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Topic: Question about TL431ACLPG shunt regulator diode (Read 1 time) previous topic - next topic

steven6282

In one of the projects I'm working on, I want to have the ability to detect when a battery is getting low on power.  I read an article about using two resistors, don't remember exactly how it worked off hand but it looked like it would be a power drain in itself to do it that way.

Since this project is trying to minimize battery usage I don't want any additional unnecessary power drains.  Then I saw this on sparfun: https://www.sparkfun.com/products/11087

It uses a TL431ACLPG which from what I can tell acts kind of like a relay and only lets power through when it is under the reference voltage.

So I have a couple questions, one would this create a power bleed off?  I'm guessing it will since it still runs the reference to ground, I think?  If it does have a power bleed, any idea how much it would be?  The datasheet for the TL431ACLPG says it has an operating range of 1 to 100ma, so I'm guessing best case scenario it would still drain 1mA continuously?

Another question I have is attempting to better understand some electrical properties of resistors and such.  Why doesn't the trim pot in the spark fun board cause the reference voltage to continually drop as the voltage from the battery drops?  It seems like the lower the voltage the battery puts out the lower the voltage that makes it through resistor would be?  Like I said this is probably simply me not understanding an electrical trait properly.

CrossRoads

I think you could also just monitor the voltage directly using an analog input pin.
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steven6282


I think you could also just monitor the voltage directly using an analog input pin.


Pretty sure that is the method I read about using two resistors. I can't find the page now, thought I bookmarked it but must've bookmarked it on my computer at work.  But reading another page it's something about using two resistors of a certain ratio as a voltage divider and one goes to an analog pin the other goes to ground.  So this does indeed create a constant bleed off in itself.  I guess I could do something like this and use a relay or something to turn on the battery check circuit only when I'm actually checking it to prevent the power drain?

Is there anything like a relay but really cheap and easy to add to a circuit and control the state of with a single pin?

Docedison

How much power do you consider a drain? any voltage divider or reference diode like the TIL431 requires power. The TIL431 is an adjustable Zener diode and does require 400 uA as a minimum cathode current for proper operation.
It and a pnp transistor could make a switch for flagging when the voltage to the device was above a "Set Point" but it would require about 1 mA for proper operation when used to measure the primary power source and there's a way with 2 transistors to enable the device that isn't very complicated and draws no current until enabled.
3 transistors 2 pnp's and an npn and about 4 or 5 resistors total maybe a .1uF cap on the '431 for stability as there is a LOT of current gain in that device.
About 100 uA to control up to 100 mA of Zener current through the diode.

Doc
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steven6282

#4
Aug 08, 2012, 01:24 am Last Edit: Aug 08, 2012, 01:26 am by steven6282 Reason: 1

How much power do you consider a drain? any voltage divider or reference diode like the TIL431 requires power. The TIL431 is an adjustable Zener diode and does require 400 uA as a minimum cathode current for proper operation.
It and a pnp transistor could make a switch for flagging when the voltage to the device was above a "Set Point" but it would require about 1 mA for proper operation when used to measure the primary power source and there's a way with 2 transistors to enable the device that isn't very complicated and draws no current until enabled.
3 transistors 2 pnp's and an npn and about 4 or 5 resistors total maybe a .1uF cap on the '431 for stability as there is a LOT of current gain in that device.
About 100 uA to control up to 100 mA of Zener current through the diode.

Doc


My goal for the project is for normal power drain to be as low as possible.  Most of the time the atmega328p and the nrf24L01 that is part of the project will be asleep.  Unless an interrupt wakes them up to transmit data.  What I'd like to have happen is during this transmission it check the battery level and transmit that as well.  Right now my total power drain while in sleep mode is 1 to 2 uA.  When it's awake and transmitting it's at around 16mA for about 3 tenths of a second.  So 1mA of continuous drain is definitely to much.  I might be ok with 10 uA of continuous drain but don't want to go much beyond that.

Like I said, I can use the resistor divider method if I can put in something like a relay to enable / disable that part of the circuit so that it doesn't drain power when I'm not measuring it.  Most relays that I've seen are to big though, plus they can be somewhat expensive for a single component in a low cost unit.  I read something about using a MOSFET switch for something like this but there are apparently a lot of different kinds of mosfet switches and not sure which one would do what I need.

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