model/16SVPF1000M
It's not just the gate voltage that is ultimately achieved. The speed at which it gets there is also important.
You may well be fine -- but you should check.
-- Tim Wescott Wescott Design Services http://www.wescottdesign.com
He can have it in parallel with the mosfet, and leave the mosfet off while C charges.
ripple spec is continuous, inrush is once now and then so can exceed ripple greatly. Just beware with tants!
its the worst case option, gives highest peak current
laekage is tiny compared to other currents, and can be completely ignored here
plus mosfet R, plus source R, plus trace R. Load current will also effect the picture a lot.
I'm not convinced you need worry about them
If you still wanted a surge handling resistor, I'd be tempted to use a resistance wire link. A filament bulb is a less elegant option. I'm not at all convinced you need it though.
NT
Why do so many people here feel the need to add personal insults to objective technical discussions? Such people clearly don't want to be helpful; they just want to insult others.
Go back to basic manners.
-- John Larkin Highland Technology, Inc picosecond timing laser drivers and controllers jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
If you mean 3.4mA through the 4.7ohm resistor, how much voltage would that be? I calculate 16mV (actually twice that because two caps are in parallel, so maybe 32mV)
A sudden change of 32mV will not be observable and cause no problem.
Do you have a schematic of your proposed circuit that you could share with us?
What are your load characteristics? What are your source characteristics? Have you thought about the fact that you have only 6m ohm of ESR in the two caps and only 1.4m ohm in your FET? What is the resistance of your wires/traces? Have you analyzed that? They might be enough to limit without additional circuitry.
This is actually an age-old situation that is easy to satisfy.
model/16SVPF1000M
You don't need a resistor that's specifically made for the service, as long as the data sheet tells you what it'll do.
Of course, if that's ALL you want the resistor for, that may be the cheapest option.
-- Tim Wescott Wescott Design Services http://www.wescottdesign.com
As low as it is, for this high capacitance value capacitor calculation shows it is a potential problem. If the resistor value is too high, as it easily can be, the ripple current specification will be exceeded.
-- To reply directly remove sj. from the domain name. This is a spam jammer.
https://en.wikipedia.or/wiki/Psychological_projection#Practical_examples
Subsection "Bully"
I am designing an inrush control circuit to prevent a pair of electrolytic bypass capacitors exceeding their rated ripple current when power is suddenly applied. When the capacitor is fully charged it will be grounded by a MOSFET. In parallel with the MOSFET there is a current limiting resistor.
I do not know what the rise time of the power supply will be. I do know I must protect against its being plugged in while turned on. So I must assume the voltage change across the capacitors is instantaneous.
The resistor value is determined by the capacitor leakage current, and maximum ripple current. Once the capacitor is fully charged its leakage current will drop voltage across the resistor. Then when the MOSFET is turned on, current will be limited only by the capacitor's ESR. So the upper limit of the resistor is the value that will cause no more than the rated ripple current to flow when the MOSFET is turned on. When the MOSFET turns on the voltage that was across the resistor will appear across the capactor's ESR value. So the upper limit is:
R = Irip * Resr / Ileak
For a safety margin it will actually be no more than half this value.
The amount of peak wattage will be high, roughly 30 Watts, but it will be a very short pulse that happens once. I know that due to the shortness of this dissipation, which is over well before thermal equilibrium is reached, it is not necessary to have a large and expensive 30 Watt resistor there. But over what amount of time should this wattage be averaged to determine the resistor's wattage rating? I know that heat capacity figures into it, but that data is not available to me.
I figure the resistor to be about 5 ohms. It will be a surface mount one.
The capacitors are a parallel pair of 16SVPF1000M
Resr 12mOhm Irip 5.4A Value 1.00mF Tolerance 20% Ileak 3.20mA
The MOSFET is a CSD17312Q5
-- To email me directly remove sj. in my email address. Ok, Sounds like you have a 12V input, 100W or I=p/e= 100/12=8A so you want to charge the 2mF cap at less than the max input current. So R=12/8 = 1.5R, so let's use a 2.0R charging resistor. Now T=R*C = 2R*2ms = 4.0mS charging time. And Ec = C*V^2/2 = 2m*144/2 = 144mJ for 4.0mS or 36W peak. Change the above to meet your requirements. Now, find a resistor that can meet these requirements. Well, I can't do everything! (Yes I found a SMD at 2 W) Harry D. Harry
Iripple^2 * Rser is the ripple-driven power dissipation in watts, where Iripple is amps RMS. That's the thing that cooks caps.
Ileak * V is the steady-state power (cap heating) caused by leakage, but will be so low that it won't contribute meaningful power dissipation. So I don't understand the equation above.
For a single event, very fast charge or discharge, the energy dumped into Rser is 0.5*C*V^2, which will be about 130 mJ for 1 mF and 16 volts, independent of the ESR value. That's tiny.
I don't think you need to do anything.
Wet aluminums start to draw more current as the voltage increases. Polymers draw very little current until roughly 1.5x rated voltage, where they suddenly fail, permanently.
-- John Larkin Highland Technology, Inc picosecond timing precision measurement jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
That was my thought. If it was higher voltage you can get some acry sparkies. (48 V - making pit's on contacts.)
12 V.. not so bad.George H.
That was my thought. If it was higher voltage you can get some acry sparkies. (48 V - making pit's on contacts.)
12 V.. not so bad.George H.
I calculate 21 kilowatts peak in Resr if you dead-short the cap at 16 volts. The mosfet will limit the peak current (the fet datasheet isn't clear how much) so the peak power will be lower, a few kilowatts internal to the cap maybe.
Still, that won't hurt the cap. There's not much energy available. The cap's internal time constant is short, about 12 us.
-- John Larkin Highland Technology, Inc picosecond timing laser drivers and controllers jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
My impression of him is that of a inexperienced recent EE grad who thinks he knows everything. It is, after all, mathematically correct that you cannot have a step voltage change across a capacitor. Everyone knows that of course. Right? :)
-- To reply directly remove sj. from the domain name. This is a spam jammer.
On Mon, 13 Jul 2015 08:47:21 -0700, Robert Baer Gave us:
Grow the f*ck up, punk.
That is an interesting part, and I appreciate your bringing it to my attention. I might use it in other designs. But in this instance it will substantially increase the series resistance of the bypass. I am going to stick with the CSD17312Q5 and a parallel resistor.
-- To reply directly remove sj. from the domain name. This is a spam jammer.
On Mon, 13 Jul 2015 09:53:09 -0700, John Larkin Gave us:
Good to see that you are trying to get back on an adult track.
On Mon, 13 Jul 2015 09:19:45 -0700, Artist Gave us:
snip
Just ignore the idiot.
Thankyou all for your comments. I have settled on a 6.8ohm, 5W, Wirewound resistor: SMW56R8JT
Initial inrush current will be 1.8A Peak dissipation will be 21W The second inrush when the MOSFET turns on will be 890mA.
The wattage rating is overkill, but it is a for sure safe one.
-- To reply directly remove sj. from the domain name. This is a spam jammer.
How is the energy stored in the capacitor, and wattage necessary to put the energy there within the charge time, relevant here?
I never insult people who ask sincere technical questions, even if they are very silly. I try to never turn a technical discussion personal. Some people here do turn technical discussions personal, example above, which starts flame wars.
-- John Larkin Highland Technology, Inc picosecond timing laser drivers and controllers jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.