# Discharging High Capacitance (6F @ 48v)

• YoungEngineering
In summary, a power resistor would not work as a solution because it would not be able to dissipate the power fast enough. You could try using a light globe but I would be surprised if it didn't go bang.
YoungEngineering
This is my first post and am curious if anybody here has any experience or advice on how to discharge a capacitor bank with 6 1F capacitors connected in parallel. I built a electric hybrid (basically electric transmission) vehicle that uses a 31hp gas engine to power a 23kW (peak) DC generator. The capacitor bank is in parallel with the generator and two VFDs and each of those is connected to a 16kW AC motor. I am currently letting the VFDs drain the capacitor bank when the engine is turned off, however that takes about 2 minutes and I am not sure if all the power is gone or the LEDs just stop blinking on the VFDs.

I am looking for a quick way to safely discharge the capacitor bank if the engine and VFDs are turned off. I was looking a powered resistor but I couldn't find one big enough. I also thought about using a relay but am worried about the instant power release. I trying to keep my power/weight ratio high so a solution (if there is one) needs to be light and mobile.

One idea would be to use an ordinary household light bulb=They run at 110 volts AC, so they should easily be able to handle 48 volts DC.

YoungEngineering
I had thought about that but my knowledge about capacitors is very limited. I originally thought about putting in a light switch with a light or heat element. My concern was when the connection was made or switch flipped, would it arc and fuse the contacts together or fry the connection due to the high amperage? Ideally I would have liked to use some type of heated wire I could place in ceramic container to keep the other components away from the heat.

The electrical resistance of a light bulb will depend upon the wattage, but will be in the range of about 100 ohms. That would give you a relatively long time constant ## \tau=RC ## of 600 seconds or thereabouts. I don't know of any good way to dissipate it much quicker, but I'm not a EE and haven't worked with that type of power levels very often.

YoungEngineering
YoungEngineering said:
My concern was when the connection was made or switch flipped, would it arc and fuse the contacts together or fry the connection due to the high amperage?
the amount of current that will flow is dependent on the circuit resistance

YoungEngineering said:
I was looking a powered resistor but I couldn't find one big enough.

you probably meant power resistor

you can work out the current that will flow using Ohms law

eg ... 48V across a 10 k Ohm resistor = 0.0048 Amps

power to dissipate = 48V x 0.0048 A = 0.2304 Watts ... let's say 1/4 Watt
so you now know you could lower the resistor value a bit to decrease the discharge time and increase the power dissipated
20W power resistors are easily availableDave

YoungEngineering
So I very checked (out of slight fear) if the capacitors were actually empty after the 2 minutes the VFDs power LED turned off. But it only took 2 minutes for that to happen. The drain of power was also from just keeping the VFDs computer on and not actually driving it, so I wouldn't think it would require much power.

But you think just putting a relay or house light switch to a 100w bulb would work?

Maybe I misunderstand the power resistors then. Because the spec sheet lists this one at 8 ohms and 20w. That wouldn't be enough resistance correct?

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YoungEngineering said:
But you think just putting a relay or house light switch to a 100w bulb would work?

I would be surprised if the light globe didn't go bang
It would depend on what the filament hot resistance value would be

YoungEngineering
YoungEngineering said:
Maybe I misunderstand the power resistors then. Because the spec sheet lists this one at 8 ohms and 20w. That wouldn't be enough resistance correct?

correct

do the Ohms law thing that I did for the other example and show your working and results

there would be a flash of light and a cloud of smoke

YoungEngineering
I have been looking for a solution for around a year now and haven't found a real solution. I don't think a power resistor is a possible solution from the products I have found. I would make a heated coil but am nervous about not getting a high enough resistance and watching something catch fire or explode.

