HV control

ut to

ght.

Id ~

t goes way

ive Id..

he gain

got 2000

that

nd you lose

l

-15V

r, the

an

ave a

and R1

g up

V for

igher

enforce

: Q1-Q2

t; the

R+C

ol loop

you have

ou

the load or

r?

o zero.

??

. e

rs

ors

l g

ith

OK, that makes sense. It's like a thermal loop with a TEC vs heater and weak link to the world. I drew this up with the push-pull driving the G-S resistor of the pfet. Then I threw away the pfet. I'll have to give it a try. BTW the HV pnp you referenced doesn't seem to be available.. there are plenty of others.

George H.

The ones I'm actually using are ZXTP01500BGTC PNP, which is a new part with 4k stock at Digikey, FJV42MTF NPN (lots of stock everyplace) and for the CE stage, MMBT6520LT1G (ditto). Only the first dissipates any power, so it's a SOT223 whereas the others are SOT23. Diodes Inc. is doing a lot of interesting things with BJTs these days--this one is a low saturation 500V 3W PNP with a nice flat beta of 100 to 300 at 1 mA, and costs 20 cents in reels.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC / Hobbs ElectroOptics 
Optics, Electro-optics, Photonics, Analog Electronics 
Briarcliff Manor NY 10510 

http://electrooptical.net 
http://hobbs-eo.com

_AoE III_ says that Figure 9.111 is a slight modification of Figure

3.111 (not available on line). Figure 3.111 is recognizable as the totem-pole mentioned on page 696. Figure 3.111 contains a stacked, non-complementary pair N-channel MOSFETS driven by a NPN CE. It also shows a terminal that connects to a +1kV supply. Perhaps a bribe will motivate someone to answer my open question about how to get started with that +1kV supply? Allow me to share a weekend crowd pleasing treat. Here's a 555 with a totem pole: :0) 73,

--
Don Kuenz, KB7RPU

_AoE III_ says that Figure 9.111 is a slight modification of Figure

3.111 (not available on line). Figure 3.111 is recognizable as the totem-pole mentioned on page 696. Figure 3.111 contains a stacked, non-complementary pair N-channel MOSFETS driven by a NPN CE. It also shows a terminal that connects to a +1kV supply. Perhaps a bribe will motivate someone to answer my open question about how to get started with that +1kV supply? Allow me to share a weekend crowd pleasing treat. Here's a 555 with a totem pole: :0) 73,

--
Don Kuenz, KB7RPU

Push-pull high voltage amp:

Using light as the base driver solves all sorts of problems.

The more obvious class B version works fine too.

--

John Larkin         Highland Technology, Inc 

lunatic fringe electronics

That's an all-NPN version, which needs a phase splitter stage (V2 in the picture). The normal "complementary symmetry" output (another old term) is NPN/PNP, and so doesn't need the phase splitter.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC / Hobbs ElectroOptics 
Optics, Electro-optics, Photonics, Analog Electronics 
Briarcliff Manor NY 10510 

http://electrooptical.net 
http://hobbs-eo.com

One simple approach is to put a bridge rectifier on the output of a CCFL backlight supply. Those are gradually disappearing, but you can still get them. Something like a BXA601 puts out 900V RMS at 30 kHz (open circuit). In stock at Digikey for $20 in onesies. For a cap you could use C330C183KGR5TA (18 nF, 2 kV, X7R, radial, 76 cents), and for the bridge rectifier, and a bridge rectifier made from SF1600-TAP fast-recovery 1.6 kV diodes.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC / Hobbs ElectroOptics 
Optics, Electro-optics, Photonics, Analog Electronics 
Briarcliff Manor NY 10510 

http://electrooptical.net 
http://hobbs-eo.com

I should add that it's probably better to use a 450-volt CCFL supply and a half-wave voltage doubler, or a 900V supply such as the above and a half-wave rectifier rather than a bridge.

Either one keeps the cold end of the secondary at ground, which is more like how CCFLs are driven. Probably only the hot end has the necessary slots, creepage distances, and so on, and the cold end may even be grounded on the board.

Do keep one hand in your pocket while playing with it.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC / Hobbs ElectroOptics 
Optics, Electro-optics, Photonics, Analog Electronics 
Briarcliff Manor NY 10510 

http://electrooptical.net 
http://hobbs-eo.com

rote:

s ]

ow

a

r9.pdf

The obvious way to make a 1kV power supply is to start off with a Baxandall Class-D oscillator (which seems to have been the basis for Jim Williams fa mous back-light driver, though Jim Williams doesn't seem to have realised t his) as describe in the Linear Technology application notes AN45, AN49, AN5

1, AN55, AN61, and AN65.

r1.htm

presents a link to Baxandall's original paper, and a couple of variations u sing MOSFET switches (which weren't around in 1959) but do work better than bipolar transistors.

