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'[AVR:] Driving ATtiny11 with 9v battery'
2004\06\29@211054 by Chetan Bhargava

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Hi All,

With the price of tiny11 falling down to a 555 I'm thinking of
replacing our starter kit for MARES Workshop
(http://www.bhargavaz.net/workshop).

Right now we use a small 555 blinker to start the 8 year old kids on
soldering. I'm planning to replace the 555 kit with tiny11 kit.

Has anyone tried AVR on a 9v battery using resistors. Reason is 9v
battery is convenient and only a battery clip is needed instead of a
holder.

Regards,

Chetan Bhargava

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2004\06\29@214300 by David VanHorn

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At 06:11 PM 6/29/2004 -0700, Chetan Bhargava wrote:

>Hi All,
>
>With the price of tiny11 falling down to a 555 I'm thinking of
>replacing our starter kit for MARES Workshop
>(http://www.bhargavaz.net/workshop).
>
>Right now we use a small 555 blinker to start the 8 year old kids on
>soldering. I'm planning to replace the 555 kit with tiny11 kit.
>
>Has anyone tried AVR on a 9v battery using resistors. Reason is 9v
>battery is convenient and only a battery clip is needed instead of a
>holder.

Add a 5.1V zener diode, and this will work.
It's not efficient, and it dosen't support much output drive, but it does work.
I'd also add a series diode to prevent reverse polarity problems.

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2004\06\29@235744 by William Chops Westfield

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On Jun 29, 2004, at 6:11 PM, Chetan Bhargava wrote:

> Right now we use a small 555 blinker to start the 8 year old kids on
> soldering. I'm planning to replace the 555 kit with tiny11 kit.

A 555 circuit uses more interesting parts.  two caps, two resistors.
A beginners kit ought to have some introduction to common parts, in
addition to being easy to put together.  Maybe.  I have an easy layout
for the traditional 2-transistor astable oscillator if you're
interested...  My daughter put one together when she was about 8.

BillW

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2004\06\30@003031 by Chetan Bhargava

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HI Bill,

Yes, I agree on that but eight year old kids don't want to know the
working of the circuit, they just want to finish the kit and see the
LEDs blinking :-) Very impatient.

It would be nice if you drop down at the workshop and see how we do.
Workshop is on Mondays 7-9pm located here:
http://www.ci.milpitas.ca.gov/citydept/planning/recreation/default.htm

Regards,

Chetan


> A 555 circuit uses more interesting parts.  two caps, two resistors.
> A beginners kit ought to have some introduction to common parts, in
> addition to being easy to put together.  Maybe.  I have an easy layout
> for the traditional 2-transistor astable oscillator if you're
> interested...  My daughter put one together when she was about 8.
>
> BillW

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2004\06\30@003645 by William Chops Westfield

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On Jun 29, 2004, at 9:29 PM, Chetan Bhargava wrote:

> Yes, I agree on that but eight year old kids don't want to know the
> working of the circuit, they just want to finish the kit and see the
> LEDs blinking :-)

Oh, I agree on not knowing the workings, but having neat looking parts
instead of just black blobs is an advantage, too.  Maybe LEDs and a
chip and some current limitting resistors are enough.  After all, you
can't get much cooler than LEDs...

BillW

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2004\06\30@024217 by hael Rigby-Jones

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{Quote hidden}

A zener and series resistor will give reasonable reverse polarity protection
by themselves.  Reverse voltage is clamped to one diode drop, whih is
generaly survivable by most IC's.

Regards

Mike




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2004\06\30@080201 by Mr MCU

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Seem like everyone forgets that zener diodes actually subtract their
voltage if placed in series.  e.g. I would use a 1N4730A in series with
the 9V.  That would give you 9V - 3.9V = 5.1V ... perfect supply up to
1W!  Although it does not regulate, you really don't need regulation in
your example as I see it.  I would also use a 1N400x series diode for
reverse polarity protection - if so just use a 1N4728A to make up the
additional voltage drop.

