can i use a 12 volt battery on a 9 volt device ?

"Ian Jackson" <ianREMOVETHISjackson_at_g3ohx.demon.co.uk> wrote in message
news:GvAqKUAuZbcKFwgH_at_g3ohx.demon.co.uk...
In message <HAhcm.73167$561.68469_at_newsfe28.ams2>, ian field
gangprobing.alien_at_ntlworld.com> writes

"Ian Jackson" <ianREMOVETHISjackson_at_g3ohx.demon.co.uk> wrote in message
news:oQqRlSBFmUcKFw3i_at_g3ohx.demon.co.uk...
In message <AC3cm.163781$bA.81816_at_newsfe16.ams2>, ian field
gangprobing.alien_at_ntlworld.com> writes

"Ian Jackson" <ianREMOVETHISjackson_at_g3ohx.demon.co.uk> wrote in message
news:RK8$DzIdHMcKFwRq_at_g3ohx.demon.co.uk...
In message <Bqfam.7029$ze1.6612_at_news-server.bigpond.net.au>, rf
rf_at_z.invalid> 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.

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.

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

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.

The cap isn't a bad idea, but 470 ohms will give you around 14mA through

In message <t-GdnS96cOSBWOzXnZ2dnUVZ_oSdnZ2d_at_giganews.com>, Zootal
usenet_at_spam.zootal.nospam.com> writes

"Ian Jackson" <ianREMOVETHISjackson_at_g3ohx.demon.co.uk> wrote in message
news:GvAqKUAuZbcKFwgH_at_g3ohx.demon.co.uk...
In message <HAhcm.73167$561.68469_at_newsfe28.ams2>, ian field
gangprobing.alien_at_ntlworld.com> writes

"Ian Jackson" <ianREMOVETHISjackson_at_g3ohx.demon.co.uk> wrote in message
news:oQqRlSBFmUcKFw3i_at_g3ohx.demon.co.uk...
In message <AC3cm.163781$bA.81816_at_newsfe16.ams2>, ian field
gangprobing.alien_at_ntlworld.com> writes

"Ian Jackson" <ianREMOVETHISjackson_at_g3ohx.demon.co.uk> wrote in
message
news:RK8$DzIdHMcKFwRq_at_g3ohx.demon.co.uk...
In message <Bqfam.7029$ze1.6612_at_news-server.bigpond.net.au>, rf
rf_at_z.invalid> 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.

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.

"Ian Jackson" <ianREMOVETHISjackson_at_g3ohx.demon.co.uk> wrote in message
news:4Lvlz7KQMdcKFwkC_at_g3ohx.demon.co.uk...
In message <t-GdnS96cOSBWOzXnZ2dnUVZ_oSdnZ2d_at_giganews.com>, Zootal
usenet_at_spam.zootal.nospam.com> writes

"Ian Jackson" <ianREMOVETHISjackson_at_g3ohx.demon.co.uk> wrote in message
news:GvAqKUAuZbcKFwgH_at_g3ohx.demon.co.uk...
In message <HAhcm.73167$561.68469_at_newsfe28.ams2>, ian field
gangprobing.alien_at_ntlworld.com> writes

"Ian Jackson" <ianREMOVETHISjackson_at_g3ohx.demon.co.uk> wrote in message
news:oQqRlSBFmUcKFw3i_at_g3ohx.demon.co.uk...
In message <AC3cm.163781$bA.81816_at_newsfe16.ams2>, ian field
gangprobing.alien_at_ntlworld.com> writes

"Ian Jackson" <ianREMOVETHISjackson_at_g3ohx.demon.co.uk> wrote in
message
news:RK8$DzIdHMcKFwRq_at_g3ohx.demon.co.uk...
In message <Bqfam.7029$ze1.6612_at_news-server.bigpond.net.au>, rf
rf_at_z.invalid> 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.

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.

Possibly. Possibly not. Murphy and his Laws move in mysterious ways. You

In message <k9ocm.127797$YA5.62631_at_newsfe08.ams2>, ian field
gangprobing.alien_at_ntlworld.com> writes

"Ian Jackson" <ianREMOVETHISjackson_at_g3ohx.demon.co.uk> wrote in message
news:4Lvlz7KQMdcKFwkC_at_g3ohx.demon.co.uk...
In message <t-GdnS96cOSBWOzXnZ2dnUVZ_oSdnZ2d_at_giganews.com>, Zootal
usenet_at_spam.zootal.nospam.com> writes

"Ian Jackson" <ianREMOVETHISjackson_at_g3ohx.demon.co.uk> wrote in message
news:GvAqKUAuZbcKFwgH_at_g3ohx.demon.co.uk...
In message <HAhcm.73167$561.68469_at_newsfe28.ams2>, ian field
gangprobing.alien_at_ntlworld.com> writes

"Ian Jackson" <ianREMOVETHISjackson_at_g3ohx.demon.co.uk> wrote in
message
news:oQqRlSBFmUcKFw3i_at_g3ohx.demon.co.uk...
In message <AC3cm.163781$bA.81816_at_newsfe16.ams2>, ian field
gangprobing.alien_at_ntlworld.com> writes

"Ian Jackson" <ianREMOVETHISjackson_at_g3ohx.demon.co.uk> wrote in
message
news:RK8$DzIdHMcKFwRq_at_g3ohx.demon.co.uk...
In message <Bqfam.7029$ze1.6612_at_news-server.bigpond.net.au>, rf
rf_at_z.invalid> 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.

