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Low Voltage Cutout
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kol...@gmail.com  
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 More options Nov 20 2008, 1:05 pm
Newsgroups: sci.electronics.design
From: kol...@gmail.com
Date: Thu, 20 Nov 2008 00:05:19 -0800 (PST)
Local: Thurs, Nov 20 2008 1:05 pm
Subject: Low Voltage Cutout
Hi

I have an access control system running off a float charged battery.
If there is a power outage of +-8 Hours, the battery drops to a level
lower than the access-control system can handle, and then it does not
recognise the tags (the readers actually lockup, but the maglocks stay
locked).

I was thinking of putting a 9v zener diode on the battery, running a
relay coil, that will cause the load to be disconnected when the
battery is at 9v. one problem i see with this is that when the load is
disconnected, the battery can 'bounce' up to above 9v again, which
will cause the relay to keep on chattering,.

Is my understanding correct, and is there an inmprovement that will
eliminate the chatter

BATTERY --------------------------|> ZENER DIODE------- RELAY COIL
              |
---------------------------------------------------------- RELAY COIL


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D from BC  
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 More options Nov 20 2008, 1:59 pm
Newsgroups: sci.electronics.design
From: D from BC <myrealaddr...@comic.com>
Date: Thu, 20 Nov 2008 00:59:29 -0800
Local: Thurs, Nov 20 2008 1:59 pm
Subject: Re: Low Voltage Cutout

I'd go for a circuit with an adjustable undervoltage trip point.
Once power returns,the charging voltage is sensed and the circuit
returns to normal.

I propose one comparator set up with hysteresis, one V reference and a
transistor to drive the relay..

Alternatively a DTL circuit could be made up.
Extra relay contacts can sometimes put to use for circuit function.

D from BC
myrealaddress(at)comic(dot)com
British Columbia
Canada


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MooseFET  
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 More options Nov 20 2008, 7:50 pm
Newsgroups: sci.electronics.design
From: MooseFET <kensm...@rahul.net>
Date: Thu, 20 Nov 2008 06:50:04 -0800 (PST)
Local: Thurs, Nov 20 2008 7:50 pm
Subject: Re: Low Voltage Cutout
On Nov 20, 12:05 am, kol...@gmail.com wrote:

If the load circuit is sure to pull the voltage down to zero quickly,
you can use the voltage on the load side to do the positive feedback.
The most extreme version would be to have a push button you need to
press to turn the system back on.


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ehsjr  
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 More options Nov 21 2008, 11:33 am
Newsgroups: sci.electronics.design
From: ehsjr <eh...@NOSPAMverizon.net>
Date: Fri, 21 Nov 2008 06:33:49 GMT
Local: Fri, Nov 21 2008 11:33 am
Subject: Re: Low Voltage Cutout

If it is a 12V battery, 9V is too low. You want to disconnect
the battery at somewhere between 10.5 and 11 volts - higher
is better for the battery.

To prevent the "bounce syndrome", hold the relay energized
through its own open point. Something like this:

                      N/O Relay
                       Contact
   Battery(+) ---+---o---
                 |       ^---o---+----> to load
                k|               |
               [D1]            [Zd]  11 volts
                 |               |
                 +--------+------+
                 | +      |      |k
              Charger  [Relay]  [D3]
                 | -      |      |
   Battery(-) ---+--------+------+

When the battery V drops below 11 volts, the relay drops and
the contact opens, isolating the battery from the load and the
zener.  D1 prevents the relay from "seeing" the battery voltage,
and D3 absorbs the inductive kick from the relay dropping out.
When the power is restored, the charger turns on and energizes
the relay, reconnecting the load.

Use a Schottky for D1, because it has a lower voltage drop
than a regular diode. If that still presents an unacceptable
voltage drop, use a double pole relay and replace the diode
with the second normally open contact.

You can use a differnt zener or a TL431 to set the drop out
voltage, but unless your battery spec says you can safely
discharge it below 10.5 volts, I recommend you use the
11 volt dropout point.

Ed


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