MGB,

- Mike



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2004\06\30@082528 by David VanHorn

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At 08:01 AM 6/30/2004 -0400, Mr MCU wrote:

>Seem like everyone forgets that zener diodes actually subtract their
>voltage if placed in series.  e.g. I would use a 1N4730A in series with
>the 9V.  That would give you 9V - 3.9V = 5.1V ... perfect supply up to
>1W!  Although it does not regulate, you really don't need regulation in
>your example as I see it.  I would also use a 1N400x series diode for
>reverse polarity protection - if so just use a 1N4728A to make up the
>additional voltage drop.

So you've removed both regulation, and reverse polarity protection as features of this design?

I like the parallel version better.

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2004\06\30@084815 by Mr MCU

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Well not really - regulation is not needed.  The additional diode
supplies the reverse polarity protection, and since it is not a shunt
regulator, there is NO power loss if you put the micro into sleep mode -
just what the micro would draw.  This method is much more efficient and
will handle varying loads without trouble.  To do so with a shunt
regulator, you will waste most of your power heating up a series
resistor.  If you want that, you can use the shunt circuit as a hand
warmer too!

- Mike


David VanHorn wrote:

{Quote hidden}

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2004\06\30@100509 by hael Rigby-Jones

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>-----Original Message-----
>From: Mr MCU [@spam@mikepKILLspamspamSGI.NET]
>Sent: 30 June 2004 13:48
>To: KILLspamPICLISTKILLspamspamMITVMA.MIT.EDU
>Subject: Re: [AVR:] Driving ATtiny11 with 9v battery
>
>
>Well not really - regulation is not needed.  The additional
>diode supplies the reverse polarity protection, and since it
>is not a shunt regulator, there is NO power loss if you put
>the micro into sleep mode - just what the micro would draw.
>This method is much more efficient and will handle varying
>loads without trouble.  To do so with a shunt regulator, you
>will waste most of your power heating up a series resistor.
>If you want that, you can use the shunt circuit as a hand warmer too!
>
>- Mike
>

It won't handle varying loads very well, a zener diodes have quite a high
dynamic resistance, especialy low voltage ones. (AKA a "soft knee").

A shunt regulator does not have to be grossly ineffecient.  If properly
designed, and the ciruit it's supplying draws a relatively constant voltage,
the current through the shunt should be quite small.

Neither circuit is good for a micropower design that may have a relatively
wide supply current range.  A micropower linear or switched regulator is the
only decent solution in this case.

Regards

Mike





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2004\06\30@175710 by David VanHorn

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>
>Neither circuit is good for a micropower design that may have a relatively
>wide supply current range.  A micropower linear or switched regulator is the only decent solution in this case.

I use as zener shunt supply on the board I use to program my 2343s, because it's cheap, and easy.

I generally, will use a 7805-ish solution for most things, where I don't need efficiency.

Buck switchers where I do, and where the current demand is relatively high, and LDO linears where I need efficiency, and the current is low.

Engineering IS the art of compromise.

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2004\06\30@192114 by Chetan Bhargava

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Also the speed of chaser can be controlled using external RC oscillator :-)

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2004\06\30@193322 by David VanHorn

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At 08:48 AM 6/30/2004 -0400, Mr MCU wrote:

>Well not really - regulation is not needed.  The additional diode
>supplies the reverse polarity protection, and since it is not a shunt
>regulator, there is NO power loss if you put the micro into sleep mode -
>just what the micro would draw.


That's a strong, and WRONG statement.

No matter what the level of current is in a series circuit, if it contains a reverse biased zener, the power converted to heat in the Zener will be I * Vz.

1mA, 5.1V zener, 5.1mW lost as heat in the zener.


If you had said, "not much", I'd agree.
Still I think the Zener shunt is simpler, and less likely to get a newbie into trouble, than what you proposed. It also provides some reverse polarity protection without adding an additional diode.

This is one that would end up in the "bad circuit" pages of "Art of Electronics", I'm afraid.

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'[AVR:] Driving ATtiny11 with 9v battery'
2004\07\01@092731 by Mr MCU
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OK, one last time on this one ... is this a perfect world? ... no, but
for the application it will work very nicely.   Everything is a
compromise.  I don't pretend to be the best engineer, but I am one of
the best designers.

