R
Roger Dewhurst
Guest
cheap.
R
R
I
Ian Jackson
Guest
In message <Bqfam.7029$ze1.6612@news-server.bigpond.net.au>, rf
<rf@z.invalid> writes
3V to drop = 5 diodes @ 0.6V per diode. No other circuitry required.
More than good enough for the job.
--
Ian
<rf@z.invalid> writes
3V to drop = 5 diodes @ 0.6V per diode. No other circuitry required.
More than good enough for the job.
--
Ian
I
ian field
Guest
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
news:RK8$DzIdHMcKFwRq@g3ohx.demon.co.uk...
current draw and rating of the diode the drop can be as low as 0.55V and as
high as 1.1V.
news:RK8$DzIdHMcKFwRq@g3ohx.demon.co.uk...
The forward conduction knee curve on diodes isn't *that* sharp, depending on
current draw and rating of the diode the drop can be as low as 0.55V and as
high as 1.1V.
R
rf
Guest
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
news:RK8$DzIdHMcKFwRq@g3ohx.demon.co.uk...
regulation.
news:RK8$DzIdHMcKFwRq@g3ohx.demon.co.uk...
Exactly one 7809 for less than a buck. No other circutry required. Perfect
regulation.
I
Ian Jackson
Guest
In message <AC3cm.163781$bA.81816@newsfe16.ams2>, ian field
<gangprobing.alien@ntlworld.com> writes
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
<gangprobing.alien@ntlworld.com> writes
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
R
rf
Guest
Roger Dewhurst wrote:
that a $.
A *few* diodes at a couple of ten cents per each. A single 7809 for less
that a $.
I
ian field
Guest
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
news
QqRlSBFmUcKFw3i@g3ohx.demon.co.uk...
draw between motor on and motor off. In the condition of low current draw
(and low diode drop) supply decoupling electrolytic capacitors can charge to
a higher voltage which is then dumped into the circuit when switched to
play.
news
QqRlSBFmUcKFw3i@g3ohx.demon.co.uk...A potential danger with a cassette recorder is the difference in current
draw between motor on and motor off. In the condition of low current draw
(and low diode drop) supply decoupling electrolytic capacitors can charge to
a higher voltage which is then dumped into the circuit when switched to
play.
I
ian field
Guest
"rf" <rf@z.invalid> wrote in message
news:5d5cm.8142$ze1.7151@news-server.bigpond.net.au...
input and output otherwise they can break into oscillation. If a 3-terminal
regulator feeds a circuit with a large supply decoupling electrolytic
(possibility of stored charge feeding backwards through the regulator when
the input voltage is switched off) its advisable to strap a diode between
the input and output terminals, otherwise the regulator can be damaged - not
conducting in the normal condition of input voltage being higher than the
output but conducts if the output tries to go higher than the input.
news:5d5cm.8142$ze1.7151@news-server.bigpond.net.au...
Not *quite* no extra circuitry, they require decoupling capacitors on the
input and output otherwise they can break into oscillation. If a 3-terminal
regulator feeds a circuit with a large supply decoupling electrolytic
(possibility of stored charge feeding backwards through the regulator when
the input voltage is switched off) its advisable to strap a diode between
the input and output terminals, otherwise the regulator can be damaged - not
conducting in the normal condition of input voltage being higher than the
output but conducts if the output tries to go higher than the input.
I
Ian Jackson
Guest
In message <HAhcm.73167$561.68469@newsfe28.ams2>, ian field
<gangprobing.alien@ntlworld.com> writes
--
Ian
<gangprobing.alien@ntlworld.com> writes
rapidly-decaying additional 3V won't hurt too much."Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
news
In message <AC3cm.163781$bA.81816@newsfe16.ams2>, ian field
gangprobing.alien@ntlworld.com> writes
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
news:RK8$DzIdHMcKFwRq@g3ohx.demon.co.uk...
In message <Bqfam.7029$ze1.6612@news-server.bigpond.net.au>, rf
rf@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
--
Ian
Z
Zootal
Guest
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
news:GvAqKUAuZbcKFwgH@g3ohx.demon.co.uk...
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.
news:GvAqKUAuZbcKFwgH@g3ohx.demon.co.uk...
It won't even be noticeable. The capacitors won't charge up that high toIn message <HAhcm.73167$561.68469@newsfe28.ams2>, ian field
gangprobing.alien@ntlworld.com> writes
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
news
In message <AC3cm.163781$bA.81816@newsfe16.ams2>, ian field
gangprobing.alien@ntlworld.com> writes
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
news:RK8$DzIdHMcKFwRq@g3ohx.demon.co.uk...
In message <Bqfam.7029$ze1.6612@news-server.bigpond.net.au>, rf
rf@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
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.
I
Ian Jackson
Guest
In message <t-GdnS96cOSBWOzXnZ2dnUVZ_oSdnZ2d@giganews.com>, Zootal
<usenet@spam.zootal.nospam.com> writes
12V. 9V at 10mA would require 900 ohms (say 1k), 90mW (so even a 1/4W
will do).
