Capacitor discharge question

"Mook Johnson" >

** The ESR figure in the table ( 1.8 ohms) is clearly stated to be a maximum value and at 120Hz.

You did not read this ??

The actual ESR figure will be lower and will fall at higher frequencies than

120 Hz - by a factor of several times.

With an ESR of say 1 ohm total for the bank, the peak surge current into

0.1 ohms will be in the order of 600 amps or 100 amps per series string.

The tantalum caps I know about will not survive that abuse, even once.

Ordinary electros are well able to survive it.

....... Phil

Reply to
Phil Allison
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?? The first page of the linked PDF (pg 73) is vs temperature; the following several pages are all ESR and Z vs frequency. Perhaps the file got munged when you downloaded it??

Cheers, Tom

Reply to
Tom Bruhns

Phil says, "The tantalum caps I know about will not survive that abuse, even once," referring to the higher current you are likely to see with the lower actual esr for real parts, and the lower esr at the higher actual frequency of a short- discharge event. Can you tell us more about your problem, and the event that's got you worried. E.g., how is it that you can place 0.1 ohms in series with the cap(s)?

Reply to
Winfield

Its a motor drive application where the phase wiresmay be shorted phase ot phase or phase to shield (ground).

I know this puts a IGBT or two in series but I'd like to be able to backup my claim that even a direct discharge into the .1ohm resistor would not be sufficient to blow it open. Another design (not mine) had this happen and everythign connected across that resistor (diff amp, +12V rail, etc) got exposed to 500V with the obvious results (BIG cloud of smooke and red faced engineer).

The PCB will also have very stout traces throught the entire current path between the 0.1 ohm resistor and ground.

Reply to
Mook Johnson

?? I have some wet-slug tantalums in my collection of parts. I just connected an SCR directly across one, a 68uF 60V part. I connected a

10k R from anode to a 58V power source, return to cathode. An HP3312 function generator set to generate pulses feeds the SCR gate through a 3.3nF cap (to keep gate drive short), with a 56 ohm termination gate to cathode. Terminal voltage on the cap decays with about 0.8usec time constant; the SCR current is low enough that it turns off within about 1.5msec after the trigger. It's been pulsing away at about 2 seconds per pulse for several minutes now, and no change in performance that would indicate the cap "not surviving." I haven't measured the ESR of the cap I'm testing; suppose I could do that without too much trouble, but not sure I can be bothered.

What I know of solid tants with normal ESR tells me that they survive this sort of thing just fine if you don't do it too often, too many times. YMMV; I suppose it depends on the quality of part you use. OTOH, it's not a good idea to do it with low ESR solid tants.

Cheers, Tom

Reply to
Tom Bruhns

Thanks for the testing, I'll be doing some spark testing in the near future but the guys at Vishay/Tansitor are not afraid of shorting their capacitors every blue moon.

Should be some interedting times in the lab.

Just out of curiosity, why use an SCR for this test? wouldn't an IGBT or MOSFET be easier? I guess its all the same but I haven't used SCRs in a design yet so I'm not comfortable with them. Any advantages to SCRs in this application?

Reply to
Mook Johnson

"Mook Johnson"

** SCRs require only a brief ( few uS ) pulse to be triggered into a low resistance state.

When the voltage across the SCR falls to circa 1.5 volts - it turns off by itself.

So, they are widely used in strobe lights and camera flashes.

........ Phil

Reply to
Phil Allison

I've used SCRs before and never had a problem - they're really simple to fire - just put the trigger current through the gate. The "advantage" is you don't have to continue to drive the gate - it latches on, like a crowbar circuit, and stays on until the forward current drops to zero. I don't know if that's an advantage in your app, but I'm guessing it's some kind of crowbar so a fault doesn't blow up the whole unit; in that case an SCR would be just the ticket. :-)

Cheers! Rich

Reply to
Rich Grise

An SCR is easy to trigger, and once triggered it turns on and stays on till the current falls below the part's holding current. I used one that I have previous experience with, and know it ramps from very low current to essentially full conduction in under 100 nanoseconds. Though a power MOSFET could be a bit faster, this should be adequately fast for the 'speriment at hand. I'm probably exceeding the rated max pulse current of this particular part in this test, but my experience has been that it can take it OK, at least for short pulses. It was easier for me to set up the test with an SCR than with a power mosfet (though I certainly could have used an IRFP4229 I have handy; rated at

180A max pulse current). With the mosfet, I'd have had to arrange a low-impedance gate drive to get high speed. With the SCR, I could use a 50 ohm output impedance generator and a simple coupling cap (3.3nF) to get short turn-on pulses.

