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'[PIC]: EE challenge PIC &&12V Battery'
2001\01\11@063629 by Germain Morbe

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Hello,

has anyone a LOW COST / LOW CURRENT idea that could help solving the
following alimentation problem?

A 16C505 or 12C508 shall be used as an ecoder for a 12V battery powered 3
button handheld transmitter. Because of the needed code completion feature
after button release, the power must be present permanently / resp. as long
as the PIC needs it.

The two approaches i see are:

a) use a very low power 5V regulator which is comparable expensive.  That
way the PIC could wakeup from sleep at button activation for as long time as
needed to complete a message after release of the button.

b) use a low cost 7805 regulator or a zener which draws to much current to
be allways on. Thus, the buttons need to switch the power directly and the
PIC needs to hold the power in the ON-state via some transistors. After
completion it could cut itself and also the stabilizing circuit from power
by releasing the transistors.

Is there a third possibility i dont think of?  I feel that from the two
mentioned either has its drawbacks in a low cost radio control? What would
the gurus advice?

Germain Morbe

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2001\01\11@070157 by Jinx

face picon face
This is asked from time to time, and you should find a few threads
in the archives

> a) use a very low power 5V regulator which is comparable expensive.

What do you call expensive ? Seiko, Linear, and National all do
ranges of low power LDO regulators. As many (including me) have
found though, you may have to buy fairly large quantities to get a
decent price

If you've got a few to make, then a circuit that requires much more
assembly than a simple regulator may cost more in labour

What is your "12V" ? Is it true 12V or a nominal 12V (perhaps 13.8V).
LP/LDO regs don't always like > actual 12V

> Is there a third possibility i dont think of?  I feel that from the two
> mentioned either has its drawbacks in a low cost radio control?

I use the LP2951, which can be turned on/off by a single pin. I also
use Seiko LDO regs when I can get them, and Seiko S81250 to
replace 78L05. How much transmitting power do you need ?

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2001\01\11@074818 by Germain Morbe

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Jinx,

we buy a seiko ldo that do 100mA, it is about 0,23$ at 10k pcs.
Its not that cheap but its main drawback is its poor availability over time.
Once you have decided for such a part you often cannot change on the fly,
especially at higher volumes.
That is why our company would love to use no specialized parts at all which
is often not a technical but a vital decision. I myself prefer state of the
art components when needed.

In generally i agree to you labour cost arguments, only because its all
surface mount, labour cost is not that high.

The battery is a small 23A type, usually its voltage not much higher than
12V.

We are allowed to radiate up to +10 dBm at 433MHz here in europe, but
usually with a poor antenna you will be happy to reach 0dBm if you put 100mW
in.

For the moment i tend to the ldo approach also. Who is building the LP2951?

Germain

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2001\01\11@090504 by M. Adam Davis

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This may not fit into your design, but there are low power switching
regulators which require fewer components than a typical switching design,
and feature shut-down modes which reduce current draw to the uA range.  A
button could supply enough power to the shutdown to start it up, and the
uC could shut it down as long as the button is not pressed.

You might try this one, for instance...
http://dbserv.maxim-ic.com/quick_view2.cfm?pdf_num=1155&Fam=DCDC_All&TREE=PowerSupplies.asp&HP=PowerSupplies.cfm

-Adam

Germain Morbe wrote:
{Quote hidden}

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2001\01\11@110018 by David VanHorn
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Ignoring the regulator, have the pushbutton supply input to the regulator.
Once the pic wakes up, it can turn on a PNP transistor, which is in
paralell with the button, to keep it's power up as long as it likes. When
it's done, just turn off the transistor.  This will take a pair of
transistors, but that's only pennies, and you won't have any availability
problems.

Now, solve the regulator as a separate problem.
Since it's input is cut off most of the time, its efficiency becomes much
less important.
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2001\01\11@110028 by Martin Hill

picon face
that's exactly what I did, using a MAX619 chip, just needs two .1uF caps.
This was to get 5v from two AA cells though, but the principle is the
same.  A button press activates it, then a pin on the pic holds it on
until it decides to turn it off.  Needs a bit of thinking about or the pic
will think the button is permanently pressed if you just drive the
same line.  Nothing a few transistors can't fix.

Martin


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2001\01\11@114553 by M. Adam Davis

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Unfortunately I couldn't find a capacitor drop down regulator, the one I
mentioned requires a coil and a diode (as well as a few caps).

As far as the button, you can use a diode to prevent the pic from
misreading the input:

vbatt+------+
           * |
             |--
           * |
           |
vneg--/\/\--+---|>|---+-------To shutdown (high=on)
           |         |
       ____|_________|_______
      |    1         2       |
      |       PIC            |

When the button is pressed it supplies enough power to bring shutdown
high, the pic wakes up, and brings I/O 2 high.  At that point it doesn't
matter whether the button is pressed, except the pic cannot turn the
regulator off if the button is pressed (in that case, bring pin 2 low, and
go to sleep - the regulator will shut off when the button is no longer
pressed).

