DC boost supply no-load voltage

Question

I am working on a Geiger counter circuit which utilizes a boost supply similar to the one shown below to polarize the Geiger tube. The tube datasheet specifies a maximum voltage of 475 V. However, I'm not really sure how to interpret this in the context of a boost supply. If the tube is an open circuit in the absence of an ionizing particle, does not the boost supply continue charging the capacitor to an infinitely high voltage in theory? When I measure the tube voltage using a 10x multimeter I get 270 V and when I measure it with 100x I get 432 V. I presume if I kept stepping up the input resistance of the multimeter that the measured voltage would continue to go up.

Is it even possible to really know the open circuit voltage of a boost supply? And if not, how do I know that I have applied the correct voltage to the Geiger tube? By the way, the tube is actually working so this is mainly an academic question at this point.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

I'm not sure what that would look like. Is there an IC linear regulator that can handle > 400 V I/O? — Qubit1028, Apr 08 '18 at 19:40
@Qubit1028 There are lots of ways of regulating the voltage. My very first Geiger counter design (almost 50 years ago) used a series chain of selected NE-2 bulbs for voltage regulation. Also, you need to seriously examine your Geiger tube's plateau region, too, if that's the region you want to operate it. — jonk, Apr 08 '18 at 21:18
@Qubit1028 Sorry. I should have said why. Some Geiger tubes use halogens to quench and others may use an organic like ethyl alcohol. The organic quenching gases tend to create a relatively flat plateau so the voltage regulation doesn't have to be as tight, once calibrated. The inorganic halogens leave a rather largish slope vs voltage which complicates interpretation, later. So you really need to know your geiger tube well, too. No escaping it. — jonk, Apr 08 '18 at 21:33
@jonk Do the tubes have any leakage current when not avalanching? Would there be any current through RL2 that might tell you when max voltage is reached? — Henry Crun, Apr 08 '18 at 23:06
@Qubit1028 You can measure the voltage with a scope, by looking at the voltage at the moment that the probe makes contact to C1. After a while (us) it drags the voltage down, but the initial contact will be at full voltage. — Henry Crun, Apr 08 '18 at 23:10
@HenryCrun I'm sure there is some, though I've never observed it (didn't ever look for it.) There is an external quenching method (so that a quenching gas isn't added) that uses something on the order of hundreds of MegOhms to form an RC filter with the GM tube itself (which has capacitance.) So the leakage is small enough that almost no voltage drop is developed across that *beast*. I'd suggest that the leakage is probably near or below a nanoamp. But obviously different designs will have different figures of merit on that score, too. — jonk, Apr 08 '18 at 23:16
@jonk Thanks for the advice on the plateau behavior of the GM tube. I'm just getting started at this and will definitely consider it going forward. — Qubit1028, Apr 09 '18 at 00:09

score 1 · Answer 1 · edited Feb 16 '20 at 12:32

In the real circuit, there are 2 implicit components and factors.

Capacitance across the switch slows rise time, absorbs energy
Slow Q1 switch off, slows rise time

The C is made up of internal (interwinding C) in the L, transistor C, diode C and stray C (but not C1).

C can limit max voltage (even when D1 is removed). Energy in L is \$L\cdot I^2\$, while energy in the implicit capacitance is \$C\cdot V^2\$. The peak possible voltage will be when all the inductor energy is transferred into C, raising its voltage until \$C\cdot V^2 = L\cdot I^2\$

The transistor switch off rate is controlled by base R and CB or Miller capacitance. As Q1 tries to turn off, the voltage shoots up, and this high slope forces current through the CB cap, keeping Q1 turned on. Thus \$\frac{dI}{dT}\$ is limited.

Either of these can be manipulated to get controlled max voltage without any explicit extra components. e.g L=47μH, I=25mA, C=10pF, => V=54V

You will note that if you Ipk is constant, then V is constant, so if you have a currentmode control circuit (i.e. Q1 switches off when I hits the threshold), the V max will be constant regardless of battery V. ( or if you have a crappy oscillator where \$T_{on} \sim \frac{1}{V_{batt}}\$, then I will also be constant)

"An engineer is a man who can do for 10 bob what any fool could do for a pound"

You can measure the voltage with a scope, by looking at the voltage at the moment that the probe makes contact to C1. After a while (us) it drags the voltage down, but the initial contact will be at full voltage.

Alternatively if you have some sort of variable high voltage supply \$[V_{psu}]\$ (or even a fixed voltage that is close to the voltage to measure), you can connect you meter between \$V_{tube}\$ and \$V_{psu}\$, and adjust \$V_{psu}\$ until the difference is 0 and then measure \$V_{psu}\$. As long as \$V_{psu}\$ is close to \$V_{tube}\$, only a small current will flow.

Note that active control of flyback is most simply done by stopping the pulses when \$V_{max}\$ is exceeded, rather than PWM varying the duty cycle.

A couple of non-obvious ways to sense the voltage for control purposes:

You could have a capacitive divider across Q1, e.g. 10pF:1nF
you could put a secondary winding around L1. If L1 is 500 turns, a single turn sense winding will make ~ 1/2V

Great suggestions. A related question I have is how do I determine the output impedance of a boost supply? I really don't know how much current the thing is capable of supplying at 400 V. — Qubit1028, Apr 09 '18 at 00:25

score 0 · Answer 2 · edited Feb 16 '20 at 12:33

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Well, the (peak) output voltage of your booster will be \$U=L \cdot \frac{dI}{dt}\$. So it depends on

how much saturating current the inductor L1 has before switching and
how "fast" your switch (Q1) is.

Lets say the inductor and the battery both have 0.5 Ohms resistance, and the switch is on for a (very) long time. This would result in a steady 9A current from the battery through the coil. If the switch turns off now, and if it had a turn off time of 100ns, this would lead to a peak output voltage of 900kV (!!). Of course this is just a theoretical answer, because the output voltage will be limited due to intrinsic capacitances of the switch and the coil.

edited Feb 16 '20 at 12:33

Ariser

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answered Apr 08 '18 at 19:56

Stefan Wyss

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This is a rubbish answer. Ever heard that the current rises from zero to some value during a certain time period for an inductor. It’s encapsulated in v equals L di/dt. – Andy aka Apr 08 '18 at 23:30
Not rubbish, but the peak current isn't going to be determined by the coil resistance -- it's actually a function of the battery voltage and the on-time of the transistor. It will also be limited by the battery itself. – Dave Tweed Apr 09 '18 at 01:23
Got a duracell part number for that 9A peak 9V alkaline? – Henry Crun Apr 09 '18 at 01:27

score 0 · Answer 3 · answered Apr 09 '18 at 01:25

On the subject of measuring with your meter.

A modern autoranging multimeter has a 1M series resistor, and low voltage switches changing the bottom divider resistors, not the top series resisotr. The input impedance is always ~1M.

An old fashioned switched range (analog) meter has a resistance of 50k/V of range. So the 1000V range is 50M ohm. So the analog meter draws 8uA, and the digital one, 400uA measuring 400V.

However this is all because the autoranging has low voltage switches. Since you multimeter has a basic input R of 1M, you can use a series R or external divider.

A good dmm might give a perfectly useful reading at 10mV - a 100,000 divide ratio from 1kV, or a 100G series resistor.

Digikey has plenty of super high R resistors and a 1G resistor is less than a dollar

Note simple series R works for DC. It does not work for AC, where reactance of stray C totally swamps the R

As for measuring the source impedance, just get your self a bunch of high voltage Rs in the Mohm range, they are cheap, and a bunch of banana sockets to make an R box.

DC boost supply no-load voltage

3 Answers3