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!
current diodes.
should be fine.
--
Ian
<usenet@spam.zootal.nospam.com> writes
Indeed. A bleed of a few mA will prevent the off-load voltage rising to"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
news:GvAqKUAuZbcKFwgH@g3ohx.demon.co.uk...
In message <HAhcm.73167$561.68469@newsfe28.ams2>, ian field
gangprobing.alien@ntlworld.com> writes
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
news
In message <AC3cm.163781$bA.81816@newsfe16.ams2>, ian field
gangprobing.alien@ntlworld.com> writes
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
news:RK8$DzIdHMcKFwRq@g3ohx.demon.co.uk...
In message <Bqfam.7029$ze1.6612@news-server.bigpond.net.au>, rf
rf@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.
12V. 9V at 10mA would require 900 ohms (say 1k), 90mW (so even a 1/4W
will do).
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!
so, no, you shouldn't parallel diodes. As you suggest, use higher
current diodes.
the LED (allowing 2V for the LED). Anything between 470 ohms and 1k
should be fine.
--
Ian
I
ian field
Guest
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
news:4Lvlz7KQMdcKFwkC@g3ohx.demon.co.uk...
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.
news:4Lvlz7KQMdcKFwkC@g3ohx.demon.co.uk...
Actually if you have two strings each having the same number of diodes andIn message <t-GdnS96cOSBWOzXnZ2dnUVZ_oSdnZ2d@giganews.com>, Zootal
usenet@spam.zootal.nospam.com> writes
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
news:GvAqKUAuZbcKFwgH@g3ohx.demon.co.uk...
In message <HAhcm.73167$561.68469@newsfe28.ams2>, ian field
gangprobing.alien@ntlworld.com> writes
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
news
In message <AC3cm.163781$bA.81816@newsfe16.ams2>, ian field
gangprobing.alien@ntlworld.com> writes
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in
message
news:RK8$DzIdHMcKFwRq@g3ohx.demon.co.uk...
In message <Bqfam.7029$ze1.6612@news-server.bigpond.net.au>, rf
rf@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.
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.
I
Ian Jackson
Guest
In message <k9ocm.127797$YA5.62631@newsfe08.ams2>, ian field
<gangprobing.alien@ntlworld.com> writes
--
Ian
<gangprobing.alien@ntlworld.com> writes
don't see many circuits with paralleled diodes (stringed or otherwise)."Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
news:4Lvlz7KQMdcKFwkC@g3ohx.demon.co.uk...
In message <t-GdnS96cOSBWOzXnZ2dnUVZ_oSdnZ2d@giganews.com>, Zootal
usenet@spam.zootal.nospam.com> writes
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
news:GvAqKUAuZbcKFwgH@g3ohx.demon.co.uk...
In message <HAhcm.73167$561.68469@newsfe28.ams2>, ian field
gangprobing.alien@ntlworld.com> writes
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
news
In message <AC3cm.163781$bA.81816@newsfe16.ams2>, ian field
gangprobing.alien@ntlworld.com> writes
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in
message
news:RK8$DzIdHMcKFwRq@g3ohx.demon.co.uk...
In message <Bqfam.7029$ze1.6612@news-server.bigpond.net.au>, rf
rf@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
--
Ian
Z
Zootal
Guest
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
newsVRIUMAqWhcKFwia@g3ohx.demon.co.uk...
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.
news
VRIUMAqWhcKFwia@g3ohx.demon.co.uk...Larger power supplies will have diodes in parallel, and they will even putIn message <k9ocm.127797$YA5.62631@newsfe08.ams2>, ian field
gangprobing.alien@ntlworld.com> writes
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
news:4Lvlz7KQMdcKFwkC@g3ohx.demon.co.uk...
In message <t-GdnS96cOSBWOzXnZ2dnUVZ_oSdnZ2d@giganews.com>, Zootal
usenet@spam.zootal.nospam.com> writes
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
news:GvAqKUAuZbcKFwgH@g3ohx.demon.co.uk...
In message <HAhcm.73167$561.68469@newsfe28.ams2>, ian field
gangprobing.alien@ntlworld.com> writes
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in
message
news
In message <AC3cm.163781$bA.81816@newsfe16.ams2>, ian field
gangprobing.alien@ntlworld.com> writes
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in
message
news:RK8$DzIdHMcKFwRq@g3ohx.demon.co.uk...
In message <Bqfam.7029$ze1.6612@news-server.bigpond.net.au>, rf
rf@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
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.
I
Ian Jackson
Guest
In message <y76dnR4FvZf0he_XnZ2dnUVZ_vudnZ2d@giganews.com>, Zootal
<usenet@spam.zootal.nospam.com> writes
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
<usenet@spam.zootal.nospam.com> writes
power supplies with diodes in parallel. I'm not saying it's never done"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
news
In message <k9ocm.127797$YA5.62631@newsfe08.ams2>, ian field
gangprobing.alien@ntlworld.com> writes
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
news:4Lvlz7KQMdcKFwkC@g3ohx.demon.co.uk...
