Capturing high frequency pulses

I hope you do not torture your eels to keep them shocking all the time =)

_nox_ said:

I hope you do not torture your eels to keep them shocking all the time =)

Click to expand...

Haha dun worry.. i will treat them well...

guys i got another query... referring to the charging circuit that i attached in my previous post... i tried using a 4700uF capacitor and use an oscilloscope to take the readings across the capacitor and i'm able to achieved 2.0V in just about 20mins. The oscilloscope was switched on throughout the duration.

If the oscilloscope was not being fixed to the circuit, the 4700uF is not able to charge up that fast probably it will only charge up to mV in 20mins.

And to make sure that it was not the oscilloscope that charge up the 4700uF cap, i fixed a discharged 4700uF across it without the charging circuit and no charge was being stored in the capacitor.

Can anyone explain why when the oscilloscope was attached to the charging circuit, the 4700uF is able to charge up much faster. Is it the principle of impedance matching??

Cheers

power from an eel?

see:
**broken link removed**

< After discharging a shock, the eel must recharge for
< nearly one hour to regain its full electric potential
< (Gerrow 2002).

The power from electric eel is limited. Even a large eel
that can pulse 600v, 1A can do so for a short burst say 10 msec and then the eel must recharge itself. see above

To measure the pulse voltage amplitude, a single fast diode with 1,000 v breakdown, can drive a 1uf mylar capacitor.

Then put other capacitors in parallel. watch the voltage
drop. remember power stored , p =1/2*C V^2

hawk2eye

Connecting the 'scope provided a ground to the charging circuit. Then the circuit's input picked up mains hum and radio waves and rectified them into DC that charged the 4700uF capacitor quicker.

hawk2eye said:

see:
The power from electric eel is limited. Even a large eel
that can pulse 600v, 1A can do so for a short burst say 10 msec and then the eel must recharge itself. see above

Click to expand...

Another site:
**broken link removed**

This one says 500V, 1A, 6msec discharge, 1 hour duty cycle => 0.8mW
Assuming 100% efficiency. Note that transfering energy directly into a capacitor without any series inductance is at most, 50% efficient, and having a big resistor in series ("to match the impedance") will burn even more power.