Approx. Ohm's law for AC?

The battery actually supplies *DC* regardless.

The load circuit may indeed generate AC.

You can't. Not from that info. You mentioned a voltage. Voltages don't take current.

Now... if you had a load across that 2V sinewave it would be possible to determine the current drawn from the battery and calculate batery life from that.

Graham

Calculations aside, what is a good rule of thumb for estimating battery life in a circuit where the vast majority of current is consumed as AC?

As a specific example, 4 x AAA cells rated at 1100 mAH's outputting a

2Vpp sinewave at 60Hz. How would one adapt a DC result from the basic ohm's law equation to estimate the appropriate resistive load for a 50 hour battery life?

Given the low frequency, can I simply upgrade the battery life by a factor of RMS, or is there more to it.

The drive circuit current requirements can be ignored for this exercise.

Thanks very much for any advice.

Robert Moore

"Robert Moore"

** ????????

** You need to know the power in the load, the percentage efficiency of the 60 Hz drive amplifier and the battery power capacity. Battery capacity depends on the acceptable end point voltage. You do not say the type of cells but from the mAH rating I take it they are NiMH - so the end point is about 1.1 volts.

The rating suggests 4 cells each with 1.2 volts nominal delivering 110 mA for 10 hours ( 5.3 Watt hours)

If discharged over 50 hours, a current of 22mA is likely = 105 mW ( at

4.8 volts)

If the drive amp is say 70 % efficient ( class B at peak level) then the AC power can be 73 mW.

A 2 volt p-p sine wave is 0.71 volts rms.

The load can consume 73 mW so solving for R

R = V squared / Power

= 6.8 ohms.

NOTE:

The assumption about the drive amp may be quite optimistic, it may have significant DC idle current and/or not be 70 % efficient when delivering 0.7 volt rms.

This will dramatically affect the available power.

....... Phil

** Yawn .......

** The OP said " 2Vpp sinewave".

** Err - the OP did say "sinewave".

You in need of new glasses ?

............. Phil

** The OP is asking for a resistance value to go with a specified AC voltage.

** That is JUST what the OP is asking for.

......... Phil

I believe 1.0V is more widely accepted.

Ted

Phil is giving you the right info. Explanation below.

This is the wrong result, but you didn't even do this division right. Looks to me like

1.1AH/50Hrs is .022 amps; then 2v/.022 amps gives 90.9 ohms. But besides that error, the 2VAC is a peak-to-peak value. The RMS value of the sine wave is .707 volts, and that's the number you need to use. And there's more to it than that, anyway.

You have to do this on an energy basis. I am going to do the calculations with some false assumptions. You say the cut-off voltage is 5V, but this is for 4 cells in series, so the cut-off per cell is 1.25 volts. If you only take the cells down to 1.25 volts, you won't get the full 1.1 Ah out of them.

See:

But, let's assume that you do. Then let's use as the average cell voltage during discharge as (6.25 + 5.00)/2 = 1.40625, or about 1.4 volts. (The real discharge curve is non-linear, but for this first crack at it, we'll assume it's linear with time.) If we assume that we'll get exactly 1.1 Ah during the 50 hour discharge, that means the current delivered by the cells for 50 hours would be 1.1/50 = .022 amps if the output voltage per cell were a constant 1.4 volts. So, if the 4 cells in series can put out .022 amps at 5.6 volts, that is a power of .1232 watts that the 4 cells can deliver for 50 hours.

This is the key to solving this problem; you have to calculate the *power* the cells can deliver as DC and then figure out what resistive load will consume this power as AC (with conversion efficiency taken into account).

The DC provided by the 4 cells has to get converted to AC somehow, and that process will happen with less than 100% efficiency. Phil picked 70% efficiency, so let's go with that number. Thus, you will be able to provide .7 * .1232 watts, or .08624 watts to your load for 50 hours. The AC voltage is 2 volts, peak-to-peak, which is .707 volts RMS. Now we're finally at a point where simple AC ohm's law can be used. Power = volts^2/resistance (P = E^2/R), so .08624 = .707^2/R, or R = .5/.08624 = 5.798 ohms.

The reason I get a slightly different result than Phil is because you told us that the cells are alkaline, which have a higher voltage during discharge than NiMH cells do.

But, I will say that you'll get substantially less than full capacity from these cells if your endpoint voltage is 1.25 volts per cell. See the curves in the PDF linked above.

Ya got me! Of course, I meant to further divide by 4 to get the per cell voltage. :-)

WattHours in - WattHourslost = Watthours out. I don't know how you can avoid calculations, even if you ignore losses. You still need to figure the watthours available from the supply, and the watthours consumed in the load.

Ed

** Depends if you ask the cell makers or electronic equipment makers.

......... Phil

discharge curve is

(6.25 + 5.00)/2 = 1.40625 ????

--DF

Thank you everyone for the replies.

Here are a few more details if anyone is inclined to comment further.

The cells are AAA alkaline, not NiMH.

Starting voltage is around 6.25. Cut-off is 5V, being the lower Vss limit of the amp which is a CA 3130.

Phil's product of 6.8 ohms seems very low to me. Can we confirm this?

I get 909 ohms for DC. 2V/(1.100AH/50Hrs). Is AC really that different in practice?

Robert Moore

"Robert Moore"

** Have you checked maker's data to verify the 1100mAH figure applies to a 1.25 volt per cell cut-off point ?

Non rechargeable cells have large operating voltage ranges.

** So this op-amp is your 2vp-p, 60 Hz source ?

It has little current output ability - you need to reconsider.

** Sorry if simple calcs have no meaning to you. ** Huh ??

Should be 90.9 ohms ?

** You do not have 2 volts DC - you have 0.71 volts rms, sine wave.

The relative power into a given load is 8 times less.

........ Phil

"Deefool"

** Yawn .....

....... Phil