Solar cell, Joule thief, and capacitor ...

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vbdev

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I am trying to use 0.5V/80mA **broken link removed** to charge a capacitor at higher voltage & then use all of the capacitor's energy emit **broken link removed**.

I do not want to power the load continuously. It may be such that as soon as the capacitor voltage rises to say 5V, the capacitors' energy will be delivered to the load until the capacitor is fully exhausted.

I am planning to use joule thief circuit to charge the capacitor at higher voltage, Will I need an another joule thief circuit at the output of the capacitor to fully utilize its charge? I have not yet started to build the projects but any thoughts are welcome. Thanks!!!

I don't want to use super-capacitors.
 

A capacitor is not a battery. Its voltage begins to drop very quickly then your sound making thing will begin loud and get less loud very quickly.
A Joule Thief circuit has a fairly low output current so it will take it a long time to charge a high value capacitor.

The alarm might not be loud because its level is 90dB to 105dB at what distance? Only if you hold it to your ear? It is small and lightweight and is powered by tiny button battery cells so it might simply make a little beep.
 

There is a chance your 0.5 V supply will provide sufficient bias to turn on a transistor, so that oscillations can get underway.

Simulation (theoretical):



The diode is necessary to prevent the capacitor from discharging through the transistor.

Where it has the switch and load, you will install a circuit which:

(a) detects when the capacitor has charged to a high enough level,
(b) turns on your sound generator,
(c) shuts off when the capacitor has discharged to a low level.
 

The simulation shows that the capacitor will take a long time to charge to 5V. Then when the alarm is powered from the capacitor it will make a little beep for only 0.05 seconds when the discharging capacitor voltage has dropped to 1.85V when the beep will be at a very low sound level if it still works at such a low voltage.
 

I will use 2 solar cells in series to get 1V. Secondly I will use a large capacitor such as 2200u/6.3V to store more energy.

A Joule Thief circuit has a fairly low output current so it will take it a long time to charge a high value capacitor.
Click to expand...

It will be okay even if it takes several minutes.

I will try using a **broken link removed** like arrangement after the diode & capacitor.
 

For each cycle of the Joule Thief oscillation, the output capacitor charges for only a moment. It might take too long for the capacitor to charge for the beeper to beep for a second burglar to trigger the beeper circuit.

Nothing limits the output voltage of a Joule Thief except its 1.8V to 3.5V LED so the beeper circuit might be destroyed if the capacitor charges to 12V or more.

Since the capacitor is not a battery its discharge voltage drops very quickly so the beeper will beep for only a moment and the short duration beep will fade away.
 

Thanks Audioguru, I don't have an issue about the long interval, the only thing I need is that the capacitor should store enough energy to make sound for at lease 2/3 seconds.

Regarding the capacitor voltage I will use 1381U which is mentioned in the solar engine. It will apply the capacitor voltage to load on reaching to 4.9V
 

vbdev said:
egarding the capacitor voltage I will use 1381U which is mentioned in the solar engine. It will apply the capacitor voltage to load on reaching to 4.9V
Click to expand...
Isn't the project supposed to be an alarm? Then you do not want the alarm to sound when the capacitor voltage reaches 4.9V, instead you want the sounder to make a sound when the magnetic sensor is triggered and you need something (a zener diode maybe?) to keep the charging voltage from exceeding 5V.
Then you do not need the 1381U voltage detector IC.
 

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