Board-mountable power supply-----help!!!1

Discussion in 'The Projects Forum' started by atariq7, Apr 4, 2008.

  1. atariq7

    Thread Starter Member

    Apr 4, 2008
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    Hello every 1 , i am new to the forum, hope to get some help. the question below.

    To design a power supply module (comprising a rectifier, reservoir capacitor, and regulator) that can be mounted as a daughter-board module on the main circuit board for which the regulated power supply is required. The input to the module is a low voltage 50Hz AC supply, and the output from the module is a DC supply.


    DC voltage = 6.3v
    DC Current = 410mA

    I will be using a full wave rectifier with reservoir capacitor as filter and for regulator i will use A zenor diode and a protective resistor. Also i have to consider the ripple voltage and all the values of the other components which will result the DC supply.

    So help!!!!!


    Now i dont know how to start.
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
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    Are you asking for help with the board layout or with the power supply circuit?
     
  3. atariq7

    Thread Starter Member

    Apr 4, 2008
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    Both, i doont know how to find the DC power supply which is given, i have already laid out the circuit design, dont know how to upload it here .
    this is how it goes

    AC + Full wave rectifier + capacitor + regulator(zenor diode and resistor ) + R load( DC supply which is given)

    Dont know wat type of diode to use, i.e wat shud be the Voltage of the zenor diode, same for the fullwave rectifier as well. How to calculate the values for capacitor and the resistor to meet the required DC supply.
     
  4. Wendy

    Moderator

    Mar 24, 2008
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    What's the voltage of the low voltage power supply?

    Is there a ground in the circuit that needs maintained?

    And last, if there is a ground, do you have a center tapped transformer?

    Here is the basics...

    [​IMG]

    The output DC voltage will be approx. 1.4 X the AC voltage.

    Amount of ripple is a function of both the power supply and the capacitor.

    A simple 3 terminal regulator will bring the voltage down to where ever you want it and dramatically reduce ripple.

    I was working on this before the 2 replies, is the zener a must?

    Is this homework by chance?
     
  5. Wendy

    Moderator

    Mar 24, 2008
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    Zeners work by dropping a constant voltage. Think of them as a resistor where the voltage is always the same, you can figure out wattages from that. The resistor that limits the current should not drop more voltage than the maxiumum load, this is a function of the DC voltage on the capacitor and the load, the zener will absorb the excess current. So the minimum current load is the maximum current the zener will have to absorb, and the maximum wattage.

    So, knowing the DC voltage under load on the cap is a must, as well as the no load voltage (they will be different). So we need the AC voltage to start off with.
     
  6. atariq7

    Thread Starter Member

    Apr 4, 2008
    10
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    thanks 1st of all, though its not clear to me yet

    No this is not homework, Its a design report which i have to have to hand in in 7hrs time.

    THere is ground, Zenor diode is a must. As for regulator i am using a Zenor diode and resistor in series.
     
  7. atariq7

    Thread Starter Member

    Apr 4, 2008
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    Also i have to assume the ripple voltage from the AC, like take few% of the AC voltage. the n uses it to find the C (capa...) i.e I/C x 1/f (which is 50HZ).

    I am not sure about this, but is this right?? Ho to calculate the value of the resistor in the regulator???
     
  8. Wendy

    Moderator

    Mar 24, 2008
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    I'll say it again, knowing the AC voltage is a must, it sets up the other voltages and calculations. We are not talking ripple at this point, that comes later.

    After we have the input AC voltage, we can calculate what the DC voltage will be. From there all else follows.
     
  9. atariq7

    Thread Starter Member

    Apr 4, 2008
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    well say if i use 240v, which is from mains supply, then wat do i do?
     
  10. Wendy

    Moderator

    Mar 24, 2008
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    You add a transformer, as shown. Then you work from the transformers specs and find out what the AC voltage from the transformer will be.

    You specified a low voltage AC though, the devil is in the details, or in this case, numbers.

    240VAC * 1.4 = 336 DC, not a workable voltage.

    A standard transformer size is 6.3VAC. Rectified and filtered that works out to be 8.8 VDC. Maybe you want to assume this voltage, but if this is a real world problem, as in it is going to be built, you need the numbers.
     
