Making a lightbulb discharger

Hello,

I'd like to make a lightbulb discharger to easily check the "healthyness" of my batteries.

I found these 200W lightbulb:

http://www.ebay.co.uk/itm/GLS-Pearl-Bulbs-100W-150W-200W-In-Bayonet-B22-Edison-Screw-E27-Cap-Fitting-/281316289122?pt=UK_Light_Bulbs&var=&hash=item417fc1ea62

And they can be connected to this holder:

http://www.ebay.co.uk/itm/Edison-Screw-ES-E27-Light-Bulb-Lamp-Holder-Socket-With-Terminal-Block-10M-Thread-/360986919304?pt=LH_DefaultDomain_3&hash=item540c7f4188

Let say now I want to discharge my 48V and 72V batteries:

- How do I know if the voltage is correct for the lightbulb? would that be too high?

- How do I make sure it discharge at the right wattage? I suppose a 100W lightbulb isn't going to discharge at 100W on a 48V and 72V battery?

- Is the blue line for positive and brown for negative?

- Can I put them in serie to increase wattage?

Thank you so much

or maybe one of these? http://www.ebay.co.uk/itm/180971553497?_trksid=p2055119.m1438.l2649&ssPageName=STRK:MEBIDX:IT

That doesn't solve my problem. I need to find the right resistance for my need. Lighbulb are resistance but I don't know how much.

This tool still need to attach resistance to it.

that kit comes with 4 x 200W power resistors.

It's suitable for fast discharging 36V and48V batteries.

http://i1163.photobucket.com/albums/q557/yanghaitao/1_conew1-25.jpg

I think I've found my answer:

http://www.practicalmachinist.com/vb/general/120v-bulbs-dc-148591/index2.html

- AC lightbulb use the same power whether it's on AC or DC.

- So I=P/V = 200W/200V = 0.9I.

- P = V*I = 48V*0.9I= 43.2W.

So each lightbulb would discharge at 43W. If I want to discharge at 400W I'd need to have 10 of them in parallel.

Because it's AC, positive or negative doesn't matter and I can plug the wire on the brown or blue either way.

Is that correct?

Thank you

that kit comes with 4 x 200W power resistors.

It's suitable for fast discharging 36V and48V batteries.

 

http://i1163.photobucket.com/albums/q557/yanghaitao/1_conew1-25.jpg

Thanks I didn't noticed the resistors. But it's going to be way more expensive than my lightbulb option and customs are going to hit me hard when they'll find these big resistors also.

The big benefit of this tool is that it can have a low voltage cut off

you need a discharge current around 4A. The light bulbs with that range of current are 12V 50W halogen bulbs.

divide your battery voltage by 12V, that will be 4 bulbs (3 not enough, the inrush current may damage your bulbs) for the 36V battery and 5 bulbs for the 48V battery.

Put these bulbs in series. You have your first array.

230v light bulbs are not good to use as a resistive load because the resistance changes so much with temperature. At 230v, they heat right up to whitevhot, but at only 72v, they probably won't even glow. The working resisistance might be about 200 ohms, but, when cold, it could be anything from 7 ohms to 2000 ohm depending on the material of the filament. It would be better to use a bank of 12v bulbs, which wiĺ get up to working temperature, or better still, the elament from one of those old electric fires that use a coil of wire around a ceramic bar. You can use a big crocodile clip to change the length and resistance. Maybe something like this:

http://pages.ebay.com/link/?nav=item.view&id=291355163362&alt=web

230v light bulbs are not good to use as a resistive load because the resistance changes so much with temperature. At 230v, they heat right up to whitevhot, but at only 72v, they probably won't even glow. The working resisistance might be about 200 ohms, but, when cold, it could be anything from 7 ohms to 2000 ohm depending on the material of the filament. It would be better to use a bank of 12v bulbs, which wiĺ get up to working temperature, or better still, the elament from one of those old electric fires that use a coil of wire around a ceramic bar. You can use a big crocodile clip to change the length and resistance. Maybe something like this:

 

http://pages.ebay.com/link/?nav=item.view&id=291355163362&alt=web

Try a mains fan heater, at 1000 watts the resistance is 57 ohms, at 2000 watts the resistance is 28 ohms and at 3000 watts the resistance is 19 ohms, so the 3000 watts setting would draw about 4 amps from 72 volt battery.

So you start at 1000 watts, then switch thru to 2000 watts, and finally to 3000 watts so avoiding a surge current.

Or... use an old 3000 watt kettle!

I use a 12v 50 watt inspection lamp to do the same for my wife's 12volt wheelchair battery which draws about 4 amps.

