Questions about capacitors, and arduino

[jehan60188]

hey all, I've been playing around with capacitors and have a few questions

Is it possible to get 5 volts over a capacitor using a AA batery? Or am I limited to 1.5 volts?
If I have more than 5 volts going into an arduino pin, that will cause damage. But I can use a voltage divider to drop that, right?

I want to use an arduino to control capacitor charging/discharging, but in order to avoid damaging the arduino, I want to use it only for controlling a relay that acts as a two position switch- one position for charging, another for discharging, and for measuring the current going into the arduino.

how can I do that second part? if I put a pin between the battery's positive terminal, and capacitor, can I have that go an analog pin for measuring?

thanks!

 

[Bob S]

A simple way to get 5 volts using a single AA battery is to charge 4 identical capacitors to 1.5 volts, then use a 4-pole double throw (4PDT) relay switch to put them in series to get 6 volts. A diode drop will reduce this to about 5.3 volts.

 

[vk6kro]

Another way is to use a solar powered garden light.

These are often discarded because the NiCd battery in them fails.

However, there is a circuit in them that converts the output of a NiCd battery (1.2 volts) to about 3.5 volts to run a white LED.

If you disconnect the solar cell, the circuit will work as if it was in darkness and give several volts out from a 1.5 volt dry cell.

 

[fgdownsups]

It seems the simplest to use the arduino itself to charge the capacitor to 5v, using a resistor if you are worried about damage to the arduino.
To measure the discharging of the capacitor, connect the negative lead of the capacitor to ground of the arduino and the other lead of the capacitor to an analog pin.

 

[jehan60188]

thanks for the info everyone, i picked 5V arbitrarily, so using the arduino's output pin isn't possible (Since I may choose 10, or 25 or whatever volts)- hence the relay (to isolate the capacitors from the arduino)

iirc, putting them in series will lower the capacitance (since they sum in inverse)
v=q/c
c goes down, q stays constant, so v increases!