18650 questions - How does internal resistance relate to voltage?

jimbo69ny

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Hello Everyone,

I recently picked up a bunch of laptop batteries and I had a few questions.

How does internal resistance relate to voltage? Does it?
I want to test to see what happens when you put a low resistance cell and a higher resistance cell, in parallel, in a circuit. Lots of people have opinions on it but I wanted to test it in a test circuit. So I tested all of my cells for internal resistance. Most of them were between 700 and 850 ohm. A couple were in the 600's. About 8 of them were near zero ohms. As my meter was connected to the near zero cells the ohm rating slowly climbed. I was surprised to see that the battery was absorbing voltage from my meter and the ohm rating was increasing. (in case you didn't know, a multi meter sends a small amount of voltage through a circuit when you are testing resistance)

I was surprised to see that internal resistance seems to change as voltage is increased. Is that normal? I thought each cell had a nominal internal resistance.

Also, does capacity relate to voltage? Capacity is inherent to the cell regardless of voltage right? For example, if you have a 3500 mah cell and a 2600 mah cell in parallel, both healthy new cells, at the same voltage, they can have different capacity right? Capacity doesn't merge when in a parallel circuit like voltage right?

Thanks!
 
The lower the internal resistance, the more amps you can pull from the cell without it overheating.

If you draw more amps than a cell can provide, there will be a voltage drop (sag), and heat.
 
I appreciate that you took the time to comment but that isn’t my question. I already knew that.

My questions are about the relation of voltage to resistance and about amp hours.

If anyone else has answers I am very curious to know what you think.
 
You can't measure the resistance of a live circuit with a multimeter. At best the measurement is meaningless, at worst it will damage the meter.
 
Again, this doesn't answer any questions. lol

I know you cant measure the resistance of a live circuit. I was measuring the resistance of a single cell. Was I not clear? Let me know if I need to reword my questions.
 
Measuring across a cell = measuring a live circuit. Literally the definition of a live circuit.

Fully charged voltage isn't related to capacity, it's determined by cell chemistry.

If you pull the same current through a high IR cell, and a low IR cell, the higher IR cell will sag to a lower voltage while it is loaded. It will also get warmer as more of it's energy is turned into heat (due to resistance), therefore it's useful capacity will be lower.

But as mentioned above, you're not properly measuring the cells IR with your multimeter, there are threads here on ES about properly measuring cell IR.
 
So measuring the resistance is creating a circuit? I guess technically maybe you are right because you are creating a path but never in any of my years of schooling did anyone claim measuring resistance was completing a circuit.

Can you send me links on these threads? If using my DMM isnt the correct way, what is the correct way to measure the resistance of a cell?

So you're saying that there is no relation to voltage and amp hour correct? That means if I have one cell that is 2600 mah and another that is 3500 mah there is no risk of damaging one cell or the other as long as voltage doesnt exceed limits.

As for internal resistance, a 2600 mah battery and a 3500 mah battery, I realize that the battery with the lower internal resistance is going to see higher amperage in a parallel circuit.
 
The average voltage while discharge a cell has influence on capacity.
Low internal resistance while discharge wil result in a higher average voltage.
Amp hours is not the correct way to measure energie,watthours is what you like to see.

A 3000mah cell with a high IR has a lower average voltage 100% to 0% 4.2V-3V:0,5C discharge ...example average voltage 3.4Vx1500mah=5.1wh
A 3000mah strong low IR cell have less voltdrop and a higher average voltage,0.5c discharge...example 3.65Voltx1500mah=5.4Wh.

Yes voltage have effect on the total watthours......average Voltage while discharge a battery.
 
jimbo69ny said:
So measuring the resistance is creating a circuit? I guess technically maybe you are right because you are creating a path but never in any of my years of schooling did anyone claim measuring resistance was completing a circuit.

Can you send me links on these threads? If using my DMM isnt the correct way, what is the correct way to measure the resistance of a cell?

So you're saying that there is no relation to voltage and amp hour correct? That means if I have one cell that is 2600 mah and another that is 3500 mah there is no risk of damaging one cell or the other as long as voltage doesnt exceed limits.

