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 Posted: Mar 13, 2010 - 02:12 PM |
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Joined: Feb 24, 2009
Posts: 140
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I am current using 5V which is the output from 7805. Is there any IC that can perform automatic power-fail detect and switch the circuitry to my battery?
I know the RTC ic contains this feature, how about any IC that can auto switch to battery for my uC, so that I won't lose any data during power failure. Once the power is restored, the IC will then switch back to the main supply, just like the RTC IC (like DS1307). |
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Posted: Mar 13, 2010 - 03:48 PM |
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Joined: Nov 22, 2002
Posts: 7176
Location: Tangent, OR, USA
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Linear Tech has a number of ICs that will do this.
Jim |
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Jim Wagner
Oregon Research Electronics, Consulting Div.
Tangent, OR, USA
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Posted: Mar 13, 2010 - 04:47 PM |
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Joined: Feb 24, 2009
Posts: 140
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| Hi Jim, can you name one please? |
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Posted: Mar 13, 2010 - 10:22 PM |
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Joined: Nov 22, 2002
Posts: 7176
Location: Tangent, OR, USA
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Posted: Mar 13, 2010 - 11:23 PM |
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Joined: Dec 15, 2005
Posts: 569
Location: Brazil
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Quote:
I am current using 5V which is the output from 7805. Is there any IC that can perform automatic power-fail detect and switch the circuitry to my battery?
What is the Voltage on the Vin of the LM7805? Other option is sense the Vin and switch to your battery before it afect your Vcc. |
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Regards,
Brunomusw
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Posted: Mar 14, 2010 - 04:18 AM |
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Joined: Aug 19, 2008
Posts: 52
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| or... if your battery voltage is a bit lower and your circuit can handle the additional 0.6V drop... a diode. |
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Posted: Mar 14, 2010 - 06:00 AM |
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Joined: Feb 24, 2009
Posts: 140
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brunomusw wrote:
Quote:
I am current using 5V which is the output from 7805. Is there any IC that can perform automatic power-fail detect and switch the circuitry to my battery?
What is the Voltage on the Vin of the LM7805? Other option is sense the Vin and switch to your battery before it afect your Vcc.
Vin is 9V DC. How to automatically switch to battery before it affect my Vcc? Sorry, I'm not good in electronics, is there such a ready circuit available or preferably just an IC? |
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Posted: Mar 14, 2010 - 06:14 AM |
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Joined: Feb 24, 2009
Posts: 140
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ka7ehk wrote:
How about this one?
the circuit looks quite simple. If my wall adapter input is supplying 5VDC, and my battery cells is 4.5VDC, and I connect my Vcc to "TO LOAD", will this circuit work? |
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Posted: Mar 14, 2010 - 06:39 AM |
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Joined: Feb 24, 2009
Posts: 140
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After many many googling, found this circuit!
According to the website, The wall adapter input can be anything between 3 to 28V DC. The battery voltage can be anything between 2.5V to 28V.
Does it mean I can put my wall adapter 12VDC, my battery 9VDC, and then "To load" connect to 7805 pin1, pin3 to my uC, and viola, my uC is 7/24 working non-stop and I don't have to worry volatile memory.
Is that so simple? If so, why people still want this eeprom thing? |
Last edited by chanseng738 on Mar 14, 2010 - 03:40 PM; edited 2 times in total
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Posted: Mar 14, 2010 - 08:27 AM |
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Joined: Aug 19, 2008
Posts: 52
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Bah in your case just use a diode like I suggested before
Adapter 12V straight to input pin of 7805
Battery 9V to anode of 1N4001 diode, cathode also to input pin
All grounds (including 7805) tied together.
Done.
Battery won't discharge when adapter is connected because of the diode. At the same time the diode prevents the battery from being charged by the adapter, otherwise you may damage it.
When voltage drops below 8.4 volts the diode starts conducting and the battery supplies the 7805.
Cost: $0.02 or less. |
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Posted: Mar 14, 2010 - 03:28 PM |
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Joined: Feb 24, 2009
Posts: 140
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Xantor wrote:
Bah in your case just use a diode like I suggested before
Adapter 12V straight to input pin of 7805
Battery 9V to anode of 1N4001 diode, cathode also to input pin
All grounds (including 7805) tied together.
Done.
Battery won't discharge when adapter is connected because of the diode. At the same time the diode prevents the battery from being charged by the adapter, otherwise you may damage it.
When voltage drops below 8.4 volts the diode starts conducting and the battery supplies the 7805.
Cost: $0.02 or less.
Like this? sorry I don't have a proper schematic drawing software, draw using Paint 
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Posted: Mar 14, 2010 - 03:37 PM |
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Joined: Feb 24, 2009
Posts: 140
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I don't understand, when +12V is conducting, why it stop battery from discharging?
Edit: ok, understand now... thanks |
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Posted: Mar 14, 2010 - 04:20 PM |
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Joined: Aug 19, 2008
Posts: 52
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| You may also want to add a second diode inline with the 12V just in case there is circuitry there that draws power from the 9V battery when it's off (indicator light for example). so 12V ---|>|--- 7805. That way no current can flow back into the power supply. |
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Posted: Mar 14, 2010 - 05:03 PM |
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Joined: Feb 24, 2009
Posts: 140
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| Thanks, Xantor. One question, how long the battery can last assuming the 12VDC supply always on. I mean battery will leak by itself even if not use right? A good Duracell 9V battery can last for how long without usage? |
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Posted: Mar 14, 2010 - 05:24 PM |
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Joined: Feb 24, 2009
Posts: 140
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Found this one from Duracell website:
Shelf Life
Up to 85% capacity remaining after 4 years of storage at 70°F (21°C).
Does that mean if my 12VDC supply always on, the battery can power the uC for 4 years in the event of power failure? |
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Posted: Mar 14, 2010 - 09:58 PM |
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Joined: Dec 15, 2005
Posts: 569
Location: Brazil
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Quote:
Does that mean if my 12VDC supply always on, the battery can power the uC for 4 years in the event of power failure?
It all depends how many power failure will happen and for how long the power failure will take.
What Duracell says is that if you don't use the battery, over 4 years it will remain up to 85% the capacity.
You could change your 7805 for a low-drop regulator, it's better when running on your battery or even a swicth regulator better yet. |
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Regards,
Brunomusw
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