Also, if you were charging a 7.5F 5.5V cap at 3.5V and ~150 mA, how long would it take to charge it to 3.5V?

I think that unless the diodes are really bad, a FULL BRIDGE RECTIFIER would be best for the quickest charging. And the rectifying diodes I have are not too terrible. I am probably going to go with one.

We begin with a first order calculation.

Charge Q = Capacitance C x Voltage V
Also Charge Q = Current I x Time t

Putting the two together

I x t = C x V
t = C x V / I = 7.5F x 3.5V / 0.15A = 175 seconds, i.e. about 3 minutes.

In reality, the charging current is not constant and diminishes as the capacitor charges. Hence the capacitor voltage follows an inverse exponential curve. Charging for three RC time constants would take about 10 minutes to charge.

 

We begin with a first order calculation.

Charge Q = Capacitance C x Voltage V
Also Charge Q = Current I x Time t

Putting the two together

I x t = C x V
t = C x V / I = 7.5F x 3.5V / 0.15A = 175 seconds, i.e. about 3 minutes.

In reality, the charging current is not constant and diminishes as the capacitor charges. Hence the capacitor voltage follows an inverse exponential curve. Charging for three RC time constants would take about 10 minutes to charge.

Thanks for the useful equations. So what would you say the best way to estimate the current going through the capacitor? Is there some constant or something that I could use to more accurately calculate the time? I have the data sheet for the capacitor, so would that help?

Also, how do I, given the capacitance and voltage it's charged to, determine how quickly it discharges? I heard that the voltage drops in a linear fashion over time, unlike a battery. Is that true not only hypothetically with ideal components, but in reality?

 

The best way to determine the current is to measure it under different conditions.

The discharge rate is a bit simpler.
If you assume that the discharge current is constant, then the time to discharge to half the initial voltage is

t = C x V / 2I

Another formula is to use the exponential decay curve.
If the discharge resistance R is constant, then the time constant = R x C is the time it would take to discharge to 37% of the initial voltage.

 

The best way to determine the current is to measure it under different conditions.

The discharge rate is a bit simpler.
If you assume that the discharge current is constant, then the time to discharge to half the initial voltage is

t = C x V / 2I

Another formula is to use the exponential decay curve.
If the discharge resistance R is constant, then the time constant = R x C is the time it would take to discharge to 37% of the initial voltage.

Is it t=C*V/(2I) or t=C*V*I/2?

Also, maybe I should start another thread for discussing the supercapaciators.

 

First suggested improvement: get some schottky diodes. You will lose a lot of the generated power in silicon diodes in such a low voltage system.https://www.mouser.co.uk/productdetail/stmicroelectronics/1n5817?qs=sGAEpiMZZMtQ8nqTKtFS/D9SVzsgHTKGsrEMHLFTAoc=light

Thanks AlbertHall. Bought this last week and really I like this schottky diodes.

 

Can you please upload the results once you are done. It will be good to know if it works and if the hand winded transformer works.

 

I ended up giving up. Some wires broke and could not be soldered.