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may work. That's like R=(14.2-1.5)/I, where 14.2V is max chargingHello,
I want to connect my 1.5VDC clock to my car battery. How do I figure the
drop resistor value? I have a DCAmp / VDC / Ohm meter.
tks,
paul
Depends on the clock. If it's purely electonic-digital a single resistor
Paul Mars wrote:
Hello,
I want to connect my 1.5VDC clock to my car battery. How do I figure the
drop resistor value? I have a DCAmp / VDC / Ohm meter.
tks,
paul
Depends on the clock. If it's purely electonic-digital a single resistor
may work. That's like R=(14.2-1.5)/I, where 14.2V is max charging voltage
and I is current in amperes.
But you could try a shunt regulation method:
Two forward biased silicon diodes(e.g 1N914B), placed in series, across
where the 1.5V battery usualy connects. Also across those same contacts
attach a capacitor of about 5 to 20mF (must be at least 20V of course).
Connect the negative of the clock to the battery negative and the
battery's positive to a 10K resistor. The other side of this resistor to
the clock positive.
Two silicon diodes together have about 1.4 - 2 volts of barrier -
functions a bit like a Zener in this circuit. 10K resistor limits the
current. The capacitor is the resovoir to support intermitant loading of
the clock's drive mechanism, if any.
Bjorn
likely has a fairly low cutoff voltage (perhaps 0.9-1.1V ?). This shouldPaul Mars wrote:
Hello,
I want to connect my 1.5VDC clock to my car battery. How do I figure
the drop resistor value? I have a DCAmp / VDC / Ohm meter.
tks,
paul
Depends on the clock. If it's purely electonic-digital a single resistor
may work. That's like R=(14.2-1.5)/I, where 14.2V is max charging
voltage and I is current in amperes.
But you could try a shunt regulation method:
Two forward biased silicon diodes(e.g 1N914B), placed in series, across
where the 1.5V battery usualy connects. Also across those same contacts
attach a capacitor of about 5 to 20mF (must be at least 20V of course).
Connect the negative of the clock to the battery negative and the
battery's positive to a 10K resistor. The other side of this resistor to
the clock positive.
Two silicon diodes together have about 1.4 - 2 volts of barrier -
functions a bit like a Zener in this circuit. 10K resistor limits the
current. The capacitor is the resovoir to support intermitant loading of
the clock's drive mechanism, if any.
Bjorn
Got to thinking, since this is probably a dry-cell battery clock it
the same.Its coming back to me now. R=E/I
Which method is better? Which pulls less battery current? And since the
engine running voltage is 13.5, I think that is what I should use.
Electronic digital LCD clock.
"John Smith" <user@example.net> wrote in message
news:7AaKg.169$I71.13@trnddc01...
Paul Mars wrote:
Hello,
I want to connect my 1.5VDC clock to my car battery. How do I figure the
drop resistor value? I have a DCAmp / VDC / Ohm meter.
tks,
paul
Depends on the clock. If it's purely electonic-digital a single resistor
may work. That's like R=(14.2-1.5)/I, where 14.2V is max charging voltage
and I is current in amperes.
But you could try a shunt regulation method:
Two forward biased silicon diodes(e.g 1N914B), placed in series, across
where the 1.5V battery usualy connects. Also across those same contacts
attach a capacitor of about 5 to 20mF (must be at least 20V of course).
Connect the negative of the clock to the battery negative and the
battery's positive to a 10K resistor. The other side of this resistor to
the clock positive.
Two silicon diodes together have about 1.4 - 2 volts of barrier -
functions a bit like a Zener in this circuit. 10K resistor limits the
current. The capacitor is the resovoir to support intermitant loading of
the clock's drive mechanism, if any.
Bjorn
Current will be most affected by the 10k. Either method will draw about
John Smith wrote:
Paul Mars wrote:
Hello,
I want to connect my 1.5VDC clock to my car battery. How do I figure the
drop resistor value? I have a DCAmp / VDC / Ohm meter.
tks,
paul
Depends on the clock. If it's purely electonic-digital a single resistor
may work. That's like R=(14.2-1.5)/I, where 14.2V is max charging voltage
and I is current in amperes.
But you could try a shunt regulation method:
Two forward biased silicon diodes(e.g 1N914B), placed in series, across
where the 1.5V battery usualy connects. Also across those same contacts
attach a capacitor of about 5 to 20mF (must be at least 20V of course).
Connect the negative of the clock to the battery negative and the
battery's positive to a 10K resistor. The other side of this resistor to
the clock positive.
Two silicon diodes together have about 1.4 - 2 volts of barrier -
functions a bit like a Zener in this circuit. 10K resistor limits the
current. The capacitor is the resovoir to support intermitant loading of
the clock's drive mechanism, if any.
Bjorn
Got to thinking, since this is probably a dry-cell battery clock it likely
has a fairly low cutoff voltage (perhaps 0.9-1.1V ?). This should allow a
shunt resistor to be used instead of a diode pair. The circuit becomes a
voltage divider. With that 10K series resistor this resistance is
10000/((14.2/1.5)-1) or about 1181 ohms. The current draw of the clock can
likely be disregarded with the 10K. When the car battery is flat, at about
11.9 volts, the clock will still see about 1.3V and should still be
running. You may also be able to go much higher with the series resistor
(matching the shunt resistance) depending on the clock's current draw
characteristics.(For calculation details, read up on on Ohm's and
Kirkoff's laws)
If both resistors were 10K the voltage would probably be too high.shunt res of 10k and series res of 10k? Please keep explanations simple.
"Bjorn" <TigerT100r@yahoo.co.uk> wrote in message
news:jjiKg.482$m36.356@trnddc02...
John Smith wrote:
Paul Mars wrote:
Hello,
I want to connect my 1.5VDC clock to my car battery. How do I figure the
drop resistor value? I have a DCAmp / VDC / Ohm meter.
tks,
paul
Depends on the clock. If it's purely electonic-digital a single resistor
may work. That's like R=(14.2-1.5)/I, where 14.2V is max charging voltage
and I is current in amperes.
But you could try a shunt regulation method:
Two forward biased silicon diodes(e.g 1N914B), placed in series, across
where the 1.5V battery usualy connects. Also across those same contacts
attach a capacitor of about 5 to 20mF (must be at least 20V of course).
Connect the negative of the clock to the battery negative and the
battery's positive to a 10K resistor. The other side of this resistor to
the clock positive.
Two silicon diodes together have about 1.4 - 2 volts of barrier -
functions a bit like a Zener in this circuit. 10K resistor limits the
current. The capacitor is the resovoir to support intermitant loading of
the clock's drive mechanism, if any.
Bjorn
Got to thinking, since this is probably a dry-cell battery clock it likely
has a fairly low cutoff voltage (perhaps 0.9-1.1V ?). This should allow a
shunt resistor to be used instead of a diode pair. The circuit becomes a
voltage divider. With that 10K series resistor this resistance is
10000/((14.2/1.5)-1) or about 1181 ohms. The current draw of the clock can
likely be disregarded with the 10K. When the car battery is flat, at about
11.9 volts, the clock will still see about 1.3V and should still be
running. You may also be able to go much higher with the series resistor
(matching the shunt resistance) depending on the clock's current draw
characteristics.(For calculation details, read up on on Ohm's and
Kirkoff's laws)
1500mAh, at a slow discharge, so your clock likely draws nearly 1mA.This sounds good:
Using a Series resistor of 10K, a Shunt resistor of 1200 could be selected.
With a "dead" battery (11.89V) you would have about 1.27V
With a charging battery (13.5V) you would have about 1.44V
What is usually taken as max in an auto (14.2V) you would have about 1.52V
This clock uses up it's1.5V AAA in about 2 months. In this time the display
gets soo dim that I need bright light and to tilt the display up close. Now
it is at this stage again and the battery is 1.02v. So the above values look
good. My only concern now is car battery drain of 1.4 mA. I think that is
ok.
Thats pretty short battery life. An AAA battery is probably about
Paul Mars wrote:
This sounds good:
Using a Series resistor of 10K, a Shunt resistor of 1200 could be
selected.
With a "dead" battery (11.89V) you would have about 1.27V
With a charging battery (13.5V) you would have about 1.44V
What is usually taken as max in an auto (14.2V) you would have about
1.52V
This clock uses up it's1.5V AAA in about 2 months. In this time the
display gets soo dim that I need bright light and to tilt the display up
close. Now it is at this stage again and the battery is 1.02v. So the
above values look good. My only concern now is car battery drain of 1.4
mA. I think that is ok.
Thats pretty short battery life. An AAA battery is probably about 1500mAh,
at a slow discharge, so your clock likely draws nearly 1mA. That's a lot
for a battery powered clock. That's enough to depress the voltage of my
crude circuit. If you replace the 1200 ohm with a 10K pot you can
compensate for this. Once you find a value you like, you can even swap the
pot out for a regular fixed resistor.
Just for reference, self discharge of Group 24 car battery is about 3mA.
With no load it will go flat in around 2.7 years (it will also be sulfated
by then as well!). Adding the clock will cut this to 1.8 years. As long as
the car is driven a bit at least once a month you should be fine.
Hello,
I want to connect my 1.5VDC clock to my car battery. How do I figure the
drop resistor value? I have a DCAmp / VDC / Ohm meter.
tks,
paul
Put high ohms pot from clock to 12vdc , adjust pot til clock starts
working just barely .
replace pot with a smaller ohms resistor .
Now place 2 silicon , small amps Diodes across the clock , band to
ground ,
diodes will limit voltage to something below 2 vdc . If voltage
tries to go
up at the clock , the diodes will take lots of amps and pull the
voltage back down
to a safe value for the clock
Or buy a LM317 at Radio Shack , or , or , or
_______________________________________________________________________-
Paul Mars wrote:
Hello,
I want to connect my 1.5VDC clock to my car battery. How do I figure the
drop resistor value? I have a DCAmp / VDC / Ohm meter.
tks,
paul
Put high ohms pot from clock to 12vdc , adjust pot til clock starts
working just barely .
replace pot with a smaller ohms resistor .
Now place 2 silicon , small amps Diodes across the clock , band to
ground ,
diodes will limit voltage to something below 2 vdc . If voltage
tries to go
up at the clock , the diodes will take lots of amps and pull the
voltage back down
to a safe value for the clock
connected betweem the clock's power input terminalsWhy do I need to limit the voltage? My car battery will never go above 14.x
volts and if it does I have greater things to worry about then a fried
clock. And yes I plan to fuse it.
What does this mean? "across the clock
one end of the diode will be marke with a band connect that end toband to ground"
1N4001 (or 1N4002 etc...) you'll have to look-up the radio shack partok. I can do this. Except please be more specific about the diodes. Please
gave me a type or a source. My local source will be radio shack and they
know less about ele components then I do.
a three terminal series voltage regulator... a device designedOr buy a LM317 at Radio Shack , or , or , or
What is a LM317 ?
---ok. I can do this. Except please be more specific about the diodes. Please
gave me a type or a source. My local source will be radio shack and they
know less about ele components then I do.
Why do I need to limit the voltage?
---My car battery will never go above 14.x
volts and if it does I have greater things to worry about then a fried
clock. And yes I plan to fuse it.
What does this mean? "across the clock , band to ground"
What is a LM317 ?
On Tue, 05 Sep 2006 18:38:39 GMT, "Paul Mars"
pmarsREMOVE@tampabay.rr.com> wrote:
ok. I can do this. Except please be more specific about the diodes. Please
gave me a type or a source. My local source will be radio shack and they
know less about ele components then I do.
Why do I need to limit the voltage?
---
Load dump.
---
My car battery will never go above 14.x
volts and if it does I have greater things to worry about then a fried
clock. And yes I plan to fuse it.
What does this mean? "across the clock , band to ground"
What is a LM317 ?
---
First, Please bottom post.
Second, do this: (View in Courier)
VBAT>---[1N4001>]---[620R]--+-------->CLOCK +
|A
[1N4001]
|
|A
[1N4001]
|
GND>------------------------+-------->CLOCK -
I'd use a 620 ohm 1 watt resistor.
--
John Fields
Professional Circuit Designer
---This is another circuit? I m about ready to shelve this project. All these
different circuits is loosing me....
On Wed, 06 Sep 2006 14:31:22 GMT, "Paul Mars"
pmarsREMOVE@tampabay.rr.com> wrote:
This is another circuit? I m about ready to shelve this project. All these
different circuits is loosing me....
---
If they'd loose you from your bad habit of top posting that'd be a
good thing.
IMO, your problem seems to be that you're clueless and you don't
want a clue, all you want is someone to give you something
_guaranteed_ to work, with no effort on your part. Why don't you
stop complaining and build some of what's been offered to you and
find out what works and what doesn't?
--
John Fields
Professional Circuit Designer
---I could buy a car clock and plug it in with all the money and time I would
spend building all those circuits. But actually, I did get my original
question answered, so I have my circuit.
Thanks all.
BTW, many threads top post. Some bottom post, and some posters tell people
to do one or the other. I do think bottom is better, but so far the majority
(from my limited exp) top post,
Then Paul Mars replied:Of all the people who responded to you here, the only ones who were
top-posting were you and werty.
---one thread is a rather limited polling.