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Battery voltage or Amps
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Jan 23, 2016 08:59:28   #
Longshadow Loc: Audubon, PA, United States
 
As batteries age, internal resistance goes up. The curve is different for each battery type.

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Jan 23, 2016 09:35:33   #
radioengineer
 
In simple terms, a battery pack is made form a series of cells, each additive. In Nicad or NiMi each cell is 1.2 volts. In a lead acid battery like in a car each cell is 2 Volts.

Often times one cell will short out reducing the total by that amount. I saw the word surface charge earlier in the posts. That is what happens with a pack having a shorted cell being recharged. The shorted cell remains dead (shorted) with the remaining good cells picking more than their designed 1.2 Volts each. A volt meter has almost no load, but put a 12V bulb across the pack while taking the voltage reading the weak pack is also providing current for the bulb which helps to determine the quality of the cells providing power. Any current draw very quickly drops the false surface charge. Find a method to have some load (current) while also checking the voltage. Volts x Amps = Watts. Watts is power.

I have found allowing a pack to get too low before recharging will cause a weak cell to become reversed charged ruining that cell.

Over charging cells to the point they become warm deceases life.

It is possible to zap one cell back to life but the pack has to be dissected and access to the bad cell exposed. After trying to zap some cells back it has been my experience they aren't dependable.

Maybe this Wiki on batteries will help https://en.wikipedia.org/wiki/Battery_(electricity)

Dave

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Jan 23, 2016 09:44:27   #
jeep_daddy Loc: Prescott AZ
 
Swayne wrote:
My question is, Can a used battery that was 11.1 volts now fully recharged to 12.6 volts have an amperage problem and cause it not to work? Do battery’s go bad and show fully charged but the Amperage is still low or no good?


Yes, batteries get weak particularly the cells can become weak. When this happens, then they have less amps. The amps is the power, not the volts. If a 6 cell had one completely dead cell, then the bad cell would cause the voltage to not be as high thus you'd know it was a bad battery. But weak cells are harder to tell so they need to be load tested.

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Jan 23, 2016 09:46:28   #
burkphoto Loc: High Point, NC
 
Does the inverter work to power the Dynalite pack with ANOTHER battery?

As I recall, Dynalite packs recycle very quickly, and the power draw is about ten amps at 120 volts. That can choke some lesser inverters, especially if their battery sources are weak!

I had good success powering Norman packs with inverters, but never had an inverter with enough kick for our Dynalite packs.

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Jan 23, 2016 09:52:15   #
BebuLamar
 
burkphoto wrote:
Does the inverter work to power the Dynalite pack with ANOTHER battery?

As I recall, Dynalite packs recycle very quickly, and the power draw is about ten amps at 120 volts. That can choke some lesser inverters, especially if their battery sources are weak!

I had good success powering Norman packs with inverters, but never had an inverter with enough kick for our Dynalite packs.


If the Dynalite draws 10A then the inverter must be drawing at least 100A at its input. With 100A drawing off the battery its voltage can't be 12V but rather my guess around 8 to 10V. But the OP noticed that the voltage doesn't drop much underload which suggested to me that the inverter wasn't drawing power from the battery. The problem could be internal to the inverter, wiring from battery to inverter but it can be that the Dynalite was defective and wasn't drawing much current from the inverter.

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Jan 23, 2016 09:52:34   #
Longshadow Loc: Audubon, PA, United States
 
burkphoto wrote:
...
As I recall, Dynalite packs recycle very quickly, and the power draw is about ten amps at 120 volts. That can choke some lesser inverters, especially if their battery sources are weak!
...


Most people don't realize about inverters, power out=power in (less the efficiency). So 120v at 10 amps out needs 100 amps in at 12v.

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Jan 23, 2016 10:43:12   #
TriX Loc: Raleigh, NC
 
joer wrote:
If remember correctly a fully charged car battery with no load applied is 14v.


A fully charged 6 cell (12v) lead acid battery is approximately 12.7 volts with no load (1.265 specific gravity)

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Jan 23, 2016 10:54:18   #
radioengineer
 
To comment on inverters and efficiency. Think of a teeter totter with two kids each 50 pounds on each end. Its balanced.

Same Watts In, Same Watts Out, is a perfect 100% efficient device.

Couple reasons inverters are used.

A transformer can not transform DC, it has to be either AC or pulsed DC.

If you want 120Volts AC in a boat or RV an inverter is used to pulse the dc voltage upward from 12V to 120V.

In the home with AC power at the socket a transformer is best choice as it can be transformed up to hundreds of volts or downward to lower volts like 12V.

Transformers are made of metal and wire and cost more money. Manufacturers starting using switching power supplies for computes, battery charges, etc.

For computers, battery chargers switching power supplies are cheaper and lighter.

In the ham radio world a linear power supply (transformer) is preferred as they don't spew out noise heard on a radio. Because of weight and added expense, along came switching power supplies (inverters).

In designing a solar panel system efficiencies down to the last millamp is calculated to conserve battery power. The best efficiency I found on DC inverters was about 90%. That meant to get the same power out (Watts) I had to supply 10% more going in than what I need out.

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Jan 23, 2016 11:34:34   #
Marionsho Loc: Kansas
 
anotherview wrote:
Ohm's Law says with the circuit resistance constant, a drop in voltage will produce a drop in amperage.

Hence, assuming a good battery, its charge to a higher voltage will produce a higher amperage.

This is a simple explanation.

I assume you refer here to a car battery.


R=ExI. If R remains the same. E goes down, I goes up.

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Jan 23, 2016 11:58:44   #
radioengineer
 
Regarding a load, such as a camera needing power, its takes voltage to produce the current needed. If the push is less (Voltage) then resistance (Ohms) being constant in the camera, the current (amps) is less.

A drop in voltage source doesn't produce more current, there is less current.

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Jan 23, 2016 12:09:22   #
Marionsho Loc: Kansas
 
radioengineer wrote:
Regarding a load, such as a camera needing power, its takes voltage to produce the current needed. If the push is less (Voltage) then resistance (Ohms) being constant in the camera, the current (amps) is less.

A drop in voltage source doesn't produce more current, there is less current.


R=ExI. 2 ohm (remaining constant.) E=1, I=2. Hense 1x2=2. Drop the voltage to .1, .1x20=2.

It would be soooo convenient if it worked the way you have described it.

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Jan 23, 2016 12:16:02   #
Apaflo Loc: Anchorage, Alaska
 
Marionsho wrote:
R=ExI. 2 ohm (remaining constant.) E=1, I=2. Hense 1x2=2. Drop the voltage to .1, .1x20=2.

It would be soooo convenient if it worked the way you have described it.

It works with Ohm's Law.

E = IR

You are using the formula for power, P = EI, where P is power, not resistance.

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Jan 23, 2016 12:16:17   #
radioengineer
 
Been doing electronics since 1958, professionally since 1961. 35 years as a broadcast engineer, Extra Class Amateur License, General Class FCC License, Certified Professional Broadcast Engineer. Still in the business.

I love photography and do not profess to offer anything to this group regarding taking pictures.

Dave

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Jan 23, 2016 12:28:21   #
Marionsho Loc: Kansas
 
Apaflo wrote:
It works with Ohm's Law.

E = IR

You are using the formula for power, P = EI, where P is power, not resistance.


You are correct. R doe's NOT = EXI. Brain fart, on my part. Thank you very much.

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Jan 23, 2016 12:35:02   #
Longshadow Loc: Audubon, PA, United States
 
Marionsho wrote:
You are correct. R doe's NOT = EXI. Brain fart, on my part. Thank you very much.


Just remember PIE (P=IxE).

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