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I found a thread in the forum about a high voltage power supply 3V to 500V DC converter and someone suggested a circuit from techlib H.V. generator for Geiger tubes:

enter image description here

However,when I tried to simulate it didn't work, the output is nearly 9V, as the input. In the schematic I drew, the only difference with the proposed circuit is that I used an equivalent of 2N4403 transistor and different diodes. I also tried reversing one of the winding connections but nothing changed. Could someone explain how this circuit works and how the output is affected by the selection of the diodes? Maybe that will also help me understand what's going wrong with the simulation.

Any suggestions?

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    \$\begingroup\$ Can you show us the circuit you're trying to simulate? There are several circuits on your links. \$\endgroup\$ – Keelan May 9 '13 at 14:44
  • \$\begingroup\$ Oh,yes.I tried to post an image but i'm not allowed. It's the first circuit on HV generator for Geiger tubes link, it says "500 Volt Geiger Counter Power supply" underneath it. \$\endgroup\$ – Chris May 9 '13 at 15:28
  • \$\begingroup\$ Okay, I edited it in for you. Once you have sufficient reputation, you can add images yourself. \$\endgroup\$ – Keelan May 9 '13 at 15:31
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Ignore (for now) the MPSA18 and the two zener diodes and 10M resistor - they are used to control the amplitude of the DC output once the circuit is producing high voltage: -

enter image description here

Upon power being applied the 10uF capacitor has a charging voltage (in red) rising from 0V and after a short while this voltage will cause the base-emitter of the 2N4403 to conduct which rapidly causes the MPSA42 to switch on via the 1k8 resistor.

The MPSA42 will switch on and immediately start dischrging the 10uF capacitor via the 1k and the 1N4007. Shortly afterwards the MPSA42 will turn off because the 2N4403 turns off due to the cap discharging.

The current that was flowing in the transformer primary has stored energy in its magnetic field and this is harvested by the secondary circuit which presumably has a higher turns ratio than the primary.

And the process starts again - MPSA42 turns on - energy is stored in magnetic field and discharges to sencondary when MPSA42 switches off.

The MPSA18 will start to conduct when approximately 240V is on the output and this will start to turn-off the 2N4403 more making the 10uF take longer to charge thus the duty cycle is reduced.

It looks to me like there will be a fixed period of a few microseconds during which the MPSA42 will conduct and an ever-increasing period where it is switched off as the output DC level gets to about 240V DC. That makes sense.

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  • \$\begingroup\$ Don't think you mentioned, that when the '42 switches on, not only does it discharge that 10uF cap, it also starts pulling current through the primary, so that's where the primary current comes from. \$\endgroup\$ – JustJeff May 11 '13 at 19:15
  • \$\begingroup\$ Anyway, I think you nailed the oscillator action, so +1 \$\endgroup\$ – JustJeff May 11 '13 at 19:15
  • \$\begingroup\$ @JustJeff - thanks dude. BTW I did mention the discharge in para 2 and para 3 (by implication) the current thru the primary. \$\endgroup\$ – Andy aka May 11 '13 at 19:44
  • \$\begingroup\$ According to the article, the transformer is a 1:1 600ohm audio isolation transformer - no voltage step up there. \$\endgroup\$ – tehwalrus Aug 7 '17 at 21:14
  • \$\begingroup\$ @tehwalrus ok point taken but it is a fly back topology and relies on back emf being generated by an open circuited inductor therefore it will ramp up the output voltage even with a 1:1 ratio. Also, given the way the primary and secondary are wired (series aiding), it will tend to act as a 1:2 transformer thus aiding the flyback process of generating a high voltage. Remember also that a single inductor used in a boost regulator can produce over 100 volts from a small battery. The point is that a flyback design is an enhancement to the standard booster. \$\endgroup\$ – Andy aka Aug 7 '17 at 21:51

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