Measuring MAX current of a battery 9 volts, AAA, AA

[BradtheRad]

I'm not so sure that battery condition and battery tests have definable boundaries.

For my own part, it's kind of amorphous. I've seen a variety of battery types, and a variety of voltages.

A new 9V battery (carbon-zinc or alkaline) used to measure 9V. Over the years, alkalines have come to read 9.6V when new.

When 9V-size rechargeables came out, they were labelled 7.2V (because of containing six 1.2 V nicads). Not near 9V. It seemed incredible that manufacturers would do this. Eventually they put in seven cells and labelled it 8.4 V.

That is one reason battery state has vague boundaries.

Rechargeable cells are such a great idea. Yet they only put out 1.2-1.25 V. Old-fashioned flashlight batteries were close to done at that voltage. So I've had to relax my standards when it comes to the state of a rechargeable.

A reason for vague boundaries.

I keep new alkaline batteries in one box. I put them in equipment which needs good 1.5 V cells.

When an alkaline gets down to 1.2 or 1.3V, I put it in a box labelled "half-good". These can be used in undemanding items such as clocks and led lights and remote controls.

I usually check a battery with my DVM. I've come to know what volt levels to expect of which types, and at what state of discharge.

I may suspect a battery is not powering a piece of equipment. That is when I connect my meter leads to the battery while it is installed in the equipment. I can watch to see if its voltage goes down during operation.

 

[danny davis]

The Battery at what state of discharge.

When a battery has a low internal resistance, there is no discharge
When a battery internal resistance raises in value , there is discharge

How do you measure the state of discharge? do you measure the voltage or the current?

But how do you know what state of discharge the battery is at when you do measure the voltage and current?

 

[betwixt]

I'm losing patience...

What resistor value would you expect to pass 25mA with 9V across it? BIG HINT: R=V/I

Brian.

 

[danny davis]

What resistor value would you expect to pass 25mA with 9V across it? BIG HINT: R=V/I

360 ohms

I'm not asking about that

I just don't understand how a brand new battery has a very low internal resistance but how does the internal resistance raise in value to get larger? what makes it grow in size?

A brand new battery has no discharge because the internal resistance is low but when the battery has a high value internal resistance there is a discharge?

When a battery has a low internal resistance, there is no discharge
When a battery internal resistance raises in value , there is discharge

How do you measure the state of discharge? do you measure the voltage or the current? does it make a difference?

But how do you know what state of discharge the battery is at when you do measure the voltage and current?

Which you know the battery is discharging from a load, but how do u know what state of the discharge the battery is in and how do u know what the internal resistance is at what value?

 

[Audioguru]

A LOAD draws current from a battery. It could be operating current or test current.
I told you already to use Ohm's Law to calculate 9V/25mA= 360 ohms.

My cheap BUT OVER-PRICED RadioShack battery tester uses 850 ohms (an 820 ohm resistor) to test a 9V battery so Ohm's Law calculates the test current to be 9V/850 ohms= 10.6mA.

"u" is not a word. You ALWAYS wrongly use the word "than" instead of "then".

 

[danny davis]

test current to be 9V/850 ohms= 10.6mA.
9V/25mA= 360 ohms.

Why are they testing the battery at 10.6mA?

What is the difference between testing at 9 volt battery at 25mA VS testing it at 10.6mA

How does this tell you the internal resistance of the battery?

Are you guys saying that at 10.6mA the internal resistance of the battery is at 850 ohms internally?
Are you guys saying that at 25mA the internal resistance of the battery is at 360 ohms internally?

 

[betwixt]

And who suggested internal resistance made the battery discharge?

I said a battery can be visualized as a perfect source with a series resistance, what makes you think a series resistance drains current? In my world, a series resistance limits current.

Brian.

 

[danny davis]

And who suggested internal resistance made the battery discharge?

What makes the battery discharge than? i thought the voltage drop was from the internal resistance getting bigger and growing larger in value which causes a voltage drop

I said a battery can be visualized as a perfect source with a series resistance, what makes you think a series resistance drains current? In my world, a series resistance limits current.

Yes true me too, but why is there a voltage drop ? when you put a LOAD on a weak battery? the battery starts to discharge

When the internal resistance gets bigger larger in value, the current draw output gets less right? but the voltage stays the SAME at 9 volts or battery voltage rating?

 

[Audioguru]

I just don't understand how a brand new battery has a very low internal resistance but how does the internal resistance raise in value to get larger? what makes it grow in size?

You have been told that some of its chemicals have been used up.

A brand new battery has no discharge because the internal resistance is low but when the battery has a high value internal resistance there is a discharge?

YOU DO NOT UNDERSTAND!!
A brand new battery has not been discharged YET. Then its internal resistance is low. Having a load discharges a battery.
When the battery is discharging then some of its chemicals get used up. Then its internal resistance becomes high.

When a battery has a low internal resistance, there is no discharge
When a battery internal resistance raises in value , there is discharge

NO!
When a battery has a low internal resistance, a discharge has not occurred YET.
When a battery internal resistance rises in value, there is discharge occurring caused by a load.
The internal resistance is in series with the battery creating a voltage divider with the load resistance.
Its load is parallel to the output of the battery but is in series with the internal resistance.

How do you measure the state of discharge? do you measure the voltage or the current? does it make a difference?

We have told you many times to use a typical load on a throw-away battery so you do not kill it and measure its voltage.
On a rechargeable battery you can measure its maximum current then re-charge it.

But how do you know what state of discharge the battery is at when you do measure the voltage and current?

You look on its datasheet. The Duracell alkaline 9V battery with a 50mA load current is almost dead when its voltage measures about 6.6V and it is half used up when its voltage measures about 7.5V.

