regulator protection diode

[starushz]

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]

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]

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]

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]

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

Is this correct for a negative regulator?

 

[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.

 

[Possum]

That's what I was thinking.

 

[Guido Tent]

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]

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]

Hope this solves the doubt...

 

[sagarverma]

i m posting the image for correctly using the diodes across the regulator ics

the diodes provide protection(they must not be reverse biased).
d1 and d2 ensure that the transients on the regulator o/ps do not drive the o/ps to a potential above their i/ps and cause damage to the regulators.
regulators may not turn on simultaneously,if such a condition occurs the o/p of the slower one may be driven to the potential of the faster one ,so d3 and d4 prevent this condition.

IN4007 or alike are sufficient.

 

[jan.didden]

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?

No. With the neg reg, if the cap discharges, the reg input is pos with respect to its output, and the diode shunts the cap discharge current away, as shown in the previous post.

Jan Didden

 

[starushz]

ok guys thanks alot. I understand now.