Need input on high voltage regulator design

Hi Graham,

Yes. Should be a nice PWM control instead. Of course, when it is cold a shunt regulator does have its merits ;-)

Why is that lousy? If the MCU has spare horse power why spend the money on a PWM chip? Of course, to be safe this would have to run reliably. Within its own interrupt routine or something like that.

Regards, Joerg

On a Geiger counter I built many years ago from scratch, I used the recommended idea of a string of series NE-2s. Worked great. I still have that Geiger counter and, aside from replacing the tube once, it's still working just fine!

Jon

You could do something like shown below with a JFET input OA. This circuit uses a resistor string to convert the HV into a current, 10uA at HV=500V, and then an operational amplifier current-to-voltage converter to produce the A/D sample. The output impedance of the circuit is 10K which most embedded A/D can handle. If you need less impedance then go to a TL082 and use the other OA as a buffer. The output clamps at 5V until the input exceeds approximately HV=400V, and for voltages greater than this the output subtracts ~25mV from 5V per volt of HV input in excess of 400V- landing you somewhere around Vout=2.5V for HV=500V.

View in a fixed-width font such as Courier.

5V | +------+-------------+ | | 9V | / / | / 100K 100K | 10K / / | / | | |TL081| | 10M 10M 10M 10M 10M | | || | HV IN >-//-//-//-//-//-+----------|- | | | | >-| V 5x 1/4 Watt | +---|+/ | out | |/| | | 1.2M | | +----//----|-------+ | | | 620K | +----//----+ | | ---

V out | V = 5V for HV < 400V 5V+--------- out | | (HV-400) 2.5| - - - - - - Vout= 5V- -------- x 1.2Meg | | 50Meg | +--------+---+----------- 500V 400V

HV-->

What does the ADC use as a reference?

How good does the regulator have to be?

Two ideas:

You could sample the primary windings voltage some time after the MOSFET is switched off. The waveform will look like this:

A B C D E ! ! ! ! ! V V V V V ....*....................... ....**...................... ....***..................... ....************............ ....***........*............ ....**.........*............ ....*..........**........... ....*..........**........... ....*..........*****........ ....*..........**..*........ ....*..........*...*........

***...............*********

(A) the MOSFET turns off and the voltage goes rocketing up, overshoots and rings a bit.

(B) the voltage flattens out at some highish level and droops slightly as we go towards (C)

(C) the energy is all out of the inductance and the voltage drops suddenly, undershoots and rings.

(D) the voltage settles down to Vcc (or maybe it doesn't get time to.

(E) the MOSFET turns on again.

The voltage you want for this sort of feedback is the one at (C). The one at (B) is almost as good. From (B) to (C) is almost a straight line.

You should be able to get fairly good regulation. The tricky bit is that the micro needs to fiddle with the ADC timing to get the right point for the feedback. The nice thing is that the load doesn't change much so the software can take a bit of time.

2nd Idea:

R1 R2 R3 ----///----///----///----+------- ! ! ! ! / --!+ R5 ! >----+-- To ADC / --!-/ ! ! ! ! GND ! ! +--//---- ! R4 / R6 ! Vref

The ADC will give a zero output up to some voltage and then go up towards full scale as the volatge increases above that point. This allows the ADCs bits to be put to better use since you know that the voltage must be near 500V

R1, R2, R3 are a string of however many resistors you need to make the impedance high enough.

(1+ R4 / R6) sets the gain for the op-amp

The divider of R5 and (R1+R2+R3..) sets the fullscale point.

Decide on a full scale.

Determine R5 and (R1+R2...) to make R5 have Vref on it at full scale

Decide on a bottom value for the ADC Lets say 450V

Figure out the voltage on R5 at 450V

The op-amp's input see the impedance of R5 in parallel with the (R1+R2)

You want the parallel combination of R4 and R6 to equal that impedance and you want them to make the op-amp have a gain of (Vref-VR5)/VR5.

The TL071,2,4 type op-amp should work for you if you don't need to accurate of voltage.

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kensmith@rahul.net   forging knowledge

Hi Scott,

Shunts are inefficient because they regulate by 'burning off' the excess. I'd just use it just as a feedback. How to do that with a TLV431 is explained in a TI app note about isolated flyback converters. I forgot the number but you should be able to find it.

