regulator protection diode

starushz · 25th January 2006, 01:49 PM

Hi guys,

How you do connect a protection diode to a 7912??...ive a high value cap at the output. I want to protect the regulator when the cap is discharging

davidsrsb · 25th January 2006, 02:01 PM

Connect between input and output of the regulator. Anode to input. Normally it will not conduct, but if the input is shorted, will discharge the output capacitor. Regulators with outputs exceeding about 7V need this protection to prevent base emitter reverse breakdown.

A 1N4004 is usually OK

starushz · 25th January 2006, 02:24 PM

doesn't really make sense. For my case whihc is a negative regulator, when the system is switched off, the cap at the o/p will discharge. However by connecting anode to input, doesnt it restrict the follow of current to the input of the IC? This might make the cap discharge into the output of the regulator thereby destroying it?

Guido Tent · 25th January 2006, 06:40 PM

Quote:

Originally posted by starushz
doesn't really make sense. For my case whihc is a negative regulator, when the system is switched off, the cap at the o/p will discharge. However by connecting anode to input, doesnt it restrict the follow of current to the input of the IC? This might make the cap discharge into the output of the regulator thereby destroying it?

The diode is to prevent reverse biasing the reg. In the case of a 7912, the anode should be at the output.

Possum · 25th January 2006, 09:56 PM

Quote:

In the case of a 7912, the anode should be at the output.

Is this correct for a negative regulator?

davidsrsb · 26th January 2006, 12:16 AM

Input is going to be around -17V, output is -12V so the bar end (cathode) should be to output so that the diode is reverse biased.

Possum · 26th January 2006, 12:27 AM

That's what I was thinking.

Guido Tent · 26th January 2006, 07:16 AM

Quote:

Originally posted by davidsrsb
Input is going to be around -17V, output is -12V so the bar end (cathode) should be to output so that the diode is reverse biased.

Yes, you are right

best

starushz · 26th January 2006, 11:12 AM

However in this cap suppose i have a large value cap at the o/p and it discharges. The diode will not protect the o/p of the regulator as it si reversed right? So doesnt it make the diode useless?

sivan_and · 26th January 2006, 02:06 PM

Hope this solves the doubt...

25th January 2006, 01:49 PM	#1
starushz diyAudio Member Join Date: Dec 2005 Location: Singapore	regulator protection diode Hi guys, How you do connect a protection diode to a 7912??...ive a high value cap at the output. I want to protect the regulator when the cap is discharging

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25th January 2006, 02:01 PM	#2
davidsrsb diyAudio Member Join Date: Dec 2005 Location: Kuala Lumpur	Connect between input and output of the regulator. Anode to input. Normally it will not conduct, but if the input is shorted, will discharge the output capacitor. Regulators with outputs exceeding about 7V need this protection to prevent base emitter reverse breakdown. A 1N4004 is usually OK

25th January 2006, 02:24 PM	#3
starushz diyAudio Member Join Date: Dec 2005 Location: Singapore	doesn't really make sense. For my case whihc is a negative regulator, when the system is switched off, the cap at the o/p will discharge. However by connecting anode to input, doesnt it restrict the follow of current to the input of the IC? This might make the cap discharge into the output of the regulator thereby destroying it?

26th January 2006, 12:16 AM	#6
davidsrsb diyAudio Member Join Date: Dec 2005 Location: Kuala Lumpur	Input is going to be around -17V, output is -12V so the bar end (cathode) should be to output so that the diode is reverse biased.

26th January 2006, 12:27 AM	#7
Possum diyAudio Member Join Date: Sep 2001 Location: Redondo Beach, CA	That's what I was thinking.

26th January 2006, 11:12 AM	#9
starushz diyAudio Member Join Date: Dec 2005 Location: Singapore	However in this cap suppose i have a large value cap at the o/p and it discharges. The diode will not protect the o/p of the regulator as it si reversed right? So doesnt it make the diode useless?