HV control

Hi all, another circuit fumble on my part. I copied this circuit fragment from whit3rd... part of thread is pasted in below. Here's the circuit I built.

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I ground the base of Q2 and added R3 so that it would start out with the HV turned off. D1 was added because I was worried about the zenering Q2's base. (opamp has +/- 15V rails.)

Here's a scope shot of the output (AC coupled Chan. 1) and the emitter (chan 2.) I'm pulsing the Fet on for ~10 usec. every ~800 usec.

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I tried adding R's and or C's here and there in hunt and peck fashion. but without much change. The cycle time is mostly set by the output cap (0.1 uF) and load resistor (100k ohm)

Sooo.... Am I doing something stupid, is there some way to fix it, or should I scrap it and start with something else?

Thanks George H.

*** previous post********

Subject: Re: Abusing the LM386 From: whit3rd

Thanks, I need pretty tight control. Some fraction of a volt out of 300. > My brute farce idea looks like this. >
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A series pass transistor is a less heat-producing option: the level translator Q1 needs HV rating, but the pass transistor Q2 is just to take off the ripple, so 100V should be plenty:

You can embellish it, of course; a collector series resistor//capacitor will share the heatload with Q1, and a source series resistor will limit shortcircuit current. Vbb* (R2/R1) is the gate drive limit.

Optos are slow, it's better to draw the milliamp or so from the HV, at 300V. I think. Different, though, if it were 3 kV.

Reply to
George Herold
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Ughh, I should have been looking at the opamp output. It's railed at +15 and then drops to -15 for ~a few us... I need a more gentle touch somewhere.

George H.

Reply to
George Herold

Snap, I put down the wrong part number for the pfet. It's a VP2450 GH

Reply to
George Herold

You need to close the ac loop and bypass the FET

So a cap from the opamp output to the inv input should make it stable

Cheers

Klaus

Reply to
Klaus Kragelund

Thanks Klaus, I'm pretty sure I tired that. I've got it working now... I ran the npn as common emitter and not common base. That 'flipped' the gain on my opamp, and I stabilized the thing with ~100pF from out to invert input as you suggest. I'm not sure why that didn't work in the other configuration.

George H.

Reply to
George Herold

Well, you don't have any active pull-down. Looks about like what I'd expect--the output droops about a volt in 250 us, so with 0.1 uF on the output, the load current seems to be about 400 uA.

Coincidentally, I'm just doing the BOM for a -250V adjustable APD supply, based on a couple of gate drivers, a 1:1:1 gate transformer, and a 4-stage Cockroft-Walton.

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
Reply to
Phil Hobbs

Assuming it's avalanche rated.

Is that a protected-gate mosfet? Some conditions could put 300 volts s-g.

--

John Larkin         Highland Technology, Inc 
picosecond timing   precision measurement  

jlarkin att highlandtechnology dott com 
http://www.highlandtechnology.com
Reply to
John Larkin

You are running the op amp wide open. Too much gain.

Remove R4 (10k) from the positive input of the op amp.

Insert it in series with the negative input.

Add an integrating cap from the output pin to the negative input.

Look at the step response.

Start with 1uF, reduce the value until it starts to fail.

Maybe add a small resistor (1k) in series with the cap if you want a faster response.

This circuit would be ideal to analyze in LTspice.

Reply to
Steve Wilson
  1. Vref has a series resistor, and you put an R+C from opamp output to -in, right? :-o
  2. You've got a gain of 1.5 with the level shifter, which is alright. As others mentioned, Q1 could stand a zener.
  3. No current limiting whatsoever. In fact...
  4. The PMOS is wide freaking open. What gain is it operating at? Who knows!

Datasheet looks like it's down in the subthreshold region, where Id ~ exp(Vgs-Vpo), maybe 100mS tops, at the current shown. But it goes way up from there at higher Id, and there's nothing to prevent excessive Id.

If the gain is 100mS and the load is, say, 200V / 10mA = 20k, the gain is around 2000.

Add to that, the single pole of 0.1uF * 20kR = 2ms, and you've got 2000 times more gain, and less than 0 degrees of phase margin, around that poor opamp! (Remember, you only have 90 degrees to spare, and you lose all of those where the 0.1uF looks like an ideal integrator.)

One more crinkle: match up the opamp's output range with the level shifter's range. If it's swinging +/-15V, divide it down to -15V instead of 0V, so that Q2's emitter range is -7.5 to 0V (or rather, the Thevenin output from the divider). Or slightly up to ensure an off-biased state.

