Using 14v or 16v battery for fishfinder?

Has anyone tried using either a 14v or 16v or higher for their fishfinder?

The bassboat guys have started doing that with high powered electronics (forward sonar and such) and it allegedly lasts longer and provides a better signal. Its probably mostly marketing, but I’m interested in the idea of a longer running ff battery since I’ve upgraded mine to a 9 inch screen.

Most new electronics can handle higher voltage inputs so I was mostly wondering about real world experiences.

Attached is a chart from a battery mfg so I have no way of certifying the info, but it’s an interesting chart.

This isn't marketing, it's Ohm's Law.  The current is directly proportional to the voltage.  If voltage goes up, current goes down.  If the device is rated for the current, then all is good. 

I = V / R where I is current, V is voltage, and R is resistance.  The resistance (load) is constant for the device, so as you increase V, I decreases.  Different battery chemistries offer different voltages.

Some of the 16V batteries are lead acid (usually AGM), some are LiFePo4.  AGM cells are ~2 VDC nominal, so they connect 8 in series to attain 16 VDC (your normal 12V battery is simply 6 cells in series).  LiFePo4 cells are ~3.2 VDC nominal, so they connect 5 of them in series to attain 16 VDC.

It comes down to the input voltage ratings of the device(s) you're powering, as you already noted.  Hope that helps!

I use Lithium batteries exclusively on my ff for over 15 years. You have many options. You can't say higher voltage equals longer output. You need to know how much power your ff needs. Sometimes ff manufacturers provide the power draw in amp. Other times they specify in watts. Power (watt) = Voltage x Current (amp). If you want a longer lasting battery, you simply need to buy a battery with a high capacity. There's no magic in having higher voltage.

This isn't marketing, it's Ohm's Law.  The current is directly proportional to the voltage.  If voltage goes up, current goes down.  If the device is rated for the current, then all is good. 

I = V / R where I is current, V is voltage, and R is resistance.  The resistance (load) is constant for the device, so as you increase V, I decreases.  Different battery chemistries offer different voltages.

Some of the 16V batteries are lead acid (usually AGM), some are LiFePo4.  AGM cells are ~2 VDC nominal, so they connect 8 in series to attain 16 VDC (your normal 12V battery is simply 6 cells in series).  LiFePo4 cells are ~3.2 VDC nominal, so they connect 5 of them in series to attain 16 VDC.

It comes down to the input voltage ratings of the device(s) you're powering, as you already noted.  Hope that helps!

Since resistance is constant, if voltage goes up, amps go down, but watts stay the same. So, 2 batteries that are rated the same in watt hours will last the same amount of time. But given 2 batteries that have the same Ah (amp hour) rating, the higher voltage one is going to last longer. Does that sound right?

This isn't marketing, it's Ohm's Law.  The current is directly proportional to the voltage.  If voltage goes up, current goes down.  If the device is rated for the current, then all is good. 

I = V / R where I is current, V is voltage, and R is resistance.  The resistance (load) is constant for the device, so as you increase V, I decreases.  Different battery chemistries offer different voltages.

Some of the 16V batteries are lead acid (usually AGM), some are LiFePo4.  AGM cells are ~2 VDC nominal, so they connect 8 in series to attain 16 VDC (your normal 12V battery is simply 6 cells in series).  LiFePo4 cells are ~3.2 VDC nominal, so they connect 5 of them in series to attain 16 VDC.

It comes down to the input voltage ratings of the device(s) you're powering, as you already noted.  Hope that helps!

Since resistance is constant, if voltage goes up, amps go down, but watts stay the same. So, 2 batteries that are rated the same in watt hours will last the same amount of time. But given 2 batteries that have the same Ah (amp hour) rating, the higher voltage one is going to last longer. Does that sound right?

That is correct. 

If I= V/R , and R is fixed, then as V increases I increases. Let R=1, then I=V

You sure you don't mean I= R/V ?

It's the power consumption of any electrical/electronic devices is what important.  The Ohm law P = V * I  proves that voltage and current are reverse proposional.

John,

tl;dr
if you skip the regulator, go with the higher voltage battery.  Battery watt hours is the most important factor.

dissertation:
those numbers look like they were all calculated, not measured - so I would not put much faith in them.  I put them into a spreadsheet and the 16V power is always 6.7% more than the 12V power, at least for the 13 I entered.

I recall you installed a voltage regulator in your battery box., and that it is a switching regulator (I think? I can't find any technical info on the nocqua regulator)  that can both lower and increase voltage.  Given you have that, all that matters for the battery behind it is how many Watt-Hours it has.  The regulator will be slightly more efficient dropping voltage vs increasing, so, if you had a higher V battery, that would be slightly better.

If the Voltage regulator is a linear regulator, and can only drop voltage - then the higher voltage battery would be better.  Also, if you wanted to skip the voltage regulator and connect directly to your FF, then the higher voltage battery would be better, as it will sink 1-2V below 'target' voltage as it drains.

I can't seem to find a good discharge chart for the Nocqua batteries.  I may need to capture this myself.

see ya,
Brian.

I would be running the battery straight to the FF, no voltage regulator wanted. Hoping that the higher voltage will keep the unit powered for longer before reaching the low voltage cutoff.