Edited February 7, 201511 yr by VictoryV

This one will give just over 2A discharge with a 72v battery or about 200w when the battery is fully charged. It would probably be a good idea to put a fan on it to keep it cool.

http://www.ebay.co.uk/itm/30R-30-Ohm-200-Watts-High-Power-Ceramic-Tube-Resistor-200W-/190836496143?pt=UK_BOI_Electrical_Components_Supplies_ET&hash=item2c6ebdd70f

230v light bulbs are not good to use as a resistive load because the resistance changes so much with temperature. At 230v, they heat right up to whitevhot, but at only 72v, they probably won't even glow. The working resisistance might be about 200 ohms, but, when cold, it could be anything from 7 ohms to 2000 ohm depending on the material of the filament. It would be better to use a bank of 12v bulbs, which wiĺ get up to working temperature, or better still, the elament from one of those old electric fires that use a coil of wire around a ceramic bar. You can use a big crocodile clip to change the length and resistance. Maybe something like this:

 

http://pages.ebay.com/link/?nav=item.view&id=291355163362&alt=web

Thanks d8veh, again you have an excellent idea.

I found this:

http://www.ebay.co.uk/itm/Nichrome-Wire-Resistance-Nickel-Chrome-Heating-Element-Hot-Cutting-Various-Sizes-/251550106470?pt=LH_DefaultDomain_3&var=&hash=item3a918e0366

It's cheap and good.. and with a resistance table!

The smallest wires 32 SWG have a resistance of 18. So if I buy 2 meter of this I'd have in serie a resistance of 36.

So, I'd discharge at 48V^2/36R = 64W.

Is that correct?

you'd need the FYD10-H for that.

It's very difficult to do a controlled discharge with light bulbs.You double the number of bulbs to halve the Amps and you have to keep an eye on the watt meter.

The nearest solution is a 2KW electric bar heater with battery croccodile clips. You move the clip up to lower the Amps.

I would not use the thermo wires, very difficult to make a bar heater with your own wires.

Is there a simple way for me... to either discharge at 50W or 1000W for my 48V battery? I need to either test at low discharge while creating a pack, then high discharge when the pack is finished

For a 48 volt battery, 4 of these in series to take up the 48 volts will provide a 50 watt load.

For a 1000 watt load u will still need the 4 in series but then 20 sets of the 4 bulbs in series, but connected in parallel !!!! (20 x 50 = 1000)

Cwah

If the device heats up the resistance will change.

So the plane wire combo won't work.

The 12 volt bulb and watt meter combo is the best.

I think I found out the solution, although not as cheap as I'd like:

- Buy 3 of these, so 9 x 100w lightbulb:

http://www.ebay.co.uk/itm/X3-100W-ES-PEARL-48V-50V-GLS-LAMP-LIGHT-BULB-LOW-VOLTAGE-/140874761935?pt=UK_Light_Bulbs&hash=item20ccca46cf

- And these holder:

http://www.ebay.co.uk/itm/10pcs-E27-Socket-Plastic-Light-Lamp-Holder-Base-AC250V-6A-/321491993135?pt=LH_DefaultDomain_3&hash=item4ada6a822f

So if I want low 50w discharge, I can just plug 2 bulb in serie. And for high discharge, I put the 9 bulb in parallel

About £45 for the whole set up

yes but 200w only, I'd need to buy 5 to achieve 1000W so back to £40 expense

And they need fan too. So additional expense.

I was thinking lightbulb may be more convenient/cheaper

Almost very good Cwah. The heating up effect comes into play again

So you cannot say two in series gives 50 watt.

Just do 1 s =100 watt and parallel up for high discharge.

So even cheaper with half the parts nessisary.

that's fine for low power discharge. For high power discharge, the battery when full outputs 54V, may blow your 48V-50V bulbs. I guess you have to try to see how it goes. So, buy one pack.

100W / 50V = 2A

The inrush current can be twice (4A each) as high as the steady current going through the bulbs when they are lit. 9 x 4A can blow the fuse of your battery or worse.I'd start with just 1 light bulb then switch in another, then another...

I just scrapped a cheap electric Heater from my workshop. Inside it is essentially four elements, wired 2s, 2p. I re-wired it to be 4p, and it draws around 6a at 36v. It gets warm, but not red hot.

Another, less destructive method is wire a four way extension lead to your battery, and plug several high load devices into (like kettles, heaters etc.)

Oh, and a watt meter or a multimeter for monitoring....

Four of these, and you make a cup of tea while you wait. You have to keep them in the water.

http://www.ebay.co.uk/itm/Hot-Water-12v-volt-heater-kettle-element-tea-coffee-NEW-/130462652665?pt=UK_Car_Accessories_Touring_Travel&hash=item1e602e18f9