[SOLVED] Voltage divider 5V to 3V to replace 2 AAA batteries with DC power supply?

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Thread Starter


Joined Jun 16, 2022
I have this portable rechargeable battery pack for charging smartphones. It is 5V, 2.4A, and 5000mAh. I would like to use it to power a wireless mouse that normally requires 2 AAA batteries, which is 3V and I'm not sure what is the A and mAh. I made a voltage divider that involves a 220Ω resistor and a 330Ω resistor so that it outputs 3V, though I don't know what the current is. However, when I connected it to the mouse, the mouse didn't turn on. Can somebody let me know what is the issue with this circuit?



Joined Feb 24, 2006
This is a very, very bad idea. The problem is that the output of a voltage divider depends on the amount of current delivered to the load. If the wireless mouse requires too much current the voltage at the junction of the two resistors will drop in such a way that the required current has two paths to ground. One path is through the bottom resistor and the other is through the wireless mouse. I'm not at all surprised the mouse did not turn on. A voltage divider is a signal level type circuit, not a power type circuit.

To see what happens apply Ohms Law to the divider without any load attached. 330 Ω + 220 Ω = 550 Ω. The power supply puts out 5 Volts with substantial current capability, but the voltage divider requires only:

\( 5\text{ Volts}/555\;\Omega\;=\;9.09...\text{ ma} \)

Assume that you connect a load that requires 5 ma. Now we have to compute the voltages and currents all over again. So we have a voltage divider and a 5 ma load. Can you figure out the current in the divider and the voltage at the midpoint?

\( (5-V_1)/220\;+\;V_1/330\;+\;5\text{ ma.}\;=0 \)
\( 5/220\;-\;5\text{ ma.}\;=\;V_1/330\;+\;V_1/220 \)
\( V_1\;=\;2.34\text{ Volts} \)

Not enough to power the mouse

The current through the 330 Ω resistor is 2.34V/330Ω = 7.09 ma.
thtotal current through th2 220 Ω resistor is 5 + 7.09 = 12.09, which can be confirmed because (5 - 2.34)/220 Ω = 12.09 ma.

There are several ways to accomplish your goal and they involve one or more type of voltage regulators. You don't happen to know how much current the mouse requires: do you?

Question: Did somebody try to prank you with this suggestion, or did it sound like a good idea until you tried it?
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Joined Mar 14, 2008
Three silicon diodes (e.g. 1N4148) in series will drop about 2V at the low current the mouse likely draws (a few mA), so that should work for you.

But what's the point in converting a wireless mouse to a wired one?
Why not just buy a wired mouse?

Thread Starter


Joined Jun 16, 2022
Ok, thanks for the information! Sorry, I'm just pretty inexperienced with circuits, so voltage dividers were the only thing I knew about. The 3 diodes worked!


Joined Mar 30, 2015
Welcome to AAC!
I'm just pretty inexperienced with circuits, so voltage dividers were the only thing I knew about.
The first "voltage regulator" they teach students is a voltage divider. Once you see how inefficient they are, you move on to more efficient methods.
The 3 diodes worked!
For higher current requirements, an LDO (Low DropOut) voltage regulator would be best. Dropping 5V to 3.3V is so common that there are 3 terminal device solutions.

Jon Chandler

Joined Jun 12, 2008
...Once you see how inefficient they are, you move on to more efficient methods....
The problem is understanding the proper use of a voltage divider. It's not meant to supply power to a load. It's for creating a reference voltage or for scaling a voltage that will measured by a high impedance input.

If you try to draw appreciable current from a voltage divider, the concept falls apart. A basic understanding of Ohm's Law makes this simple to understand.
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