I am thinking that I could use capacitors in parallel too my 11.1v 3s2p 15000mAh 100c Continuous 200c burst Lipo battery for my spotwelder, so that I don't abuse the battery as mutch. Only problem is that I know very little about capacitors and don't know how many Farads I would need for this too work. As the battery has 1500A/3000A output atleast what specs say, doubt that they count shortcurcuit current though. The burst rating might be the same as shortcurcuit as it is the max it can handle. But I digress, So the caps would need too be able too handle that current and as the battery is 11.1v 3000A max then that is 33300w. But I can't find any info on how many farads I need too keep the same output. The closest I came was with an online calculator wich sayd 3000F for this too work. But I am also scared that the caps will push more than the battrry and blow my welder so please someon help me here as I don't want too blow things upp.
Parallel use of capasitors.
- Thread starter Unitox
- Start date
Chalo
100 TW
3000 farads? That's not happening. Are you talking microfarads or millifarads?
In your situation, I'd experiment and add capacitance in stages until I got the results I wanted, rather than trying to guess and then rolling with whatever number I came up with.
Why can't you use cells that are big enough to do the job you have in mind? Then you wouldn't have to mess around with welding lipsticks together.
Capacitors control voltage spikes. Your original post appear to be related to current spikes.
A-DamW
1 kW
I use 6 of these in series, paralleled to a small lead acid, to start a 5.9L cummins diesel engine (yes, 3000 farads):
You could connect 6 in series, of the above capacitors, paralleled to the lipo's.
Inductive kickback, causing voltage overshoot, is typically what kills mosfets.
Inductive kickback comes from supply voltage sagging under load, then spiking when the load is removed.
The less voltage sag you have, the less voltage spike you will have.
This also known as "Supply Ripple"
Imagine you are pulling a load, someone in a wagon, but you are pulling the wagon with an elastic rubber band,
and the rubber band gets released at full stretch/load(voltage sag), the energy stored in the rubber band has to go somewhere, ouch.(voltage spike)
Now pull the wagon with a steel rod, the elastic deformation(stored energy, voltage sag) of the steel rod is so low, you don't have to worry about the whiplash.
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Chalo
100 TW
I use 6 of these in series, paralleled to a small lead acid, to start a 5.9L cummins diesel engine (yes, 3000 farads):
I would love to see what 6 of those in series would do to a vape pen cell.
A-DamW
1 kW
Well, since this is a place of learning, let's have some thought experiments.
If the 6 capacitors in series were charged to 3.7 volts, and the vape pen cell was at 3.7 volts, then electrically paralleled:
nothing, nothing would happen, there is no potential difference to equalize or vaporize anything.
If the capacitors where charged to 16 volts, then connected reverse or same polarity to the vape pen cell at 3.7 volts:
well, we don't know, because I probably died trying it.(hint, lots of fire)
If the capacitors where charged to 11.1volts, and a 3s2p 15000mAh 100c Continuous 200c burst Lipo battery was charged to 11.1volts,
then electrically paralleled, and connected to a spot welder of reasonable construction,
then connected to one end of a vape pen cell through nickle tabs for a few milliseconds:
a spot weld!
@Unitox , What kind of spot welder do you have?
Chalo
100 TW
Assuming the spot welder is a simple timer circuit, and can handle whatever current the caps can deliver, it could blow a crater in the cell casing.
6 of those caps (not necessary, because 5 would do) would reduce their capacitance to 500F. At 11.1V, that's around 30,800 joules, or enough to heat 74ml of water from 0C to 100C. So that timer circuit had better be fast.
CamLight
10 kW
No sagging needed. The voltage spike is created when the current flow stops but wire/circuit inductance wants to force the current flow to continue. The voltage rises in an attempt to do this, creating a spike.
It’s all about the inductance, not the sag. More inductance and/or more current = larger spike.
if the spike is larger when there’s more sag it’s not due to the magnitude of the voltage sag. It’s due to the higher current.
This is completely different. In a switching supply the input or output filter caps help smooth the discontinuous current flow. But it takes a crazy amount of capacitance to get it down to almost zero so some voltage/current ripple is always left. The circuit designer can balance the cost and size of the caps with the ripple requirements. It’s not related to circuit inductance.
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