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'power switch'
2000\01\12@211254 by Gordon Varney

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I am looking for a way to power up a circuit from a 1.5V AAA cell battery.

I have a AAA battery, a switch mode power supply running 5V powering a PIC.
(Total current 150 ma on the battery, circuit current is only 40 ma.)

I have tried an N channel Mosfet switching the ground side of the circuit.
(won't work - at 150 ma it takes over 2.5 Vdc to enhance the gate.)
I tried a P channel mosfet on the high side. (won't work - P channel
requires -2.5 Vdc at with a +1.5 Vdc on the source at150 ma)
PNP and NPN transisters are out of the question (the .7 volt drop will bring
the battery voltage below .8 volts which is the minamum voltage for my
SMPS.)

Mechanical switches won't work (I need to hold the power on untill the
microcontroller has completed its cycle then power off.)
Note: the cycle is started by a push button, which supplies 1.5V battery to
"?what ever?"

I have never heard of such a device, have you?

Gordon

2000\01\12@213501 by Erik Reikes

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At 08:08 PM 1/12/00 -0600, Gordon Varney wrote:
>I am looking for a way to power up a circuit from a 1.5V AAA cell battery.
>
>I have a AAA battery, a switch mode power supply running 5V powering a PIC.
>(Total current 150 ma on the battery, circuit current is only 40 ma.)
>
>I have tried an N channel Mosfet switching the ground side of the circuit.
>(won't work - at 150 ma it takes over 2.5 Vdc to enhance the gate.)

Place your pushbutton switch in parallel with the n-channel FET (or p
channel) and drive the fet from the PIC (at 3v or 5v or whatever).

User presses button, switcher fires up and pic starts running and your SW
turns on the FET.  User releases the button and the PIC continues to hold
the FET 'on' .  I'm not sure, but I think all of this should take something
less than 10-20 ms.  I don't think a user would really be able to press the
button faster than that.

Let me know if this works?

Erik Reikes
Software Engineer
Xsilogy, Inc.

spam_OUTereikesTakeThisOuTspamxsilogy.com
ph : (858) 535-5113
fax : (858) 535-5163
cell : (858) 663-1206

2000\01\12@222646 by Ken Webster

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>PNP and NPN transisters are out of the question (the .7 volt drop will
bring
>the battery voltage below .8 volts which is the minamum voltage for my
>SMPS.)


VCE for a saturated bipolar junction transistor is typically around .3 volts
and can be significantly smaller if the transistor is driven deep into
saturation.

I think Erik's suggestion is better than using a BJT anyway, but,
nevertheless, I thought you may like to know that a BJT typically drops much
less than .7V from collector to emitter when saturated.  The .7V drop is
from base to emitter.

Cheers,

Ken

2000\01\12@224418 by Gordon Varney

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   Your right, however, To drive it that hard will take about 5 ma on the
base. That will translate to about 20 ma on the battery after stepping it up
to 5 Vdc.
   I agree Eric's idea is almost perfect, and solves most of the turn on
problems. Jost one more detail, I was hoping to be able to monitor the
switch to determine when the user releases the button. Of course that would
be an ideal situation, and we all know that Murphy would not allow this....
:-)

Gordon

> >PNP and NPN transisters are out of the question (the .7 volt drop will
> bring
> >the battery voltage below .8 volts which is the minamum voltage for my
> >SMPS.)
>
>
> VCE for a saturated bipolar junction transistor is typically around .3
volts
> and can be significantly smaller if the transistor is driven deep into
> saturation.
>
> I think Erik's suggestion is better than using a BJT anyway, but,
> nevertheless, I thought you may like to know that a BJT typically drops
much
> less than .7V from collector to emitter when saturated.  The .7V drop is
> from base to emitter.
>
> Cheers,
>
> Ken

2000\01\12@224836 by Gordon Varney
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Eric,
   I belive that this will solve the switching problem, can you now tell me
how to monitor the switch to determine when the user lets go of the button.
(I am trying to make this difficult :-) I only want the world and own it to.

