Charging supercapacitor using peltier module

Using peltier module and with additional circuitery i was able to achieve a dc voltage of 5V and current 5mA. I need to charge a supercap but the current is too low for that. Any idea about how can i charge a supercapacitor with such low current in say 4-5 hours?

• What is the additional circuitry? It will help people answer if you provide a schematic, as there could be something in this 'additional circuitry' that could be amended to achieve a better result – MCG Jun 2 '17 at 12:37
• Use a smaller value of supercapacitor! If 5mA at 5V is all you have, that's all you have – Dirk Bruere Jun 2 '17 at 12:40
• Use a bigger TEG? Or apply a greater temperature difference across your TEG? Or use a more efficient boost(?) circuit? – The Photon Jun 2 '17 at 15:37

The rate at which any capacitor charges is given by the formula:

t = RC

where t is the time in seconds, R is the value of resistance in ohms, and C is the value of capacitance in Farads. This will take the capacitor to a ~63% charge. To take the capacitor to nearly full charge, it will take 5 times that long.

Knowing that there are 3600 seconds in an hour, you can then re-arrange the formula to calculate the resistance needed for full charge in 4 hours:

R = 14400*C/5

So if you have a super capacitor of 1.0 Farad, you will require a 2.88k ohm resistor in series with the capacitor.

Then you want to check that you will not exceed your power supply current. The maximum current occurs when the capacitor has no charge and the charging cycle begins. At this point, it is as if the capacitor is short circuited so we can use Ohm's law to calculate the maximum current:

I = E/R = 5 volts / 2880 ohms = 1.7 mA

So this would certainly work within your power supply current limitations.

• Err... No? I=C*dU/dt so It=CU of you start from zero and with a constant current of 5 mA for 5 hours, you end up with 18 F to reach 5 V. If you have a constant power of 5 mA * 5 V for 5 hours, you end up with 450 joule or E=CU^2/2 so C=36 F. – winny Jun 2 '17 at 18:19
• On the other hand, it is a constant voltage supply...so yes my formula is correct. – Glenn W9IQ Jun 2 '17 at 18:56