I was incorrect in stating NO power loss, what I should have said is
that the current draw is ABOUT the same on both sides of the zener.
e.g. if the micro is using 1ma (when the LED is off) it will draw APPROX
1ma from the 9V battery.  When the LED is on, (let say you want to use a
cheap one and be somewhat bright so you set it up to draws 6Ma) then 7Ma
(1ma + 6ma) is drawn from the battery.  If you put the micro into sleep,
maybe it draws 100uA or so about from the battery.  Try that with your
shunt regulator!  How much current would the shunt use from the 9V in
all three of these conditions?

I made an assumption as to how the 'product' would work - that is using
a tact switch connected to a port line to turn the unit ON/OFF as well
as work speed and/or brightness - instead of a slide in series with the
9V to simply turn it ON/OFF - not only more function but at a lower
cost.  You could even replace the tact switch with pcb artwork to be
bridged by one's finger ... at NO additional cost - except for a couple
of resistors (for the input spike protection).  Gee, you could go for
broke adding another 'touch switch' or more to do even more functions.

As for voltage regulation, as an LED blinker, there will not be any
significant difference if the micro sees 3.5 volts or 5, other than the
brightness of the LED and possibly the speed of the blinking.  If they
were controlled via the touch switch - no one would notice.

Sure a low Q linear regulator or a switcher would work, but at a much
greater expense and NOT needed for THIS project.  A zener is about
$0.03, 1N400x diode about $0.007, so for less that 4 cents you have an
ALTERNATE low Q supply.  How is the series zener any more 'complicated'
that the shunt?  In your concept, one would save maybe $0.004 (cost
savings of using a resistor instead of the 1N400x) at a cost of MUCH
greater current draw from the 9V.

As far as getting a newbie into trouble, I assume that the kit comes
with instructions AND a description of operation.  If he gets into
trouble, its because the author did a bad job of writing or the newbie
didn't read it!

The ART of design is IDENTIFYING the problem and working out the best
solution, coming in at the lowest possible cost without compromising
function or longevity, and doing so in a unique way.  I have been doing
such for 20+ years with 10's of thousands of products in the field - to
state "bad circuit" pages of "Art of Electronics" - I am afraid that you
know not of where you speak!

Good luck in your endeavors and keep doing things the same old way - its
safer and doesn't require much creativity!

- Mike




David VanHorn wrote:

{Quote hidden}

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2004\07\01@094355 by David VanHorn

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At 09:26 AM 7/1/2004 -0400, Mr MCU wrote:

>OK, one last time on this one ... is this a perfect world? ... no, but
>for the application it will work very nicely.   Everything is a
>compromise.  I don't pretend to be the best engineer, but I am one of
>the best designers.
>
>I was incorrect in stating NO power loss, what I should have said is
>that the current draw is ABOUT the same on both sides of the zener.

??? Kirchoff's law, the current everywhere in a series circuit is exactly the same.
(DC circuits, AC is different.)

>e.g. if the micro is using 1ma (when the LED is off) it will draw APPROX
>1ma from the 9V battery.  When the LED is on, (let say you want to use a
>cheap one and be somewhat bright so you set it up to draws 6Ma) then 7Ma
>(1ma + 6ma) is drawn from the battery.  If you put the micro into sleep,
>maybe it draws 100uA or so about from the battery.  Try that with your
>shunt regulator!  How much current would the shunt use from the 9V in
>all three of these conditions?

Of course the shunt zener actually regulates the voltage, where the series does not.
Vout = Vin-Vz, and the series zener wastes nearly as much power as a 7805-ish solution would, the difference being only the quiescent current of the regulator.

If you're going to have a significant load, then a 7805-ish linear would be a much better choice.


>As for voltage regulation, as an LED blinker, there will not be any
>significant difference if the micro sees 3.5 volts or 5, other than the
>brightness of the LED and possibly the speed of the blinking.  If they
>were controlled via the touch switch - no one would notice.

Clock speed varies with voltage, when using the internal RC osc.

>Sure a low Q linear regulator or a switcher would work, but at a much
>greater expense and NOT needed for THIS project.

Where was "THIS" project clearly defined?



> How is the series zener any more 'complicated' that the shunt?

It isn't, it just doesn't offer any significant advantage, and it has significant disadvantages. If you're so concerned over cost, then do an emitter follower with a 2N3904 on the shunt, and get moderate current output, and reverse polarity protection for another penny.