Gordon

> At 08:08 PM 1/12/00 -0600, Gordon Varney wrote:
> >I am looking for a way to power up a circuit from a 1.5V AAA cell
battery.
> >
> >I have a AAA battery, a switch mode power supply running 5V powering a
PIC.
> >(Total current 150 ma on the battery, circuit current is only 40 ma.)
> >
> >I have tried an N channel Mosfet switching the ground side of the
circuit.
> >(won't work - at 150 ma it takes over 2.5 Vdc to enhance the gate.)
>
> Place your pushbutton switch in parallel with the n-channel FET (or p
> channel) and drive the fet from the PIC (at 3v or 5v or whatever).
>
> User presses button, switcher fires up and pic starts running and your SW
> turns on the FET.  User releases the button and the PIC continues to hold
> the FET 'on' .  I'm not sure, but I think all of this should take
something
> less than 10-20 ms.  I don't think a user would really be able to press
the
> button faster than that.
>
> Let me know if this works?
>
> Erik Reikes
> Software Engineer
> Xsilogy, Inc.

2000\01\13@003249 by Peter Wintulich

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Just an idea
If the dc/dc will run up on a lower voltage ~1.2V then you may be able to use a low V drop  diode (gemanium?) in series with the push button (on the dc/dc side).
once the pic is up it can switch on the fet, bypassing the diode.
Then run two resistors from the junction between the switch & diode  ~10K each, one to ground 0V, the other to an input pin on the pic.
The ascii art is poor (also in a proportional font sorry).

                    +-----<   to pic o/p (may need a large value pull down?)
                     |
           ___ - - - _____
           |                     |
           |    S1             |
           |    __             |
+1.5 --+----    ---+--|>|--+----[dc/dc]-------- V+ o/p
                       |
                       |
 0V ----[10kE]--+-----[10kE]--------- to pic i/p

Regards Peter


>>> Gordon Varney <.....varneyKILLspamspam@spam@CLAS.NET> 01/13 12:38 PM >>>
I am looking for a way to power up a circuit from a 1.5V AAA cell battery.

I have a AAA battery, a switch mode power supply running 5V powering a PIC.
(Total current 150 ma on the battery, circuit current is only 40 ma.)

I have tried an N channel Mosfet switching the ground side of the circuit.
(won't work - at 150 ma it takes over 2.5 Vdc to enhance the gate.)
I tried a P channel mosfet on the high side. (won't work - P channel
requires -2.5 Vdc at with a +1.5 Vdc on the source at150 ma)
PNP and NPN transisters are out of the question (the .7 volt drop will bring
the battery voltage below .8 volts which is the minamum voltage for my
SMPS.)

Mechanical switches won't work (I need to hold the power on untill the
microcontroller has completed its cycle then power off.)
Note: the cycle is started by a push button, which supplies 1.5V battery to
"?what ever?"

I have never heard of such a device, have you?

Gordon

2000\01\13@010127 by Ken Webster

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face
>    I agree Eric's idea is almost perfect, and solves most of the turn on
>problems. Jost one more detail, I was hoping to be able to monitor the
>switch to determine when the user releases the button. Of course that would
>be an ideal situation, and we all know that Murphy would not allow this....
>:-)


Not without spending a little more money and making the circuit a tad more
complex ...

Is a dual-pole pushbutton out of the question?


If so, how about the following:
(I am assuming that your switching power supply has a common V- in/out which
is the same as GND, V+ in = battery, V+ out = Vdd)

Battery [-] to GND

Battery [+] to N-channel MOSFET [drain]

Battery [+] to PNP #1 [emitter]

N-channel MOSFET [source] to SPS [V+ in]

PNP #1 [collector] to SPS [V+ in]

PNP #1 [base] to 68 ohm [1]

68 ohm [2] to pushbuton [2]

pushbutton [1] to GND

pushbutton [2] to 22k [1]

22k [2] to PNP #2 [base]

PNP #2 [emitter] to Vdd

PNP #2 [collector] to PIC [sense input]

PIC [sense input] to 10k [1]

10k [2] to GND

SPS [+5 out] to Vdd

SPS [GND] to GND

N-channel MOSFET [gate] to PIC [power control output]


------
The PNP transistors may be 2N3906 or similar.

The PIC [sense input] will be at Vdd when the pushbutton is pressed (and the
circuit is powered, of course) and will be at GND otherwise.

