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I want to test my car battery with as little investment as possible. I also want to understand the reliability and working principle behind such a test.

I have access to a multimeter with a min max function. The battery is now fully charged and at 12.6 V. According to this video if the voltage does not drop below 9.6 V when starting the car; it is usable. Does this 9.6 V apply to a diesel as well?

The cheapest stand alone car battery testers are of this type:

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They are typically rated as 100 amps and the display seems to show the voltage. So I am guessing they draw 100 amps and measure the voltage across the load? If so they are merely a more convenient way of doing the multimeter test?

More advanced (conductance) testers like this one:

enter image description here

seems to report both the cold crancking ampere (CCA) and the internal resistance of the battery. How do they do this? Are such more advanced testers more reliable than the older type?

Clearly a high voltage drop under load indicates a larger internal resistance. Is there some way I can calculate the internal resistance of the battery (and the CCA) from this voltage drop (from 12.6V to 9.6V)?

My battery

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A car battery tester is not just a voltmeter; it also contains a heavy duty temporary load. Effectively, the internal resistance of the battery can be determined from the voltage open-circuit, and then under load.

A simplified case: The tester has a 0.1 Ω load, the battery shows 12.6 VDC open-circuit, and 6.3 VDC under load. In this fictitious example, 6.3 VDC is dropped inside the battery, i.e., half, showing its internal resistance is equal to that of the load, 0.1 Ω, and that it can supply 63 A current, but at a terminal voltage likely too low to start an automobile engine (perhaps it might start a lawn tractor).

However, as you state, you can use the starter motor as the load resistor, which is actually more realistic than a one-size-fits-all resistor built into the meter. The 5.2 liter V12 in an Aston Martin Superleggera probably needs a little bit more current to turn over the engine than the 0.841 liter, straight 3-cylinder, two-stroke Shrike engine in a SAAB 96, so observing the voltage under the actual load is more meaningful. BTW, an inexpensive analog voltmeter such as this will give a better reading than a digital (DVM), because the D’Arsonval meter movement smooths the display, while numbers jump too rapidly on most DVM's to be read.

Also note that it is harder to start a Diesel engine than an Otto-cycle engine, because the diesel compression ratio is higher. Some Diesel's have 24 volt systems for extra oomph.

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  • \$\begingroup\$ Diesels do not compress a fuel-air mixture, they compress the air then the fuel is injected as max pressure is reached. \$\endgroup\$
    – Solar Mike
    Commented Jun 6, 2023 at 12:49
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More advanced (conductance) testers seem to report both the cold cranking ampere (CCA) and the internal resistance of the battery. How do they do this?

Well, the answer could fill a book, but essentially they sample the voltage drop due to a known resistance load like imagined and calculate them.

Are such more advanced testers more reliable than the older type?

Not necessarily more reliable than an older type, but likely more accurate than a simpler type of tester. The first tester has a small analog meter, so it would be hard to determine if the reading was 320A or 350A, whereas a digital unit would read out the exact value, say 335A.

Clearly a high voltage drop under load indicates a larger internal resistance. Is there some way I can calculate the internal resistance of the battery (and the CCA) from this voltage drop (from 12.6V to 9.6V)?

If your load resistor is 0.03Ω and connecting this across the battery causes it to drop from 12.6V to 9.6V, then:

  • You have 9.6V across the 0.03Ω load resistor.
  • Ohm's Law states that E=I*R, so I=E/R.
  • 9.6/0.03 = 320A (can't perform this test for very long because things will melt.)
  • 320A * 9.6V = 3kW. Resistor power rating is generally sized to be twice the operating power, so a 6kW resistor would be needed to dissipate this indefinitely.
  • When pulsing a resistor, a reduced duty cycle means a reduced power rating requirement. i.e., a 3kW resistor would be needed if pulsed at 50%, 1.5kW at 25%, etc. So for very short pulse times (ms), a relatively small 100W resistor can be practical.
  • Not all resistors are rated for pulse applications. You couldn't use carbon or metal-film as they would heat up too quickly and vaporize. Generally, resistance wire or wirewound ceramic might be used - something robust and massive, able to absorb the heat generated. A fan might also be used, to keep the resistor temperature low and stable.

A sophisticated tester may pulse this load to get some readings without melting anything, then average the data for more accuracy. It could also try an even lower resistance, almost a dead-short (perhaps 0.005Ω or 5mΩ, whatever the cables and PCB resistance equal) and use that for the CCA reading. But it becomes challenging to get a good enough connection to the battery for such high-current tests. Every piece of wire, alligator clip, solder joint etc. all add some resistance and thus become a weak-point for overheating and failure.

But my guess is that the CCA reading is extrapolated from the normal load-test data. Depending on how the tester was designed, this could be as simple as doubling the value as an estimate of the CCA, or it could attempt to fit the data to a curve and extrapolate a hypothetical 0Ω current value that way.

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