Arghhh! don't get in the habit of "trying to measure the current from a battery" by putting an ammeter across the terminals.

It's just a bad idea.

 

Arghhh! don't get in the habit of "trying to measure the current from a battery" by putting an ammeter across the terminals.

It's just a bad idea.

Yeah. I swear I'll never short circuit when I measure current again!

But putting a 10ohm resistor in series with the ammeter WORKED! Yay! No more short circuit protection.

 

I'm trying to make a life logging camera because the products sold for life logging is rediculously overpriced.
Basically, a lifelogging camera is just a normal camera with a rediculously long battery life and storage.

So I bought a Keychain "Spycamera" from China and I'm considering removing the 3.7V lipo battery and replacing it with a USB Power Bank(5V, 2A, 10Ah) so the camera would last a VERY long time. I'd put in a 16gb microsd and I calculated for the low resolution, I'd be 4GB/day or ~1.5TB/Year or ~100TB-150TB/Lifetime.....pretty good to be able to have a record of an entire life.

I NEED the 2A or at least 1A...

You say you need 2 A and the power bank you are using is 10 Ah. Even with a 100% efficient switcher to get to 3.7 V that's at best 6.7 hours of operation, and that's assuming you can deplete the power pack. I don't know how much they are designed to keep in reserve before shutting down (or perhaps the 10 Ah rating reflects the useable capacity, but knowing marketing types, I doubt it).

 

You say you need 2 A and the power bank you are using is 10 Ah. Even with a 100% efficient switcher to get to 3.7 V that's at best 6.7 hours of operation, and that's assuming you can deplete the power pack. I don't know how much they are designed to keep in reserve before shutting down (or perhaps the 10 Ah rating reflects the useable capacity, but knowing marketing types, I doubt it).

Oh by 2 amps, I meant I had measured the 3.7v lipo battery of the camera current short circuited. I wanted to measure the voltage and current of the included lipo battery which was 3.7v, 2A. And then I would mimick that with the power bank.

Because I thought just knowing the voltage of the lipo battery was not enough since all 3.7V doesn't necessary mean the current is the same.
The battery short circuit gives 2A but I think the camera itself uses much less because it's lipo battery is only 200mah but it lasts 1 hour.

In other words, I did not measure the current draw of the camera itself, just the current output of the lipo battery so I could find a corresponding battery

Here's a question I've been having for a while.
Do all batteries/power supply the Same Current with the same voltage?
Does a 5V arduino supply the same current, as 5V USB, as 5V lipo, as 5V li-ion, etc.
I'm assume it doesn't due to difference resistence, so I was trying to know what this lipo's current was.

I assumed if the current was too low, it wouldn't power on, and if it was too high, it would burn out even if the voltage is the same.

 

The battery short circuit gives 2A

the battery pack is liked rated 2a max - fairly typical of usb chargers.

a lipo pack can produce current substantially higher than 2a.

 

"Intelligent" USB powered devices go through a process called "enumeration" when they are first plugged in to an "intelligent" host. That process uses signalling on the data+ and data- leads. Only when the enumeration process is completed does the "intelligent" host supply power to the "intelligent" device. When powering an "intelligent" USB device with a "dumb" USB power source (one that does not use the enumeration process,) the data+ and data- leads should be shorted together in order to power the "intelligent" device.

 

the battery pack is liked rated 2a max - fairly typical of usb chargers.

a lipo pack can produce current substantially higher than 2a.

The meter says 2A for the lipo, 3.8V.
My Power bank is 2A, 3.8V (Even though it's supposed to be 5V..but confirmed on multiple meters).

Anyways the powerbank successfully substitutes for the lipo, and I've been recording hours of video with it clipped to my shirt
It's mounted using Lego technic haha.

 

Oh by 2 amps, I meant I had measured the 3.7v lipo battery of the camera current short circuited. I wanted to measure the voltage and current of the included lipo battery which was 3.7v, 2A. And then I would mimick that with the power bank.

That approach can get you in a lot of trouble -- try measuring the short-circuit current of your car battery and you will immediately understand what I mean. (Hint -- DON'T actually do this!).

Here's a question I've been having for a while.
Do all batteries/power supply the Same Current with the same voltage?
Does a 5V arduino supply the same current, as 5V USB, as 5V lipo, as 5V li-ion, etc.
I'm assume it doesn't due to difference resistence, so I was trying to know what this lipo's current was.

The current capacity of a power source is pretty much independent of its output voltage. A 13,000 V neon sign power supply might be only capable of outputting a few tens of milliamps while a 13,000 V power distribution line might be able to deliver a few hundred amps. One 2 V lead-acid cell might be able to deliver a few amps and another might be able to deliver a few thousand amps.

I assumed if the current was too low, it wouldn't power on, and if it was too high, it would burn out even if the voltage is the same.

The first part is basically correct -- if you try to draw more current from a power source than it is capable of delivering, the voltage will drop. Exactly what happens after that depends on the voltage source. If it's a battery, then often times the voltage will drop until the demand at the lower voltage matches the current that it IS able to deliver. For other batteries, the extreme current draw causes them to heat up and go into thermal runaway resulting in an explosion of fire. For more active power supplies, they might go into a current-limiting mode or they might shut down completely.