Using the same principle (matching operating ranges), Q1 should have a source resistor (fixing its gain ~constant, except near cutoff), and R1 could have some idle bias in it so there's minimal deadband coming up from cutoff.

As long as you don't mind a fairly high dropout voltage (about 10V for R1-R3 as shown), current limiting is also automatic: it's never higher than (V(R1)

  • Vgs(th)) over (source resistor). A zener, to enforce V(R1) maximum, provides further assurance of this.

With these changes, it's easy to see the block diagram equivalent: Q1-Q2 is a transconductance source, current proportional to opamp output; the 0.1 and the load is the "plant", a simple single-order one, so an R+C compensation network allows full freedom to set a 2nd order control loop of desired damping.

Oh, check the gate-R1 time constant, too. If it's too slow, you have another pole in the loop (and don't forget Miller effect!), and you won't be able to stabilize it anywhere near as fast as the output capacitance would suggest. Make sure this is less than 1/3 the load or controller time constant.

Tim

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

"George Herold"  wrote in message  
news:f086a724-cd0e-45dc-ad46-7cfa3bc8cc45@googlegroups.com... 
> 
> Hi all, another circuit fumble on my part. 
> I copied this circuit fragment from whit3rd... part of thread is pasted in 
> below.  Here's the circuit I built. 
> 
> https://www.dropbox.com/s/skaspzjafoqmcfl/HV-control.JPG?dl=0 
> 
> I ground the base of Q2 and added R3 so that it would start out with the 
> HV turned off.  D1 was added because I was worried about the zenering 
> Q2's base.  (opamp has +/- 15V rails.) 
> 
> Here's a scope shot of the output (AC coupled Chan. 1) and the emitter 
> (chan 2.) I'm pulsing the Fet on for ~10 usec. every ~800 usec. 
> 
> https://www.dropbox.com/s/eoszsjj16dsn86r/Vout-E.BMP?dl=0 
> 
> I tried adding R's and or C's here and there in hunt and peck fashion. 
> but without much change.  The cycle time is mostly set by the output 
> cap (0.1 uF) and load resistor (100k ohm) 
> 
> Sooo....  Am I doing something stupid, is there some way to fix it, or 
> should I scrap it and start with something else? 
> 
> Thanks 
> George H. 
> 
> 
> 
> *** previous post******** 
> 
> Subject: Re: Abusing the LM386 
> From: whit3rd  
> 
>> Thanks,  I need pretty tight control.  Some fraction of a volt out of  
>> 300. 
>> My brute farce idea looks like this. 
>> https://www.dropbox.com/s/mfqhln20x7tkfk3/10-300V.JPG?dl=0 
> 
> A series pass transistor is a less heat-producing option: the level 
> translator Q1 needs HV rating, but the pass transistor Q2 is just to 
> take off the ripple, so 100V should be plenty: 
> 
>  
> 
> You can embellish it, of course; a collector series resistor//capacitor  
> will 
> share the heatload with Q1, and a source series resistor will limit 
> shortcircuit current.   Vbb* (R2/R1) is the gate drive limit. 
> 
> Optos are slow, it's better to draw the milliamp or so from the HV, at  
> 300V.  I think. 
> Different, though, if it were 3 kV.
Reply to
Tim Williams

This is exactly the kind of thing well suited for LTspice. You can try different configurations much faster than with hand soldering. You can see the open loop gain and margins which you cannot see on a breadboard. You can view the step response which is difficult on a breadbard. You can look for overstressed components which may be difficult with an oscilloscope. You can see glitches that may be impossible on a scope.

LTspice is an amazing tool. It has a huge learning curve, but there is an abundance of help available. It is an excellent way to document circuits so

6 months later when you have forgotten how something works, you can refresh instantly. You can also post your circuits to sed and gain significant suggestions on improvements or changes. You can solve problems much faster and gain a deeper insight into how a circuit works than with any other method.
Reply to
Steve Wilson

A pull down where? Not the gate? Driving home I was thinking that my working common emitter had 1k or degeneration, maybe some base resistance? (the npn base was soldered right to ground.)

Huh, well color me unsurprised, if your design is bucket loads cheaper and better than mine. :^) talking to the APD people (Laser components) it seems they are hot these days... self driving cars and such.. Which is fine by me I'm happy to surf along. (I paid a bunch for the good one they told me I wanted.)

