- #1
Venus Marie
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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?
Capacitance for Voltage Multiplier
- Thread starter Venus Marie
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Discussion Overview
The discussion revolves around calculating the capacitor value for a voltage multiplier circuit and how to achieve voltage doubling across a 20 ohm load resistor. Participants explore theoretical and practical aspects of voltage multipliers, including circuit design considerations and component specifications.
Discussion Character
- Technical explanation
- Mathematical reasoning
- Debate/contested
Main Points Raised
- Some participants inquire about the specific schematic of the voltage multiplier circuit and whether it operates from AC mains or an oscillator, emphasizing the need for starting and ending voltage specifications.
- There is a discussion about the relationship between voltage drop across a resistor, current flow, and Ohm's law, with references to the Cockcroft–Walton generator as a voltage multiplier example.
- One participant suggests an assumed capacitor value of 10000uF but expresses uncertainty about how to calculate the appropriate capacitor value based on the circuit parameters.
- Another participant explains that the peak voltage on the reservoir capacitor is crucial for determining the output voltage and discusses the impact of the rectifier diode's voltage drop on achieving the desired voltage doubling.
- Calculations are presented for determining capacitance based on the peak capacitor voltage, average DC voltage, load current, and time between peaks, with examples illustrating different scenarios based on diode types and their voltage drops.
- Concerns are raised about the resistance in the circuit affecting performance, and suggestions are made to consider different rectifier technologies to minimize voltage drop.
Areas of Agreement / Disagreement
Participants express various viewpoints on the calculations and assumptions regarding the voltage multiplier circuit, with no consensus reached on the exact capacitor value or the best approach to achieve voltage doubling.
Contextual Notes
Limitations include the need for specific circuit parameters such as supply frequency, voltage, and load characteristics, which are not fully defined in the discussion. The impact of circuit resistance and diode characteristics on performance remains unresolved.
- #2
berkeman
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Welcome to the PF.Venus Marie said:
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?
- #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.Venus Marie said:
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.
- #4
Venus Marie
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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
- #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...Venus Marie said:
Can you write out your full assignment statement? What exactly do they want you to do?
- #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 modeled 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.
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 modeled 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.