>As far as getting a newbie into trouble, I assume that the kit comes
>with instructions AND a description of operation.  If he gets into
>trouble, its because the author did a bad job of writing or the newbie
>didn't read it!

Students frequently make mistakes.
What <I> would do, is not worry about minimum power consumption, and instead design for a robust supply, and protect the micro from typical student mistakes, like inverted supply polarity, or disconnecting the ground before Vin.

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2004\07\01@104935 by Russell McMahon

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> Seem like everyone forgets that zener diodes actually subtract their
> voltage if placed in series.  e.g. I would use a 1N4730A in series with
> the 9V.  That would give you 9V - 3.9V = 5.1V ... perfect supply up to
> 1W!  Although it does not regulate, you really don't need regulation in
> your example as I see it.  I would also use a 1N400x series diode for
> reverse polarity protection - if so just use a 1N4728A to make up the
> additional voltage drop.

The supply is a NOMINAL 9 volt battery. These can start out with as much as
9.6v (maybe even more at the very start) and down to say 6v as a reasonable
end point. Even at 9v to 6v that's a 3 volt swing which is quite demanding
for a microprocessor to handle. Say 2v to 5v or 2.5 to 5.5. And somewhat
more than that in reality. While that may be OK for a circuit that was being
built with utterly minimum parts as the main objective, it's not really
practical for powering real world circuits in this case.

A shunt zener supply is also pretty poor with such a wide input voltage
swing. Say you wanted at least 10 mA at 6v, then you must dimension your
dropping resistor such that you'd have 45 mA at 9.5v - and most of the power
is wasted. This is because you must dimension the dropper resistor for
maximum current at minimum voltage so that you draw far more as the voltage
rises to maximum. When Vin closely approaches Vout the results get
horrendous.

In this situation the cost of a switcher is hardly justified (unless you
want to use an eg Black Regulator and show what engineers can really do !
;-)
That's not a totally silly idea - the dearest part is the inductor - and it
would be a superb technology demonstrator.

But a linear regulator is probably the most practical choice. This could be
an (ugh) 78L05 with very poor dropout voltage, or an expensive LDO OR a roll
your own 1 transistor plus zener and 1 resistor (as has been suggested) (and
probably a filter cap). The zener and resistor form a shunt regulator at
about 5.6v BUT at low power, and this supplies the base of an emitter
follower with collector at "9v" and emitter = output. This is not an ideal
circuit as you want to run the zener at low current to reduce quiescent
power and the knee will be a bit soft there, but the results are entirely
bearable for this application and largely superior to the alternatives.

Lets see: Say target Iout is 50 mA max (allow some load capability at 5v.
Say emitter follower beta is 50 (should be easy enough) so  base current is
1 mA.
BUT if battery endpoint is 6v then min resistor drop is (6-5.6) = 0.4V.
So max base current at 9.5V is (9.5-5.6)/(6.5-5.6) x 1 mA =~ 10 mA
That's a pretty ugly quiescent current for a 50 mA regulator.
For less than 50 mA or a slightly higher battery endpoint (say 6.5V) it gets
much better.

eg 6V, 10 mA = 2 MA quiescent.
6.5V, 50 mA =~3 MA
6.5V, 10 mA = minimal

You COULD use clever tricks to get the quiescent current lower, but it's
hardly worthwhile.

SO

NPN transistor (I'd use BC337 but use what suits).
Base to +9V = 1k
Cap 10 uF base to ground.
5v6 zener base to ground.
Emitter = output.

Iq at 9v =~ (9-5.6)/1k = 3.4 mA
Max I at 9v = 170 mA (beta = 50)

Iq at 6v = 0.4 mA
Max I at 6v = 20 mA (beta = 50)

Iq at 7v = 1.4 mA
Max I at 7v =  70 mA (beta = 50)

A reasonably acceptable solution I think.

If Vout is 4.5v or 4v or less you get FAR better results as the ratio of
voltages across the zener feed resistor is much lower.

Rough design equations:

   Iq ~~= (Vin-Vout)/Rz
   Rz = (Vin_min-Vout)/ Imax
   Iout = Iq x beta

or rearrange as required.

       eg Iqmin = Iout/beta
       etc

In all cases you'll want an output filter capacitor.


