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AC constant current source

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Dr. Dark Current

Thu Jun 19 2008, 08:53PM

Registered Member #152 Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384

Is there any inductive (ie. no loss) component which can provide constant current output from AC voltage input?

Steve Ward

Thu Jun 19 2008, 10:07PM

Registered Member #146 Joined: Sun Feb 12 2006, 04:21AM
Location: Austin Tx
Posts: 1055

How constant does the current need to be?

An inductor alone will have some impedance, which can limit the current of a shorted load, but otherwise the current is based off of the complex impedance of the inductor combined with whatever your load is.

A true constant current source isnt particularly easy. It can be done with solid state switching methods, or with some types of reactors where the inductance is controlled to change the impedance, controlling the current. The response time is typically slow, at least for the ones ive seen, which were mechanically controlled.

More information about your intended application might turn up some better solutions for you.

Dr. Dark Current

Fri Jun 20 2008, 02:23PM

Registered Member #152 Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384

Well Im playing with hv sparks (as always

) and I have a theory that if you feed your transformer with constant current instead of constant impedance, the arcs should be more than 2x as long with the same short circuit current.

I was thinking along the lines of some magnetic component (reactor) which can vary its inductance based on the voltage drop it sees...

Dr. Slack

Mon Jun 23 2008, 08:07AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

Yes there is. You can have a 2 port device that you connect to an AC input. It takes a constant voltage in, and generates a constant current out. I would have tried it myself, but frankly I have neither the time nor the balls.

I got the idea sharing a few shandies with Richie the evening before Derek's Nottingham Gaussfest. The seed was the behaviour of matching components in a class-E drive. Now all RF engineers know that you can invert a short circuit into an open circuit using a quarter wave length of transmission line, the so-called quarter wave transformer. What is current at one end is voltage the other other, related by the impedance of the line. Most RF component matching is done by simulating a transmission line at a single frequency with lumped components. So the question is not "does it work scaled up?" but "why has no-one tried it before?" It's not like the resonant MOT arcs, as they lack the output capacitor.

First of all, a caution. The wall socket can deliver constant volts, at any current up to infinity (breakers permitting). The output of the impedance inverter will deliver constant current, at any voltage up to infinity (electrical breakdown of output components permitting, and losses limiting Q). That means that if you run this and there is any possibility of the output going open circuit (duh, certainty more like), something is going to flash over. Make sure that something is an anti-fuse (a fuse goes open circuit at over-current, an anti-fuse goes short circuit at over-voltage). A simple fixed spark gap would be sufficient in the short term but would get hot as with limited current, the gap voltage will not drop to zero. A spark gap with a shorting bar held off by some candle-wax would work to limit the gap temperature rise.

Finally to the circuit. All you need to synthesise a transmission line is series Ls and parallel Cs, a minimum of 3, either as a pi network of CLC, or as a T network of LCL. I am going to do all the sums for 50Hz and 240v. Anyone who dares to run this inverter will be capable of computing the components for any arbitrary frequency and voltage. The components must be tuned to 50Hz. The output current is given by the input voltage divided by the line impedance. It turns out that regardless of whether you use a pi or T network, the impedance is sqrt(L/C), and the frequency is 1/2.pi.sqrt(LC), where L and C are the values of the components used.

So if we want 1 amp output, we need 240ohm line, which at 50Hz you can make from 0.76H and 13.2uF. The choice of pi or T is arbitrary, but it you want to run the input from a MOT rather than direct from mains, the LCL configuration would allow you to absorb some of the MOT leakage inductance into the network. A CLC configuration can use a pre-existing power factor correction bank as the input C, or you can dispense with the input C altogether if you don't worry about getting to zero input current at s/c output. A CLC configuration has only one L, so will probably be cheaper to implement than the LCL.

The series L and the output C need to be rated for your intended maximum output voltage, that is twice the nominal set voltage of your output anti-fuse.

