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I'm trying to make a 400V boost converter for a Geiger counter, but the circuit is drawing far too much current than what I would expect.

I'm using the following circuit:

enter image description here

My Analog Digital Discovery (cheap oscilloscope + function generator) shows a current delta of about 130-150mA when the converter is on versus idle, when I should expect around 10x less. Obviously this means that I have pretty significant losses, but I can't seem to figure out why. This happens at all frequencies and inductor/capacitor values.

Is there a way to reduce power losses? I know I can replace the diode with a MOSFET, but if possible I'd like to see if I can reduce the losses just with component selection/topology.

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  • \$\begingroup\$ Do you have a load other than the 10M resistor? If so, what is the impedance of this load? \$\endgroup\$ Nov 19, 2022 at 22:19
  • \$\begingroup\$ I'm using my multimeter as the load. The geiger tube would act as a switch, and the Geiger counter circuits I've seen use a couple megaohms in series with the tube. \$\endgroup\$ Nov 19, 2022 at 22:51
  • \$\begingroup\$ Is the actual circuit as unregulated as it looks? I have no idea how sensitive Geiger-Müller tubes are, but I'd be concerned about this outputting far too high a voltage. \$\endgroup\$
    – Hearth
    Nov 20, 2022 at 17:24
  • \$\begingroup\$ By the way, here's an old thread on Geiger-Müller power supplies: 3V to 500V DC converter \$\endgroup\$ Nov 20, 2022 at 18:11

3 Answers 3

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This is expected with the component values that you've chosen.

Let's assume that your circuit is actually able to produce 400V (which it likely isn't, but let's pretend it is). This means that the drain of your IRF840 will have a 400Vpp square wave on it. The IRF840's output capacitance is 310pF according to the datasheet and your converters switching frequency seems to be about 50kHz.

With this data, we can calculate the losses caused by the IRF840 alone:

P = V² * C * f = (400 volts)² * 310 pF * 50kHz = 2.5W

This means that there's no way to build a 400V step-up with an IRF840 operating at 50kHz that doesn't at least dissipate 2.5W as heat in the transistor. It gets even worse when you take the diode into account as well.

TL;DR: Those 0.7W of power that your circuit draws are all being dissipated in the transistor because a single-stage boost converter is simply not suited for stepping the voltage up by a factor of 80 in a low-power application. You need to use a different circuit altogether.

You can build a flyback converter to get higher efficiency. (A flyback will have lower losses because the MOSFET's drain doesn't experience such an extreme voltage swing.)

Given the low output power requirements of your application, I'd go for a DCM flyback converter with peak-current mode control. The transformer turns ratio should be about 1:40. The primary inductance will have to be rather high, which means that you might want to use an ungapped core, which is rather unusual for flyback converters. Something on the order of 1mH primary inductance should work. You will also have to find a diode with extremely low junction capacitance. You also won't be able to operate the converter in open-loop mode due to varying diode losses - or in other words, you'll need a feedback loop and an actual flyback controller.

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  • \$\begingroup\$ My circuit is actually able to produce more than 400V IRL. I use an FR207 diode instead of the UF4007 because the FR207 doesn't exist in Multisim. Still, I see your point about the MOSFET having very high losses. \$\endgroup\$ Nov 19, 2022 at 22:55
  • \$\begingroup\$ Very good point, +1. However I believe capacitive power is only a half of what you wrote, since MOS is only the discharge path of its own DS capacitance. Charging takes place through the inductor and hence it's lossless. There could actually be some more MOS loss at the end of current, if DCM applies, when ringing periodically switches-on the body diode. Don't think this is significant but I'm not sure. \$\endgroup\$
    – carloc
    Nov 20, 2022 at 7:03
  • \$\begingroup\$ The current in Q1 and D1's parasitic capacitance is not automatically a power loss; to some degree, it forms an LC circuit, and for the appropriate operating frequency/duty cycle some of the energy in that LC is scavenged in the next cycle. @carloc: it's not sure "ringing periodically switches-on the body diode"; and even if it did, that power loss is very low: it's UI with U<1V the diode's forward voltage, for I=0 most of the cycle. Otherwise said, most of the energy stored in L1 when Q1 conducts evacuates thru D1. \$\endgroup\$
    – fgrieu
    Nov 20, 2022 at 9:04
  • \$\begingroup\$ @fgrieu yes, I agree, that's part of what I was trying to highlight \$\endgroup\$
    – carloc
    Nov 20, 2022 at 9:40
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    \$\begingroup\$ if this is a one-off project you may find a suitable flyback transformer in a junk "phone charger" \$\endgroup\$ Nov 20, 2022 at 11:03
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You have a very light load, according to your schematic. Low efficiency is typical of the boost converter topology (and many other DC-DC converter topologies) at very light loads.

