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Integrating LVD circuit in 12V system

Posted on 2014-12-07
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Last Modified: 2014-12-08
I'm trying to build a Low Voltage Disconnect circuit with ultra-low power consumption and came across this particular circuit on the net. The circuit is shown in the attached image "LVD circuit".

My challenge is to modify it to suit my purposes, and this might be a little bit out of my league, so I definitely need some help. The modified circuit should have the following properties:

-Dimensioned for control of a 12V battery package
-More or less zero current draw when off
-Less than 250 microamperes current draw when on

When integrated in my project (a battery-driven lamp) this circuit also needs to behave as follows:

-Shut down when I turn of the main lamp switch
-Turn on (start monitoring battery voltage) when I turn on the main lamp switch and keep doing so unitl the battery voltage becomes too low or the main switch is turned off.

For the on/off functionality, I've come up with a sketch. See attached image "LVD circuit modified". In this diagram I'm assuming 5V battery voltage. I've removed switch S2 and added a by-pass capacitor to kick-start the circuit without permanently bypassing the relay. But: what happens when the main switch is turned off? How can I make the capacitor discharge quickly enough without adding another resistor that continuously steals current?

As for the 12V aspect, using a voltage regulator to feed the circuit appears problematic since all normally sized (through-hole mounted) regulators are guzzling >1 mA. This is about a thousand times more than the original circuit claims to consume, and consequently a bad solution.

That was probably 10 questions in one post, so bear with me. I'm a newbie when it comes to this type of electronics, but very eager to learn. If someone could point me in the right direction, I would really appreciate it.


Best regards,

Einar
D--LVD-circuit.png
D--LVD-circuit-modified.png
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Question by:EISTO
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6 Comments
 
LVL 83

Expert Comment

by:Dave Baldwin
ID: 40486086
First, the two ICs can only take 5.5VDC max.  Second, your proposed modifications do nothing to help the circuit.  Third, when you turn off the main switch in your version, there is no path for the cap to discharge although adding a resistor around the relay isn't a problem since it only takes effect when the main switch is off.  Inputs to the MAX917 are not supposed to exceed the power supplies so you can't use it to monitor a 12 volt lamp.  http://datasheets.maximintegrated.com/en/ds/MAX917-MAX920.pdf

Since any lamp you might use will require many times the current consumption of the LVD circuit, trying to use a micropower circuit to monitor it is not necessary.
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Author Comment

by:EISTO
ID: 40486413
Hi Dave, thanks for answering!

First: I know the ICs can only handle 5.5V. That's why I asked how I could modify the concept to suit 12V applications.
Second: OK. Care to share why the modifications won't work as well? I know the resistor R5 won't use current on a regular basis and that the capacitor won't discharge. That's why I commented upon it in the text. It was perhaps a bit unclear, but my point was whether I would need to add a resistor between the + side of the capacitor and GND to discharge the capacitor when the switch is turned off. If so, this resistor would have a continuous current, adding to the power loss.

Finally: Well, that depends. It's a LED lamp. If the circuit consumes 5% of the total current, that's a bit too much. If there are no alternatives I will of course have to settle for that. The problem is that this lamp will be used at a summer house without access to electricity, so charging should take place as seldom as possible.

The modified circuit only incorporates the switching functions. I haven't drawn the voltage regulator since I want to avoid using it. Perhaps a voltage divider could do the job? That will of course depend on how much power is required. If the voltage divider resistors have to be below say 5-10k I might as well use a voltage regulator.

Einar
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LVL 27

Assisted Solution

by:d-glitch
d-glitch earned 500 total points
ID: 40486891
>>  As for the 12V aspect, using a voltage regulator to feed the circuit appears problematic

You are missing some very good regulators.  The MCP17039 is available for $0.39 in a thru-hole TO-92 package and operates at 2uV.
     http://www.microchip.com/wwwproducts/Devices.aspx?product=MCP1703
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Author Comment

by:EISTO
ID: 40487121
Nice!
Being a newbie I just searched for "Linear voltage regulators". LDO regulators are apparently a separate category at Mouser.com.
Thanks for the tip!

Thats one problem solved. Any ideas on how I can make the whole thing work as specified with a single main switch?


Einar
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LVL 27

Accepted Solution

by:
d-glitch earned 500 total points
ID: 40487273
The latching relay is a critical component for this design, since it can hold its position without any power.
You do need two switches to control it, but both switches can be controlled by a single actuator.

A DPDT Switch with (ON)-OFF-(ON) action would work like a pair of NO Push Buttons.  

The ( ) indicate MOMENTARY operation.  The handle normally sits in the middle.  You lift it up to turn on the light.  You push it down to turn it off.  In each case it returns to the center when you let go.

    http://www.mouser.com/ProductDetail/CK-Components/7205P3YZQE/?qs=%2fha2pyFaduh89A3Drs4ISWf%2firMxnTtBMaM2mMT7gsvhnDVqOGlCtg%3d%3d
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Author Comment

by:EISTO
ID: 40487397
Thats brilliant. I had completely forgotten that there are mechanical SPDT switches as well.
Thanks for two good solutions!

Einar
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