AC to DC voltage doubler not working (1 Viewer)

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electricalguy

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I have an AC to DC voltage doubler circuit. I am using a full wave bridge rectifier that uses two diodes and two capacitors that are 470uFarad each. My supply is a 50 watt transformer that outputs 13.9 volts and 3.790 amps. The supply frequency is 60 hz. The output frequency is 120 hz from the rectifier. The voltage output is 3.72 volts DC at 1.650 amps DC. I already know that the power supply is too low for the load I have which has a resistance of 2.3 ohms. My question is how do I calculate the power required to be able to allow the doubling effect to happen?
 

davenn

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I have an AC to DC voltage doubler circuit. I am using a full wave bridge rectifier that uses two diodes and two capacitors that are 470uFarad each. My supply is a 50 watt transformer that outputs 13.9 volts and 3.790 amps. The supply frequency is 60 hz. The output frequency is 120 hz from the rectifier. The voltage output is 3.72 volts DC at 1.650 amps DC. I already know that the power supply is too low for the load I have which has a resistance of 2.3 ohms. My question is how do I calculate the power required to be able to allow the doubling effect to happen?

1) show us your schematic
2) show us a photo or 2 ( sharp and well lit) of your construction
 

tech99

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I notice that 28 volts across 2.3 Ohms is 340W which is too much for the transformer.
 

davenn

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I notice that 28 volts across 2.3 Ohms is 340W which is too much for the transformer.
yeah, it's a wonder the resistor didn't vaporise in a puff of smoke
 

Baluncore

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It will not double the voltage if there is any load. It will instead pump a DC current through the load. The voltage doubler involves a series capacitor. That is what limits your output current.
You know that C = Q / V; and Q = I · t; So C = I · t / V; or I = C · dv / dt

One cycle of 60 Hz takes 16.667 ms, so half a cycle takes 8.33 msec = dt.
Secondary voltage 13.9 Vrms has a Vpp of 13.9 * 2 * Sqrt(2) = 39.3 Vpp = dv

C · dv / dt = Iout; 470 uF * 39.3 V / 8.333msec = 2.21 amps
That ignores voltage drops across diodes and in the secondary winding.
It also explains why your output current is limited.

There are several voltage doubler topologies possible. Please provide a circuit diagram.
Drag and drop a circuit .jpg or .png onto your next post.
 

CWatters

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What happens if you test it with no load?
 

Henryk

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The transformer output voltage is RMS, I suppose, therefore the 13.9 VRMS translated into 19.6 VPEAK. Allowing for 0.6 V drop across the diodes, the doubler would produce open circuit voltage of 38 V. At that voltage, the current across 2.3 Ohm resistor would 16.5A and the power would be 628 W. Way out of the transformer specs.
Why do you need a doubler?
What power do you need to deliver to the resistor?
 

davenn

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and yet another OP @electricalguy who hasn't returned to the thread :frown:


Please check in and respond to questions and comments :smile:
 

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