Thats a neet device there. I wonder if the internal resistance increases as the cell ages/gets used alot.

electricjelly wrote:

Thats a neet device there.

Very neat, I did post a review of it: http://budgetlightforum.com/node/56582

I think they are taking advantage of the principle behind these equations.
BTW, sorry that my symbol for R-internal looks a lot like R-1

The circuit represents a Voltage source (cell) with it’s internal resistance hooked up to 2 different resistors, with 2 different current draws.
The value of each resistor is unknown, nor does it matter

What is measured is the voltage drop across R-1 and R-2, the load resistors, and the current thru them.
The equations show that the R-internal is calculated from the change in voltage drop divided by the change in current.

The value calculated does not depend on the values of the 2 load resistors
Instead of using 2 load resistors, they are using variable voltage and current, AC, and doing the math.

The reason why this is so good, I believe, is because any contact resistance would be part of the load resistance, and wouldn’t matter, because the resistance of the load resistor is not used in the calculation.
.
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dchomak wrote:

I think they are taking advantage of the principle behind these equations.

No, as you can see in my review it uses AC and 4 terminal connection.
The AC also means the value will be lower than chargers measures, because they use DC.

HKJ wrote:

dchomak wrote:

I think they are taking advantage of the principle behind these equations.

No, as you can see in my review it uses AC and 4 terminal connection.
The AC also means the value will be lower than chargers measures, because they use DC.

The circuit I drew up would use 2 meters at the same time. One measuring current and one measuring voltage, for a total of 4 separate terminal connections.
I still think that is the principle behind what they are doing.

dchomak wrote:

The circuit I drew up would use 2 meters at the same time. One measuring current and one measuring voltage, for a total of 4 separate terminal connections. I still think that is the principle behind what they are doing.

They are measuring current and voltage at the same time, but with AC you do not change current or even draw current (I checked this with an oscilloscope)

This is the standard way to measure internal resistance.

nice device.
i need to get one.
the dick smith esr works well on batteries but lacks resolution on the very low end where it is most desired.

HKJ wrote:

dchomak wrote:

The circuit I drew up would use 2 meters at the same time. One measuring current and one measuring voltage, for a total of 4 separate terminal connections. I still think that is the principle behind what they are doing.

They are measuring current and voltage at the same time, but with AC you do not change current or even draw current (I checked this with an oscilloscope)

This is the standard way to measure internal resistance.

“They are measuring current and voltage at the same time”

Yes, I think what is being done is that AC is used and they are calculating the delta V and delta I at each sample point of the Sine Wave
No, the peak current will not change from cycle to cycle, but it would all along each cycle.

It could be that they are calculating the instantaneous change in voltage with respect to the instantaneous change in current along that sine wave. Dividing the 2 will give the internal resistance.

Now, who am I to say how they are actually doing it. All I am saying is that they are calculating the internal resistance based on the principles of my circuit and equations.

snakebite wrote:

nice device. i need to get one. the dick smith esr works well on batteries but lacks resolution on the very low end where it is most desired.

Yes, very nice indeed!
I want one too. $58 on AliExpress

dchomak wrote:

Now, who am I to say how they are actually doing it. All I am saying is that they are calculating the internal resistance based on the principles of my circuit and equations.

As long as you circuit can measure AC only need to divide voltage with current, no need to use the DC formulas.
You can find it all in my article about resistance/impedance: http://lygte-info.dk/info/Internal%20impedance%20UK.html

Pardon my ignorance, will the internal resistance measurement be affected if the battery is fully charged or fully drained?

I notice most of the cells above are measured with DC voltage at around 3.5-3.6v

Wow, that Samsung 21700 30T looks like it could deliver the highest current of any cell I’ve seen. I see articles saying it’s a Panny/Tesla 21700 competetor cell. Looking at the reviews, very impressive.

lawallac wrote:

Wow, that Samsung 21700 30T looks like it could deliver the highest current of any cell I’ve seen. I see articles saying it’s a Panny/Tesla 21700 competetor cell. Looking at the reviews, very impressive.

That 30T cost me a battery holder during test, my holder could not handle 50A

d_t_a wrote:

Pardon my ignorance, will the internal resistance measurement be affected if the battery is fully charged or fully drained?

It will increase when the battery is close to empty.

HKJ wrote:

d_t_a wrote:

Pardon my ignorance, will the internal resistance measurement be affected if the battery is fully charged or fully drained?

It will increase when the battery is close to empty.

Will the difference be significant for that same battery? (eg. measuring the same battery at 3.0v vs 3.6v vs 4.2v)
or will there be a big variance depending on the type/chemistry/age of the battery?

I saw the sudden cut out in the graph. 50A just wow. That explains it. Thanks for testing it though!

For flashlight use the relevant measurement for battery resistance is the DC IR. As mentioned this meter measures something different which might or might not be useful.

All these measurements seem very close. Some batteries listed I would have thought to have higher resistance. So these are very accurate numbers? Shows perception does not mean much.