30 kV 10 mA power supply

I am sorry, but if you connect them in series, the twice higher voltage between secondary and primary may exceed their safe insulation capacity.

Adding to what "Ignoramus" (who seems to not be one) already posted... a serious problem when you get to high voltages in transformers is corona. If there is corona around organic insulation (which is almost always used in transformers), the insulation will eventually break down. It may not happen right away, but could instead be a field failure built into every unit.

On the other hand, what's wrong with using a voltage doubler? The transformer is rated, apparently, at high enough power for your application, so why waste money on two of them? 12kVrms should give you over 17kV peak, more than enough to get your 30kV with appropriately sized capacitors.

Be prepared to use corona rings, lots of space, etc., to keep electric field strength under control...

Cheers, Tom

At someone lower current this one might do the job. I do not know the seller but he has been trying to sell it for a while. Like he says, this thing is lethal. When hooked up to sizable cap you will not have the time to smell the burning flesh. What is it that you are trying to do?

Regards,

Boris Mohar

Got Knock? - see: Viatrack Printed Circuit Designs (among other things)

void _-void-_ in the obvious place

Thgese are NOT for running off the Ac line, NOR for putting in series.

The secondary is apparently center tapped, with the center grounded, so it's better explained as a 6000-0-6000 volt transformer. NOT something that can be used as a 12,000 volt transformer, and definitely not in series with another such ilk.

And you can't drive it form the AC line, as it only has about six turns on the primary. You have to drive them with what they allude to as their "power supply, schematic included".

Have you ever built a 400 watt switching power supply, from scratch, before? Not as easy as you might think.

I suggest trying something less likely to result in smoke, fire or death, like dismantling land mines.

Do your parents or life-insurance agent know what you're planning?

Thanks for your safety concerns. I plan to put the transformer in oil anyway.I guess that will solve the insulation problem. I'm aware that I cannot use AC directly. There must be some high frequency switching involved. The output of the secondary will be rectified. The voltage doubler idea seems interesting instead of using 2 transformers in series. But how much space will it occupy?

Tom Bruhns yazdi:

On 28 Nov 2006 00:53:47 -0800, "booth" Gave us:

A voltage doubler is two HV capacitors and two HV diodes. It would likely take up less space than the xfmr footprint, and likely cost less as well.

booth wrote:

Amplifying on what JoeBloe wrote: Assume one side of the transformer secondary is grounded. The other side drives one end of a capacitor, C1. The other side of that capacitor goes to the cathode of diode D1 and the anode of diode D2. The anode of diode D1 is grounded. Thus, it prevents the "diode" end of C1 from going significantly more negative than ground. The cathode of diode D2 connects to the output terminal, and between the output terminal and ground is C2. C2 must be rated to withstand the peak output voltage, of course. C1 must be rated to withstand the peak transformer voltage, roughly half the peak doubler output voltage, and in addition, must be insulated from its surroundings so that when the diode end reaches the circuit output voltage, there will not be unwanted arcs. Each diode must be rated to withstand the full output voltage when reverse biased. The diodes must not have so much stored charge that they won't recover quickly relative to the switching rate at which you are driving the transformer. The value of the capacitors must be such that they do not discharge significantly over the course of a cycle. You can determine how much charge is removed from C2 by the load during one cycle, and that charge must be replaced each cycle by C1. Note that the peak diode current will be considerably more than the load current, since the diodes conduct in relatively short pulses. The amount of charge and the capcitance will tell you how much ripple voltage there is, and how much the output voltage is below the peak-to-peak transformer voltage. Of course, you also must consider diode drops and drops in the transformer impedance.

Cheers, Tom