Cheers, Tom

Reply to
Tom Bruhns

Check the datasheet for how much trigger current to use. Too much is obviously bad. Too little is bad too if the anode current rises quickly. What happens when you use a low trigger current is that you only trigger part of the device and then the conduction spreads out until the whole device is on. If the anode current rises quickly, a small portion of the whole device tries to take the whole current.

There is a small "holding current" number that the current must fall below before the device will turn off. The current doesn't have to really stop. The turning off process is slow. If the anode current only briefly stops (lets say 1uS) the device may not turn off.

When fired an SCR looks very much like a normal Si diode that is conducting. The forward drop follows the same sort of rule as a rectifier of about the same size would.

I don't know if that's an advantage in your app, but I'm

Reply to
MooseFET

They usually recommend a trigger of the initial pulse of high current followed by a pedestal of lower current. It is even better to trigger the SCRs by a burst of short pulses of high current to the gate. This allows for the quickest transition of the whole structure into the conductive state without risk of damaging the gate. The PWMs on some of the motor control MCUs are supporting for this mode of operation.

I'd say 1uS would be a good time for the ordinary SCR to turn on. The relaxation time is at the order of hundreds of microseconds.

There are two forward biased junctions hence the drop is twice as big as that of a diode. Usually around 1.5..2 Volts.

Vladimir Vassilevsky DSP and Mixed Signal Design Consultant

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Reply to
Vladimir Vassilevsky
[....]

That depends a lot on the type of SCR and how it gets down to the holding current. If the decrease is rapid, there is a huge stored charge and it will be a longer time than if the decrease is slow.

This isn't true. Observe Figure 9 in

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and figure E6.19 in

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Compare with Figure 1 in:

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The simple model of the SCR with two transistors is not accurate but lets consider it:

-------------+------- Anode ! e\\! PNP !----- /! ! ! ! ! ! ! !/ Gate----+-----! NPN !\\e ! -------------------+--- Cathode

Notice that the EB junction drop is in series with the collector of the other transistor in each case. There are two diodes in there but they are more like in parallel than series. This makes the on voltage far less than two diode drops and a lot more like the forward drop of just one diode.

In actual fact the SCR when it is on is not really like the two transistors any more. Internally there are 4 layers like this:

PPPPPPPPPPP < heavy doping PPPPPPPPPPP nnnnnnnnnnn < light doping nnnnnnnnnnn ppppppppppp < light doping ppppppppppp NNNNNNNNNNN < heavy doping NNNNNNNNNNN

Unlike a real transistor each on in the model has a lightly doped collector. When the SCR in in heavy conduction, there are many more carriers in the lightly doped parts than would normally be the case.

Reply to
MooseFET

You are right.

I am aware of the voltage drop of thyristors; what I did not expect is that there are some diodes with the voltage drop that high. Thank you for the references. I checked some other similar diodes and they are rated at Vf =

1...1.2V at max. current. Would it be the correct assumption that the diode drop is considerably less than that of thyristor?

V = Vbe + Vce. Not like one diode, but neither like two diodes.

[...]

I see. Thank you for the detailed explanation.

VLV

Reply to
Vladimir Vassilevsky

The package size and ability to get rid of heat are what sets the maximum power of semiconductor parts. As a result, the rated maximum current of a schottky will tend to be higher than for a normal rectifier but the voltage at the rated curent not as much lower as you would expect.

In a given package size, the power loss in a SCR will be about the same as the rectifier's. This makes the curves end up being scaled to me more alike.

Reply to
MooseFET

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