Why bother to buy a milliohm ESR polymer cap, and then put a 6.8 ohm resistor in series with it?
You don't need any resistor.
-- John Larkin Highland Technology, Inc picosecond timing laser drivers and controllers jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
They also don't like it much when you reverse the voltage. It takes them a while to fail.
-- Best regards, Spehro Pefhany Amazon link for AoE 3rd Edition: http://tinyurl.com/ntrpwu8 Microchip link for 2015 Masters in Phoenix: http://tinyurl.com/l7g2k48
The only case that I've ever seen of adding a resistor in series with a cap to protect it is MnO2 dry tantalums. High dV/dT, namely high peak currents, can literally detonate them. Of course, there's not much use for a big cap with ohms of added ESR.
People do make tantalums with fuses inside, basic fire protection.
-- John Larkin Highland Technology, Inc picosecond timing precision measurement jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
Excuse me, but very little of that makes sense.
Got a schematic?
-- John Larkin Highland Technology, Inc picosecond timing precision measurement jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
As whit3rd pointed out, it is just a matter of energy.
So you need to charge 2000uF to 16V, yes? The energy that the power source will need to supply is c*v^2 = .002 * 144 = 0.288 joules. The capacitors will get half of that and the resistor will get the other half. The resistor then needs to withstand .144 joules.
Now, you said your resistor selection is a 4-1879233-1 which is a 5W WW. The data sheet shows that particular resistor is capable of 5 times the power rating for 5 seconds. So, that would mean that the resistor is good for 25*5 or 125 joules! Where is the problem?
Because the power supply will rise with an unknown slew rate. There is also the possibility it will be plugged in live. This rapid voltage change will cause the ripple current rating for the capacitor to be exceeded. It is common practice in this situation to design in an inrush control circuit.
Once the capacitor is charged as much as can be with the series 6.8 ohm resistor it is shunted out by the parallel MOSFET, which remains on for the remainder of the power up.
I recommend a web search for "inrush control circuit".
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It is a gate driver designed to drive MOSFET gates. MIC4420YM TR
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Sorry if this has already been suggested, but why wouldn't you simply buy a resistor with a guaranteed surge rating that meets your requirements?
DIY qualification of components may be justified sometimes, but if you can buy what you need off-the-shelf it's almost always cheaper and better.
-- Best regards, Spehro Pefhany Amazon link for AoE 3rd Edition: http://tinyurl.com/ntrpwu8 Microchip link for 2015 Masters in Phoenix: http://tinyurl.com/l7g2k48
Phil Hobbs wrote:
I looked up a suitable "pulse-withstanding" resistor. This is the CRM2512-JW-6R8ELF
It is a much lower cost than the one I selected, and the Pulse Load Characteristic" curve shows my circuit would be within its limits. Unfortunately if I want them I have to buy 4000 of them. This is not feasible for prototype quantities. The cost to acquire 4000 of them greatly exceeds the cost of twelve of the wirewound resistor I selected. Also they are not stocked. I do not have time to wait for a factory order. I appreciate knowing this type of resistor exists though. I might use it later, most especially if it is ever offered in lesser quantities.
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There is no problem because a resistor similar to it is the one I finally selected.
When I started this thread I was going to use a thick or thin film one that would have eventually failed according to DecadentLinuxUserNumeroUno. I changed to wire wound on his recommendation.
The energy absorption is not the only criteria. The resistance has to be low enough so as not to cause ripple current to be exceeded when the MOSFET is turned on.
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capacitors/os-con-aluminum-polymer/series/svpf/AAB8001/model/16SVPF1000M
capacitors/os-con-aluminum-polymer/series/svpf/AAB8001/model/16SVPF1000M
So pick a different one. There are thousands of them listed on that page. For instance, the Stackpole RPC2512JT6R80 is rated for 400W for 1 millisecond. Digikey sells them for 26 cents in hundreds, and has
11,500 in stock.Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
The Guidelines and Precautions document says those caps can take 10x ripple rating as a rapid charge/ discharge limitation. All of your assumptions ar e borderline hysteria. You don't say squat about the voltage source but a s urge limiting resistor is probably not needed.
I have opened a tech support request at Panasonic that asks if the ripple current rating can be exceeded in a one time initial event at power that is very short in duration. I will post the answer in this thread here.
The power supply is a 20A supply. It is a Power Stream PWS310-110-12.
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Current limit works but the output voltage crowbar is 32.0+/-1.0 Volt, which means your 16VDC capacitor is a bad choice. You would want to use a 35VDC minimum. Too bad there are no 1000uF in that voltage, back to the drawing board as they say.
model/16SVPF1000M
model/16SVPF1000M
Digikey has a checkbox in their search page that hides anything that's not regularly stocked.
It's the first thing I click on when I'm doing searches.
-- www.wescottdesign.com
This is a 12V 20A supply. This is sufficiently below the 16V rating of the capacitor.
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Yes, until a part fails and it becomes a 32v psu.
NT
How is the energy stored in the capacitor, and wattage necessary to put the energy there within the charge time, relevant here?
Stephen,
The energy stored in the cap equals the energy dissipated in the resistor in series with the cap. Most of that energy is dissipated in R*C so now you must find peak power in the resistor. so Ppk = Edr/RC = C*V^2/2*RC = V^2/2*R +/-10%
Now, prove me wrong.
Cheers, Harry
How is the energy stored in the capacitor, and wattage necessary to put the energy there within the charge time, relevant here?
Stephen,
The energy stored in the cap equals the energy dissipated in the resistor in series with the cap. Most of that energy is dissipated in R*C so now you must find peak power in the resistor. so Ppk = Edr/RC = C*V^2/2*RC = V^2/2*R +/-10%
Now, prove me wrong.
Cheers, Harry
Ok, ok, So some smart math jockey out there should be able to come up with this much needed exact equation of Peak Resistor Power as a function of V, C, R to charge a cap. No integrals please.
Cheers, Harry
The board has a Vishay Semiconductor 5KASMC12AHM3/57 TVS Diode across the power input. Its stand off is 12V.
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