The PIC [power control output] is raised to Vdd to hold the power on and
dropped to GND to shut the circuit down (MOSFET must be a logic-level
enhancement mode device)

Note that PNP #1 will have its base-emitter junction reverse-biased by
(Vdd - VBE[Q2] - Vbattery+) when the circuit is powered and the pushbutton
is released.  This is ok since most transistors have a BVEBO of 6 to 8
volts.

I would be interested in hearing if this works if you build it (I think it
will work but haven't "debugged" it).

Cheers,

Ken

2000\01\13@043033 by paulb

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face
The idea of using the power switch to turn on the circuit directly,
then hold it on with a FET in parallel is quite straightforward.

 If however, you want something different, such as switching on
remotely or via a high impedance control line, or keeping the pushbutton
grounded but switching the positive line, then you just use a transistor
to perform the initial switch-on via the pushbutton, then a FET in
parallel with this to provide a lower "on" resistance for continuing
operation.

 Generally speaking however, the SMPS itself contains a switching
device and can be controlled on or off by interrupting the bias on this.

 Now, as to the matter of sensing the pushbutton - think laterally!
The SMPS contains a capacitor on the output.  Simply switch off the FET
or whatever is in parallel with the switch and measure the input to the
SMPS (using an I/I line and isolating resistor).

 If it falls, the pushbutton has been released.  As soon as this has
been discerned, switch the control device back on.  This can be polled
about 10 to 50 times a second depending on what resolution is desired,
and each resulting interruption will only last a few µs, not affecting
the SMPS output significantly, if at all.

 Even this polling need not continue after the pushbutton is released,
unless further presses need to be detected.

--
 Cheers,
       Paul B.

2000\01\13@082332 by jkitchen

picon face
What's wrong with a simple 2N3904 (NPN) or 2N3906 (PNP) transistor?

Gordon Varney wrote:

{Quote hidden}

2000\01\13@083200 by Richard Dungan

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face
On Wed, 12 Jan 2000 20:08:50 -0600, you wrote:

>I am looking for a way to power up a circuit from a 1.5V AAA cell battery.
>
>I have a AAA battery, a switch mode power supply running 5V powering a PIC.
>(Total current 150 ma on the battery, circuit current is only 40 ma.)
>
>I have tried an N channel Mosfet switching the ground side of the circuit.
>(won't work - at 150 ma it takes over 2.5 Vdc to enhance the gate.)
>I tried a P channel mosfet on the high side. (won't work - P channel
>requires -2.5 Vdc at with a +1.5 Vdc on the source at150 ma)
>PNP and NPN transisters are out of the question (the .7 volt drop will bring
>the battery voltage below .8 volts which is the minamum voltage for my
>SMPS.)

A well saturated bipolar transistor will have a VCE of only about 0.2 volts,
even though the VBE is 0.7 volts.

Richard

--
------------Richard Dungan-------------
Radix Electronic Designs, Orpington, UK
 Email: richardspamKILLspamradix-design.co.uk
---------------------------------------

2000\01\13@142118 by Erik Reikes

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At 11:02 PM 1/12/00 -0700, you wrote:
>>    I agree Eric's idea is almost perfect, and solves most of the turn on
>>problems. Jost one more detail, I was hoping to be able to monitor the
>>switch to determine when the user releases the button. Of course that would
>>be an ideal situation, and we all know that Murphy would not allow this....
>>:-)
>
>
>Not without spending a little more money and making the circuit a tad more
>complex ...
>
>Is a dual-pole pushbutton out of the question?
>

If you went dual pole, I don't think you'd need all the transistors and
stuff.  Just have 5V routed to the other pole (actually it would have to be
dual pole, dual throw... do they make pushbuttons like that?) wired so that
when it is released it closes a circuit to a PIC input pin.  This way you
use only 1 DPDT pushbutton, one FET, and two PIC IO pins.  One IO to drive
the FET and one to sense the fact that the button has been released.

