Gen 1 HV BATTERY REBUILD WITH 2G CELLS

Do you know the age and odometer reading of the 2G donor vehicle? It's not reassuring to see those modules leaking slightly.

A flashlight load is not enough to stress the battery module. A flashlight bulb might consume 4W at 12V. Since the module produces 7V, the bulb will probably consume 2W of power or 300 mA of current.

If you want to apply a decent load, I would use a couple of halogen headlight bulbs. They consume 55W each at 12V, so that would result in ~10A drain if you put two modules in series to produce ~14V to light the headlight bulbs. That's enough to stress the modules and allow you to sort the chaff from wheat.

 

If this is a little air compressor that can plug into your aux 12V socket, it probably uses 5-6A. That would be a pretty good load.

 

Actually I was thinking of turn signal, backup, dome or stop lights. The idea was seeing them glow gives a visceral feel of the module SOC and health. But lights are notorious for not having a linear load but very nice for sinking heat. <grins>

The compressor would be a fine substitute and more likely to present a consistent load.

The perfect answer is 2 ohm, high wattage resistor:

7.2 V / 2 ohm = 3.1 A. about 1/2 C.

3.1 A x 7.2 v ~= 23 w SO I'd probably want a 50 W., 2 ohm resistor. But this gets a little pricey for a one-time survey.

GOOD LUCK!
Bob Wilson

ps. If you can, shrink your photos to ~640 width. They'll be a lot smaller and easier to view. But you've got the right idea!

 

Hello UK, Since you seem to be new in Electronics, I do hope that when working around these serious current draw batteries you bought a pair of HV gloves. Also Pat and Bob are using Ohms law equations, that is EIR, (E=I X R) and any variation. I=E/R and R=E/I. Same with the Power formula, which is PIE. Power, in Watts = current, in Amperes times voltage. All this can be really useful when you'll want to catorize and differentiate between modules and put the best combination together. I see you are using a analog meter, which is alright, but a digital readout would be more accurate. As for a practical load, outside of the vehicle, a 220V, filament bulb will work, to test the whole restored pack. The more bulbs you put in series, the greater the load. In the case of Bob Wilson and probably Hobbit too, they have no doubt fabricated a load bank and keep it around for testing.On the analog meter, don't forget and leave it in ohms position and then check a voltage, you'll blow it out immediately ( unless it's protected) I see also you may be working in a apartment with the three modules, I hope you have a partner around, for safety. Watch out for those inadvertant SHORTS. Keep us posted! Andy:cheer2: Sorry, I did not see your little yellow digital meter.

 

I agree with the use of bulbs if it is desired to test the fully-assembled battery outside the car, but the bulbs should be wired in parallel to increase the load. If in series, the load will diminish since the total resistance of the bulbs in a series string will increase.

 

Sorry, many of us have dealt with electronics for a long time and often forget that not everyone has a background in it. I'll explain some of the techno-babble below but I am curious, you have a VOM. Have you studied electricity in school or physics? It isn't ego as much as I want to post things that make sense without 'going down the rabbit hole.' <grins>

Yes, the compressor works great because it will look like a constant load and you'll hear the feedback from how much noise it makes. We do want the voltage across the battery while the compressor is running. The ONLY risk is the compressor motor can sometimes induce small spikes in the voltage. Electric motors are notorious for these electrical spikes. If the needle is steady, there is no problem.

As for image sizes, it is partially a problem with different browsers. I liked your photos for the detail but they were kinda big. Sometimes folks can use photo editing software to reduce the apparent size to something more easily handled.

