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Author Topic: back current..  (Read 485 times)
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CR2032 Battery
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« on: February 02, 2010, 12:58:00 PM »

nagkakaproblem po kami eh..
nagkakaroon kasi ng back current.
pag shinoshort namin yung transistor, nashoshort na din yung SCR kahit wala pang input..

sa word ko na lang ginawa.. Grin
pagpasenyahan na..

thanks..
http://www.sendspace.com/file/3ffokm

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« on: February 02, 2010, 12:58:00 PM »

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zer0w1ng
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« Reply #1 on: February 02, 2010, 01:12:22 PM »

Paano mo isino-short ang transistor?

Nilagyan mo ng 1 sa base?
Kung ganoon, base sa diagram ang base-to-emitter junction ang nagpa-ilaw sa LED.
Try mo i-move ang transistor sa ground and the LED sa cathode ng SCR (ipalit ang location ng NPN at LED).
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« Reply #2 on: February 02, 2010, 01:41:13 PM »

ang problema po talaga..yung pagkakashort ng SCR which is di naman sana dapat ganun..

thank you po..

pag-aaralan po ulit namin..
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TinTopHack
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« Reply #3 on: February 02, 2010, 08:24:52 PM »

Your circuit has several problems.
1. Your LED is connected to ground while the transistor that drives it is on top. This is not recommended because the turn on voltage of the LED which can vary widely from device to device can affect the base drive required to turn on the transistor. Just like what zer0wing said, the base current could turn on your LED since your have no base resistor. It is better to put the LED on top and the switches (transistor and SCR at the bottom.
2. You have no resistors from base to ground and from gate to ground. This can allow the base or gate to pick up noise and cause unwanted turn on of transistor or SCR. Take note that when your SCR turns on it can latch in the on condition so it will appear shorted until you remove the current through it.
3. You have no current limiting resistors to your base and gate. That can result in overdrive condition that could damage the base junction of the transistor or the gate of the SCR.
4. You have no current limiting to your LED. Depending on your supply voltage, you could damage the LED when the SCR and transistor turn on.

I suggest your revise your circuit as follows:
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