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I'm currently constructing a power supply for a small helium neon laser tube, based on this circuit:

Helium neon laser power supply

The output of a centre tapped transformer (small neon sign transformer), is rectified by two high voltage diodes, smoothed by around 1uF of capacitance and fed through a ballast resistor which regulates current to the laser tube itself (due to it's negative resistance characteristic). I know this isn't the most efficient power supply design, with significantly more energy being lost as heat in the resistor than is consumed by the tube itself, however, the entire system into which this laser is being integrated is intended to resemble quite accurately, the early helium neon lasers of the 1960's, which as mentioned in the below link, used a power supply similar to this.

This brings me to my question. On the site where I found the original schematic (here), the author mentions a voltage of 7.07kv across the capacitors (5*1.414), but it seems to me, as both taps are being utilized, the voltage would actually be 14.14kv, the peak voltage of the 10kv input, is this correct, or am I missing something?

Thanks in advance for any help!

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  • \$\begingroup\$ What are the turns ratios? \$\endgroup\$ Aug 22 '15 at 1:24
  • \$\begingroup\$ Of the two halves of the secondary? 1:1, they both produce 5kv open circuit. Of the primary to the secondary in total? 1:42, as 240v input, 10kv open circuit output. \$\endgroup\$
    – CoffeeCrow
    Aug 22 '15 at 1:29
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Full wave rectifier is just two opposite phase half wave rectifiers in parallel.

This is obviously only half the end-to-end voltage:

schematic

simulate this circuit – Schematic created using CircuitLab

I can use the unused end to have another 5kV power supply.

schematic

simulate this circuit

And if I connect both of them in parallel, I get full wave rectification:

schematic

simulate this circuit

If I want 10kV, I would have to connect it like this:

schematic

simulate this circuit

Now the center tap is not used and the full voltage of the transformer is on the cap.

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Suppose there are 10 turns on the primary and 10 turns on the secondary (5, then the center tap, then 5 more). In that case the top and bottom of the secondary relative to the grounded center is half of the input. Only one of the taps is ever conducting into the load at one time giving a waveform like this.

There are probably more than 10 turns these are presented for relative interpretation only.

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