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Author Topic: Trying to understand battery power (math)  (Read 355 times)

KC3EDP

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Trying to understand battery power (math)
« on: December 10, 2019, 02:50:36 PM »

    Can anyone please send me to a link or youtube to try and grasp battery Amp hours and such? The PDF I have linked sound a little like it makes sense.

   Something for beginners, and not rocket science.  ;D

https://www.vhfclub.org/pdf/Emergency%20Power%20Training%20feb%202016.pdf
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KC3EDP

AA4PB

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Re: Trying to understand battery power (math)
« Reply #1 on: December 10, 2019, 05:07:57 PM »

The article in your link looks pretty accurate to me. One thing they forgot to take into account is that the battery AH ratings are for the battery being drained down to a certain voltage (typically 10V). That means an 80AH battery will supply 80AH until the battery reaches 10V. Unfortunately many radios begin to have problems when the voltage reaches 11V so you can't get the whole 80AH for those radios. Additionally, the AH rating is specified for a certain current drain. If you exceed that then you get a little less AH output. However, the process described in your link should get you pretty close to the actual operating hours.
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Bob  AA4PB
Garrisonville, VA

K5LXP

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Re: Trying to understand battery power (math)
« Reply #2 on: December 10, 2019, 05:24:54 PM »

The linked presentation covers a lot of the basics including how to calculate run time based on loads and capacities.  What other information are you looking for specifically?

Mark K5LXP
Albuquerque, NM
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W9IQ

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Re: Trying to understand battery power (math)
« Reply #3 on: December 10, 2019, 05:32:28 PM »

In addition to Bob's comment, the author makes a consistent error. For example, he states:

Required AH= (Transmit current x .35) + (receive current x . 65)

The result of the formula is amps, not amp hours. If he were to multiply the right side of the equation by 1 hour, he would then have amp hours.

If you divide the battery amp hour rating by his amp hour units, it would produce a unit-less result. But when you divide amp hours by amps, the result is hours which is what the author was apparently intending but did not demonstrate.

It is also worth noting that most Ah ratings for deep discharge batteries are based on a 20 hour level current discharge. So a 100 Ah deep discharge battery generally means a 5 amp discharge rate for 20 hours. If you discharge this battery at 10 amps, it will last for less than 10 hours.

- Glenn W9IQ
« Last Edit: December 10, 2019, 05:52:52 PM by W9IQ »
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- Glenn W9IQ

God runs electromagnetics on Monday, Wednesday and Friday by the wave theory and the devil runs it on Tuesday, Thursday and Saturday by the Quantum theory.

K5LXP

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Re: Trying to understand battery power (math)
« Reply #4 on: December 10, 2019, 08:28:52 PM »

most Ah ratings for deep discharge batteries are based on a 20 hour level current discharge.

Since many hams vapor lock on figuring out the mode and TX/RX duty cycles I'm not sure you'll get as far as including peukert constant.  I use a spreadsheet to input load, minimum voltage and battery type variables and it will crunch the run time figuring peukert.  But even with that, the result is still pretty speculative because it assumes the load model to be exactly correct and in real life it rarely follows it exactly.  Useful for what if's and to see how close you are, then go from there to throw some margin in for the actual field deployment.  To get an accurate peukert constant you need to run a couple discharge profiles on the very battery you intend to use which exceeds the attention span of most, so for a horse shoes/hand grenades number you're probably close enough just using the rated Ah and adding some margin based on what the load is vs the 20 hour rate.

Mark K5LXP
Albuquerque, NM
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KC3EDP

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Re: Trying to understand battery power (math)
« Reply #5 on: December 11, 2019, 03:09:34 AM »

The linked presentation covers a lot of the basics including how to calculate run time based on loads and capacities.  What other information are you looking for specifically?

Mark K5LXP
Albuquerque, NM

   Thanks,  I understand that. I'm just not getting it.
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KC3EDP

W9IQ

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Re: Trying to understand battery power (math)
« Reply #6 on: December 11, 2019, 03:15:42 AM »

The challenge with trying to apply Peukert's law is that the exponent is not a constant. It varies by battery chemistry and even by product within a chemistry. It is better to pick a conservative exponential value for the particular chemistry involved.

The other fundamental issue is that current averaging is not a valid use of the equation.

