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Thread: Driving ATtiny11 with 9v battery
face BY : Russell McMahon email (remove spam text)

> 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

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.


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

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

       Russell McMahon

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