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Critical damping versus diode protection?

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LAN_403

Bonehead

Fri May 15 2009, 04:49PM

Registered Member #2046 Joined: Sun Mar 22 2009, 01:12PM
Location: Stockholm, Sweden
Posts: 23

I've been pondering this for a few days.. How to get the best waveform when firing a coilgun.
I did a few simulations with Barry's RLC sim and found out that if you simply discharge directly into the coil you get a very bouncy wave whilst if you get it critically damped you don't have to deal with the negative "backflows"..

But this brings the question, how would the discharge look if I used diode protection?
My very non-scientific theory is that it'd be similar to the critically damped one except possibly with a longer discharge but would there be any way to calculate just how long if this is the case?

Edit: spelling fixes

big5824

Fri May 15 2009, 05:15PM

Registered Member #1687 Joined: Tue Sept 09 2008, 08:47PM
Location: UK, Darlington
Posts: 240

I think the wave would just fall like normal, then when it hits 0 it would just stop.

Backyard Skunkworks

Fri May 15 2009, 06:26PM

Registered Member #1262 Joined: Fri Jan 25 2008, 05:22AM
Location: Maryland, USA
Posts: 451

Exactly, the discharge will be no longer, because there is no increase in resistance in the path from the cap bank to the coil.

The negative side of the ring down will still be there to a very small extent, with the peak reverse voltage being equal to the diode foreward voltage.

klugesmith

Sat May 16 2009, 01:54AM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1715

Have you read in the HvWiki about coilguns and Quenching?

You ask "what would the discharge look like if we use diode protection?" and show us simulated current waveforms for underdamped and critically damped cases. But your circuit with the diode has two loops, so -three- different current waveforms to consider. Let's call them Ic, I, and Id (for the cap, inductor, and diode currents).

You might also want to sketch in the voltage waveform. That leads the current by about 90 degrees. Very roughly, the voltage crosses zero when the current reaches its maximum, at which point the diode (if present) turns on. Without the diode, the voltage reaches a minmum (of about -500 volts) when the current reaches zero at 2.92 ms.
Finer analysis depends primarily on where the 255 milliohm resistance goes in your model -- there will be about 200 volts across that resistor when the diode turns on, and the meaning of "voltage waveform" depends on which pair of nodes is measured.

Suppose we lump the 255 mO together with the 700 uH, and take the diode to be ideal (not a bad approximation).
Then in the first quarter cycle, Id = 0 and Ic = I = same as "simple coil" case.
At about 1.5 ms the diode turns on and conducts about 800 amps while the capacitor current suddenly stops.
I and Id will then exponentially decay with a characteristic time of L/R = 2.75 ms.
(The "braking" voltage L * di/dt comes entirely from the parasitic resistance.)
I think this is not a wildly unrealistic model, given the values in your example.
You can put a resistor in series with the diode to speed up the decay, at the cost of more voltage reversal at the capacitor.

Suppose we separate the 255 mO and move it to the "bottom" leg of the schematic, leaving (in parallel with ideal inductor) a diode with constant 1V forward drop. In this case, after the current switches from the capacitor loop to the diode loop, it would decay linearly to zero in about 1/2 second. (The "braking" voltage is only the diode drop.) This is not a physically realistic model, not so much because of the diode as because your amateur, non-superconducting, air-core coil will have L/R much less than 1/2 second.

Hope that helps. You can get some insight by learning to use a more general circuit simulator such as Spice.

Rich

Barry

Sun May 17 2009, 05:58PM

Registered Member #90 Joined: Thu Feb 09 2006, 02:44PM
Location: Seattle, Washington
Posts: 301

Good questions that strike at the heart of performance design.

Diodes ... At one time I started to add a diode to my RLC simulator but never quite finished. The reason is that performance-wise the protective diode is a bad idea. Once the diode turns on, the inductor current basically decays with a time constant of L/R. Since the coil resistance R is as small as we can make it, this is a slow decay rate.

Critical damping ... This seems like such a great idea to eliminate the diode. If there's no ringing then there's no need for a protective diode. Great. So I built a Mark 4 demonstrator coilgun with exactly this idea. Several inches of straight wire let me easily tune the resistance to achieve critical damping.

The Mark 4 results were okay but not very impressive. The measured efficiencies were around 0.25%.

I ran some equations to compare the peak current for critical damping with that of undamped R=0. It turns out that, in theory, a critically damped circuit will have peak current only about one-third (36%) of the maximum possible with an undamped RLC circuit. Ouch.

As a result, I think I'll look at all sorts of other solutions for capacitor protection, but not diodes or critical damping. Some likely possibilities include a V-switch to quench the coil quickly, or a diagonal half-bridge to recover some coil current. Or I might just let it ring down and see how long the capacitor bank will survive.

Cheers, Barry
I enjoy using the comedy technique of self-deprecation - but I'm not very good at it.

MachineShop

Fri May 22 2009, 02:13PM

Registered Member #2107 Joined: Mon May 04 2009, 05:02AM
Location:
Posts: 12

when using diodes whats the best way to ensure circuitry safety? like one at the begining and end of the coil to ensure no back flow or....

Wessel

Fri May 22 2009, 02:35PM

Registered Member #2127 Joined: Wed May 20 2009, 03:35PM
Location:
Posts: 8

You could ground it and use a full bridge rectifier? You could also protect the bank with diodes and use 2 additional banks with diodes to absorb the rest of the current.

MachineShop

Fri May 22 2009, 06:54PM

Registered Member #2107 Joined: Mon May 04 2009, 05:02AM
Location:
Posts: 12

how about protecting the circuit as a whole? is there some particular way to keep the energy from goin right back into the circuit and smacking it silly?

big5824

Fri May 22 2009, 07:49PM

Registered Member #1687 Joined: Tue Sept 09 2008, 08:47PM
Location: UK, Darlington
Posts: 240

Not too sure what that last post really meant, but the only really important diode is the beefy one in antiparallel to the coil. If you really want to you can try tracking down some huge diodes to put in series with your bank.....but to be honest i think youd just be adding unnecessary resistance to your circuit and wasting a lot of money.

MachineShop

Fri May 22 2009, 08:26PM

Registered Member #2107 Joined: Mon May 04 2009, 05:02AM
Location:
Posts: 12

big5824 wrote ...

Not too sure what that last post really meant, but the only really important diode is the beefy one in antiparallel to the coil. If you really want to you can try tracking down some huge diodes to put in series with your bank.....but to be honest i think youd just be adding unnecessary resistance to your circuit and wasting a lot of money.

understood!

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