- Glenn W9IQ
« Last Edit: December 11, 2019, 03:26:38 AM by W9IQ »
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- Glenn W9IQ

God runs electromagnetics on Monday, Wednesday and Friday by the wave theory and the devil runs it on Tuesday, Thursday and Saturday by the Quantum theory.

W9IQ

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Re: Trying to understand battery power (math)
« Reply #7 on: December 11, 2019, 03:20:12 AM »

Thanks,  I understand that. I'm just not getting it.

As Mark mentioned, a lot of hams struggle with this.

The first part involves calculating the average amperage that your radio will draw.

The second part is using this average amperage to calculate how long a battery would last.

At what part are you stuck?

- Glenn W9IQ
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- Glenn W9IQ

God runs electromagnetics on Monday, Wednesday and Friday by the wave theory and the devil runs it on Tuesday, Thursday and Saturday by the Quantum theory.

KC3EDP

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Re: Trying to understand battery power (math)
« Reply #8 on: December 15, 2019, 04:47:22 PM »

Thanks,  I understand that. I'm just not getting it.

As Mark mentioned, a lot of hams struggle with this.

The first part involves calculating the average amperage that your radio will draw.

The second part is using this average amperage to calculate how long a battery would last.

At what part are you stuck?

- Glenn W9IQ

Thanks!
Glenn, lets try to break this down to as simple terms as possible.
1) Let's try two numbers. I'll run @40w and if necessary, 45w. How do I calculate ave amps?

2) average amperage to calculate how long a battery would last?
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KC3EDP

W9IQ

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Re: Trying to understand battery power (math)
« Reply #9 on: December 16, 2019, 03:18:33 AM »

The first step is to determine how many amps your radio draws during transmit and receive - two separate numbers. Generally the receive current is lower than the transmit current. You can often get these numbers by looking at the specification page in your manual. If you cannot find this information, then mention the make and model of your radio.

Also comment on your modulation mode - AM, FM, SSB, CW, FT8, digital, etc.

Then you need to estimate the percent of time you will be listening (receiving) vs transmitting when you are on battery power. If you are on a VHF net, you might be listening 95% of the time and transmitting 5% of the time, for example. Or if you are on an active HF band on FT8, you may be receiving 50% of the time and transmitting 50% of the time, for example.

If you post back with this information, we can work through the math for the first step using your data.

- Glenn W9IQ
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- Glenn W9IQ

God runs electromagnetics on Monday, Wednesday and Friday by the wave theory and the devil runs it on Tuesday, Thursday and Saturday by the Quantum theory.

KC3EDP

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Re: Trying to understand battery power (math)
« Reply #10 on: December 16, 2019, 06:55:51 AM »

The first step is to determine how many amps your radio draws during transmit and receive - two separate numbers. Generally the receive current is lower than the transmit current. You can often get these numbers by looking at the specification page in your manual. If you cannot find this information, then mention the make and model of your radio.

Also comment on your modulation mode - AM, FM, SSB, CW, FT8, digital, etc.

Then you need to estimate the percent of time you will be listening (receiving) vs transmitting when you are on battery power. If you are on a VHF net, you might be listening 95% of the time and transmitting 5% of the time, for example. Or if you are on an active HF band on FT8, you may be receiving 50% of the time and transmitting 50% of the time, for example.

If you post back with this information, we can work through the math for the first step using your data.

- Glenn W9IQ

   Thanks for your time with this.
   I'll guesstimate with a Emcom situation. 30% TX- 70% RX.. Digital modes

 From the manual Yaesu 857D
Current Consumption: squelched 550 mA (aprox)
                     receive: 1A
                     transmit: 22A

transmitter RF power output
160-6M SSB/CW/FM 100W     AM CARRIER 25W

I couldn't find anything Dig. Power TX. I use "user-U" upper side band for Dig

I included a link to the whole manual.

https://www.hamradio.hanklambert.com/Downloads/FT857/FT-857DManual.pdf


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KC3EDP

W9IQ

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Re: Trying to understand battery power (math)
« Reply #11 on: December 16, 2019, 09:43:50 AM »

Let's work through the average current draw using your numbers. This isn't a simple average but rather a weighted average with the weighting coming from the percent of transmit and receive time:

Average Current = (Receive Amps * Receive Percent) + (Transmit Amps * Transmit Percent)
Average Current = (1 amp * 70%) + (22 amps * 30%)
Average Current = (0.7 amps) + (6.6 amps)
Average Current = 7.3 amps

Note that this assumes a transmit mode where peak and average current are the same. For most digital modes, this is the case. If this were voice SSB, we would need to apply an additional weighting factor for the transmit current otherwise this formula would overstate the amps drawn during transmit.

