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tazntex
- 18th August 2010, 14:07
Just curious, is there a simple efficient way to add on a 9V battery powered project ,feeding a LP2950ACZ-5.0 regulator (5V) with a 16F628A, processor some kind of protection that when the battery goes down to about 6.2V it will not provide power to the regulator?

Thanks for any suggestions.

languer
- 18th August 2010, 19:01
why 6.2V? will the regulator be unpowered? If you want to "feed" two power supplies into the regulator, and want the highest one to always power it, a blocking diode is the solution (it can be more complicated than this, but the idea is the same).

tazntex
- 18th August 2010, 19:34
Thanks for the reply, what I am trying to do is to prevent any damage to any components because of a low battery. What I mean is if I am using a 9V battery to feed the lp2950 5v regulator and forget to turn if off over a period of time the battery will be discharged and and I would like to "disconnect" the battery from the circuit that way I could prevent damaging any components like voltage regulator and anything things else connected that requires 5v. I having been experimenting with a 6.2v zener but something around that voltage would allow me to get the maximum use of my 9v battery but would like it to have a very low current. Anyhow I am just looking for a suggestion, I've been searching the forum but haven't found anything of interest on a low battery condition but after few batteries I figured I needed to do something.


Thanks

rsocor01
- 18th August 2010, 20:33
What I mean is if I am using a 9V battery to feed the lp2950 5v regulator and forget to turn if off over a period of time the battery will be discharged and and I would like to "disconnect" the battery from the circuit that way I could prevent damaging any components like voltage regulator and anything things else connected that requires 5v

I have never seen any component getting damaged because of a low voltage. I usually test my circuits by let them run until the battery completely drains to get a battery lifetime plot, and never had any problems. Let's see what other people in the forum has to say about it.

Robert

Acetronics2
- 18th August 2010, 21:21
Hi,

The 16F628 has built in Comparators ... so, it's easy to get an interrupt if an input voltage goes under a choosen reference ...

till the regulator has sufficient supply, you can get some 5v absolute reference ...

Why not do it simple ??? :rolleyes:

Note a MC33064P5 can also do it ...

Alain

Tywais
- 19th August 2010, 03:27
A LTC1440 can do it and is what I use for one of my systems with 9V battery operation. It can operate with supply voltage from 2-11 volts. 2 resistors to set the trigger point and a on/off output that can be used to drive an LED or disconnect your power source from the circuit using either a relay or MOSFET.

languer
- 19th August 2010, 09:01
The IC Alain recommends, MC33064P5, is as simple as it gets. You still need to have some pass-thru device (transistor) between the supply and regulator, and the supervisor IC will enable/disable it based on the battery voltage. You can do the same with the MCU and external components (i.e. use MCU's comparator, external Vref, and pass-thru device).

tazntex
- 19th August 2010, 16:41
Thanks for all the suggestions. Using the comparator or the MC34064 both are great ideas, but wouldn't the just keep reseting? What I mean is applying the 9V battery to the input of the LP2950 5V regulator, and the powering the processor with the regulator having the MC34064 or the the processor on power up, code would start, compare the 9v battery is it above 6.2v or whatever if so, continue program, if not well, turn off a FET or transistor that would be in series with the 9v positive lead or ground leg, then once that happen let's see, wouldn't it start up again and do the same thing continuously?

Thanks again.