Or, better yet I just thought if you use DPST you could have a pull up on
the 'button down' side  when it is released you lose a voltage to the PIC.
Yes, that's better.  You wire 5 volts to the other pole of the switch, and
the closed circuit supplies voltage to a PIC input pin with a pull down.
When it is released voila 0 volts (or the other way, use ground and a pull
up at the PIC, whichever seems best.).

Not too bad for a software guy, huh?

-Erik Reikes
Erik Reikes
Software Engineer
Xsilogy, Inc.

.....ereikesKILLspamspam.....xsilogy.com
ph : (858) 535-5113
fax : (858) 535-5163
cell : (858) 663-1206

2000\01\13@150037 by Gordon Varney

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Eric,
   Very good, However The switch I am using is only a SPST and they only
make it that way. I am looking for another type of switch.

Thank You
Gordon Varney


> At 11:02 PM 1/12/00 -0700, you wrote:
> >>    I agree Eric's idea is almost perfect, and solves most of the turn
on
> >>problems. Jost one more detail, I was hoping to be able to monitor the
> >>switch to determine when the user releases the button. Of course that
would
> >>be an ideal situation, and we all know that Murphy would not allow
this....
> >>:-)
> >
> >
> >Not without spending a little more money and making the circuit a tad
more
> >complex ...
> >
> >Is a dual-pole pushbutton out of the question?
> >
>
> If you went dual pole, I don't think you'd need all the transistors and
> stuff.  Just have 5V routed to the other pole (actually it would have to
be
> dual pole, dual throw... do they make pushbuttons like that?) wired so
that
{Quote hidden}

2000\01\13@155443 by Ken Webster

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part 0 1116 bytes
>>>    I agree Eric's idea is almost perfect, and solves most of the turn on
>>>problems. Jost one more detail, I was hoping to be able to monitor the
>>>switch to determine when the user releases the button. Of course that
would
>>>be an ideal situation, and we all know that Murphy would not allow
this....
{Quote hidden}

...

Exactly.  That is what I had in mind.  The circuit is only for the case
where you don't want a DPDT switch (because they cost too much, don't have
the right "feel", etc.).  Sorry about the hard-to-read text circuit
description .. Jinx's post gave me the right idea:  Attached is a .GIF of
the circuit.

Ken

Attachment converted: wonderland:pushbutton_power.gif (GIFf/JVWR) (00014CA6)

2000\01\13@162748 by Gordon Varney

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Ken,
   In your drawing, you are using a P channel FET on the highside. The
problem is that the Gate must be approx., 3.5 Vdc bellow the Source to
conduct. If the Source is 1.5 Vdc then the Gate must be -2.0 Vdc. If I had 5
Vdc on the source then your circuit is great. (this is great help, thank you
guys for your input)

Gordon


> Erik Reikes wrote:
>
> >>>    I agree Eric's idea is almost perfect, and solves most of the turn
on
> >>>problems. Jost one more detail, I was hoping to be able to monitor the
> >>>switch to determine when the user releases the button. Of course that
> would
> >>>be an ideal situation, and we all know that Murphy would not allow
> this....
> >>>:-)
> >>
> >>
> >>Not without spending a little more money and making the circuit a tad
more
> >>complex ...
> >>
> >>Is a dual-pole pushbutton out of the question?
> >>
> >
> >If you went dual pole, I don't think you'd need all the transistors and
> >stuff.  Just have 5V routed to the other pole (actually it would have to
be
> ...
>
> Exactly.  That is what I had in mind.  The circuit is only for the case
> where you don't want a DPDT switch (because they cost too much, don't have
> the right "feel", etc.).  Sorry about the hard-to-read text circuit
> description .. Jinx's post gave me the right idea:  Attached is a .GIF of
> the circuit.
>
> Ken

2000\01\13@164637 by Ken Webster

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Gordon Varney wrote:

>Ken,
>    In your drawing, you are using a P channel FET on the highside. The
>problem is that the Gate must be approx., 3.5 Vdc bellow the Source to
>conduct. If the Source is 1.5 Vdc then the Gate must be -2.0 Vdc. If I had
5
>Vdc on the source then your circuit is great. (this is great help, thank
you
>guys for your input)


No, actually that is an N-channel MOSFET (in the shorthand notation the
arrow points out for an N-channel .. the longhand with the substrate shown
has the arrow pointing inward .. this is a bit confusing).

