Here's a schematic of the junkbox HV supply I built.
It's working okay. The only disappointing thing is that the efficiency is a rather unimpressive 65% at 12 volts in, ~180 volts out at 20 mA current. Frequency of the switching waveform is about 60 kHz.
Is there anything I can do with the design to improve this? There's some ringing at the drain connection...here's the relevant scope waveform on my trusty Kikusui:
It would probably help a lot to use a decent MOSFET driver. And much depends on the quality of the inductor, capacitors, and rectifier. 60 kHz is rather high for a scratch built project. Layout of wiring can be fairly critical. Do you have a picture of the "beast"?
Paul
The circuit looks mostly fine to me, but there are potentially significant i^2r losses with those components.
From your numbers, I estimate you'd be putting at least 1A peak through your ~4 ohm FET if your circuit were operating in 'boundary' mode, and a lot more if it is operating as discontinuously as the 'scope waveform suggests.
That means you're dropping >4V (peak) of your 12V drive in the FET.
So, a lower-resistance FET would help.
Changing your oscillator / controller to keep the inductor current either continuous or less discontinuous would help a lot too. Those choices lower peak FET and inductor current, cutting i^2r loss.
You can do that with a fixed FET 'on' time and making the 'off' time last until the inductor rings off (i.e, after it has dumped its boost energy into the load).
DCR of the inductor would be another place to look, and Trr of your junkbox rectifier, too. Switching time of the FET is another place to look, but I'd suspect these other factors first.
Cheers, James Arthur
There's almost 2 mA dumping into the 100K feedback divider, over 300 milliwatts.
-- John Larkin Highland Technology, Inc lunatic fringe electronics
True. I can sort that...
Absolutely, and a decent MOSFET as well. This would have higher Ciss and exacerbate the lack of a decent driver. It's important to quickly turn off the MOSFET, otherwise it wastes a high Id*Vds as its drain flies up. You won't have a MOSFET driver in your junkbox, but you can use a PNP transistor driven from your CD4093's output, to rapidly pull its gate down, with a diode to pull it up. A serious high-current PNP like a 2N4403 is called for.
--
Thanks,
- Win
Is the bottom of the divider connected to something? (I don't see how the feedback works, does the tl431 have a fixed input impedance?)
George H.
Yep, it's grounded (just a sloppy schematic)
There's another ~20 mW in the 4.7K resistor.
-- John Larkin Highland Technology, Inc lunatic fringe electronics
That's 10% right there.
(20mA @ 180V) / (12V * n=65%) = .46 amps average input current @ 12V.
iL ramps from zero, so has to hit at least 2x average at its peak, which means the 4 ohm FET is dropping >4V, peak.
Crude approximation: integrating i^2r di yields ~[(1A^3)/3] * 4 ohms = 1.3W conduction loss in the FET for 93% duty cycle, continuous or boundary mode (worse for discontinuous).
Win's idea about a pnp turn-off is a good one too.
Cheers, James Arthur
Yes. Pull a UC3842 out of your junkbox.
Or if you well and truly don't have one, construct one with an LM393, CD4013 and a complementary emitter follower driver (2N4401/3, say).
And put a tap on that damn inductor!
Tim
-- Seven Transistor Labs, LLC Electrical Engineering Consultation and Contract Design Website: http://seventransistorlabs.com
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.