"Ian Jackson" wrote in message
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In message , Zootal
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"Ian Jackson" wrote in message
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In message , ian field
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"Ian Jackson" wrote in message
news In message , ian field
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"Ian Jackson" wrote in
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In message , rf
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Roger Dewhurst wrote:
You can get simple to use regulator chips that drop the voltage
down, you only need a couple of components to make a working
voltage
regulator. Ask on News:sci.electronics.basic - the folk on there
are
usually very helpful and should explain all you need.


Why not just drop the voltage through a few diodes? Very simple.
Very
cheap.

A *few* diodes at a couple of ten cents per each. A single 7809 for
less
that a $.

More like "a *few* diodes at a couple of cents per each".

3V to drop = 5 diodes @ 0.6V per diode. No other circuitry required.
More
than good enough for the job.
--
Ian

The forward conduction knee curve on diodes isn't *that* sharp,
depending
on
current draw and rating of the diode the drop can be as low as 0.55V
and
as
high as 1.1V.

For most 'normal' Si diodes, that isn't really the case. The actual
voltage drop does, of course, increase with current, but at 'sensible'
currents, you can reckon on around 0.65V per diode. How much current
is
the Tardis toy going to take? 1A max? 4 or 5 1N4000-series diodes
should
work fine in this application. I've used this non-elegant 'KISS'
technique
on several occasions, and haven't found any problems.
--
Ian

A potential danger with a cassette recorder is the difference in current
draw between motor on and motor off. In the condition of low current
draw
(and low diode drop) supply decoupling electrolytic capacitors can
charge
to
a higher voltage which is then dumped into the circuit when switched to
play.

True, true. But I reckon that a momentary short burst of a
rapidly-decaying additional 3V won't hurt too much.
--
Ian

It won't even be noticeable. The capacitors won't charge up that high to
start with, and they don't "dump" into the circuit, they just quickly
discharge down to the lower voltage that is present at the output of the
last diode - how fast this happens depends on the size of the caps and the
load. I wouldn't even call it a surge. A resistor from the last diode to
ground will prevent them from charging more then a few tenths of a volt
and
is a good idea.

Indeed. A bleed of a few mA will prevent the off-load voltage rising to
12V. 9V at 10mA would require 900 ohms (say 1k), 90mW (so even a 1/4W will
do).

And the capacitor doesn't need to be that big.

Which capacitor do you mean?

I would *not* use 1n400x diodes. ~1 amp will make them hot and susceptible
to failure. Use 2 or 3 amp diodes, they are cheap and readily available.
They are bigger and will run cooler and won't fail as easily.

1A through diodes dropping 0.6V means 600mW per diode. 1N4000-type will
run a bit warm, so maybe a physically larger (higher current) diode might
be better. But it depends on how much current the Tardis takes!

Or use two
strings of 1n400x in series, that is good enough for an app like this.

So - 12v - diode - diode - diode - diode - ~9.2v Output

That's only one string of diodes in series. Did you mean parallel? If so,
no, you shouldn't parallel diodes. As you suggest, use higher current
diodes.

Actually if you have two strings each having the same number of diodes and
put the two strings in parallel it doesn't matter, when you have a few or
more diodes in series as the variation in Vf for each diode averages out.