Which you know the battery is discharging from a load, but how do u know what state of the discharge the battery is in and how do u know what the internal resistance is at what value?

The curves on the datasheet show the state of discharge and Ohm's Law can be used to calculate the internal resistance.

Simple stuff.

 

[danny davis]

Ok, I'm getting it know

We have told you many times to use a typical load on a throw-away battery so you do not kill it and measure its voltage.
On a rechargeable battery you can measure its maximum current then re-charge it.

Oh ok , so what you guys are saying is that a "Non-Rechargeable Battery" you can't measure the maximum current because it will kill it or use up all the chemicals?

When the battery is discharging then some of its chemicals get used up. Then its internal resistance becomes high.

This is what i'm confused about

Why does the internal resistance becomes HIGH when the chemicals get used up? plus when you measure the voltage of the battery it's still measures 9 volts without a load

The internal resistance is in series with the battery creating a voltage divider with the load resistance.
Its load is parallel to the output of the battery but is in series with the internal resistance.

When you SHORT the battery terminals together, the voltage will measure ZERO volts, but the current is at the max. because the ESR internal series resistance is converting the battery voltage into a current

This internal series resistance is maybe up of chemicals or charges

Why does the internal series resistance get larger in value when the chemicals get used up? or there is discharge?

 

[Audioguru]

so what you guys are saying is that a "Non-Rechargeable Battery" you can't measure the maximum current because it will kill it or use up all the chemicals?

Correct.

Why does the internal resistance becomes HIGH when the chemicals get used up?

Because of the chemical reaction slowing down when most of the chemicals are used up.

when you measure the voltage of the battery it's still measures 9 volts without a load

You still don't understand that the internal resistance is in series with the battery and the load forms a voltage divider that reduces the voltage at the junction of the resistors. Without a load the internal resistance is MUCH lower than the resistance of the meter so you measure the full voltage of the battery.

When you SHORT the battery terminals together, the voltage will measure ZERO volts, but the current is at the max. because the ESR internal series resistance is converting the battery voltage into a current

Don't be silly. I showed you the datasheet for a Duracell 9V alkaline battery that says its internal resistance is 1.7 ohms when new.
Do the math with Ohm's Law and the shorted current is 9V/1.7 ohms= 5.3A for less than 2 seconds then the battery is dead. 9V x 5.3A= 47.7w so the battery will begin getting very hot.

This internal series resistance is maybe up of chemicals or charges

"maybe up" is bad English.

Why does the internal series resistance get larger in value when the chemicals get used up? or there is discharge?

You do not understand electronics and chemistry and "discharge".

 

[danny davis]

the shorted current is 9V/1.7 ohms= 5.3A for less than 2 seconds then the battery is dead. 9V x 5.3A= 47.7w so the battery will begin getting very hot.

So every 9 volt battery in general outputs 5 to 6 Amps of current?

Because of the chemical reaction slowing down when most of the chemicals are used up.

So slowing down chemical reaction cause the internal resistance to get larger in value? i still don't get as to why

Internal resistance gets larger in value
1.) Chemical reaction slowing down
2.) Chemical being used up

 

[BradtheRad]

There is more than one reason a battery's chemistry declines.

A) A battery's can develop high internal resistance sitting on the shelf for years. The electrolyte has a tendency to dry out. A meter reading might be 1.5V (just like when it was new), but if we attach even a slight load, the voltage drops.

B) A brand new battery drained quickly by a heavy load. It put out high amperes as its voltage went down gradually. We disconnect the load. A meter reads 1.2V with no load. Its chemistry is no longer able to produce 1.5V. But it still has low internal resistance.

The battery in case B is not necessarily at the end of its useful life. Some types can be recharged to a certain extent (such as carbon-zinc and alkaline). However it is not always problem-free to do so.

 

[betwixt]

Danny try this VERY SIMPLE experiment:

Start with a good battery, any voltage will do, you can even use a regulated power supply if you want. For sake of argument I'll say here it is a 9V one because thats what you have been mentioning.

Connect TWO resistors, one end of each resistor to the + side of the battery. Leave the other end of each resistor disconnected for now. One resistor should be 10 Ohms, the other 10K Ohms.

PRETEND those resistors are inside the battery, we are using them to emulate the batteries internal resistance. 10 Ohms for a good battery and 10K for a bad battery.

Use your DVM to measure the voltage between the negative side of the battery and the free ends of each resistor. Tell us what voltage your DVM says.

Now add a third resistor to the circuit. We are going to use this to emulate a 10mA load so it's value should be (R=V/I) 9/.01 = 900 Ohms. It doesn't have to be exatly that value, something nearby will do.

Connect one end to the negative side of the battery and the other end to the free end of the 10 Ohm resistor. Measure and tell us the voltage across the 900 Ohm resistor.
Now disconnect the 900 Ohm resistor from the 10 Ohm resistor and reconnect it to the 10K resistor instead. Again measure the voltage across the 900 Ohm resistor and tell us what your DVM reads.

So you should have four results for us - voltage from negative to the free end of the two resistors without the 900 Ohms load and the voltage from the negative to the same points WITH the 900 Ohms load connected to them.

Report the results here please.

Brian.

 

[Audioguru]

So every 9 volt battery in general outputs 5 to 6 Amps of current?

Of course not. Only a brand new Duracell 9V ALKALINE battery. The cheap Chinese ones in The Dollar Store are probably a few years old and use obsolete carbon-zinc chemicals. They might be able to produce 100mA if they do not leak too much "beer".

So slowing down chemical reaction cause the internal resistance to get larger in value? i still don't get as to why

Learn chemistry to find out why.

Internal resistance gets larger in value
1.) Chemical reaction slowing down
2.) Chemical being used up

Yes.