  11. atariq7

    Thread Starter Member

    Apr 4, 2008
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    thanks again, i am confused my self

    the question says that i will have to use a low voltage 50HZ AC supply. it doesnt specify wat voltage to use.
     
  12. atariq7

    Thread Starter Member

    Apr 4, 2008
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    The other parts of the question after the part i mentioned above in 1st thread.

    You may use either (i) a half-wave, or (ii) a full-wave, rectifier circuit with the reservoir capacitor, and for the regulator you may use either (a) a Zener diode plus protective resistor, or (b) a transistor, Zener diode, and resistor, in an emitter follower arrangement. For both the rectifier and regulator sub-systems you should evaluate the two alternatives, and recommend a particular combination in order to meet DC Supply voltage/current specification.
     
  13. atariq7

    Thread Starter Member

    Apr 4, 2008
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    The other part of the question


    You may use either (i) a half-wave, or (ii) a full-wave, rectifier circuit with the reservoir capacitor, and for the regulator you may use either (a) a Zener diode plus protective resistor, or (b) a transistor, Zener diode, and resistor, in an emitter follower arrangement. For both the rectifier and regulator sub-systems you should evaluate the two alternatives, and recommend a particular combination in order to meet DC SUpply voltage(6.3v)/current(410mA0 specification.

    I have to use a low voltage , 50HZ for AC supply
     
  14. Wendy

    Moderator

    Mar 24, 2008
    20,766
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    OK, this give you some wiggle room, and you can make assumtions. If you were designing from scratch this would likely bite you in a most sensitive region, but as it is you can mention in your paper you needed the info and it wasn't given, so you got to make them up as you went along.

    Lets start with worst case on the resistor / zener senario. You are drawing 410ma off a 8.8VDC power supply. BTW, if you have a chance to check it I bet you'll find the assignment specified 6.3AC, like I said it is a standard.


    8.8VDC 0----\/\/\/\(R1)-----/\/\/\/(Rload)-----Gnd

    The load has 6.3 VDC on it. It's resistance will be 6.3V/.41A = 15.4 ohms

    R1 is dropping 8.8V - 6.3V = 1.5V @ .41A, so it will be 3.65 ohms. We will round this off to 3.3 ohms, a standard resistor value.

    V^2/R is power, so 1.5^2/3.3 = .68 Watts, you will need a 1 W resistor.

    OK, the zener in a no load condition is getting .41A, and dropping 6.3V, so it's max power will be 6.3*.41A = 2.5Watts (pretty large, but 5W zeners exist).

    This circuit will be pulling the full .41A continously.

    Lets looks at the emitter follower design...

    [​IMG]

    It's big advantage is you can bring the currents down to 1 ma on the zener and resistor, and the transistor will conduct no more current than the load requires, making for a much efficient circuit.

    As for ripple, I'll let you worry about that, full wave bridge rectifiers will generate less ripple than half wave by a lot. Both regulators will effectively remove ripple. I didn't take the diode drops into account in any of the designs, so you will want to recalculate the values. Hope it helped.
     
  15. atariq7

    Thread Starter Member

    Apr 4, 2008
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    thanks dude, u have already done half of it.

    I c u have used a transistor, do i have to use it??
     
  16. SgtWookie

    Expert

    Jul 17, 2007
    22,182
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    Along with the output voltage, you'll also need the wattage or volt amp rating for the transformer. Being a linear supply, you'll have a lot of power dissipated in heat.

    I don't think you'll get there with just a Zener and limiting resistor unless your load is pretty static. You will probably need a pass transistor. Perhaps you should look at low-dropout regulators.

    [eta] OK, now I see Bill's posted an emitter follower circuit, which is one of the directions I was heading.

    R1 will actually need to be fairly low unless the emitter-follower has a lot of gain; perhaps a Darlington. Otherwise, you'll have a lot of "droop" in the output as load increases.

    The output should also have caps on it to supress transients, like a 0.1uF ceramic and 10uF to 100uF tantalum or electrolytic.
     
  17. atariq7

    Thread Starter Member

    Apr 4, 2008
    10
    0
    SgtWookie, u r making it so complicated man.

    Do i have to use a transformer?? what if i use low voltage like in the range of 15-25V as AC supply.

    What happens if i do not use transistor?? i mean use Resistor and Zener diode as regulator and then the R load.
     
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