10A you'd be dissipating 108W, which is *very* close to the maximum power rating of the transistor, 115W. You don't want to run close to this or your transistor will not last long at all. Usually, run components within half their maximum rating to give you a safety margin and to extend service life; that means, for 5V out and 12V in you'd be limited to 8.2 amps maximum.

    Don't forget, to output 10 amps, you will need some really heavy duty wiring. Thin wires don't cut it. I think that 10 amps is way more than what you will ever need. What are you using it for?
     
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  15. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    I collected all the needed components and mechanical parts, but still looking for a switch with suitable current rating and 6 pins (3 and 3.. two contacts).. A wooden box will prepared tomorrow. I'll build the circuits and finish everything tomorrow.

    [​IMG]
     
  16. wayneh

    Expert

    Sep 9, 2010
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    That's a very ambitious plan. I recommend taking your time and checking and double-checking your work as you go. Patience is a virtue.
     
  17. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    Actually the circuit is not that hard, it's easy and doesn't take much time to build. Most of the time I'll take is in external shape and adding extras to the power supply (ex. zener voltage tester) and in a good inside arrangement...
     
  18. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    OH. There is a problem... 1 to 1.5 should flow through that 22 Ohm resistor at the voltage regulator input. So??? I'll not use a 50W resistor!
     
  19. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    I see it should be a 0.47 ohm resistor, or something between 0.47 and 0.68 ohm. I'm not sure, but 0.7V/1.5A = 0.4666 Ohm and 0.7V/1A = 0.7 Ohm. The resistor should be 0.47 Ohm 2W.
    Right?
     
  20. tom66

    Senior Member

    May 9, 2009
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    Power through a resistor is P = I^2 * R, so a 0.47 ohm resistor with 1A flowing through it dissipates 0.47W and has a voltage of V = IR or 0.47V across it. Choose a resistor at least double the rating. 2W is adequate for high reliability.
     
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