You may have to tie the shutdown low - many regulators have it tied high
internally through a very weak resister.

-Adam

Martin Hill wrote:
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2001\01\11@130625 by Roman Black

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Germain Morbe wrote:
{Quote hidden}

I'm not a remote guru, but it does seem the max
priority is to conserve the battery.

I would use the buttons to switch the entire
power to the PIC (and regulator), and use a fet
or transistor to keep the power on under PIC
control until it decides it is finished doing
the task.

For the 3 button dilemma, I would use 3 diodes,
one from each button to the PIC power Vdd, this
is the common for all three buttons of course.
Then each button is also connected to one PIC
input pin. This will allow any button to power
the PIC, and also activate its own input pin.

The diodes will drop 0.6v, assuming cheap diodes,
but you only have to get the PIC running long
enough for it to turn the transistor on, and
then you have full power anyway. :o)
-Roman

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2001\01\11@132048 by Don Hyde

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The PIC is pretty tolerant of voltage variations, so it could be supplied
via a zener.  I imagine that the radio is a little more critical, so it
could be supplied by a regulator that is switched on by the PIC (easiest is
a PNP transistor with collector to +12V, emitter to regulator, base to
open-drain PIC output pin).  That way the regulator which supplies the radio
can be a simple 78L05, which has lousy quiescent current but is cheap.

> {Original Message removed}

2001\01\11@144201 by steve

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> Is there a third possibility i dont think of?  I feel that from the two
> mentioned either has its drawbacks in a low cost radio control? What would
> the gurus advice?

I don't know what it costs, but is PIC16HV540 worth considering ?
It may end up being cheaper to use that and no regulator.

Steve.
======================================================
Steve Baldwin                Electronic Product Design
TLA Microsystems Ltd         Microcontroller Specialists
PO Box 15-680, New Lynn      http://www.tla.co.nz
Auckland, New Zealand        ph  +64 9 820-2221
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2001\01\11@145830 by James Paul

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Germain,

Just an observation on my part and a bit of a tangent from your
original question, but regarding the RF power output of your unit.
You state that you are allowed a mx of +10dbm which is 10mw, correct?
And that you'd be happy to get 0dbm, which is 1mw correct?  And that
this could be had with only 100mw input.  Is this input to the entire
unit, or only to the RF section?  If to the whole unit, then what
portion of this input goes to the RF section?   If this is to the RF
section only, then you must have a very ineffecient RF section to be
sure.  It seems to me that even the most ineffecient transmitter RF
power output driver/amplifier can convert at a reat of 30% or so.
At best, you are getting about 10% and at worst about 1%.  Is this
correct, or am I miscalculating the numbers here?  Please advise.

                                              Regards,

                                                Jim





On Thu, 11 January 2001, Germain Morbe wrote:

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2001\01\11@150045 by M. Adam Davis

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I hadn't even thought of that!  This might be the best way to do things,
actually.

It has a wake up on pin chaange, so you can sleep the entire time without
using the watch dog to wake you up for polling.  Not only that, but it has
high voltage outputs which might help you with your transmitter design.
You could power your transmitter from the pic.

-Adam

Steve Baldwin wrote:
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2001\01\12@140257 by Peter L. Peres

picon face
You can save on transistors and switchoff logic by using a capacitance
multiplier with a single transistor. This saves a PIC pin (required to
hold the power on) (I suspect you are using a 8-pin PIC). The required
capacitor will be small, too.

Peter

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2001\01\12@140312 by Peter L. Peres

picon face
The poor efficiency of some rf tx modules is dued to the bandwidth
requirements and to on board PLL (if any). The devices run in class A or
AB at best. Class C is not usable at those power levels without using
expensive filters afterwards. Getting 1mW with 100mW input is about
standard. 100mW input is 5V, 20mA if you are wondering. If you think that
this is inefficient, compare to a PC computer or to a laser ;) Of course
one can make expen$ive modules with 60% or better efficiency.

Peter

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2001\01\13@022716 by Roman Black

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Peter L. Peres wrote:
>
> You can save on transistors and switchoff logic by using a capacitance
> multiplier with a single transistor. This saves a PIC pin (required to
> hold the power on) (I suspect you are using a 8-pin PIC). The required
> capacitor will be small, too.

Very clever! I had thought of a way to get
the PIC to pulse a line capacitor coupled to
the transistor, to give a turn on period and
get rid of the bias problems when just using
one transistor for the power.

Your way is better, and has less parts count
and much simpler.

What about getting rid of the transistor
all together, just charging a large cap with
the button, which then takes X time to discharge
powering the zener reg and PIC?? Assuming the
PIC draws regular amount of current it would
give predictable time to go down. Hmm. Maybe
battery voltage could cause problems as
battery starts getting flat?.. But this would
be even simpler. :o)
-Roman

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2001\01\14@171733 by Peter L. Peres

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>What about getting rid of the transistor
>all together, just charging a large cap with
>the button, which then takes X time to discharge

Only one small snag: capacitor size. When I do things like this I fit them
in a keyfob usually ;-)

Peter

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