YoungEngineering said:
I have been looking for a solution for around a year now and haven't found a real solution. I don't think a power resistor is a possible solution from the products I have found. I would make a heated coil but am nervous about not getting a high enough resistance and watching something catch fire or explode.
you haven't done the maths that I suggested to see the difference between the two resistor values
I showed you my resistor value would work, but that you could lower the value somewhat and increase the
discharge rate

YoungEngineering
I understand the math that you showed and that would work. However a product with those specs does not exist.

48v * 2.2 ohms = 105.6 amps
48v * 105.6 amps = 5068 w
5068w > 1000w rated power resistor = Not possible

davenn
YoungEngineering said:
I understand the math that you showed and that would work. However a product with those specs does not exist.

48v * 2.2 ohms = 105.6 amps
48v * 105.6 amps = 5068 w
5068w > 1000w rated power resistor = Not possible

48 / 500 Ohms = 0.096 A

48V x 0.096A = 4.6W

http://au.mouser.com/Passive-Components/Resistors/Wirewound-Resistors/_/N-7fx9f?P=1z0x8b9Z1z0wn9mZ1z0x6d8

48V / 150 Ohms = .32 A

48V x .,32 A = 15.4 Watts

http://au.mouser.com/Passive-Components/Resistors/Wirewound-Resistors/_/N-7fx9f?P=1z0x8ajZ1z0wn9mZ1z0x6d8

finding a suitable resistor isn't your issue ! it's looking like a 100W 110V light globe as Charles suggested may well work
Im just concerned with it's significantly varying resistance with temperatureDave

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YoungEngineering
A light bulb works really nice because it shows you when discharge has finished.

You don''t want to be switching a hundred amps with everyday switches
but ten or so is no problem.

There exist 48 volt headlamps for golf carts, seems like a natural fit for your EV
search turns up plenty of them

but it would be a lot cheaper to use an everyday 500 watt 120 volt halogen lamp, like those used in torchiere floor lamps.

http://www.homedepot.com/p/Philips-...ded-4-7-in-Light-Bulb-2-Pack-415729/202800575
you could buy a socket in the lamp parts section of your hardware store, or peruse local yard sales and curbside rubbish piles for a whole lamp .

davenn
Face Palm... I get so excited thinking about the project I jump steps and messes me up. Thanks Dave!

I would really like to have this light/resistor be controlled from my computer controller that also takes in joystick position, key switch,... etc. Do you guys think a simple relay would be able to handle it?

I have always been under the impression that the capacitor bank was basically a bomb when it discharged and I needed something heavy duty to control the explosion.

YoungEngineering said:
I have always been under the impression that the capacitor bank was basically a bomb when it discharged and I needed something heavy duty to control the explosion.

only if you short circuit the terminals ... but that is the same as a car battery etc

using a resistor or lamp ( appropriate values) gives a controlled discharge

YoungEngineering and jim hardy
davenn said:
using a resistor or lamp ( appropriate values) gives a controlled discharge

Incandescent lamps when cold have resistance about 1/10 of hot value

so inrush can be significant

a 500 watt 120 volt lamp should be around 28 ohms hot, 3 ohms cold
at 48 volts that'd be 16 amps inrush to a cold filament.

YoungEngineering said:
Do you guys think a simple relay would be able to handle it?

Microwave ovens and coffee percolators use a ten amp relay that should do the job, often with a 12 volt DC coil - take a dead one apart and study the relay, find its datasheet.

Have fun ! old jim

YoungEngineering and davenn
jim hardy said:
a 500 watt 120 volt lamp should be around 28 ohms hot, 3 ohms cold
at 48 volts that'd be 16 amps inrush to a cold filament.

that's what I was concerned about ... hence why I prefer the power resistor approach

YoungEngineering and jim hardy
I just looked at the relay board i salvaged from a junkyard microwave oven, for a project
has a 12Volt power supply and a handful of relays on it.

Main relay contacts are marked 20 amps 120V~, 17A 250V~. coil is 12VDC. It's TYCO 0MIF-S-112LM

http://pdf1.alldatasheet.com/datasheet-pdf/view/228047/MACOM/OMIF-S-112LM.html

it should do the job. But it's not rated to turn the lamp off against your full 48 volts DC, so only use it for discharge duty when incoming power is off, or protect the contacts..