So let me run this idea by you. These Spads, are going to be used by students. From what I understand if you turn the room lights on while the detector is biased (above breakdown), you break the diode. "bvvvt" (my guess at the sound, there may not be any v's :^) (~$1-2k to replace) So I was going to try an lnd150 and ~10k* g-s resistor as a current limit in the HV line. With the 'right' amount of C after... I'm not sure how big the PD's are. (thinking thermally here)

George H.

*from one measurement 10k is ~0.1mA, it costs about a volt. I could put in 100k, 10 uA... Oh, each event discharges the whole PD capacitance.... ~1pf (I'm not sure of the photon flux... I've seen mostly low numbers in papers, 10k max) so say 10^6, Q=C*V, the voltage depends on how high it's biased above break down, ~1V but maybe more... I think there's a sweet spot in there somewhere, well so I'm hoping. GH
Reply to
George Herold

All my pass bits are ~450+ volts. I want the the output to go from 0-300V.. (it might be only 200V max.) I haven't tried the HV yet.. So far the HV line has been connected to +15V

George H.

Reply to
George Herold

Since the feed back is positive I tried an integrating cap from output to the non-inverting input. Didn't work. that R can't hurt. I agree that I'm not sure it does any good.

Oh as Tim observes below Vref is a voltage reference feeding a 10 k pot... So I've got some R in the Vref lead. I feed back C to there but it didn't help. (I might have made some silly mistake.)

Right, I thought I could just do solder. Then I can attach a real device to it.

Thanks, George H.

Reply to
George Herold

nd.)

The slew is asymmetric because you have a strong pull-up and a weak pull-do wn. A class-B totem pole would make it more symmetrical.

Something like a ZXTP01500 would do a good job, I expect. (That's what I'm using as the main pass transistor in my negative supply.)

Other folks have commented on the frequency-compensation issue. IME it work s best if you make each stage as fast as reasonably possible and apply loca l feedback to turn it into a well-behaved first-order block, and then let t he slowish outer loop look after the remaininf discrepancies.

Nah, horses for courses. Mine runs at 2.2 MHz, which causes gross losses in the transformer and the diode capacitances, but satisfies the customer's D C-2MHz

on while the detector >is biased (above breakdown), you break the diode.

Reply to
pcdhobbs

Yeah, 10k pot (from 10V ref) feeds the inverting input.

Where?

Right. I looked at the opamp and it was in the most sorry of states. mostly railed positive, but bashed to the negative rail for ~10usec.

Oh boy, Tim, thanks. I think I understood most of that... I'm mostly an idiot, saved by negative feedback. :^)

George H.

Reply to
George Herold

requirement.

A fast foldback sounds like the ticket. Make sure that the output bypass isn't big enough to blow up the diode by itself.

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
Reply to
Phil Hobbs

ng

ground.)

l-down. A class-B totem pole would make it more symmetrical.

I'm using as the main pass transistor in my negative supply.)

works best if you make each stage as fast as reasonably possible and apply local feedback to turn it into a well-behaved first-order block, and then l et the slowish outer loop look after the remaininf discrepancies.

and

s in the transformer and the diode capacitances, but satisfies the customer 's DC-2MHz

ghts on while the detector >is biased (above breakdown), you break the diod e.

Ahhh, is a fast foldback, the depletion fet plus R? Picking the size of the last bypass cap is what I need to guesstimate... I should reach out to some engineer's at the company, they must have blown some up.

George H.

Reply to
George Herold

A foldback makes the current limit a function of the voltage across the load as well as the output current. It's often done by summing a voltage divider between the output and ground with the voltage drop of the current limiting resistor.

The usual wisdom is to keep the high-voltage current limit above a quarter of the zero-volt value in order to avoid latch-up states.

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
Reply to
Phil Hobbs

if all he needs is a couple mA output, why not do a shunt regulator?

Reply to
bitrex

Well, that would explain it not working as originally sketched out: the PMOS was intended to turn ON with circa 10Vgs, and the base-emitter-emitter resistor was intended to take up to a full op amp swing (12V or so), but there's not enough volts to do all that... so perhaps the Q2 transistor is saturating.

The rail-rail behavior would indicate not enough negative feedback, so it might also pay to put a speedup capacitor across R5 (to swamp the input capacitance of the op amp). Too much capacitance and you get output ramping (OK for a regulator, not for a pulser).

Reply to
whit3rd

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