       Russell McMahon

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2004\07\01@145610 by Ben Hencke

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I like the series zener idea. It just subtracts the unwanted voltage
without burning excess current. This is as simple as it gets.

>
> I was incorrect in stating NO power loss, what I should have said is
>...

The power loss in the zener is the zener voltage times whatever the
circuit draws. If the circuit draws nothing, the zener wastes nothing.
The only thing more efficient would be an actual power converter (ie
switching regulator).


>
> As far as getting a newbie into trouble, I assume that the kit comes
>...

Just make sure newbies understand that a zener is supposed to be
pointing the "wrong" way ;-)

- Ben

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2004\07\01@150225 by David VanHorn

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>
>The power loss in the zener is the zener voltage times whatever the
>circuit draws. If the circuit draws nothing, the zener wastes nothing.
>The only thing more efficient would be an actual power converter (ie
>switching regulator).

However, there is NO regulation at all.
Any voltage variation on the supply is passed direct to the micro.

For that matter, if efficiency and low cost are the issue, why are we talking about a 9V battery?  Four AA cells and a series diode would give better regulation than the 9V/Zener system, and reverse protection as well. An LDO would be better, but I could argue for this system in a product design, in some cases.

Caps are needed in all cases, I figured those as a "gimme".


>>
>> As far as getting a newbie into trouble, I assume that the kit comes
>>...
>
>Just make sure newbies understand that a zener is supposed to be
>pointing the "wrong" way ;-)

It's not the zener that I'm worried about, it's the main supply.

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2004\07\01@155525 by Ben Hencke

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Thats why you also add a regular diode and compensate with a lesser
value zener. I believe this was already mentioned. But in assembly, it
is easy for a newbie to be confused about how a zener works
differently than a regular diode.

Yes there is no regulation, but is it required? It depends on the
ciruit and I don't think that a blink-a-led needs regulation. PICs do
not always need regulation, especially with such a wide operating
voltage range of most PICs. I have found the INTRC to be very
forgiving of supply variation and have not had a PIC lock up due to
supply ripple.

9v is the de-facto standard power supply for newbies. What could be
easier for a newbie to diagnose and replace?

my $1/50

- Ben



On Thu, 1 Jul 2004 14:02:32 -0500, David VanHorn <spamBeGonedvanhornspamBeGonespamcedar.net> wrote:
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2004\07\01@160155 by Chetan Bhargava

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I have a bag full of old si diodes that look like 1N4001 but have a
greater voltage drop. Connecting a few in series should drop the
voltage.

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2004\07\01@223317 by William Chops Westfield

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On Jul 1, 2004, at 1:02 PM, Chetan Bhargava wrote:

> I have a bag full of old si diodes that look like 1N4001 but have a
> greater voltage drop. Connecting a few in series should drop the
> voltage.
>

Yeah.  Is it still simpler than your 555 blinky circuit?  I've done
this sort of thing: "hey! I can use a micro instead of a 555."  "but
wait, it needs ~5V.  And a decoupling cap.  And it doesn't do +/-200mA
output.  And...  Hmm; maybe that 555 wasn't so bad after all."

BillW

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2004\07\01@223729 by William Chops Westfield

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On Jul 1, 2004, at 7:49 AM, Russell McMahon wrote:

> The supply is a NOMINAL 9 volt battery. These can start out with as
> much as 9.6v (maybe even more at the very start) and down to say 6v as
> a reasonable end point. Even at 9v to 6v that's a 3 volt swing which
> is quite demanding for a microprocessor to handle. Say 2v to 5v or 2.5
> to 5.5.

Since the tiny11 is spec'ed at 2.7 to 6V, that sounds like a pretty
good match...

BillW

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2004\07\01@223937 by David VanHorn

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At 12:54 PM 7/1/2004 -0700, Ben Hencke wrote:

>Thats why you also add a regular diode and compensate with a lesser
>value zener. I believe this was already mentioned. But in assembly, it
>is easy for a newbie to be confused about how a zener works
>differently than a regular diode.

