My dad has acquired an outdoor anemometer that’s cleverly built around a bicycle computer. The bicycle computer uses a couple of coin batteries (total voltage, 3V), and he wants to convert it to plug-in so he doesn’t have to keep buying batteries for it. I’ll be visiting in a couple of weeks, and he asked me to dig through my junk heap to find a 3Vdc power supply to bring with me.
Crap: the closest I could find was a 5Vdc power supply.
Question: is it likely that I will be able to power this bicycle computer from a 5Vdc power supply without frying it?
Diodes, great idea, thanks; wish I’d thought of that.
It looks like 1N4001 has about 1.1 volts drop. Wouldn’t I just want two of them, so as to drop 2.2 volts and leave me just under three for the computer?
…
wait a minute…
…
I don’t know what my dad’s got, but estimating from the reported battery life of the Cateye Urban bicycle computer, and the capacity of its CR2032 batteries, it looks like the current flow to a bicycle computer is on the order of a milliamp. As I look at Figure 2 on this PDF data sheet, I see that for such tiny currents, the drop is more like 0.7 volts, which explains why three in series is appropriate.
Voltage dividers using simple resistors only work when the load is constant. If the load varies, then its effective impedance varies and the voltage division likewise varies.
You have to be careful with diodes. A diode doesn’t provide a constant voltage drop. At low currents the diode won’t drop the voltage much at all. A more accurate representation of a diode is a constant voltage drop plus a resistor. In reality the voltage/current curve is exponential so it doesn’t exactly match up with that either.
This is what they invented voltage regulators for.
A lot of battery powered devices have a voltage regulator built into them and can actually handle a fairly wide range of input voltages. You don’t really know for certain though unless the manufacturer’s specs say so or you open it up yourself and take a look.
I would worry more about the current than the voltage. The small CR2032 will be limiting the current due to its internal resistance, the large power supply (even with the diodes) may not.
I would use a potentiometer that will drop the voltage to 3v at the halfway point (approx) and supply the 1 mA current.
So - a simple calc gives you R = V/I = 2/1mA = 2k Ohms at the half point - so 4k Ohms overall. This is an approximation. Approximating to a 5k Potentiometer - I’d buy something like this.
I would start with the lowest voltage to your circuit and keep raising it until your anemometer works fine and leave it there.
Just wire up an adjustable linear voltage regulator. They’re incredibly straightforward to configure. That way you can have exactly 3 volts over a range of loads.
Another thing to note is that if you take a normal adapter type power supply, the voltage is at the rated current output (at least for a decently reputed manufacturer). 1mA is pretty much open circuit voltage which can be easily 20% to 50% more. Then you get into rectification issues, power surges etc.
I would recommend just going to 2 AA battery supply which should last plenty (would still recommend a current limiter).
