Im planning to use a simple blocking oscillator with a CCFL transformer and Cockcroft-Walton diode multiplier to supply HV for a Geiger-Muller tube. This requires 1600V but at miniscule current.

I wish to make this draw as low a current from the supply as possible, ideally below 1mA from a 9V supply.

What is responsible for limiting the current (other than the load) in such a circuit? Is it just the back-EMF in the transformer primary?


The impedance of the voltage multiplier plays a part in limiting the current from the high voltage output.

I've had a look at the impedance of the Cockcroft-Walton voltage multiplier before, and found the following equation:

$$ E_{out} = 2nE_{pk} - \frac {I_{load}}{2 \pi fC} (4n^3 + 3n^2 - n) - 2nV_{f}$$

What that does is to compute your output voltage from several parameters:

  • \$E_{pk}\$ the peak voltage of the AC input (not the peak to peak voltage, but the simple peak voltage.)
  • \$I_{load}\$ - the load current.
  • \$f\$ - the frequency of the AC input
  • \$C\$ - the capacitance of your individual capacitors (in farads.)
  • \$n\$ - the number of stages in your multiplier.
  • \$V_{f}\$ - the forward voltage of your diodes.

That applies to the commonly used half wave Cockcroft-Walton voltage multiplier.

The relationship between the number of stages and the impedance is not obvious just from looking at the circuit.

The output impedance is this part:

$$\frac {4n^3 + 3n^2 - n}{2 \pi fC} $$

The maximum output voltage of the multiplier is given by this part:

$$ E_{out} = 2nE_{pk} - 2nV_{f}$$

Figure your maximum voltage, then use the impedance to figure the voltage drop at the supply current your tube needs. Vary the capacitance, the number of stages, and the frequency to get a multiplier that delivers 1600 VDC at whatever current your tube needs.

The current drawn from the low voltage supply will be higher than the current out of the high voltage side.

I never measured it while I was doing my experiments, so I don't know for sure how much higher.

I'd expect the input current to be at least \$N \times I_{out}\$ where \$N\$ is the number of stages of your multiplier. Since capacitors leak, the input current will be higher.

I almost expect the oscillator and the driver for the transformer to draw more than 1mA.

  • \$\begingroup\$ cheers, thanks. The G-M tube current is really hard to define, as outside of an ionization event, the tube is effectively open-circuit, and the only current then drawn by the HV side is from leakage. Even in a 'worst-case' scenario for this particular tube, the anode resistor limits the current to a few hundred uA. It is more the LV and oscillator side im looking at - whether there is anything there that can draw more current than is really needed, and can be 'designed out' to reduce the current draw, such as by limiting the base current on the oscillator transistor. \$\endgroup\$ – Martin G7MRV Dec 15 '19 at 13:05
  • \$\begingroup\$ If I can drive the CCFL transformer to an output of 400V or more, then I should only need two CW stages, so worst case current becomes about 800uA (allowing for some losses plus current in a regulator chain of zeners). 1mA does look too optimistic for this (I can do 400V supplies that low), 30-40mA looks more probable for this supply. \$\endgroup\$ – Martin G7MRV Dec 15 '19 at 13:12
  • \$\begingroup\$ @MartinG7MRV: Is that 400V RMS, or 400V peak, or 400V peak to peak? \$\endgroup\$ – JRE Dec 15 '19 at 13:28
  • \$\begingroup\$ Good point! I forgot I was going from AC (well, pulse) to DC. I require 1600V DC for the G-M tube. Presuming half-wave rectification, thats about 2.5kV rms, so probably about 650Vrms out of the transformer. A lot will depend on what I can drive the salvaged CCFL transformers to. But I believe they are designed for 3kV pk-pk when driving CCFL tubes \$\endgroup\$ – Martin G7MRV Dec 15 '19 at 13:50
  • \$\begingroup\$ I ask because the number of stages you need depends on the peak voltage of the AC input to the multiplier. If you have 400V peak, then you need 2 stages to reach 1600VDC. If you have 400V peak to peak, then it'd be 4 stages. If its 400V RMS, then a single stage would get you 1100VDC and two stages would be 2200VDC. \$\endgroup\$ – JRE Dec 15 '19 at 13:55

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