I wouldn't make it a requirement to understand how the power supply works.
Designing power supplies isn't really all that trivial.
Where I was going, was a reliable, simple, and inexpensive power supply for running student projects. Such a supply will need to tolerate mistakes that students might make, and not introduce any "flakiness" that will complicate the student's experience when learning about the processor.

>Yes there is no regulation, but is it required? It depends on the
>ciruit and I don't think that a blink-a-led needs regulation. PICs do
>not always need regulation, especially with such a wide operating
>voltage range of most PICs. I have found the INTRC to be very
>forgiving of supply variation and have not had a PIC lock up due to
>supply ripple.

The only valid question here, is do you go outside the spec?

>9v is the de-facto standard power supply for newbies. What could be
>easier for a newbie to diagnose and replace?

They are rather expensive on a $/Watt basis.

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2004\07\02@033310 by Russell McMahon

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> > The supply is a NOMINAL 9 volt battery. These can start out with as
> > much as 9.6v (maybe even more at the very start) and down to say 6v as
> > a reasonable end point. Even at 9v to 6v that's a 3 volt swing which
> > is quite demanding for a microprocessor to handle. Say 2v to 5v or 2.5
> > to 5.5.

> Since the tiny11 is spec'ed at 2.7 to 6V, that sounds like a pretty
> good match...

No - the series zener concept will not work across a full working battery
range with a tiny11.
Actually (or at least according to the Atmel tiny 11 & tiny12 data sheet Rev
1006D--AVR-07/03) the voltages are

2.7 - 5.5    ATtiny11L-2 & L-4        2.8v range
4.0 - 5.5    ATTiny11-6                    1.5v range

Note that the bulk buy part is liable to not be the L version. It would work
only until the battery went from new to 8.1V !!!

To guarantee that datasheet spec is met for the L part (and guaranteeing
that datasheet spec is met is an essential part of acceptable design), you
could allow the battery to go from 9v6 down to 6v8 IF the zener knee was
100% square. In practice the zener is operating from a varying current load
and you probably should allow another 0.2v on knee variability. This gives
battery range of from new down to 7v. You can't use the processor for a UART
with RC control with such a supply. Other timing applications using RC clock
are also essentially impractical.


In fact, while in the above I said "even at 9v to 6v" (= 3v range), I also
made it clear that the real range is liable to be 9.6 - 6 = 3.6v range.
Both of these are outside the maximum Vcc swing of 2.8v for the L part and
1.5v swing for the normal temperature part.

Even if the processors operating range could JUST be made to fit the 3.6v
battery swing (eg 5.6 - 2v operation) I would tend to recommend against it
for all but very very very dedicated jobs that could tolerate it, as such a
wide Vcc swing in operation is liable to cause "issues" in general
applications. While it MAY be OK to run a blink-a-LED program with such a
widely varying supply, the savings of a few cents in the power supply is
exposing beginners to a range of other factors which may confuse and
mislead. A keen beginner will consider playing with the program and giving
them an approximately stable clock is highly desirable. Even having the tone
of a doorbell or code practice oscillator vary with battery voltage is
liable to be unacceptable.

Using Digikey 100 pricing,

BC337    $0.14
Zener       $0.04
Resistor   $0.02 say

$US0.20 buys you the 3 parts needed to build quite a reasonable regulator.

Note that Digikey prices for 78L05 in 100 quantity are as low as $US0.30
OR you can get the superior 7805 for the same money!
But, then you'd have to use them :-)


       Russell McMahon

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2004\07\02@085055 by David VanHorn

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>
>To guarantee that datasheet spec is met for the L part (and guaranteeing
>that datasheet spec is met is an essential part of acceptable design), you
>could allow the battery to go from 9v6 down to 6v8 IF the zener knee was
>100% square.

Zeners have bad knees.
Worse, as the current goes down, the knee gets worse, and the Vz changes.


> While it MAY be OK to run a blink-a-LED program with such a
>widely varying supply, the savings of a few cents in the power supply is
>exposing beginners to a range of other factors which may confuse and
>mislead.

Bingo.

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2004\07\02@103253 by Chetan Bhargava

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Found Fairchild LM78L05 for 10 cents at Arrow. I think, I'll use that.

