# Discharging circuit problems for Pulsed plasma thruster

1. Jan 4, 2012

### lomaikai

I have been working on a circuit (similar to car ionizer circuit) for a pulsed plasma thruster, which is able to charged up a capacitor to voltage as high as 1.8kV. I have a diode which is able to block reverse voltage up to 2kV interfaced between the charging circuit and the discharged capacitor.

This capacitor is then connected to two electrodes (air gap) that discharge when the breakdown voltage is reached.

The problem is, whenever the capacitor discharges, a random component in my circuit will be spoilt. What must i do to overcome this problem?

Last edited: Jan 4, 2012
2. Jan 4, 2012

### DragonPetter

Do you have a schematic?

3. Jan 4, 2012

### lomaikai

Hi there, attached is the schematic. Hope it helps =D

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4. Jan 5, 2012

### lomaikai

Questions on car ionizer circuit (a.k.a discharge circuit)

hi all, i'm coming from a mechanical background and was tasked to carry on an electronic project.

Attached is the schematic of the circuit, i tried looking up the net on the respective components and understand how each component works.

I also understand the part after the transformer which is a voltage multiplier circuit, which steps up the 240v from the transformer to several folds.

However, i do not understand how the circuit before the transformer actually works. Greatly appreciate any help! Thanks !

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5. Jan 6, 2012

### Staff: Mentor

Re: Questions on car ionizer circuit (a.k.a discharge circuit)

These cascade voltage multipliers work by charging the capacitors in parallel, then connecting them in series while they discharge. They are a clever design, and are most useful for low-current applications. http://en.wikipedia.org/wiki/Voltage_multiplier

I notice you have not recorded a value for the caps in your voltage multiplier.

6. Jan 7, 2012

### Mike_In_Plano

Re: Questions on car ionizer circuit (a.k.a discharge circuit)

It looks like a small high voltage generator of the sort that was used in air ionizers that plugged into a cigarette lighter.

It has a diode in the input to prodect it from the power supply voltage being reversed, and an LED to tell you that it's connected to power.

Below that are two transistors that are connected in what's termed a Darlington configuration. This allows them to work as one transistor which has much more gain than a normal transistor.

Some current passes through a resistor to start the transistors turning on. As they draw current, it passes through a transformer that couples even more drive to the transistors, ensuring they turn on completely.

While the transistors are on, the current through the transformer to the transistors will increase according to di/dt = V / L.

Eventually, the current will reach a peak where either due to the transistor's limitations, or perhaps the resistance in the transformer's winding, the current will no longer be able to increase at this point, V across the transistor will go towards zero, the transistors will be turned off by a winding of the transformer, and this will be regenerative.

A diode prevents the drive from the transformer from placing a substantial negative voltage on the base of the first transistor. This is to protect the transistor.

During the next portion of time, the transistors are off, and the energy stored in the transformer will cause it to "ring" the voltage at the collectors of the transistors will go higher than the power supply voltage during this time.

After a short time, the energy in the transformer will be depleated and the voltage on the collectors of the transistors will fall. During this time, energy may be stored in the various leakage capacitances across the transformer. This energy will tend to make the transformer ring under the power supply voltage and aid in turning the transistors on again. Thus the cycle repeats.

On the output, the combination of diodes and capacitors serve to multiply the output voltage from the transformer until it's high enough to cause ionization of air.

I believe the oscillator is termed a blocking oscillator, and the output circuit is a voltage multiplier.

Mike

7. Jan 7, 2012

### Mike_In_Plano

Let me guess...
One of the rectifiers in the voltage multiplier typically dies?

When a voltage multiplier undergoes a hard discharge, it seends high currents through the diodes of the multiplier.

The solution is to place a physically large, high-surge resistor between the output of the multiplier and the output capacitor. This will alleviate the torture on the diodes.

Also, if you are intending to use this in any sort of high performance application, the suggested circuit is a poor choice.

8. Jan 8, 2012

### lomaikai

hi Mike ! really appreciate for the help. Yes, most of the time, the rectifiers died on the discharge. I will try out your solution and see if it improves the situation !

yea very much agree with you, but due to limited knowledge of electronics, this is the closest i ever come to ! =S

Last edited: Jan 8, 2012
9. Jan 8, 2012

### Staff: Mentor

(Moderator note -- two similar threads merged)

10. Jan 8, 2012

### Mike_In_Plano

For the basic power supply, I like the cold cathode fluorestent lamp (CCFL) power supplies. Small ones can be purchased through computer hobby stores for about $10. They're robust and efficient, but they don't have the voltage multiplier on the output. This has to be built seperate. To help your supply survive discharges, I suggest an Ohmite, SM104032003FE. At 200k-ohm and 10kV withstand, this resistor will stand off the peak currents between your voltage multiplier and output capacitor. Mouser stocks these parts for about$5 as well as some high speed, high voltage rectifiers, and high voltage capacitors.

I'm curious, do you ever want to test this in a vacuum? Interesting things happen when your high voltage projects are placed in a vacuum.

