Powering a Device with AAA Batteries and USB

AI Thread Summary
To power a device that uses two AAA batteries with USB, it's essential to understand the voltage and current requirements. AAA batteries provide 1.5 volts each, totaling 3 volts, while USB supplies 5 volts at a maximum of 500 mA. To adapt the USB output, using four silicon diodes in series can drop the voltage to around 2.6 volts, suitable for the device. It's crucial to measure the device's current draw with a multimeter, as USB can only supply up to 500 mA, and exceeding this limit could cause issues. Properly configuring the power supply ensures the device operates efficiently without frequent battery replacements.
nikolai850
Messages
2
Reaction score
0
so i have a device which uses 2 AAA batteries. i don't like charging batteries all the time so instead i want to power the device via usb.

a AAA battery: 1.5 volts and 300-1000 mAh (milliampere hours)
usb has 5 volts and max current is 500-900 mA

how do i do this?
 
Engineering news on Phys.org
What current does the device need? Does it fit USB specification?

Knowing internal resistance of the device you may be able to add a resistor in series, so that 3V are for the device and 2V are lost on the resistor.
 
Since you know the amp-hour capacity of your batteries, I guess they must be NiCd or NiMH and not alkalines, because alkalines do not have this written on them.

So, you need about 2.4 volts - 2.6 volts

You will need to know the current that your device draws. I think the limit for USB is 500 mA, so if your device draws more than that, you should not use USB.

Probably the easiest way of doing this is to put 4 silicon diodes (1N4007s or similar) in series from the +5 volts. These diodes can handle up to 1 amp. This is more than USB can deliver.

This will drop about 2.4 volts leaving you with about 2.6 volts which is similar to the output of two rechargeable AAA cells. This 2.4 volts is fairly constant but it does increase with greater current.
 
how do i find out how much current the device draws?
 
What is it?

If you want to do any electronics, you are going to need a multimeter. Have a look at these ones:
http://www.harborfreight.com/catalogsearch/result?category=&q=multimeter
There is one there for less than $4 which would be quite OK for this.

You can get similar digital meters at other Electronics stores or get them on Ebay like this:
http://cgi.ebay.com.au/New-LCD-Digital-Voltmeter-Ammeter-Ohm-Multimeter-DT830B-/260517261237?pt=AU_B_I_Electrical_Test_Equipment&hash=item3ca809efb5
Posted to your door for about $4.

To measure current, you have to put the meter (which is set to measure current, on the highest range to start with) in series with one side of the battery.
 
USB hosts will supply at least 100 mA at a nominal 5V. Call it greater than 4.6 to 5.0V. The 500 mA capability a potential USB host can supply is determined by negociation with the host. You're not going to design your own USB device capable of communicating with the host, so presume 100 mA for low power devices with USB hosting. However, you can presume that the USB port can supply 500 mA out of something like a PC but crash something less powerful than a PC.

If you just want to charge from a PC assume you can get 4.6-5.0 volts drawing 500 mA maximum current.
 
Last edited:
Very basic question. Consider a 3-terminal device with terminals say A,B,C. Kirchhoff Current Law (KCL) and Kirchhoff Voltage Law (KVL) establish two relationships between the 3 currents entering the terminals and the 3 terminal's voltage pairs respectively. So we have 2 equations in 6 unknowns. To proceed further we need two more (independent) equations in order to solve the circuit the 3-terminal device is connected to (basically one treats such a device as an unbalanced two-port...
suppose you have two capacitors with a 0.1 Farad value and 12 VDC rating. label these as A and B. label the terminals of each as 1 and 2. you also have a voltmeter with a 40 volt linear range for DC. you also have a 9 volt DC power supply fed by mains. you charge each capacitor to 9 volts with terminal 1 being - (negative) and terminal 2 being + (positive). you connect the voltmeter to terminal A2 and to terminal B1. does it read any voltage? can - of one capacitor discharge + of the...
Back
Top