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.

Possibly. Possibly not. Murphy and his Laws move in mysterious ways. You
don't see many circuits with paralleled diodes (stringed or otherwise).
--
Ian

"Ian Jackson" <ianREMOVETHISjackson_at_g3ohx.demon.co.uk> wrote in message
news:PVRIUMAqWhcKFwia_at_g3ohx.demon.co.uk...
In message <k9ocm.127797$YA5.62631_at_newsfe08.ams2>, ian field
gangprobing.alien_at_ntlworld.com> writes

"Ian Jackson" <ianREMOVETHISjackson_at_g3ohx.demon.co.uk> wrote in message
news:4Lvlz7KQMdcKFwkC_at_g3ohx.demon.co.uk...
In message <t-GdnS96cOSBWOzXnZ2dnUVZ_oSdnZ2d_at_giganews.com>, Zootal
usenet_at_spam.zootal.nospam.com> writes

"Ian Jackson" <ianREMOVETHISjackson_at_g3ohx.demon.co.uk> wrote in message
news:GvAqKUAuZbcKFwgH_at_g3ohx.demon.co.uk...
In message <HAhcm.73167$561.68469_at_newsfe28.ams2>, ian field
gangprobing.alien_at_ntlworld.com> writes

"Ian Jackson" <ianREMOVETHISjackson_at_g3ohx.demon.co.uk> wrote in
message
news:oQqRlSBFmUcKFw3i_at_g3ohx.demon.co.uk...
In message <AC3cm.163781$bA.81816_at_newsfe16.ams2>, ian field
gangprobing.alien_at_ntlworld.com> writes

"Ian Jackson" <ianREMOVETHISjackson_at_g3ohx.demon.co.uk> wrote in
message
news:RK8$DzIdHMcKFwRq_at_g3ohx.demon.co.uk...
In message <Bqfam.7029$ze1.6612_at_news-server.bigpond.net.au>, rf
rf_at_z.invalid> 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.

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.

Possibly. Possibly not. Murphy and his Laws move in mysterious ways. You
don't see many circuits with paralleled diodes (stringed or otherwise).
--
Ian

Larger power supplies will have diodes in parallel, and they will even put
the output trannies in parallel. You can get away with it with diodes but
you have to be carefull to use the same type, preferably from the same
batch. Otherwise you end up with some saturated and others barely turned on,
and you get a high failure rate.

During 50 years 'in electronics', I can't immediately recall seeing any

In message <y76dnR4FvZf0he_XnZ2dnUVZ_vudnZ2d_at_giganews.com>, Zootal
usenet_at_spam.zootal.nospam.com> writes

"Ian Jackson" <ianREMOVETHISjackson_at_g3ohx.demon.co.uk> wrote in message
news:PVRIUMAqWhcKFwia_at_g3ohx.demon.co.uk...
In message <k9ocm.127797$YA5.62631_at_newsfe08.ams2>, ian field
gangprobing.alien_at_ntlworld.com> writes

"Ian Jackson" <ianREMOVETHISjackson_at_g3ohx.demon.co.uk> wrote in message
news:4Lvlz7KQMdcKFwkC_at_g3ohx.demon.co.uk...
In message <t-GdnS96cOSBWOzXnZ2dnUVZ_oSdnZ2d_at_giganews.com>, Zootal
usenet_at_spam.zootal.nospam.com> writes

"Ian Jackson" <ianREMOVETHISjackson_at_g3ohx.demon.co.uk> wrote in
message
news:GvAqKUAuZbcKFwgH_at_g3ohx.demon.co.uk...
In message <HAhcm.73167$561.68469_at_newsfe28.ams2>, ian field
gangprobing.alien_at_ntlworld.com> writes

"Ian Jackson" <ianREMOVETHISjackson_at_g3ohx.demon.co.uk> wrote in
message
news:oQqRlSBFmUcKFw3i_at_g3ohx.demon.co.uk...
In message <AC3cm.163781$bA.81816_at_newsfe16.ams2>, ian field
gangprobing.alien_at_ntlworld.com> writes

"Ian Jackson" <ianREMOVETHISjackson_at_g3ohx.demon.co.uk> wrote in
message
news:RK8$DzIdHMcKFwRq_at_g3ohx.demon.co.uk...
In message <Bqfam.7029$ze1.6612_at_news-server.bigpond.net.au>, rf
rf_at_z.invalid> 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.