In message <t-GdnS96cOSBWOzXnZ2dnUVZ_oSdnZ2d@giganews.com>, Zootal
usenet@spam.zootal.nospam.com> writes
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
news:GvAqKUAuZbcKFwgH@g3ohx.demon.co.uk...
In message <HAhcm.73167$561.68469@newsfe28.ams2>, ian field
gangprobing.alien@ntlworld.com> writes
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in
message
news
In message <AC3cm.163781$bA.81816@newsfe16.ams2>, ian field
gangprobing.alien@ntlworld.com> writes
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in
message
news:RK8$DzIdHMcKFwRq@g3ohx.demon.co.uk...
In message <Bqfam.7029$ze1.6612@news-server.bigpond.net.au>, rf
rf@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
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
I
ian field
Guest
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
news:M9lIelA4fpcKFwB6@g3ohx.demon.co.uk...
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.
news:M9lIelA4fpcKFwB6@g3ohx.demon.co.uk...
A company I used to work for used paralleled rectifiers in ultrasonicIn message <y76dnR4FvZf0he_XnZ2dnUVZ_vudnZ2d@giganews.com>, Zootal
usenet@spam.zootal.nospam.com> writes
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
news
In message <k9ocm.127797$YA5.62631@newsfe08.ams2>, ian field
gangprobing.alien@ntlworld.com> writes
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
news:4Lvlz7KQMdcKFwkC@g3ohx.demon.co.uk...
In message <t-GdnS96cOSBWOzXnZ2dnUVZ_oSdnZ2d@giganews.com>, Zootal
usenet@spam.zootal.nospam.com> writes
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in
message
news:GvAqKUAuZbcKFwgH@g3ohx.demon.co.uk...
In message <HAhcm.73167$561.68469@newsfe28.ams2>, ian field
gangprobing.alien@ntlworld.com> writes
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in
message
news
In message <AC3cm.163781$bA.81816@newsfe16.ams2>, ian field
gangprobing.alien@ntlworld.com> writes
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in
message
news:RK8$DzIdHMcKFwRq@g3ohx.demon.co.uk...
In message <Bqfam.7029$ze1.6612@news-server.bigpond.net.au>, rf
rf@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
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.
I
Ian Jackson
Guest
In message <XQBcm.127794$qz1.125762@newsfe12.ams2>, ian field
<gangprobing.alien@ntlworld.com> writes
Was there any current equalisation (obvious, or 'hidden')?
--
Ian
<gangprobing.alien@ntlworld.com> writes
Was there any current equalisation (obvious, or 'hidden')?
--
Ian
I
ian field
Guest
"Ian Jackson" <ianREMOVETHISjackson@g3ohx.demon.co.uk> wrote in message
news:jLZjTSFLtucKFw2f@g3ohx.demon.co.uk...
small square slab of aluminium that was bolted to an elongated cube of more
aluminium that pressed into the middle of a forced air cooled heatsink
assembly, each pair of rectifiers was linked together with a triangle of
16SWG TC wire, the 2 ends that came together were crimped into a solder tag
whose "eye" was the solder point for the big fat wires from the mains
transformer.
news:jLZjTSFLtucKFw2f@g3ohx.demon.co.uk...
Absolutely none - the (pair of) 2 press fit rectifiers were pressed into a
small square slab of aluminium that was bolted to an elongated cube of more
aluminium that pressed into the middle of a forced air cooled heatsink
assembly, each pair of rectifiers was linked together with a triangle of
16SWG TC wire, the 2 ends that came together were crimped into a solder tag
whose "eye" was the solder point for the big fat wires from the mains
transformer.
Z
Zootal
Guest
Having diodes from the same batch is something we do at home when we make
our own klugey experimental stuff. Of course they aren't going to do this in
production equipment!
I only have 25 years against your 50, but I've seen so many things I can't
immediately recall seeing a lot of the things I've seen
. At one time I was doing the work on our power supplies, and we had a lot. And of course
there were resistors in series with each diode, the purpose of which is to
minimize variations in the operating parameters of the diodes themselves. I
never bothered to test them to see how equally they shared the load, but the
things worked. I've never had to replace one of the diodes themselves. It's
possible that the diodes would turn on such that one didn't turn on until
another approached saturation. So long as they don't burn out I guess it
would work OK. Hmm...Let's use big doides mounted in a huge heat sink
because some are going to be near saturation while others are idling. I've
seen enough klugey designes that I would not be surprised if that really was
how it worked
A
Any one
Guest
Ian Jackson wrote on 29-Jul-09 14:40 :
-if- junction temperature remains constant.
-if- load current remains constant.
-if- each selected diode returns precisely 0.6v @ 'load current'.
-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?
BWHAHAHAHAHAHAHAHA!!!! *cough* *wheeze* *wheeze*
-if- junction temperature remains constant.
-if- load current remains constant.
-if- each selected diode returns precisely 0.6v @ 'load current'.
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?