Regards


{Quote hidden}

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2004\07\02@111025 by artin Klingensmith

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On Fri, 2 Jul 2004 19:32:42 +1200, Russell McMahon
<TakeThisOuTapptechEraseMEspamspam_OUTparadise.net.nz> wrote:
> Actually (or at least according to the Atmel tiny 11 & tiny12 data sheet Rev
> 1006D--AVR-07/03) the voltages are
>
> 2.7 - 5.5    ATtiny11L-2 & L-4        2.8v range
> 4.0 - 5.5    ATTiny11-6                    1.5v range
>
> Note that the bulk buy part is liable to not be the L version. It would work
> only until the battery went from new to 8.1V !!!

Hi Russell,
The part we're getting is the -6 version just to let you know.

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2004\07\02@172107 by Russell McMahon

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> > Using Digikey 100 pricing,
> >
> > BC337    $0.14
> > Zener       $0.04
> > Resistor   $0.02 say

> Found Fairchild LM78L05 for 10 cents at Arrow. I think, I'll use that.

Certainly the easiest solution.
You can also probably find suitable transistors and zeners at eg Arrow for
equivalently less as well for a total cost of maybe 5 cents in parts - at
which point it becomes not worthwhile ;-)

The 78L05 has far better regulation than a zener based arrangement but is
still inferior in a few ways. These probably don't bother you in this
application but are worth noting in passing.

Quiescent current is 3 mA typical and 5 mA worst case. The simple transistor
with zener arrangement someone suggested can have a quiescent current
designed around your maximum current drain and actual transistor beta and
will usually be lower than the 78L05 for many applications. But this is
probably not too important here.

Worse though, the dropout voltage is over 1.5v at 25C at 1 mA and more like
1.8v at 25C at 70 mA (it gets worse at lower tempeatures). This means the
battery voltage can not be less than about 6.6 volts. The 1 transistor
regulator does better in this respect. Again though, it's not the end of the
world in such a simple application if you can't use the tail end of the
battery power without regulator dropout.

I'd say the 78L05 at $US0.10 is a good compromise. To put the quiescent
current in perspective - using an Alkaline 'PP3' 9V transistor battery with
no laod at all you'd get about 100 to 150 hours of battery life with the
regulator alone as load.



       Russell McMahon






{Quote hidden}

regulator.
> >
> > Note that Digikey prices for 78L05 in 100 quantity are as low as $US0.30
> > OR you can get the superior 7805 for the same money!
> > But, then you'd have to use them :-)
>
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2004\07\02@175850 by Chetan Bhargava

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Thanks Russell,

Low temp problem would be a concern if the kids go on a Alaska
vacation and take their LED chaser with them. :-)

I'm running out on number of holes allowed on the PCB so instead of
using discrete components regulator, I'll use a 78L05.

Regards,

Chetan


On Sat, 3 Jul 2004 09:20:28 +1200, Russell McMahon
<RemoveMEapptechspamTakeThisOuTparadise.net.nz> wrote:
{Quote hidden}

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2004\07\02@181550 by William Chops Westfield

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On Jul 2, 2004, at 7:52 AM, Martin Klingensmith wrote:

>> 4.0 - 5.5    ATTiny11-6                    1.5v range
>>
oops.  i was obviously getting the AVRs confused with the PICs,
which have the wide operating range on their "default" parts.
Chalk up another advantage to PICs.

BillW

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2004\07\05@040825 by hael Rigby-Jones

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>-----Original Message-----
>From: William Chops Westfield [westfwEraseMEspam.....MAC.COM]
>Sent: 02 July 2004 22:08
>To: EraseMEPICLISTspamMITVMA.MIT.EDU
>Subject: Re: [AVR:] Driving ATtiny11 with 9v battery
>
>
>On Jul 2, 2004, at 7:52 AM, Martin Klingensmith wrote:
>
>>> 4.0 - 5.5    ATTiny11-6                    1.5v range
>>>
>oops.  i was obviously getting the AVRs confused with the
>PICs, which have the wide operating range on their "default"
>parts. Chalk up another advantage to PICs.

PIC's are typicaly only rated from around 4.0 to 5.5volts for the standard
part (actualy voltage varies with the part), and 2.0-5.5 for the low voltage
"LF" parts.  The maximum operating clock frequency is reduced significantly
at lower voltages as well.

Regards

Mike




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