Basically a divider is hooked up to the high voltage so that the TLV starts to pull current when Vref is reached. That signals the MCU or whatever is on the generating side to throttle down. The TLV431 needs a minimum 'cathode' current of 100uA. Since your application appears not to be an isolated one this feedback circuitry will become much simpler than in the TI note.

Regards, Joerg

Well.... for next to no current draw you could choose a simple inverter. Simply choose the turns ratios to suit the supply and take into account Vce sat on the primary driver. Wind it so the primary takes next to no magnetisation current.

Should do the trick. And you could still tickle it with your MCU !

Graham

Only if it's a multi kW supply !

;-)

That's the sort of thing I meant.

Graham

The TL082 is still available from Radio Shack- as well as 10Meg 1/4W carbon film 5-packs - so that's a plus, and why I chose that one.

If you go with the TL082 and don't need the buffer, then a good use for it would be as low-pass filter and buffer for the 5V shown in the schematic...this will eliminate your 5V switching noise from the MCU and other digital switching you have going on there.

"Scott Miller" schreef in bericht news: snipped-for-privacy@corp.supernews.com...

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HV

Could you replace the D10 diode with a transistor, emitter to ground? Then you get a digital signal that tells if the voltage is above or below 500V, which you can use to increase/decrease the pwm.

I assume you are trying to keep current consumption to a minimum, running everything from a 9V battery?

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Thanks, Frank.
(remove 'x' and 'invalid' when replying by email)

Hi Graham,

One of the folks here on s.e.d., I believe it was Chris, once mentioned that many higher end battery-to-mains inverters are run with a uC being the PWM controller. Unfortunately there isn't much out there in terms of application notes.

Regards, Joerg

Hi Frank,

You'd also need a resistor from base to emitter, to create a real voltage divider.

Then R9 and R10 would need to be very large values. Unless I am wrong I believe Geiger counters only draw current when radiation triggers them so in dormant mode this sense circuitry might be the only significant load on the HV side.

Regards, Joerg

Hi Scott,

That shunt would drain your battery rather quickly. Q1 keeps pumping in energy and the varistors just burn that off. Also, please note that the metal oxide variety has a finite lifetime. They are like a bank account, after so many joules they go kaputt.

Regards, Joerg

Hi Scott,

I'd just go with a resistive divider and feedback. BTW, 10mA for uC and LCD display sound a bit high as well. That alone would empty a regular

9V battery in under 24 hours. I guess it's too late now but next time I would look into lower power uCs like the MSP430.

Regards, Joerg

Hi Scott,

With a serial controller on there? That is one heck of a deal. Much more $$$ out here in Northern California :-(

Or dump the regulator and run it with three or four cells.

Regards, Joerg

Yeah, it used power. My supply for it, back then, was two 45V telephone battery packs (bunch of 1.5V AA types in a package.) So I had 90V to work with and for my needs the battery lifetime was just fine -- memory is thin and I've got it packed away so I cannot easily verify, but I think it ran for about 8-10 hours. I used it intermittently for testing rock samples (autunite, for example), so I could be off by quite a bit.

Jon

You can clean up the 5V feed into the op-amp like so:

View in a fixed-width font such as Courier.

9V 5V | 100K | +------+---------+-//-+ | | | |* | / / | === / 100K 100K | | 10K / / | --- / | gnd | |TL081| | 10M 10M 10M 10M 10M | | || | HV IN >-//-//-//-//-//-+----------|- | | | | >-| V 5x 1/4 Watt | +---|+/ | out | |/| | | 1.2M | | +----//----|----------+ | | | 620K | * +----//----+ 0.1u||0.01u | | --- gnd

Which is where the compromise is.

Yeah, I forgot you had a mosfet there.

You're running the primary from +9V ( regulated ? ). You need a 55:1 ratio step up TX. and the output will be 500 V.

There's a bit more to it than that but basically you don't need feedback regulation if the supply voltage is regulated. Certainly not for next to no load.

And the transformer needs to be in the opposite phase. Your current design is flyback which can only regulate by feedback or ( as in your case ) shunt regulation.

Google 'single transistor forward converter' and ignore anything about feedback

- the turns ratio sorts it.

You might google 'inverter' too for good measure.

CCFL backlight transformers are designed as inverter transformers btw.

Graham

Given the device physics, that would have to be made from 89 5.6V zener dies in series. :>) (If a zener reference was desired, it would make more sense to attenuate the HV down to 5.6V, compare and regulate.)

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 Thanks,
    - Win