Mike

11. Jan 8, 2012

### lomaikai

I checked the input to CCFL power supply, and the lowest I've seen is about 11vac. Will my current circuit be sufficient to supply that 11vac to kick start the CCFL ?

Ah yes that will be interesting, but it is an air prototype at the moment, most likely won't be tested in a vacuum.

12. Jan 8, 2012

### Mike_In_Plano

The input to the CCFL supply is typically 12 volts, but the open circuit voltage must be high enough to ionize the gas in the lamp. Thus these little inverters typically output about 1kV RMS through a low value capacitor which serves to limit the current after the lamp ionizes.

With a voltage doubler (two rectifiers and two capacitors), you can probably get about 2 - 2.5 kV DC from a low cost inverter.

I have to admit I've been curious about this manner of thruster as it regards the ultrasmall (10 cm^3) satallites. Such a tiny satallite has no room for plumbing, but could still benefit from attitude correction.

Extreamely tiny, electrically actuated thrusters would be great.

13. Jan 8, 2012

### lomaikai

hey Mike, really appreciate the help so far! With your suggestion, I managed to find an inverter by JKL components, part no. BXA-502, which is able to step up 5vdc to 650vrms. shld be able to step up to around 3kv after a series of voltage multiplier.

Yes, we are actually looking this type of thruster for these nano-sats, but most likely only suitable for micro-sats. The size of the capacitor needed to discharge sufficient energy across the teflon (our fuel for the thruster) gets really big and unsuitable.

For these nano-sats, the same concept can be applied but will need the help of MEMs technology to achieved this. Cheers

14. Jan 10, 2012

### lomaikai

I bought a bxa-502 inverter by jkl component. it has an output of 650vrms with frequency between 70-90khz. I used a low speed/freq. capacitor for the voltage multiplier circuit but only able to achieve a mere 200vdc.
how should I calculate the required high speed/freq capacitors and diodes for implementation so as to achieve a dc voltage around 3000.

15. Jan 12, 2012

### Mike_In_Plano

Hiya,

I'm glad to see your well after it. Firstly, don't trust what most meters will tell you about the output. This is twofold:

1 - most meters are not intended to read such high frequency AC voltage. So, they read low.

2 - Many meters have capacitance across the input that will load down the circuit.

Not to worry though. No harm, no foul. These inverters are tough.

Chances are, you're closer to your 3kV goal than you think. The reason lies in the behavior of the CCFL lamp. when cold, the lamp typically takes about 3x it's operating voltage in order to ionize the gas within.
The inverter is designed to provide this voltage, but must then reduce the voltage, or excessive current will flow. It does this by placing a small value capacitor in series between the lamp and inverter. this will generally be a sky-blue colored capacitor near the lamp's terminal.
This capacitor will limit your output, but I'd leave it in at first, because it will help protect your parts. Later, I might double or triple the value.

In any case, with no load you'll have much more than 900 V peak at the output :-)

You may want to simply purchase a voltage multiplier. I found a number on ebay, including this one:
http://www.ebay.com/itm/M3KV6-posit...695?pt=LH_DefaultDomain_0&hash=item3f1264d51f

If you wish to build it from scratch, a Rectron R5000F-B will likely be a good rectifier. With that, a 2.2nF HV capacitor, such as a Murata DECE33J222ZC4B.

HV can be creepy when you don't have a way to discharge it, so you probably want a large value (ie 22meg), high voltage (4kV), resistor to place across the output of the multiplier.

If you choose to go Frankenstein with the output, you may want to place the whole thing in oil or pot it. I've found that I can trick silicon caulking into making a nice conformal layer by dissolving it in a heaver solvent, such as MEK. Then dip the part, allow it to dry, and set it aside a night for the silicon to harden. Don't rush the process with heat!

Best of Luck!

Mike

16. Jan 13, 2012

### lomaikai

hi, sorry but i have met with another problem. Quite a serious one. Inverter circuit works like a charm, a couple of times, including discharging via a capacitor. But the next day when i did the same thing, the discharge actually spoilt the inverter circuit together with the power supply.

Is it possible that during the discharge, the huge voltage drop caused a surge in current to replace that voltage drop, thus spoiling the DC supply. If that is so, how should one avoid this problem?

Many Thanks.

17. Jan 13, 2012

### Staff: Mentor

Your 5v dc supply is damaged? You don't know what is wrong, except you are not now getting any dc output from it? I'm wondering whether you might have accidently included the 5v supply in the discharge path when you shorted the high voltage capacitor?

18. Jan 15, 2012

### Mike_In_Plano

Assumming the 5 volt supply works when disconnected from the inverter, it's probably ok. Otherwise, it sounds as if your capacitors discharged through a path that led though the 5V supply. Sometimes, the arcs can find sneak paths that would otherwise be open. I find these more easily by discharging it in a darkened room.

19. Jan 16, 2012

### Mike_In_Plano

Ouch! I'm sorry to hear that.

Are you working with the original inverter with the added resistor? Or the new inverter?

20. Jan 17, 2012

### lomaikai

yea, was working with the BXA-502 inverter and added resistor before the discharge capacitor. spoilt the circuit together with the power supply when the capacitor discharges.