Here is a CircuitLab simulation of your circuit, using a 18 kHz, 50% duty cycle, 0-16 V square wave.

schematic

simulate this circuit – Schematic created using CircuitLab

This is only a simulation, so the reality may be different. Nevertheless, it will point us to how the circuit will change in its performance as the parameters of the circuit are changed.

Here is how the circuit performs as is:

Output Voltage enter image description here

Inductor Current enter image description here

It might be useful for discussion to show a zoomed in version of the inductor current.

Inductor Current (Zoomed) enter image description here

It can be seen that the inductor current is 0 for part of the cycle. This means that the converter is operating in discontinuous conduction mode (DCM). This is what we would expect with such a light load. When the switching element (the MOSFET) conducts, the current in the inductor rises gradually. Then when the switching element is turned off, the inductor current falls rapidly. The falling current reaches zero, and then goes past zero. This is due to the reverse recovery current of the diode. Another diode, or a synchronous switching MOSFET, might reduce this effect, but we will not consider such changes in this answer.

One thing to take note of is that if we ignore the reverse recovery current, and treat it as zero, (which it clearly is not), then about half the time the inductor is conducting, and about half the time it is not. That means that the average current for a whole cycle will be about half the current of the half cycle in which it is conducting. But the average current in the half cycle in which it is conducting is about half of the peak current. Thus, we can use 1/4 of the peak current as an estimate of the average current.

Is there a way to reduce power losses? I know I can replace the diode with a MOSFET, but if possible I'd like to see if I can reduce the losses just with component selection/topology.

Here is the same circuit with the value of the inductor increased by a factor of 100, the frequency reduced by a factor of slightly more than 10, and the capacitor increased by a factor of 10.

schematic

simulate this circuit

The voltage is pretty much unchanged.

Output Voltage enter image description here

The peak inductor current, however, is dramatically reduced -- from about 900 mA to about 100 mA.

Inductor Current enter image description here

The peak inductor current has been dramatically reduced. But is the average current still approximately 1/4 of the peak current? Just to verify, let's again zoom in on the inductor current, to see if the waveform is significantly different.

Inductor Current (Zoomed) enter image description here

We see that the waveform has retained its shape, and again, the average inductor current is approximately 1/4 of the peak inductor current. (BTW, this is a consequence of a) the steep decline in inductor current when the switching element turns off, which is a result of the light load, and b) the 50% duty cycle of our switching signal.) So, very roughly, the average current fell from about 225 mA to about 25 mA. Of course this is only a simulation, but it points towards how improvements may be made.

Normally, the duty cycle and frequency are controlled by a chip designed for that purpose, using feedback from the output voltage. In your circuit, as given, the duty cycle and frequency are fixed, so you will need to adjust your signal generating source "manually" or "externally". Alternatively you could use a more complex circuit that generates its own PWM signal and duty cycle and uses feedback from the output voltage. If you fail to properly adjust the duty cycle and/or frequency, you may accidentally cause the output voltage to rise above the 400 V you specify. Obviously, this could destroy the MOSFET as well as other components.

Conclusion: using your topology, you should be able to substantially decrease the power consumed by your boost converter by a) increasing the inductance, b) decreasing the switching frequency to restore the voltage, and c) increasing the capacitance to restore the smoothness of the output.

The changes require larger valued components, which are quite likely more expensive. For that reasons, it is definitely worth considering alternative topologies, such as fly-back, as suggested by Jonathan S. in his answer. On the other hand, if this is just a project for fun, and you have the larger inductor and capacitor on hand, as well as an appropriate signal generator, then your circuit topology can be made to work more efficiently than in your original circuit, even though from an economic standpoint, it may not be the best choice.

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Apparently, my AD2 doesn't give the actual current draw from the voltage source. I hooked up an ammeter to the voltage source, and with a 4x voltage multiplier, I was drawing about 9mA from the source with a 5.05M\$\Omega\$ load. This means that my efficiency was:

$$ \eta = \frac{\frac{V_{out}^2}{R_L}}{I_{Source} \cdot V_{Source}} = \frac{\frac{400V^2}{5.05M\Omega}}{9mA \cdot 5V} = \approx 70\% $$

I would think that for a 80\$\mu A\$ load current that's not terrible, and is much better than what I thought it was before. Adding an extra 2 stages to the multiplier does not increase efficiency - most likely because of the capacitor/diode losses from too many stages.

I was running the PWM signal at 2kHz with a 46% duty cycle, capacitors were all 1uF. This combination seems to yield the best results.

I was able to reduce the input current to about 50 mA (from 150 mA) by adding a Cockcroft–Walton voltage multiplier to the output. Since I don't need the output to sustain high current, this is a good enough solution. I think that both reducing the frequency and increasing the inductor size will further reduce (as "Math Keeps Me Busy" said in their answer) the current through the MOSFET.

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