You'll probably have to wind your own ferrite cored transformer - RM cores work fine, and you can buy plastic coil formers to go with any size of core you choose. John Larkin hates winding special purpose coils or getting the m wound, but it's not that demanding.

There are off-the-shelf inductors that can work in the circuit, but winding an inductor isn't much extra work.

--
Bill Sloman, Sydney

utput to

lright.

re Id ~

t it goes way

essive Id..

, the gain

ve got 2000

nd that

, and you lose

evel

to -15V

ther, the

ure an

d have a

), and R1

ming up

10V for

r higher

to enforce

ent: Q1-Q2

tput; the

an R+C

ntrol loop

ow, you have

d you

put

/3 the load or

ator?

of

n to zero.

ent??

e.g.

mize

stors

istors

will

cing

t with

in

I'm still waiting for an excuse to try your HV amp. (I'd certainly start with class B.) LT's 6090 is a nice 140 V opamp.

George H.

Right, I was trying to check it out open loop, and then spent too much time f'ing around getting the bias right for the first stage. (I just needed help from the opamp/feedback gods. :^)

George H.

to

t.

~

goes way

e Id..

gain

t 2000

at

you lose

15V

the

n

e a

d R1

up

for

her

force

Q1-Q2

the

+C

loop

ou have

e load or

zero.

s

Hi all, I had some time to try Phil's idea this morning. Here's my current circuit,

I had to add R2, otherwise I got these big long tails (~milli sec.)

It's still ratty. Here's the step response of the output.

I'm not sure I'm making any progress. :^)

Anything stupid or obvious?

TIA George H. (Oh I have no HV pnp's (on order) so I can't try this at HV. There is also a diode to ground on Q1's base so the opamp won't zener it.)

You completely missed my post. You've committed an even bigger sin: Q1 has wide open gain, so not only is the opamp's loop gain completely nuts, it varies exponentially with bias!

This is on top of Q2 still being wide open.

The ramping output is somewhat characteristic of an integrator controlling a slew-limited pole at too much gain. A zero in the control loop, at the pole, makes it settle much faster. TLDR: put a resistor in series with the

33p.

Tim

--
Seven Transistor Labs, LLC 
Electrical Engineering Consultation and Contract Design 
Website: https://www.seventransistorlabs.com/

Hi Tim, I did read your post. (I'll admit I didn't understand all of it.) And then I latched on to the idea of getting the charge off the gate faster.

RC in integrator loop... I'll try. Thanks George H. (I'll go read it again.)

Get rid of Q2 entirely--the emitters are the output of the regulator. The point is to have a totem-pole output. Also you'll need to put some degeneration on the CE stage, or maybe feedback from the output, to give the stage a well-defined gain. At the moment its gain and bandwidth are a strong function of output voltage.

Then you need to draw a Bode plot and fix the frequency compensation.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC / Hobbs ElectroOptics 
Optics, Electro-optics, Photonics, Analog Electronics 
Briarcliff Manor NY 10510 

http://electrooptical.net 
https://hobbs-eo.com

Didn't help much. But then I recalled whit3rd's advice of bypassing the 3 meg ohm FB divider. 10 pF in parallel.

I should reduce the 3 meg to 1 meg

I'm winning :^)

Oh I went back to the original circuit. No push-pull at the top.

George H.

ut to

ght.

Id ~

t goes way

ive Id..

he gain

got 2000

that

nd you lose

l

-15V

r, the

an

ave a

and R1

g up

V for

igher

enforce

: Q1-Q2

t; the

R+C

ol loop

you have

ou

the load or

r?

o zero.

??

. e

rs

ors

l g

Hi Phil, yeah I tired that first, but it was a bit of a train wreck. (I do still have 1 k ohm of degeneration in the CE.. not shown on schematic )

It worked better with R2 *collector resistor* at 10 k ohm, but it can't be that small.

Hmm, now that I've sorta discovered (thanks to whit3rd) that it's my feedback R's that need bypassing, the push pull thing might still work.

Anyway for the moment I'm going to work with the PFet circuit. (and put an order in for some HV pnp's) Maybe I can revisit this in the future. I really need to move on and start counting photons.

(I need to make a low photon flux light source.)

Or just bang on it and blob caps in till it looks OK. (I've got to check it at higher voltages.)

Thanks again for all the help.

George H.

It's awfully complex already, but adding a resistor in the emitter of Q1 would linearize it some.

Trade it for the 10K in series with FB.

How about this?

--

John Larkin         Highland Technology, Inc 
picosecond timing   precision measurement  

jlarkin att highlandtechnology dott com 
http://www.highlandtechnology.com

Thanks John, I scribbled your previous depletion fet circuit into my note book and was going to try that. (I've got some DN2540's) But now I've bypassed the FB divider R.. and reduced it by a factor of ~3 I've got a nice looking step response. (no wiggles and ~100 us rise time) And it works up to 250 V with no change. (Oh this is back to something like the original circuit. no push-pull)

George H.