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Voltage doubler and low load values

  1. Apr 11, 2012 #1
    How can I double the voltage through a 3 ohm load with a voltage doubler when the output voltage usually collapses due to low load resistances. I have a feeling it is due to capacitance values and frequency. Can someone please explain.
     
  2. jcsd
  3. Apr 11, 2012 #2
    Are you using a DC source with discrete switching? Like you said, your output current is dependent on the size of capacitors you use and how quickly you switch them. The effective switched capacitor resistance goes down with increased frequency.

    You will have ripple or like you said complete discharge if the switching frequency or capacitance is not high enough. The problem with this is the surge currents you will need to keep your capacitor charged can be damaging to your switching components, and all of these factors limit the practical load a voltage doubler can source.

    The attachment shows a relationship between frequency, capacitor value, and how much current you can source.
     

    Attached Files:

  4. Apr 11, 2012 #3
    The source I am using is 60 Hz AC power from a transformer. I have been using a Delon voltage doubler with 2200 Micro Farad capacitors. With the load being 3 ohms the voltage is half that of a bridge rectifier.
     
  5. Apr 11, 2012 #4

    vk6kro

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    What transformer voltage are you using and what output voltage do you want?

    What is your load? Is it just a resistor?
     
  6. Apr 12, 2012 #5
    The transformer voltage is 19 volts AC. On the output through the 3 ohm resistor the DC voltage from a bridge rectifier is 3.5 volts. From the voltage doubler it is 2 volts.

    Just so that I understand with lower frequencies on the input to the voltage doubler the capacitor takes longer to charge, but with a low resistive load the capacitor discharge relatively quick compared to charge time. So the output voltage would have a ripple in it with highs and lows and the multimeter reads an average lower voltage. Making the voltage doubler not effective.

    So is it right to say that the voltage doubler would need a higher value capacitor with a larger input current to keep the capacitors charged to reduce the ripple voltage on the output.
     
  7. Apr 12, 2012 #6

    vk6kro

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    It looks like the transformer is not capable of supplying enough current for the circuit to work properly.

    I did a simulation of this and it indicates that the output should be about 18 volts if the transformer was perfect, but I get your results if I add about 16 ohms resistance in series with the power transformer secondary.

    Charging currents of about 25 amps (peak) are needed to charge up that capacitor.
     
  8. Apr 13, 2012 #7

    NascentOxygen

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    Does the transformer have a current rating printed on it?
     
  9. Apr 14, 2012 #8
    The transformer does not have a current rating on it but with the secondary winding connected in series with the 3 ohm resistor the voltage from the secondary is 17.66 volts AC at 0.385 amps AC.
    What size of capacitor would I need to make this circuit work properly with a transformer that can output 8 amps AC? I do not have a transformer capable of supplying 25 amps.
     
  10. Apr 14, 2012 #9

    vk6kro

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    That doesn't seem right. 17.66 volts AC at 0.385 amps must be across 45.87 ohms. So, you can't have the 3 ohm resistor directly across the transformer secondary.

    If you did have 18 volts across 3 ohms there would be 6 amps flowing.and it would be dissipating 108 watts.

    The capacitor is OK although it would give some ripple voltage. However, the transformer seems to be the main problem.
    You are getting 3.5 volts after the bridge rectifier but 19 volts from the transformer (presumably without a load), so the drop is either in the transformer or the bridge rectifier. The transformer is a lot more likely.
     
    Last edited: Apr 14, 2012
  11. Apr 15, 2012 #10

    NascentOxygen

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    I think you had better test each diode, one at a time, as just a half-wave rectifier using your transformer and a suitable load, to make sure you can measure the expected average (i.e., DC) voltage.
     
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