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Capacitance for Voltage Multiplier

  1. Mar 16, 2017 #1
    Can someone help me on how can I get the value of capacitor on voltage multiplier and how can i double the voltage through a 20 ohm load resistor?
     
  2. jcsd
  3. Mar 16, 2017 #2

    berkeman

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    Welcome to the PF.

    Please post the schematic for the voltage multiplier circuit you are referring to. Are you wanting it to run off of AC Mains at 50/60Hz? Or do you have an oscillator circuit that will be pumping this multiplier? What are your starting and ending voltages for this circuit? Where is the 20 Ohm load? At the doubled output? How much power is that?
     
  4. Mar 17, 2017 #3

    Baluncore

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    A voltage drop occurs across a resistor. A current flows through the resistor. Ohms law says that the current flowing, i, will be equal to the voltage drop, v, divided by the resistance, r.
    i = v / r.

    If you build a voltage multiplier by using capacitors and diodes such as;
    https://en.wikipedia.org/wiki/Cockcroft–Walton_generator
    For a 20 ohm load, the capacitor value will be dependent on the voltage and frequency of the AC supply.

    What type of voltage multiplier were you considering. The capacitance values can be calculated when you specify the resistor current and the supply frequency and voltage.
     
  5. Mar 19, 2017 #4
    Here is the schematic we simulated,but we aren't sure on how we will calculate the capacitor value of the circuit. so we assume 10000uF capacitor upload_2017-3-19_23-48-54.png
     
  6. Mar 19, 2017 #5

    berkeman

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    What is the output impedance of that signal generator? In real life, they are typically 50 Ohms. That would affect this circuit quite a bit...

    Can you write out your full assignment statement? What exactly do they want you to do?
     
  7. Mar 20, 2017 #6

    Baluncore

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    What seems like a simple question can be very confusing unless approached from the right direction. This question comes down to the peak voltage on the reservoir capacitor, Vcpk, at the end of the short charging period. That is determined by the AC peak voltage, less the voltage drop across the rectifier diode when it conducts.

    Doubling the average DC voltage across the resistor will double the average current through the resistor, and so will dissipate four times the power. The doubler can be modelled as two separate half-wave supplies connected in series, one positive, the other negative. Each needs an average output voltage of 2.00 VDC, so as to double the total voltage across the resistor.

    If the rectifier dropped so much voltage that Vcpk was less than 2.000 volts then it will never be possible to double the load voltage and current. If the rectifier dropped no voltage then Vcpk, above or below zero volts would be 2 * sqrt(2) = 2.828 volts.
    So we now know that the peak capacitor voltage will be somewhere in the range between 2.000 and 2.828 volts.

    Given a value for the peak capacitor voltage, Vcpk, we can compute the capacitance needed.
    We know the average DC voltage will be 2.000 volts so it will start with Vcpk and fall through 2.000 volts at half time, to a minimum of (2 * 2.000) – Vcpk, before the next capacitor charge.

    Charge will flow from the each reservoir capacitor through the load at a rate of 4V/20 ohm = 0.2 amp.
    We know the time between peaks, dt, will be; 1/60Hz = 16.667 msec.
    We know the difference voltage, dv, will be; 2* (Vcpk – 2.000V).
    Capacitance is defined as C = Q / V; Charge, Q = I · time; ∴ C = I · t / v;
    The answer to your question is therefore C = I · dt/dv.

    As an example, if Vcpk was 2.3 volts then dv = 2 * (2.3 – 2.0); ∴ dv = 0.6 volt.
    Then C = 0.2A / ( 60Hz * 0.6V ) = 0.2 / 36 = 0.00555 farad = 5.555mF

    When you are rectifying low voltages, the critical factor is the voltage drop across the diode for the short time that it conducts near the peak voltage. You have specified a simple silicon junction diode that will have a drop of about Vj = 0.8V while conducting and so require two very large capacitors. You can do better by using a Schottky power diode which will have a voltage drop of half that. The latest technology for low voltage rectifiers would use MOSFETs that were switched synchronously, which would drop less than Vj = 100mV.

    Here is a summary that justifies the use of more expensive rectifiers, or a different circuit.
    Vj = 800mV. Silicon PN junctions. Vcpk = 2.028; C = 0.119047 farad = 120,000 mF.
    Vj = 400mV. Schottky rectifier diodes. Vcpk = 2.428; C = 0.007788 farad = 7,800 uF.
    Vj = 100mV. Synchronous MOSFETs. Vcpk = 2.782; C = 0.004262 farad = 4,700 uF.

    As Berkeman noted; you must get the AC supply, diode and capacitor circuit resistance very low to make it work.

    I have ignored that resistance so far. You must add the voltage drop across the circuit resistance to Vj to calculate Vcpk.
    Worst of all, the time available for conduction of the rectifier is so very short that the peak rectifier circuit current will be maybe 100 times Iout, which will drop much higher voltages in the rectifier resistance.

    You may wish to change your voltage doubler circuit topology.
     
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