Now that we have the weighted average current, we can convert this to amp hours (Ah), which is a unit of electrical charge, simply by multiplying by the number of hours of operation. So, for example, if you were to operate your radio this way for 5 hours, it would be:

Amp Hours = 7.3 amps * 5 hours = 36.5 Ah

The next step is to convert this to the necessary battery capacity but I will pause here for any feedback or questions from you.

- Glenn W9IQ

« Last Edit: December 16, 2019, 10:00:22 AM by W9IQ »
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- Glenn W9IQ

God runs electromagnetics on Monday, Wednesday and Friday by the wave theory and the devil runs it on Tuesday, Thursday and Saturday by the Quantum theory.

ND6M

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Re: Trying to understand battery power (math)
« Reply #12 on: December 16, 2019, 01:58:12 PM »

...edit...
In addition to Bob's comment, the author makes a consistent error. For example, he states:

Required AH= (Transmit current x .35) + (receive current x . 65)

The result of the formula is amps, not amp hours....

- Glenn W9IQ

Actually, the ".35" and the ".65" figures are hours, He clearly explained them in the article.

He just didn't write it in the equation.
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W9IQ

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Re: Trying to understand battery power (math)
« Reply #13 on: December 16, 2019, 02:57:57 PM »

Actually, the ".35" and the ".65" figures are hours, He clearly explained them in the article.

He just didn't write it in the equation.

I don't wish to distract from the current direction of the thread, but he clearly did not describe them as hours. He simply referenced them as a percent:

Quote
[slide 4]Determine transmit duty cycle (Typical):
– 35% for net control
– 10% for net participant

[slide 5]At a 50 watt transmit power for a net control station the
power requirement in ampere hours will be:
Required AH= (Transmit current x .35) + (receive current x . 65)
Required AH= (10 amperes x .35) + (0.5 x . 65) or 3.83 Ampere Hours

The result would be amps, not amp hours (much less "AH").

- Glenn W9IQ
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- Glenn W9IQ

God runs electromagnetics on Monday, Wednesday and Friday by the wave theory and the devil runs it on Tuesday, Thursday and Saturday by the Quantum theory.

KB8VUL

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Re: Trying to understand battery power (math)
« Reply #14 on: December 16, 2019, 05:13:30 PM »

First thing you need to understand is just what an AMP-HOUR is. 
simple version is if a battery is rated for 1 AH (amp-hour) and its fully charged that you can draw 1 amp for one hour and the battery will be depleted.
The second thing is what's depleted,,, that is when the manufacture says the battery is dead.  It's some loaded voltage, typically 11 volts.
Then there is the base line math.
1 AH means 1 amp load for 1 hour
it also means that a .5 amp load at 2 hours, or a 2 amp load to .5 hour.
So that is the really basic level explanation.
Now into the real world.
A 50AH battery will support a 10 amp load for 5 hours.
But we are talking about a radio that is not being talked on constantly.
The radio draws 1 amp when it's receiving, so if it never transmitted, it would run for 50 hours before the battery was flat. 
But since you are talking on it, it will draw 10 amps when you are doing that.  So the discharge rate is much higher when you are transmitting.
Figuring out exactly how long the battery will last is difficult, and in truth, can only be a good estimate as you cant tell how long you will be talking on the radio as opposed to how long you will be listening to it.
So saying a 50AH battery will last until 5PM tomorrow running radio X is near impossible to honestly state. 
There are some ratios that get thrown around, but you are never going to accurately predict the point that a battery will go flat with a varying load.  And if you add a charger to the mix, the current available from the charger gets into the math.  If a 5 amp charger in running, you are putting 4 amps into the battery and one amp is going to the radio.  But if the radio is transmitting, you are supplying 5 amps to the radio and the battery is supplying 5 amps and is discharging during transmit.  Then it goes back to receive and we go back to the 1 and 4 amp feeds. 
I am hoping that sort of helps. And it shows you that trying to make an accurate calculation is very difficult, and while you can get reasonably close it's still sort of a guessing game.

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