Does it work? It will take only a couple of minutes to whack the few components into the simulator of your choice (hint - Simmetrix) and prove it to yourself. Assuming the CLC configuration, then we can see what happens with a) s/c output, b) o/c output and c) matched output. With a short circuit on the output, the output current is given by the current through the series inductor from 240v to ground, which has an impedance of roughly 240.j.ohm, so 1 amp. However, the input current is the sum of that through the inductor, and what you can think of the input power correction cap, which is -1 amp, so zero input current (neglecting losses of course). For an open circuit output, the series L resonates with the output cap, and the amplitude of the output builds cycle to cycle to infinity (do a transient simulation to see this behaviour), while the input current also climbs to inifinity. For a matched output, the 240ohm load appears as 240ohm at the input, so draws 1 amp.

If you use this circuit ahead of a transformer, then the transformer leakage reactance will appear in the sums (probably need the LCL configuration), and the high output voltage could break the transformer. Also, getting a good anti-fuse is much more difficult at lower volts where a spark gap doesn't work as well, might need an transorb/triac crowbar.

A more subtle output protection method would be to note that the input current flows mainly in the series inductor. The inductor could be sized to saturate, therefore lose inductance and de-tune the network, before the output voltage has risen to a dangerous level. The network should then be self-protecting.

Enjoy

Steve Conner

Mon Jun 23 2008, 08:49AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

The grid-intertied inverter I built was basically that, a 50Hz AC constant current source. It used an H-bridge with current-mode PWM.

I don't expect it would make sparks 2x longer: the length of the spark is determined by the maximum real power available for it to consume. By stretching the arc until it goes out, you're adjusting its resistance to find this maximum power point of your driver.

With an ideal current source driving a step-up transformer, the maximum power would still be limited because the voltage the transformer can handle is limited by saturation. Likewise, Neil's clever network described above will have its output voltage limited by saturation of the inductor. If I started designing out the iron core to avoid saturation, and optimizing for longest sparks, I'd end up with an SSTC.

Well, the volt-second product is limited, so you can get round that by increasing the frequency. De-shunted resonant MOT stack driven by a 400Hz inverter, anyone?

Dr. Dark Current

Mon Jun 23 2008, 09:08AM

Registered Member #152 Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384

Thanks Dr Slack, that was an interesting read. However I'm not sure if the circuit you drew would be any usable for hv transformer drawing arcs, because the output looks capacitive and capacitive arcs are not arcs but sparks. Maybe it would help to add another series L on the output of the network.

Dr. Conner wrote ...

I don't expect it would make sparks 2x longer: the length of the spark is determined by the maximum real power available.

Yes, and with constant current source the maximum power available is 4x higher that of a constant impedance source (if you calculate upon the same short circuit current, that is, to keep maximum copper losses the same).

The constant impedance source will deliver maximum power at output voltage V/2 and current I/2, P=VI/4. The constant current source will reach its maximum power point at output voltage V (limited by saturation if we are talking about real situations) and current I, that is P=VI.

Dr. Conner wrote ...

Well, the volt-second product is limited, so you can get round that by increasing the frequency. De-shunted resonant MOT stack driven by a 400Hz inverter, anyone?

Do you think a MOT can hold up 16kV?

Dr. Slack

Mon Jun 23 2008, 10:16AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

because the output looks capacitive

sounds like a very strong vote for the LCL configuration then

Steve Conner

Mon Jun 23 2008, 01:22PM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

Dr. Kilovolt wrote ...

Do you think a MOT can hold up 16kV?

I bet you guys could have a lot of fun finding out

BTW, didn't Neil say that his circuit could be inverted so that it had a series capacitor and two shunt inductors to ground? That would solve the capacitive arcs problem.

GeordieBoy

Mon Jun 23 2008, 01:31PM

Registered Member #1232 Joined: Wed Jan 16 2008, 10:53PM
Location: Doon tha Toon!
Posts: 881

Ahhh, I remember discussing this with Neil and Paul Benham in Nottingham over a few pints.

I agree that the "T" variant (LCL) of this "impedance-inversion" circuit would probably work better that the pi-model (CLC) with an arc as the load. The pi model would be apt to discharge its output capacitance straight into the arc's negative resistance in a blinding flash and bang!

-Richie,

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