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.

Possibly. Possibly not. Murphy and his Laws move in mysterious ways. You
don't see many circuits with paralleled diodes (stringed or otherwise).
--
Ian

Larger power supplies will have diodes in parallel, and they will even put
the output trannies in parallel. You can get away with it with diodes but
you have to be carefull to use the same type, preferably from the same
batch. Otherwise you end up with some saturated and others barely turned
on,
and you get a high failure rate.

During 50 years 'in electronics', I can't immediately recall seeing any
power supplies with diodes in parallel. I'm not saying it's never done
but, if it is, there should also be some current-balancing resistance in
each path. This could be low-value resistors, or even the resistance of
the secondary windings of paralleled transformers which feed each set of
diodes. I don't think that having to use diodes from the same batch is a
very good design criterion!
--
Ian

A company I used to work for used paralleled rectifiers in ultrasonic
cleaner generators up to 1500W, I suspect reliability might not have been
all it could have been but they weren't exactly dropping like flies.

That's nearly enough to power a REAL Tardis, let alone a toy one!

In message <XQBcm.127794$qz1.125762_at_newsfe12.ams2>, ian field
gangprobing.alien_at_ntlworld.com> writes



A company I used to work for used paralleled rectifiers in ultrasonic
cleaner generators up to 1500W, I suspect reliability might not have been
all it could have been but they weren't exactly dropping like flies.

That's nearly enough to power a REAL Tardis, let alone a toy one!

Was there any current equalisation (obvious, or 'hidden')?
--
Ian

Larger power supplies will have diodes in parallel, and they will even put
the output trannies in parallel. You can get away with it with diodes but
you have to be carefull to use the same type, preferably from the same
batch. Otherwise you end up with some saturated and others barely turned
on,
and you get a high failure rate.

During 50 years 'in electronics', I can't immediately recall seeing any
power supplies with diodes in parallel. I'm not saying it's never done
but, if it is, there should also be some current-balancing resistance in
each path. This could be low-value resistors, or even the resistance of
the secondary windings of paralleled transformers which feed each set of
diodes. I don't think that having to use diodes from the same batch is a
very good design criterion!
--
Ian

In message <Bqfam.7029$ze1.6612_at_news-server.bigpond.net.au>, rf
rf_at_z.invalid> 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 Jackson wrote on 29-Jul-09 14:40 :
In message <Bqfam.7029$ze1.6612_at_news-server.bigpond.net.au>, rf
rf_at_z.invalid> 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.

BWHAHAHAHAHAHAHAHA!!!! *cough* *wheeze* *wheeze*

-if- junction temperature remains constant.
-if- load current remains constant.
-if- each selected diode returns precisely 0.6v @ 'load current'.

More than good enough for the job.

AHAHAHAHAHAHAHA!!!

-if- source voltage remains constant
-if- there aren't any ICs that could suffer over-voltage damage
-if- you're as cheap and gullible as a Republicunt

lemme guess -- you're a close relative of teh chucktard?

In message <d5udnXVAg8R3sO7XnZ2dnUVZ_vydnZ2d_at_earthlink.com>, Any one
writes
Ian Jackson wrote on 29-Jul-09 14:40 :
In message <Bqfam.7029$ze1.6612_at_news-server.bigpond.net.au>, rf
rf_at_z.invalid> 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.

BWHAHAHAHAHAHAHAHA!!!! *cough* *wheeze* *wheeze*

-if- junction temperature remains constant.
-if- load current remains constant.
-if- each selected diode returns precisely 0.6v @ 'load current'.

More than good enough for the job.

AHAHAHAHAHAHAHA!!!

-if- source voltage remains constant
-if- there aren't any ICs that could suffer over-voltage damage
-if- you're as cheap and gullible as a Republicunt

lemme guess -- you're a close relative of teh chucktard?

Lemme guess....
You have a degree in Electrical Over-Engineering?
--
Ian

In message <d5udnXVAg8R3sO7XnZ2dnUVZ_vydnZ2d_at_earthlink.com>, Any one writes
Ian Jackson wrote on 29-Jul-09 14:40 :
In message <Bqfam.7029$ze1.6612_at_news-server.bigpond.net.au>, rf
rf_at_z.invalid> 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.

BWHAHAHAHAHAHAHAHA!!!! *cough* *wheeze* *wheeze*

-if- junction temperature remains constant.
-if- load current remains constant.
-if- each selected diode returns precisely 0.6v @ 'load current'.

More than good enough for the job.

AHAHAHAHAHAHAHA!!!

-if- source voltage remains constant
-if- there aren't any ICs that could suffer over-voltage damage
-if- you're as cheap and gullible as a Republicunt

lemme guess -- you're a close relative of teh chucktard?

Lemme guess <backhand