So, the gate must be aprx 3.5 VDC above the source .. thus, 5V with a fresh
battery, and less than 5 if the battery is older.  Just make sure your DC-DC
is really putting out 5V and not 4.5!  (or get one with a control input like
Paul suggested)

Cheers,

Ken

2000\01\13@170444 by Ken Webster

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part 0 855 bytes
>No, actually that is an N-channel MOSFET (in the shorthand notation the
>arrow points out for an N-channel .. the longhand with the substrate shown
>has the arrow pointing inward .. this is a bit confusing).
>
>So, the gate must be aprx 3.5 VDC above the source .. thus, 5V with a fresh
>battery, and less than 5 if the battery is older.  Just make sure your
DC-DC
>is really putting out 5V and not 4.5!  (or get one with a control input
like
>Paul suggested)


Actually, I just had a look at a textbook.  My notation has gotten a bit
sloppy -- having the line for the gate come in at one side kinda makes it
halfway between the "official" shorthand and longhand versions.  Sorry about
that.  A corrected version is attached.

Cheers,

Ken

Attachment converted: wonderland:pushbutton_power.gif 1 (GIFf/JVWR) (00014CB4)

2000\01\13@173144 by Gordon Varney

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Ken, Eric,
   This will do nicely. Yes an N channel will work under this condition.
Thank you both for your input and to all the others that offered ideas
"Thank You"

Gordon

> I wrote:
>
> >No, actually that is an N-channel MOSFET (in the shorthand notation the
> >arrow points out for an N-channel .. the longhand with the substrate
shown
> >has the arrow pointing inward .. this is a bit confusing).
> >
> >So, the gate must be aprx 3.5 VDC above the source .. thus, 5V with a
fresh
> >battery, and less than 5 if the battery is older.  Just make sure your
> DC-DC
> >is really putting out 5V and not 4.5!  (or get one with a control input
> like
> >Paul suggested)
>
>
> Actually, I just had a look at a textbook.  My notation has gotten a bit
> sloppy -- having the line for the gate come in at one side kinda makes it
> halfway between the "official" shorthand and longhand versions.  Sorry
about
> that.  A corrected version is attached.
>
> Cheers,
>
> Ken
>

2000\01\13@181724 by Ken Webster

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>Ken, Eric,
>    This will do nicely. Yes an N channel will work under this condition.
>Thank you both for your input and to all the others that offered ideas
>"Thank You"
>
>Gordon


Glad I could help.  Thanks to Eric for suggesting the MOSFET in the first
place.  Thanks to others for various ideas.  Please let us know if it works,
eh?  This ckt may come in handy for other applications.

Ken

2000\01\14@104640 by Wagner Lipnharski

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face
Sometimes the use of a simple latching relay (5Vdc) solve this problem.

See, an Omron/Aromat 5Vdc (or less) subminiature relay drains a pulse of
aprox 10mA during 50ms or less to latch, and the "unlatch" is zero power
since it uses the capacitor charged during the latching as the energy
source.

A simple (start) switch contact starts the step up 1.5 --> 5Vdc, that
starts the processor and a port pin drives (via a capacitor) the
latching relay that closes its contact in parallel to the start switch
(no power consumed by the relay while it is latched).

At the time the processor wants to power off, it simple reverts the port
pin level, it will drain the capacitor (in series to the latching relay
coil) and it will disengage, cutting power, easy and simple.

If the user still holding the start button when the processor issued
power off to the relay, nothing will hapens, since the processor would
be sleeping already (infinite loop waiting for lost of power) and the
relay would be disengaged already, so when the user releases the button,
power will finally goes down.  I don't see much Murphy's law here
(except if the battery is disconnected while the processor is on - the
relay will still latched and ).

(A pull up resistor could be necessary to supply enough current to the
capacitor + coil at the latching up time.)

For more info about the TK2 (1.5Vdc - 24Vdc) Aromat relays (4x10x9mm):
http://ctlgserv.mew.co.jp:80/ctlg/acg/eng/relay/mech_eng/@Generic__BookTextView/139715;td=2;cs=dynaweb1.wv;uf=0#X

Wagner Lipnharski

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