YoungEngineering
davenn said:
it's looking like a 100W 110V light globe as Charles suggested may well work
Im just concerned with it's significantly varying resistance with temperature
A light globe is self protecting if selected for a voltage greater than or equal to it's voltage rating.

The resistance of a tungsten filament is proportional to absolute temperature. From 300°K at room temperature to operating at about 3000°K there will be a factor of 10 change in resistance.

As an example of the computation. A 115 V globe rated at 60 watt will have a “bright” current of 60W / 115V = 0.522 amps. The hot resistance is then 115V / 0.522A = 220 ohm. When cold it will have a minimum resistance of 22 ohm.

When first turned on to the full rated voltage there should be a very short current peak reaching about 10 times the final steady current as the filament heats. As the capacitor voltage falls the filament will cool, so the resistance will fall, but the current will also be falling due to the lower voltage.

YoungEngineering
It's OK that you want to discharge it fast, but this caliber of capacitors often has some 'maximal discharge current'. Was this parameter considered in this task?

YoungEngineering said:
I understand the math that you showed and that would work. However a product with those specs does not exist.

48v * 2.2 ohms = 105.6 amps
48v * 105.6 amps = 5068 w
5068w > 1000w rated power resistor = Not possible
V=IR
so I = V/R
Your current calculation is a bit out of kilter!

Can you find a suitable heating element of ~ 2 Ohms? There are cheap 110V heating elements for cook tops. I don't think those have as much in-rush current as a light bulb, not such a great temperature delta. Water heater elements also, but if used near their rating for more than a few seconds, will need to be in a water bath. Or just a suitable length of ni-chrome wire?

Or 100' of 22 AWG wire is ~ 1.6 ohms (correction, I had 16 at first). So 125' for 2.0 ohms. Only rated for one amp continuois, but check fusing currents - it might take much higher for a short time. Your V (and therefore I) will be decreasing exponentially with time. I see 22 AWG rated at 41.5A fusing, so that could work (check the numbers from sophiecentaur above I = V/I).

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This thread has been drifting for a while now without specification of the maximum permitted capacitor current. What is the make and model of the capacitor? How many capacitors are used in parallel?

jim hardy
sophiecentaur said:
V=IR
so I = V/R
Your current calculation is a bit out of kilter!

sophiecentaur
Baluncore said:
This thread has been drifting for a while now without specification of the maximum permitted capacitor current. What is the make and model of the capacitor? How many capacitors are used in parallel?

He did state the cap arrangement in the OP. " 6 1F capacitors connected in parallel". And 6F @ 48V is in the title.

But you are correct, we don't know if these can handle the peak discharge currents. OP - give us a part number and/or spec sheet.

## 1. What is "discharging high capacitance"?

Discharging high capacitance refers to the process of releasing stored electrical energy from a capacitor with a large capacitance value.

## 2. What is the significance of 6F @ 48v in this context?

The value of 6F @ 48v represents the capacitance (6F) and voltage (48v) of the specific capacitor being discharged. This information is important for determining the amount of energy being released and the potential hazards involved.

## 3. Why is it necessary to discharge high capacitance?

High capacitance capacitors can store a large amount of electrical energy, which can be dangerous if not properly discharged. Discharging the capacitor reduces the risk of electrical shock and damage to equipment.

## 4. What methods can be used to discharge high capacitance?

There are several methods for discharging high capacitance, including using a resistor to slowly release the energy, using a discharge tool to quickly drain the energy, or short-circuiting the capacitor with a wire.

## 5. Are there any safety precautions to take when discharging high capacitance?

Yes, it is important to follow proper safety precautions when discharging high capacitance, such as wearing protective gear and using appropriate tools. It is also important to discharge the capacitor slowly and carefully to avoid any potential hazards.

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