Power supply desiginig Qs

Discussion in 'Homework Help' started by AIN, Apr 15, 2011.

  1. AIN

    Thread Starter New Member

    Apr 11, 2011
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    Hi

    I am designing a power supply and i have some Qs:

    What kind of pcb board is suitable?

    should I build the circuit directly on a pcb? because debugging after soldering is not easy i think.

    How to choose the appropriate fuse/circuit breaker and where to put them?

    if i am using a big transformer and a bridge rectifier to derive an unregulated voltage for my first stage, what safety procedures I should make?


    Thnaks
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
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    282
    1. FR4 is stronger than phenolic, but that probably does not matter.

    2. Making an untested design first is not smart. You can simulate the circuit or use a breadboard.

    3. Fuse the line into the transformer. For extra protection, you can fuse the regulator output.

    4. Define "big", in volts and amps.
     
  3. AIN

    Thread Starter New Member

    Apr 11, 2011
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    Thank you for the fast reply beenthere.

    The transformer is 250VA rated for 10 amps secondary current.
     
  4. beenthere

    Retired Moderator

    Apr 20, 2004
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    A 35 amp 400 volt bridge - http://www.web-tronics.com/mb354.html - is about as cheap as any.

    This is no fun having to drag every detail out, one at a time. Will the PS be regulated? If so, what voltage/s and current/s do you expect to output?
     
  5. AIN

    Thread Starter New Member

    Apr 11, 2011
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    somehow yes. I have to provide power to three blocks.

    1st block requires 5 amps/16 volts. Since it can handle unregulated voltage, the instructor suggested to connect it directly to the bridge rectifier.

    The other two blocks both need 200mA/5 volts. So, I will also connect a regulator to the bridge to provide the blocks the needed power.
     
  6. beenthere

    Retired Moderator

    Apr 20, 2004
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    That transformer is way larger than you need. Is that a single secondary?

    As this is an assignment, this thread will get moved to the Homework section.

    At this point, you should be able to calculate the voltage on the filter capacitor, as well as the size of that capacitor. How close to 16 volts will that voltage be?
     
  7. AIN

    Thread Starter New Member

    Apr 11, 2011
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    Yes it is a single secondary. Do you recommend having 24V T 5 A instead? I just thought it's good to have some margin.

    The ripple Voltage is given by this formula: VR = (Secondary AC Rating×1.414-2VD) × (1÷(2×f×RL×CL))

    How can I determine RL? Since the the rectifier will be connected to the main block and a regulator that will derive the 5 V to the other two blocks.

    Is the right way way to determine the capacitor size is by letting 1000uF for every 10WRMS ? I have been reading on tutorials and found these approximate information.

    thanks
     
  8. beenthere

    Retired Moderator

    Apr 20, 2004
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    With Ohm's law. You have given the characteristics of the first block (whatever that may mean) as:
    What is R given that I and V?

    The other two blocks get figured the same way.

    You have also given:
    What does that suggest about the transformer output?
     
  9. AIN

    Thread Starter New Member

    Apr 11, 2011
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    So for the 1st block:

    RL =16/5 = 3.2 ohms from the first block.

    Since the power of the 1st block is 16*5 = 80 Wrms , CL = 8000 uF

    And using the transformer rated 24 V @ 5 Amps:

    and VR = (24×1.414-2*0.53) × (1÷(2×60×3.2×8000*10^-6)) = 10.7018 Volt

    I couldn't interpret that. Does that mean I'm off by about 5.3 Volts?
     
  10. AIN

    Thread Starter New Member

    Apr 11, 2011
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    So the cap size should be reduced to 5000 uF to achieve 17.12 Volts. Is that right?
     
  11. AIN

    Thread Starter New Member

    Apr 11, 2011
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    I was confused. Vripple has to be minimum. is a good value in mV?
     
  12. beenthere

    Retired Moderator

    Apr 20, 2004
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    No, you can't adjust a rectified output through changing a capacitor. The DC is ACVrms X 1.414. That will be a peak voltage of the unfiltered pulsing DC, or the smoothed DC using a filter cap. There will always be some residual ripple.

    There are some problems with this. You can't use VR, as there is no regulation (assuming VR is to identify a regulated voltage).

    In (24 X 1.414-2*0.53), where does the 2*.53 term come from?

    The other formula is somewhat mysterious. If you need to calculate ripple voltage, see this link - http://forum.allaboutcircuits.com/showthread.php?t=17114

    Given that your filtered DC voltage is greater than the ACrms of the transformer secondary and the statement that:
    What does this suggest?
     
  13. AIN

    Thread Starter New Member

    Apr 11, 2011
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    I see ..

    VR is meant to be the voltage ripple. I have got that equation from some tutorial!

    * So from the link you have provided: Vripple = I/2fc = 5/(2*60*8000uF) = 5.2 volts.

    * 0.53 is the voltage drop of the schottky diode I am planing to use: http://www.mouser.com/ProductDetail...GAEpiMZZMuIUjt4yeP9czzQat3OYqJfoh%2bczmlJOMg=

    * The filtered DC voltage = ACrms*1.414 = 24*1.414 = 34 Volts

    I guess that suggest I better chose a smaller transformer, right?
     
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