# All the power in the world and not enough battery!

Discussion in 'Amateur Radio News' started by VK6FLAB, Mar 2, 2019.

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1. ### WA6VVCHam MemberQRZ Page

I was thinking the same thing with the diodes. Add a switch across the diodes when your down to the end and you'er on reserve power time. Seems to me it would provide a little more voltage headroom for some of those sensitive less that 12 volt rigs. NO???

2. ### K3KICHam MemberQRZ Page

That's not how you calculate efficiency. You need to calculate the total energy used and deduct the amount lost in the diodes.

3. ### KD6VXIHam MemberQRZ Page

The rig pulls 22 amps.

At 1.2 volts drop, that's 24ish watts.

For a rig that pulls 22 amps at the given 16 volts we are looking at 352 watts of input power from the batteries, give or take.

My simplistic formula was still pretty damn close, and he's still in the 90s.

Being pedantic doesn't make you right.

Even if you wanted rig input power alculate the power used that's 22x13.8= 304 watts. And the 22 watts wasted by the diodes still puts him in the upper 80s this way.

And you still don't have the problems with a DC to DC cheap chicom switcher.

If you have a problem with the math, show where it's wrong so people can learn. Please don't just sideswipe with a fairly useless comment.
--Shane
KD6VXI

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4. ### K3KICHam MemberQRZ Page

You need to calculate Watt hours or similar units to determine which approach is more efficient as both battery voltage and current draw is dynamic during discharge.
Your response included the comment that math was easy for some of use and then you proceeded in using the wrong analysis.
The numbers you provide above are an instantaneous calculation of losses at the start. Not across the full discharge cycle.
KF5LJW stated that 15% of the total power would be lost using the diode approach. Without running the numbers my educated guess is that he is right. I just would have presented that thought differently. (I've done many battery discharge tests for projects at work.)

Last edited: Mar 7, 2019
5. ### KD6VXIHam MemberQRZ Page

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I used an apples to apples analysis.

I was comparing the efficiency of the DC to DC converter vs his diode drop. I also stated that it was being simplified by thinking constant carrier.

I didn't touch on the dynamic draw, duty cycle, battery VDrop over time, etc.

His DC to DC converter didn't take any of that into account, either.

Please don't try to change my intent of the post to fit an argument.

--Shane
KD6VXI

6. ### K3KICHam MemberQRZ Page

KL5LJW referenced battery capacity. Neither of your calculations related to capacity. I think knowledgeable readers of the thread would have noticed this as did I.
I merely pointed it out because of your quirky comment.
As someone who operates remotely I wouldn't throw capacity away when I'm on a mountaintop and worked hard to haul that weight up the hill. So I wouldn't use diodes to drop the voltage.
I just hook up the battery with the understanding that it's a tad over the voltage spec for a while.

7. ### 4Z1UGPlatinum SubscriberPlatinum SubscriberQRZ Page

Could you use one of the Chinese buck converters that keep the output voltage constant as the battery voltage starts above the 13.8 then falls below it?

8. ### N7KOHam MemberQRZ Page

Thank you, Very good information, and well written. You shed some light in that Rabbit Hole.