As for internal resistance, a 2600 mah battery and a 3500 mah battery, I realize that the battery with the lower internal resistance is going to see higher amperage in a parallel circuit.
Firstly, you need more/better schooling on electrical theory such that you can understand simple basics like measuring the resistance of a passive circuit or component is very different to trying to measure the resistance of an active circuit or battery cell.
Capacity of a cell is only a function of its voltage for that one specific cell, you cannot compare different cells by voltage and assume they are the same capacity.
Ask google or use this sites search function, or read some reference books.
 
jimbo69ny said:
How does internal resistance relate to voltage? Does it?

When there is no load, there is no relationship.

Once a load is applied, a battery with a higher internal resistance will have a lower voltage under load, than a battery with a lower internal resistance.

A battery with internal resistance is the equivalent of an ideal power source with an external resistor in series as part of the circuit. The overall result depends on what else is on the circuit.

jimbo69ny said:
I want to test to see what happens when you put a low resistance cell and a higher resistance cell, in parallel, in a circuit. Lots of people have opinions on it but I wanted to test it in a test circuit.

If they are at the same voltage and no load, absolutely nothing.

If they are not at the same voltage, then current will flow from the higher voltage cell, to the lower voltage cell, at a rate limited by the internal resistance of both cells added together.

If a load is applied, the battery with the lower resistance will supply more power, in accordance to Ohm's law. After the load is removed, the battery with lower internal resistance will now have a lower resting voltage. Power will flow from the higher resistance battery to the lower resistance battery as described in the paragraph directly above this one.

jimbo69ny said:
So I tested all of my cells for internal resistance. Most of them were between 700 and 850 ohm. A couple were in the 600's. About 8 of them were near zero ohms.

A battery that had an internal resistance of 700 ohm would be near useless. As soon as you applied a load, the voltage would drop to near zero. Even a battery with 700mOhm (0.7Ohm) would be near useless. A new high C rate LiPo cell should have an internal resistance of around 0.001 ohms. Even a high rate 18650 should be 0.01, and a poor quality 18650 should be 0.1. Anything greater than 1 would be useless for anything except maybe running a wrist watch.

A 1 Ohm internal resistance, would mean that a load of 1A will drop the battery voltage 1v. So your fully charged 4.2v battery would sag to 3.2 Ohm, and trip your eBike's low voltage cutoff. At 1A - not even enough to spin a wheel unloaded. Imagine how useless a 700ohm battery would be.


jimbo69ny said:
As my meter was connected to the near zero cells the ohm rating slowly climbed. I was surprised to see that the battery was absorbing voltage from my meter and the ohm rating was increasing. (in case you didn't know, a multi meter sends a small amount of voltage through a circuit when you are testing resistance)

That's not how internal resistance meters work, which, together with other people's comments here suggest to me that you didn't actually measure the internal resistance.

A DC IR meter measures the voltage with no load, the applies known loads to record voltage drop. It can calculate the internal resistance from there.

An AC IR resistance meter sends a pulse of electricity through at 1000Hz, and sees the reactance of the cell to estimate resistance.

jimbo69ny said:
I was surprised to see that internal resistance seems to change as voltage is increased. Is that normal? I thought each cell had a nominal internal resistance.

It's unlikely you conducted the experiment properly, but the technically correct answer is yes, that's normal. A LiPo battery will have a higher internal resistance 3.8v-4.2 and 3.2-3.6v, than between 3.6 and 3.8v, where most the capacity is. I'm not sure if this is correlated, or causal. perhaps someone with better knowledge of battery design can assist here.

jimbo69ny said:
Also, does capacity relate to voltage? Capacity is inherent to the cell regardless of voltage right? For example, if you have a 3500 mah cell and a 2600 mah cell in parallel, both healthy new cells, at the same voltage, they can have different capacity right? Capacity doesn't merge when in a parallel circuit like voltage right?

Capacity does not relate to voltage. A 2600mah cell and a 3500mah cell will both have the same operating voltages.

When you parallel cells of different capacity, they have an additive relationship, but internal resistance is the sum of the inverse of resistance (meaning the combined battery has a better, lower internal resistance). Those two cells together will amount to a 6100mah, with a lower internal resistance than either battery. (1/Rt = 1/R1 + 1/R2)

All good questions, but I think built on the wrong premise that you could measure internal resistance with a multimeter. Hope the answers help though. A test might be worth a thousand opinions, but only if it's performed correctly. Me? Whenever there is a difference between a test and the theory, I wonder what I did wrong with the test, never the theory.
 
Wow! Thank you Sunder!! You answered all of my questions in beautiful detail! Thank you so much!
 
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