internal resistence?

[r2]

Will someone please explain internal resistence?

I know that alkalines have higher internal resistence than lithiums or NiMH and it means that they can't deliver high current as well. I don't quite understand how this relates to normal resistence put inline between batteries and an LED.

From what I understand, resistence lowers the voltage. If you have 4 alkaline cells in series driving an LED, you can put enough resistence in to pull the voltage down to a safe level, but the current draw is still largely determined by what the LED draws at the particular voltage. If the batteries are incapable of supplying that much current then the voltage drops further until the whole system reaches equalibrium. That much makes sense to me.

If this is accurate, then it seems different than what I read about internal resistence in the batteries, which doesn't seem to figure into the voltage (we still rate them at 1.5V/cell even though the internal resistence is fixed for the cell) as it would (?) if it were a resister external to the cell, but rather it limits the current without affecting the voltage directly? This is the part I don't understand.

This is just out of curiousity, but I'd appreciate it if someone could clear things up a bit.

Thanks,

Russ

 

[MrAl]

Hello there,

If i understand you right, you pretty much already
know what internal resistance in a battery is.
The only unusual thing to think about is that this
resistance DOESNT affect the battery voltage until
a load is applied, so we end up calling it a 1.5v
cell even though it might be much lower after a
heavy load is applied.

The internal resistance doesnt affect the open circuit
voltage, so the cell is still called a 1.5v cell.

The in-circuit voltage might only be 1.3 volts, due to
the internal resistance dropping 0.2 volts.

Good luck with your LED circuits,
Al

 

[INRETECH]

ESR is very important in batteries, for example - the ESR of a AA is much higher than a D

When you draw current out of a battery, the internal resistance of the battery causes HEAT, that is - trying to draw out large amounts of current out of a small battery is impossible, and "bad" for the battery since the internal resistance of the battery will convert the voltage across the internal resistance into HEAT

When designing circuits using batteries, or for that matter - a CAP, its important to know the ESR of the device during charge/discharge cycles

That is why you can destroy a cap without exceeding its working voltage simply by putting too high a ripple current thru the part

 

[php_44]

[ QUOTE ]
INRETECH said:
ESR is very important in batteries, for example - the ESR of a AA is much higher than a D


[/ QUOTE ]

This is a good topic for discussion!

Not to nit pick, but non-rechargable AA's seem to be designed with very low internal resistance (and in many cases it is lower than larger cells - like "D" cells). This is why when feeding a high performance light you need to know your batteries - it's not always intuitive which battery has the lowest internal resistance.

For example compare the internal resistances of the following Energi*er cells when fresh:

<font class="small">Code:</font><hr /><pre>

Size Alkaline Alk E2 Lithium NiMH
---- -------- ------ ------- ----
AAA 0.205 0.156 0.1
AA 0.146 0.127 ~0.25 0.03
C 0.324 0.171 0.011
D 0.173 0.234 0.011
123 ~0.25
</pre><hr />

For alkalines, the internal resistance goes up significantly as the cell discharges. When an alkaline cell is 90% discharged the internal resistance is ~5x greater than fresh.

I couldn't find good info in the lithiums, but I'm estimating based on other data that their internal resistance increases less dramatically as the cell discharges.

For NiMH/NiCd the internal resistance can double when the cell is at 1/2 capacity, but it starts out so low that it is not a significant factor unless you're drawing a lot of amps out of them.

Again - none of this is an issue when drawing a few dozen mA from a cell to power a few 5mm LEDs. However, if you're trying to draw >1/2 of an amp from a cell, then the internal resistance of the batteries you intend to use should be factored in. If you don't you'll probably be wasting a lot of your battery $$$ on heating up your batteries rather than generating light from them.

 

[Doug S]

php-44. Thanks for taking the time to post the above data. I have noticed in the past that many people have had the misconception that the C and D cells have significantly lower Ri than AA cells in the alkaline chemistry. People should especially note the Ri of the non-premium C alkaline cells. For high drain applications they suck. With C cells you get to pay the same price as for D cells but only get half the capacity and high rate performance worse than D or AA cells [worse in the sense of terminal voltage droop].

 

[pwell]

[ QUOTE ]
r2 said:
From what I understand, resistence lowers the voltage. If you have 4 alkaline cells in series driving an LED, you can put enough resistence in to pull the voltage down to a safe level, but the current draw is still largely determined by what the LED draws at the particular voltage.

[/ QUOTE ]

This is not correct
Think of resistance as limiting current. Remember V = IR

In a simple circuit with a batt connected to a resistor, the voltage accross the resistor never changes (assuming the battery has no internal resistance), but the current can be varied by changing the resistor.

To calculate the resistance required to limit the current in an LED circuit use the following formula

R = voltage accross resistor/20mA
= (batt Voltage - LED voltage)/20mA

The LED voltage will remain the same, whether it be 0.7V or 2.2V or whatever, no matter what resistor you use.

A LED WILL ALWAYS draw as much current as you let it. It will happily draw 20A or more of current but only for half a second (unless its for high power use), until the magic smoke comes out. Try connecting an LED to 2 series 3V lithiums and see how long it lasts. (dont use a good LED).

[ QUOTE ]
r2 said:
If the batteries are incapable of supplying that much current then the voltage drops further until the whole system reaches equalibrium. That much makes sense to me.


[/ QUOTE ]

The greater the curent draw, the greater the drop in battery voltage. The effect varies for different cell chemistries.

NiMH dont like being discharged at greater then 3C
NiCD ... 10C !

NiCD are still the king of high current draw applications.
Go the Nicad!


My 0.02

Sean

 

[AilSnail]

What does the C stand for?

 

[B@rt]

C stands for the capacity of the cell. /ubbthreads/images/graemlins/wink.gif

 

[AilSnail]

I still don't get it (I think). Does that mean a 1.2Ah Nicd battery can deliver 12 Amps? sounds like a lot.

 

[Doug Owen]

Yup, in simple terms a 1.2 AH battery can deliver 1.2 Amps for an hour, 12 Amps for 6 minutes ( a tenth of an hour), .12 Amps for ten hours and so on.

The oft quoted 3C and 10C limits mean 'don't try to drain the full charge in less than 20 or six minutes respectively. In a real world, lower rates (less current) actually run a little 'extra long' for a number of reasons. But 'simple math' is close enough for jazz in most cases.

Doug Owen