Need circuit explanation

iONic

Joined Nov 16, 2007
1,650
Can anyone explain how the 150V, 160uF cap gets charged using the step-up
transformer? I believe the primary to Secondary ratio is aprox. 1:200.

Thanks

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iONic

Joined Nov 16, 2007
1,650
The circuit is actually part of a flash circuit from a disposable camera. I removed the flash unit and the trigger coil to set the flash off.

Tahmid

Joined Jul 2, 2008
343
Hi,
The - terminal of the capacitor receives - charge through the diode. The + terminal of the capacitor is charged through the 470p cap. This is a half wave rectifier. One part of the capacitor receives AC. This passes through the 470p capacitor when this part of the AC waveform is positive. During the same cycle, the negative travels through the diode. During the next cycle the cap is not charged as positive charge does not go through the diode in your diagram. During this cycle, even though the + terminal of the capacitor receives positive charge, it does not explode as there is nothing on the - terminal.

Thanks.

iONic

Joined Nov 16, 2007
1,650
So there is an oscillation taking place. This is the part I am interested in. In other words you could use a 555 to create an oscillator. Is there any Advantages of using a higher voltage source, say 3V - 5VDC!

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studiot

Joined Nov 9, 2007
4,998
In principle yes, but the original had a tuned circuit which produces a sinusoidal oscillatioon voltage. This works best with transformers so your output may well be rather less than you expect with a nonsinusoidal waveform. A 555 oscillator can produce square waves or sawtooth waves. Your suggested relaxation oscillator produces sawtooth waves. Don't forget the 'form factor' for sawtooth and triangle waves is onlly 1/3. The average (equivalent DC) level is the form factor times the peak so you won't have gained anything.

iONic

Joined Nov 16, 2007
1,650
Thought that might complicate things a bit. So what if I were to change the source voltage to 3V from 1.5, would this increase the charge rate? or would I have to change the 150K resistor or the 470pF cap.

And also what would change if I replaced the 160uF/150V cap with an 80uF/150V cap?

thingmaker3

Joined May 16, 2005
5,084
Thought that might complicate things a bit. So what if I were to change the source voltage to 3V from 1.5, would this increase the charge rate?
The charge rate (percent of full charge per unit time) would remain the same, as it is governed only by capacitance and resistance - not by voltage. http://www.allaboutcircuits.com/vol_1/chpt_16/2.html

The ammount of charge (coulombs) will be reduced. 1 Farad = 1 Coulomb/Volt, so 1 Coulomb = 1 Farad * 1 Volt. What happens when you reduce the Volts by half?

And also what would change if I replaced the 160uF/150V cap with an 80uF/150V cap?
What would change in the above equation if you reduced the number of Farads by half?

SgtWookie

Joined Jul 17, 2007
22,210
I'm curious as to what you're actually attempting to accomplish?

The 2SD965 is a cool find. Beats a 2N2222 to heck and gone. Great gain, amazing current carrying capabilities in a TO-92 package.

But what are you really trying to do with this?

You can decrease the charge time by decreasing the size of the cap. However, you won't have as much energy available. The cap voltage rating is nearly the voltage applied; don't expect it to last very long.

If you want it to last and to charge faster, use a cap rated 2x the voltage being applied.

iONic

Joined Nov 16, 2007
1,650
My thoughts were to make a shocking circuit to keep the squires away from my bird food. I would of course be cutting the current way way down with a series resistance and the charged capacitor, maybe 200k to 300k. Maybe I don't even need a battery but just a small solar cell to keep the cap charged.

studiot

Joined Nov 9, 2007
4,998
Won't it also zap the birds?

Audioguru

Joined Dec 20, 2007
11,249
A Squire is a knight??? Are they bothering you???

The shocking circuit doesn't care if the victim is a bird or a squirrel (or a squire).
Goodbye birds and squirrels (and squires).
Maybe a black bear will survive.