Now to make sense of the techno-babble:

• Volts - the pressure of the electrons or how strongly they want to move forward. A higher voltage means it will conduct more current through a resistor, like a human or light bulb, than a lower voltage. As a general rule of thumb, voltages above 45-50 V will begin to shock humans by conducting an electric current through dry skin. Wet the skin and a lower voltage can be felt.
• Current - the actually current flow measured in amps, which we usually identify as electrons moving. There is a 'shadow' of an electron called a positive charge carrier but this becomes metaphysical ... the 'hole where an electron should be.' Regardless of how the current is expressed, it is the part that does the work. The more current, the more work it can do.
• Resistance - the property of matter to resist current flow. The higher the resistance, the more voltage is needed.
• C - is a common battery term that is expressed as Amp-Hour, the stored current capacity of a battery. Our Prius batteries are rated at 6.5 Ahr which is 1 C for these batteries. A typical car battery would be say 50 Ahr, so 1 C would be 50 amps of current for one hour before the battery is considered discharged. So when we calculate the current drawn from a battery, one shorthand is the ratio of current to Amp-Hour.
• Power - the watts, is the product of the voltage times the current, and measures the amount of energy in a current flow. So a 100 W light bulb in a 200 V power line would be drawing 0.5 amps: 100 W/ 200 V = 0.5 Amps.

Does this help? Wiki and Google can lead you to other web sites that may more easily explain Volts, Current, Resistance, Power and battery capacity.

Bob Wilson

 

Maybe our terminologys are different Pat, I was using lightbulbs, resistors, compressors as all representing a load on the circuit. Resistors are additive in a series circuit, 3 ea. 1 ohm resistors = a 3 ohm load. 1000 1 ohm resistors = a load of 1Kohm. Now parallel IS totally different as the load seen by the battery pack would be smaller in identical sized resistors In non-identical sized resistors the load is seen as something less than the smallest resistor. I think you may have mixed up series and parallel. Or were you referring to Power in watts as load?

 

Hi Andy,

Suppose you have a battery producing 200V and a bunch of 200V light bulbs whose resistance is 200 ohms each. If you hook up one light bulb across the battery, the current will be 1A and the power consumed will be 200W, based upon Ohm's Law:

Voltage = Current x Resistance
Power = Voltage x Current, or (Current)^2 x Resistance

If you agree with the above, then let's move to an example where you have two light bulbs in series. This means that the cumulative resistance of the two bulbs will now be 400 ohms. The current flow will be 0.5A and the power consumed will now be 100W. Therefore, the series light bulb connection results in a smaller load applied to the battery.

Finally, if you connect two bulbs in parallel, the combined resistance of the two bulbs will be 100 ohms, the current flow will be 2A, and the power consumption will be 400W.

 

Hi Pat, You are correct, I was only speaking of a resistive load in a load bank of light bulbs. I was not equating resistance with the PIE formula. The more bulbs connected to a series circuit the dimmer the overall brightness of the lightbulbs would be as more voltage is dropped at the individual resistances (bulbs) This would be the visual effect. Bob was speaking of using a compressor, which is also good for an aural effect. At some point the 220V light bulbs should be at the same brightness as UK is used to for his normal 220V, AC house current. I think now we would get into ,707 effective value of AC versus pure DC power. Anyway UK should probably still continue to check individual modules, it's safer too.

 

Individual module testing is critical to rebuilding a traction battery pack so it has to be done anyway. Furthermore, whole bank testing is not trival since the packs have an open, safety relay, that breaks them into two banks.

To test all modules at once, a jumper would have to be installed and that would make me give pause due to the safety aspects. It is similar to removing one safety on a system that has lethal hazards. It enables whole pack voltage at each end and leads to a temptation to measure or connect to these exceptionally hazardous voltages.

Bob Wilson

 

Good point.

 

UK, I hope we havn't scared you away with Ohms law, we are all truly interested in your progression. Andy.

 

Yes! You need an RC charger that with the ability to handle:

• 6 cells, NiMH series battery pack (mandatory) - each module has six cells.
• dV charge detector (mandatory) - this means the charger is smart enough to recognize the module has been fully charged and STOPS!
• automatic discharge-charge (mandatory) - the ability to carefully discharge the module to a safe voltage, ~1 V per cell, and then at a constant current, charge until it detect the peak voltage swing, the dV. It should tell you what the Amp Hour (Ahr) capacity of the modules so you can pick the good ones.
• temperature probe (optional) - a backup overcharge protection, this ensures the battery won't thermal runaway

Did you get a chance to see this work?

It is more important to work safely and carefully rather than in a rush and turning stuff into junk.

Bob Wilson