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High frequency current transformer....reducing sec current by phase cutting?

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How can I implement a variable phase-cutting rectifier on a 50KHz current transformer secondary?

I have a current transformer with a split secondary which is charging a battery from its secondary. The primary current is a fixed sine wave.

Schematic: Full wave rectifier on current transformer output………

…As you see, there is a full wave rectifier which puts DC into the battery from the secondary as follows…..

Full Wave Current waveform into battery:….

The primary current is fixed and is a sinusoidal 50KHz.

I now wish to reduce this current by phase cutting the rectified sinusoidal output. I can do this by periodically shorting out the rectifier diodes.

Here is the waveform that I now require into the battery……(0.3 duty cycle)………

…as you can see its phase cut version of the initial waveform and has a 0.3 duty cycle.

I also want to be able to do any duty cycle from 0.1 to 0.9…….so for example, I may sometimes want the following waveform to charge the battery………

Current waveform into battery (0.7=duty cycle)

Here is a rough version of the phase cutting circuit…..

…as you can see, the controlled voltage source switchs the FETs on and off periodically via the FET drivers, shorting out the rectifier diodes as it does so.

Though my question is, how can I most simply provide this “phase cut” waveform?
What circuitry can most simply do this job?

…on any particular product I only need one duty cycle…and it might be anywhere between 0.1 to 0.9…..(so that makes things easier)

You must explain to me how shorting out a rectifier reduces the current with out blowing up the source. I would replace D3 with a FET, and switch it off when no current is required. Use a current sensing resistor in the positive lead to the battery and a fully smoothed low voltage power supply (you could use the battery!) to power an opamp as a comparator. Use a CR time constant to change the incoming square wave into a ramp and smooth the DC from the current sensing R as the comparator inputs.
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The power is supplied by a current source.

If you short a current source, you do not get a high current , because its a current source.

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