"Ian Jackson" wrote in message
...
In message , ian field
writes

"Ian Jackson" wrote in message
news
In message , ian field
writes

"Ian Jackson" wrote in message
...
In message , rf
writes
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. And the capacitor doesn't need to be that big.

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. 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

From the output run a 470 ohm half watt resistor through a standard 1/4" LED
to ground (I like the LED so you can see when the circuit is on). In
parallel to this connect a capacitor, say 100uF. Presto, ~9.2V, adequately
regulated, and minimal voltage increase when the load is off.