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'[EE]: Detecting low input voltage'
2003\04\10@091259 by Olin Lathrop

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part 1 981 bytes content-type:text/plain; (decoded 7bit)

I responded to an earlier post asking how to detect low input voltage to a
LM7805 regulator, and posted a schematic with a "fancy" solution.  That
was a cut and paste from another design, and I realized after hitting send
that some of the constraints in the original design don't apply to the
simple question that was asked.

Attached is a simpler design that may be good enough for the purpose.  It
has a softer knee, but that may be fine depending on the application.

I didn't reply to my other post because it seems to have vanished into
never never land again.  I probably forgot to restore the [EE]: tag that
Outlook Express stripped off the reply.  Argh!  I know this isn't the
list's fault, but Outlook Express is such a common email client.  Is there
anything that can be done about this?

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2003\04\10@100129 by fred jones

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Hi,
I'd like to ask a couple of questions and I apologize if they are "dumb" but
I'm pretty weak with analog electronics.  Can you explain what is meant by a
softer knee?  and is there a reason you used the transistor you did or could
I use a 2n3906 instead?
Thanks,
FJ






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2003\04\10@102141 by Eduardo Garcia

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The MC34064 solve your problem?
MC34064 is a brown-out circuit that output 0v when voltage is below to 4,6V.
It is cheap and little circuit (a TO-92 and a resistor).

What you say about it?



{Original Message removed}

2003\04\10@130847 by Olin Lathrop

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> I'd like to ask a couple of questions and I apologize if they are
> "dumb" but I'm pretty weak with analog electronics.  Can you explain
> what is meant by a softer knee?

The output will no instantly click from high to low as the input voltage
drops.  There will be a transition region around the input threshold where
the output will be at intermediate levels.  The size of the input voltage
transition region is larger in this circuit than in the previous.

> and is there a reason you used the
> transistor you did or could I use a 2n3906 instead?

Just about any small signal PNP transistor will do.  The 2N4403 happens to
be the PNP "jelly bean" transistor I keep around.


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2003\04\10@164846 by Andrew Warren
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Olin Lathrop <KILLspamPICLISTKILLspamspammitvma.mit.edu> wrote:

> I didn't reply to my other post because it seems to have vanished
> into never never land again.  I probably forgot to restore the
> [EE]: tag that Outlook Express stripped off the reply.  Argh!  I
> know this isn't the list's fault, but Outlook Express is such a
> common email client. Is there anything that can be done about
> this?

   You mean, besides switching from that POS non-standard email
   client to a real one?  Probably not.

   I use Pegasus Mail, by the way; it's absolutely free (not even
   shareware) and is available from http://www.pmail.com .

   -Andy

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2003\04\11@031458 by hael Rigby-Jones

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> -----Original Message-----
> From: Olin Lathrop [SMTP:TakeThisOuTolin_piclistEraseMEspamspam_OUTEMBEDINC.COM]
> Sent: Thursday, April 10, 2003 2:13 PM
> To:   RemoveMEPICLISTspamTakeThisOuTMITVMA.MIT.EDU
> Subject:      [EE]: Detecting low input voltage
>
> I responded to an earlier post asking how to detect low input voltage to a
> LM7805 regulator, and posted a schematic with a "fancy" solution.  That
> was a cut and paste from another design, and I realized after hitting send
> that some of the constraints in the original design don't apply to the
> simple question that was asked.
>
> Attached is a simpler design that may be good enough for the purpose.  It
> has a softer knee, but that may be fine depending on the application.
>
I don't understand how this circuit works.  Assuming (dangerous!) that the
regulator will always have at least it's dropout voltage between it's intput
and output, (typicaly 2 volts for a generic 780x) the transistor will
continue to be biased into conduction even the the regulators output is very
low.  The UNREG_OK signal will simply track the output.

Please tell me where I'm going wrong in this analysis...

Mike


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2003\04\11@141421 by Dwayne Reid

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At 08:09 AM 4/11/03 +0100, Michael Rigby-Jones wrote:
> >
>I don't understand how this circuit works.  Assuming (dangerous!) that the
>regulator will always have at least it's dropout voltage between it's intput
>and output, (typicaly 2 volts for a generic 780x) the transistor will
>continue to be biased into conduction even the the regulators output is very
>low.  The UNREG_OK signal will simply track the output.
>
>Please tell me where I'm going wrong in this analysis...

Assume low drop-out regulator, then all should make sense.  You have to add
1 extra resistor between E-B if you want to use a 7805 type regulator.

dwayne

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2003\04\11@152201 by Olin Lathrop

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> I don't understand how this circuit works.  Assuming (dangerous!) that
> the
> regulator will always have at least it's dropout voltage between it's
> intput
> and output,

The point was to test whether the input voltage is too low.  If the
regulator always has enough voltage to work with, then there isn't a
problem.  This circuit will start raising UNREG-OK when the drop on the
regulator gets to around 1.4V.  UNREG-OK should be fully high when the
drop is about 1.9V.  You can add another diode in series with D1 if you
want the thresholds to be a little higher 600mV to 700mV higher.


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2003\04\11@173010 by Olin Lathrop

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> Assume low drop-out regulator, then all should make sense.  You have to
> add 1 extra resistor between E-B if you want to use a 7805 type
> regulator.

That would work.  You could instead add another diode in series with the
emitter.  That would raise the threshold point to about 2V while
maintaining a sharper response at a slightly higher cost.


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2003\04\11@223002 by Thomas N

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part 1 2005 bytes content-type:text/plain; format=flowedHi everyone,

This is the circuit that I have (see the attachment).  Since the turn-on of
the transistor is based on its gain, the circuit still suffers from the
transistor's gain variation (correct me if I am wrong!).  The diode is there
to compensate for the transistor Hfe changes due to temperature.  This
circuit has steeper turn on/off slope than the one Olin suggested.

What do you think?
Regards,
Thomas

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2003\04\12@083603 by Olin Lathrop

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> This is the circuit that I have (see the attachment).  Since the
> turn-on of the transistor is based on its gain, the circuit still
> suffers from the transistor's gain variation (correct me if I am
> wrong!).  The diode is there to compensate for the transistor Hfe
> changes due to temperature.  This circuit has steeper turn on/off slope
> than the one Olin suggested.
>
> What do you think?

You need to define what this circuit is supposed to do, else it's
impossible to judge if it does it successfully.  However, the diode most
likely doesn't belong there.


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2003\04\14@041822 by hael Rigby-Jones

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> -----Original Message-----
> From: Olin Lathrop [SMTP:@spam@olin_piclist@spam@spamspam_OUTEMBEDINC.COM]
> Sent: Friday, April 11, 2003 8:20 PM
> To:   spamBeGonePICLISTspamKILLspamMITVMA.MIT.EDU
> Subject:      Re: [EE]: Detecting low input voltage
>
> > I don't understand how this circuit works.  Assuming (dangerous!) that
> > the
> > regulator will always have at least it's dropout voltage between it's
> > intput
> > and output,
>
> The point was to test whether the input voltage is too low.  If the
> regulator always has enough voltage to work with, then there isn't a
> problem.  This circuit will start raising UNREG-OK when the drop on the
> regulator gets to around 1.4V.  UNREG-OK should be fully high when the
> drop is about 1.9V.  You can add another diode in series with D1 if you
> want the thresholds to be a little higher 600mV to 700mV higher.
>
The point I was trying to make is that the drop accross the regulator is
pretty constant once it goes out of regulation.  For a 7805 this is over 2
volts so the circuit wouldn't work, but as already explained by someone,
this application is for an LDO.

Regards

Mike


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