connecting 2200 mfd capacitor on 7805 voltage regulator output

i am connecting a 2200mfd capacitor on output of 7805 regulator. i am doing this because i need 2 seconds power in my circuit when power fails. my doubt is can i connect 2200 mfd 50v capacitor on output of 7805 will it affect the ic life

No, it wont. But, whether you do get your +5 volts for 2 seconds after losing power would depend on your load circuit.

Have a look at LM78xx/MC78xx datasheets for where to put the diode when using large output capacitors, I understand that size capacitor could damage the regulator IC, given that 100uF can be detrimental without the protection diode.

I believe you'll have a longer 'On' time if you put the large cap on the input side.

A capacitor draws surge current on startup, until it charges. If you put it on the output of the regulator IC, it will try to draw surge current through the regulator.

BradtheRad said:

I believe you'll have a longer 'On' time if you put the large cap on the input side.

Click to expand...

This is definitely the better solution.
If you double the input voltage to the regulator, the same capacitor can store FOUR times the energy. Its also a lot more gentle on the 7805.

If doing this, the 7805 itself might draw more power than your circuit.
If so, check out some of the very low power five volt regulators such as LP2954 and LM2936.

Hi, correction (maybe), checked a couple of datasheets for a paragraph, the 1084 on page 12 discusses protection diodes and says 1000uF upwards if input shorted. However, having too many regulator app notes and datasheets to look through all evening to find specific paragraph I was looking for, I'm certain protection diode is good practice and even "low" uF can be detrimental.

https://www.ti.com/lit/ds/symlink/lm1084.pdf

ST 78xx series datasheet says on page 34: "Application with high capacitance loads and an output voltage greater than 6 volts need an external diode (see Figure 22 on page 31) to protect the device against input short circuit. In this case the input voltage falls rapidly while the output voltage decrease slowly. The capacitance discharges by means of the base-emitter junction of the series pass transistor in the regulator. If the energy is sufficiently high, the transistor may be destroyed. The external diode by-passes the current from the IC to ground."

Not sure how that wouldn't also apply to a 5V regulator and 2200uF output cap, maybe it doesn't, but hope info of use, and seems to tie in with Brad's comment.

Hi,

Connect the capacitor at the input, as suggested before.
This is the better solution with longer back up time.

Klaus

now i changed my circuit i am using lm2576 12 dc-dc converter (circuit refereed by lm2576 data sheet) at the output of dc-dc converter i am connecting a series 1 amps Schottky diode 5819 , at the cathode end i am connecting a 3300 mfd 25v capacitor then connecting this line to 7805 input. is this will be safe? kindly explain.

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i am using lm2576 12v ic because i need 12 also.

Hi,

...sorry raman00084:

A picture speaks a thousand words, 100 words do not always paint an accurate or clear picture. Failing a schematic, a link to the actual datasheet or taking the minute it takes to upload it would help anyone unfamiliar with the LM2576, friend.

This isn't just you, so sorry to say it here, but you would think eda members are datasheet encyclopedias and clairvoyant sometimes, given how often references are made to something only the poster can see

To answer, don't know - sounds okay, what does the 2576 datasheet say about output capacitance?

raman00084 said:

i am connecting a 2200mfd capacitor on output of 7805 regulator. i am doing this because i need 2 seconds power in my circuit when power fails. my doubt is can i connect 2200 mfd 50v capacitor on output of 7805 will it affect the ic life

Click to expand...

Please state the full design requirements with Max/min input range, output range and worst case loads for all outputs and acceptance criteria for decay time vs specified voltage tolerance.

Hi,

I wonder why you connect the capacitor to the output, while we suggest to connect t it to the input....

Klaus

I think what he is trying to say, is he is using a lm2576 switching regulator to make 12vdc and after that he is putting a diode and 3300uF cap to feed a 7805. It appears he only needs the 2 seconds of power from the 5vdc supply.

There still needs to be 2 seconds of power available to supply the IC...

And 2 seconds of power in front of that...
Etc.

Hi,

Yes maybe. It's like so often. We have to guess.

If we had a schematic, a description, mA values...and so on then we could calculate and answer all most probable with a single post.

Klaus

SunnySkyguy said:

Please state the full design requirements with Max/min input range, output range and worst case loads for all outputs and acceptance criteria for decay time vs specified voltage tolerance.

Click to expand...

If you cannot define your task with all variables, then please close your question.

Gentlemen, sometimes we need to consult our crystal balls.

raman00084 said:

i am connecting a 2200mfd capacitor on output of 7805 regulator. i am doing this because i need 2 seconds power in my circuit when power fails. my doubt is can i connect 2200 mfd 50v capacitor on output of 7805 will it affect the ic life

Click to expand...

Power loss in regulator Pd= (Vin-Vout)*Iout. Loss > 1W normally takes a large heatsink. with low Rth['C/W] computed by T rise = Pd * Rth {W * 'C/W}
The IC shutdown protects itself from failure so no worry about IC life, but it may not be best choice of design if this happens frequently.

I assume when you say 2200 mfd that you mean 2200 µF
as "m" means milli and µ ( or u) is micro.

A 2 second holdup time may be excessive unless your load * 2200uf = 20 seconds. meaning low drop in 2 seconds. with Cap on input with diode to cap. above 7V perhaps 9V source.