So i am going to show you how to power nearly anything off a USB port
This is for the USB competition
USB runs at 5v. The max current you can draw is 500ma. Therefore the max load is 5v x 0.5A=2.5. Watts. (W=VxI) If you try and draw more than 500mA, you may overload the port which will cause it to break
UPDATED PICTURES :)
Just a WARNING to look what your circuit needs if it needs a higher Voltage/current this can break your ports make sure you have got the correct details of the circuit WARNING
Remove these ads by
Signing UpStep 1: How Many Volts ?
Looking at what battery is being used
4 x aaa = OK
4 x aa = OK
USB runs at 5v. The max current you can draw is 500ma. Therefore the max load is 5v x 0.5A=2.5. Watts. (W=VxI) If you try and draw more than 500mA, you may overload the port which will cause it to break
Just a warning to look what your circuit needs if it needs higher this can break your ports
Or looking at a the transformer and see how many votls go in to the circuit ( if your 5 or under your ok )
and you can power of the Port :)
Most electronics nowadays are under 10 V and most of them Under 5V which is good for USB







































Visit Our Store »
Go Pro Today »




Voltage is controlled by the supply (the port in this case)
Current is controlled by the device connected to the supply
The port SUPPLIES 5 V
The device has to be rated at 5 V or ABOVE (if it is rated too high, the device won't function as the port won't supply enough voltage. If the device is rated lower than 5 V, you risk damaging the device, unless you limit the voltage before the device)
The device must DRAW no more than 500 mA at the voltage supplied. Remember that Power (voltage x current) is always constant to the device. This is important.
Say you have a device rated at 7 V, 500 mA, 3.5 W
If you supply it with 5 V, assuming it can handle the voltage variation, it will still draw 3.5 W.
So, P = VI, therefore I = P/V
I = 3.5 / 5
I = 0.7 A (700 mA) which is too high for your USB port.
So to check if you can power something:
First check the Voltage rating of the device is 5 V or OVER.
If it is 5 V exactly, check the current rating is 500 mA (0.5 A) or LESS
If the voltage rating is over 5 V, use the power consumption (calculated from the RATED voltage and the current if necessary) to calculated the new current draw at 5 V. If this is under 500 mA, then you can plug in :)
You go some way to explaining this, but saying that the device needs to be 5 V or under is incorrect.
I am looking for an instructable that shows how to modify a "mains" to usb wall plug down to 3.7 volts so as to replace the LiPo/Lion batteries with constant power.
Can you go into more detail about how to implement this voltage modification, or point me to an instructable if you know of one?
(i.e. if 2 diodes drops the 5v to 3.8 would that work?, where do the belong in the circuit?)
Thanks
I've been doing some research since I read your post, I've been looking into your other suggestion of using regulators.
A linear regulator seems to be the best candidate from what I read (for it's size and simplicity, and low voltage)
What would be the difference in using a resister instead of the diodes?
I am open for suggestions of course. In fact I'd be willing to start from scratch if a better suggestions was found.
My current idea is the AC to USB adapter is 95% there.
NOTE: the load will be < 150ma so I'm going for the SMALLEST solution available.
The adapter above is only 300-400ma, so pretty close to my specs.
??? How do those photo LED night lights deal with the mains AC?
Do all of the components operate on full line current? or is it converted somehow?
I tried to read the voltage across the LEDs but didn't have any success.
Here is a picture of what I mean: