Simple Low Voltage Shut Down

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I have a little circuit consisting of an LM358 and BC547 emitter  
follower. It runs off 4 x AAA alkaline batteries and consumes about 50mA.

What is the simplest way, without consuming additional power, of  
disabling the output from the LM358 when the battery voltage drops below  
around 4V?

In other words, I would like to have the circuit deactivate itself  
before a distorted wave form is output.

Prefer all passive components.

Robert Miller

Re: Simple Low Voltage Shut Down
Robert Miller wrote...
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.
. --+-----E   C----
.   |       B
.  6.2k     |  ZTX718
.   |      470
.   |       |
.   |       K
.   +-----R    TL431
.   |       A  $0.40
.  10k      |
.   |       |
. --+-------+------


--  
 Thanks,
    - Win

Re: Simple Low Voltage Shut Down
Winfield Hill wrote...
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 A little hysteresis would be good.

.   4.0-volt low-V shutoff.
.              
.              ZTX718
. --+-----E   C-----+----
.   |       B       |
.  6.8k     |      120k
.   |      470    5% hyst
.   |       |       |
.   +------ | -|<|--'
.   |       K
.   +-----R    TL431
.   |       A  $0.40
.  10k      |
.   |       |
. --+-------+------

 Omit the diode, if a little OFF leakage
 from 127k of resistance can be tolerated.


--  
 Thanks,
    - Win

Re: Simple Low Voltage Shut Down
On Thu, 30 Mar 2017 10:20:54 -0700, Winfield Hill wrote:

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Hysteresis will just make the thing oscillate slower, at least if you're  
turning the battery off all the way.

--  

Tim Wescott
Wescott Design Services
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Re: Simple Low Voltage Shut Down
On 30/03/17 23:13, Winfield Hill wrote:
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That's a neat little circuit, but I was hoping for something passive.

Maybe a dumb question, but, given the low current, would it be possible  
to power the amp through a zener diode?

Robert Miller

Re: Simple Low Voltage Shut Down
On 3/31/2017 4:42 PM, Robert Miller wrote:
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What would that accomplish other than losing a lot of power and voltage  
to the amp?

--  

Rick C

Re: Simple Low Voltage Shut Down
On 01/04/17 09:03, rickman wrote:
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The amp would be powered through the zener until the supply voltage is  
too low to cause breakdown. Then the zener would stop conducting.

The effect of dynamic resistance may need to be factored in.

I am willing to accept the standard diode voltage drop.

Robert Miller

Re: Simple Low Voltage Shut Down
On 3/31/2017 7:35 PM, Robert Miller wrote:
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The Zener would drop a constant voltage as the supply voltage drops  
leaving very little for the amp to operate from.  As the voltage drops  
to the Zener breakdown voltage the current will tend to zero, but  
largely because the voltage on the amp will tend toward zero.


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What standard diode voltage drop?  Try drawing the circuit and calculate  
the voltages.

--  

Rick C

Re: Simple Low Voltage Shut Down
Robert Miller wrote...
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 In engineering terms, passive means no transistors.
 Really?  How could that work?  My simple circuit is
 passive in the sense that no other power is required.
 But I do suggest the hysteresis version, with more
 hysteresis to force shutoff after the battery dies.


--  
 Thanks,
    - Win

Re: Simple Low Voltage Shut Down
On 03/31/2017 08:56 PM, Winfield Hill wrote:
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Since this is a battery-powered circuit, maybe a BSS209 PFET ($0.07 @
1000 pcs) and a TLV431 ($0.16 @ 1000 pcs, 80 uA min), with 33k or so
between source and gate?  Still equally impassive, of course. ;)

Cheers

Phil Hobbs

--  
Dr Philip C D Hobbs
Principal Consultant
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Re: Simple Low Voltage Shut Down
On Thu, 30 Mar 2017 18:21:15 +1100, Robert Miller

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What does the emitter follower do? If it's a linear circuit and the
emitter follower has a passive pull-down, that might waste a lot of
power.

Got a schematic?


--  

John Larkin         Highland Technology, Inc

lunatic fringe electronics  


Re: Simple Low Voltage Shut Down
On Thu, 30 Mar 2017 18:21:15 +1100, Robert Miller wrote:

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First, there's a lot of useful energy you can get from an alkaline cell  
between 1V an 0.9V (NiMH, too, but not as much).

Second, you want some hysteresis, or at a certain level of charge the  
thing will start madly turning on and off.  Without hysteresis the  
current will get shut off, the battery voltage will rebound, the circuit  
will turn on, the battery voltage will get pulled low, and the cycle will  
repeat.

I'd design the circuit to work well down to 0.9V/cell, and I'd design the  
turn-off circuit to latch off -- maybe design the thing with an "on"  
pushbutton and an "off" pushbutton, with "off" coming either from the  
user or from the low-voltage cutout.

--  
Tim Wescott
Control systems, embedded software and circuit design
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Re: Simple Low Voltage Shut Down
Tim Wescott wrote...
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 True, but a 1-volt battery with no load doesn't
 rebound very much.  If the hysteresis is, say,
 0.25V, it should stay off.


--  
 Thanks,
    - Win

Re: Simple Low Voltage Shut Down
On Thu, 30 Mar 2017 13:18:33 -0700, Winfield Hill wrote:

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An Internet acquaintance of mine who used to be a battery engineer for GE  
claimed that nearly-fully-discharged NiCd cells would rebound to 1.2V if  
you gave them enough time, even if they couldn't deliver any significant  
charge.

But -- I'd have to test that theory on alkalines.

--  

Tim Wescott
Wescott Design Services
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Re: Simple Low Voltage Shut Down
Tim Wescott wrote...
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 Alkalines are rather different, in a useful way.
 Their open-circuit voltage drops from 1.55V to
 1.0 or 0.9V, etc, as they're depleted, providing
 a convenient way to evaluate their condition.


--  
 Thanks,
    - Win

Re: Simple Low Voltage Shut Down
On 3/30/2017 2:31 PM, Winfield Hill wrote:
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Is there a simple explanation for why this happens?
It's a chemical reaction with voltages determined by
chemistry/physics.  I can understand why the series
resistance would increase as the reagents are "used up".
But, why does the open-circuit voltage change?

I don't mean to hijack the thread, just a few words in
layman's terms or a link will suffice.
Thanks,

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