Powering camera using USB power

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Skedgy Sky

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I want to power my camera using USB power.

The camera is a Canon S95 and takes 3.7V (1000mAh) batteries which read an output voltage of about 4.1 when charged.

I want to keep the camera on for a long period of time so is it safe for me to use a few resistors (which I did) and decrease the USB output voltage to 4.1 and power the camera that way?

Will this cause damage to the camera?

Do milliamps also play some role in this that I haven't yet considered?
 

Ubrales

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Capacity is in Amp-hours. 100 Amp-hours could be 1 Amp for 100 hours OR 100 Amps for 1 hour (not a linear relationship).

Somewhere on your camera or in the manual, the Watts may be stated. This is a better indication for the Amp draw. In DC circuits, Watts = Amps X Voltage. From this you can do the calculations.
 

Skedgy Sky

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I kind of already figured that.

I'm just wondering if powering a camera using an external power source means that a certain amperage is needed and if using more than the amperage of the batteries would cause damage to the camera.
 

Ubrales

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When you know it is not working, stop! Clearly, more amps are required than the USB port can supply. Why find out whether damage could happen?
 

Skedgy Sky

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Well I don't really want to know if it'll get damaged; I don't want to damage it.

I'm trying to find out how many amps the power source needs to be able to power the camera (since I want the camera on for long periods of time).

If the batteries for this camera have a capacity of around 1000mAh (and output around 4v), does that mean I need a power supply that outputs around 4 volts at 1 amps? Does amperage work the same way for batteries as it does for regular dc power supplies?
 

Ubrales

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Capacity is in Amp-hours. 100 Amp-hours could be 1 Amp for 100 hours OR 100 Amps for 1 hour (not a linear relationship).

Somewhere on your camera or in the manual, the Watts may be stated. This is a better indication for the Amp draw. In DC circuits, Watts = Amps X Voltage. From this you can do the calculations.
 
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