Any electronics experts out there? Would appreciate help with a simple
circuit:

to light an LED when current in a 12vdc circuit goes above
approximately 10 amps (adjustable from about 8 - 13 amps would be
nice) and drop out when current goes below that set point. Maximum
current is about 20 amps.

Any ideas?

Thanks for the help

On 24 Feb 2004 11:48:17 -0800,
Norm Freedman wrote:
Any electronics experts out there? Would appreciate help with a simple
circuit:

to light an LED when current in a 12vdc circuit goes above
approximately 10 amps (adjustable from about 8 - 13 amps would be
nice) and drop out when current goes below that set point. Maximum
current is about 20 amps.

Any ideas?

Thanks for the help


Try Maxim,

http://www.maxim-ic.com/appnotes.cfm/appnote_number/2015/ln/en

They are usually quite happy to send a sample of an IC. Or you can order
them from the usual suspects like Mouser. Note that the appnote above,
also cuts off the load in the event of overcurrent, but you can cut that
part out and ignore it.

If you'd rather roll your own, you'll need a (couple of) precision low
value resistor(s) about 0.01 Ohm or so, then rig up a voltage drop
measuring circuit around that, adjust for expected current levels. Most
of the circuit cookbooks will have examples you can copy. With 0.01 Ohm,
passing 10A through it, results in 0.1V drop, which is a bit small for
some circuits to reliably measure, but only dissapates 100mw, and if the
current goes to 20A, your resistors are only dissapating 200mw, which is
probably ok for most things. Although it's a couple of amp hours a day
if it runs that long.



You might be able to get away with using a magneto-resistive system, but
with DC, the results are likely to vary widely depending on what other
magnetic fields are around, and are likely to change when you start or
stop the engine, or other unrelated magnetic sources.

--
Jim Richardson http://www.eskimo.com/~warlock
If you can modify it via software, then it doesn't count as hardware.
-- Lionel, in the Monastery

On 24 Feb 2004 11:48:17 -0800,
Norm Freedman wrote:
Any electronics experts out there? Would appreciate help with a simple
circuit:

to light an LED when current in a 12vdc circuit goes above
approximately 10 amps (adjustable from about 8 - 13 amps would be
nice) and drop out when current goes below that set point. Maximum
current is about 20 amps.

Any ideas?

Thanks for the help


Try Maxim,

http://www.maxim-ic.com/appnotes.cfm/appnote_number/2015/ln/en

They are usually quite happy to send a sample of an IC. Or you can order
them from the usual suspects like Mouser. Note that the appnote above,
also cuts off the load in the event of overcurrent, but you can cut that
part out and ignore it.

If you'd rather roll your own, you'll need a (couple of) precision low
value resistor(s) about 0.01 Ohm or so, then rig up a voltage drop
measuring circuit around that, adjust for expected current levels. Most
of the circuit cookbooks will have examples you can copy. With 0.01 Ohm,
passing 10A through it, results in 0.1V drop, which is a bit small for
some circuits to reliably measure, but only dissapates 100mw, and if the
current goes to 20A, your resistors are only dissapating 200mw, which is
probably ok for most things. Although it's a couple of amp hours a day
if it runs that long.



You might be able to get away with using a magneto-resistive system, but
with DC, the results are likely to vary widely depending on what other
magnetic fields are around, and are likely to change when you start or
stop the engine, or other unrelated magnetic sources.

--
Jim Richardson http://www.eskimo.com/~warlock
If you can modify it via software, then it doesn't count as hardware.
-- Lionel, in the Monastery

There are electronics groups better suited to discuss your need.
But here's a thought, any way.

A current sense is often provided by a low value series resistor on
the + rail across the base/emitter junction of a transistor which
turns on when the voltage drop exceeds about 0.6 volts
A collector resistor limits the output current to about 10 milliamps
and activates a LED connected to the - rail when active.

This is a low count solution, but probably not best suited
to your purpose, because it drops a volt and wastes some power.

You might consider a VERY low value series resistor - i.e. the voltage
drop across a length of the positive rail cable feeding the two inputs
of an op amp, whose output feeds a dropper resistor of 1.2Kohms say,
to a LED on the negative rail.

The adjustability is provided by a 25Kohm pot connected to the op amp
output at one end, whose wiper connects to the + op amp input though
a 200 ohm resistor, and the other end of the pot connects to the most
positive end of the + rail .
The - input of the op amp connects to the least positive end of the
positive rail through a 200 ohm resistor.

Not just any op amp will do for this: the key is an op amp type that
does NOT latch on an input as high as the + rail.

Brian Whatcott Altus OK

On 24 Feb 2004 11:48:17 -0800, (Norm Freedman)
wrote:

Any electronics experts out there? Would appreciate help with a simple
circuit:

to light an LED when current in a 12vdc circuit goes above
approximately 10 amps (adjustable from about 8 - 13 amps would be
nice) and drop out when current goes below that set point. Maximum
current is about 20 amps.

Any ideas?

Thanks for the help

There are electronics groups better suited to discuss your need.
But here's a thought, any way.

A current sense is often provided by a low value series resistor on
the + rail across the base/emitter junction of a transistor which
turns on when the voltage drop exceeds about 0.6 volts
A collector resistor limits the output current to about 10 milliamps
and activates a LED connected to the - rail when active.

This is a low count solution, but probably not best suited
to your purpose, because it drops a volt and wastes some power.

You might consider a VERY low value series resistor - i.e. the voltage
drop across a length of the positive rail cable feeding the two inputs
of an op amp, whose output feeds a dropper resistor of 1.2Kohms say,
to a LED on the negative rail.

The adjustability is provided by a 25Kohm pot connected to the op amp
output at one end, whose wiper connects to the + op amp input though
a 200 ohm resistor, and the other end of the pot connects to the most
positive end of the + rail .
The - input of the op amp connects to the least positive end of the
positive rail through a 200 ohm resistor.

Not just any op amp will do for this: the key is an op amp type that
does NOT latch on an input as high as the + rail.

Brian Whatcott Altus OK

On 24 Feb 2004 11:48:17 -0800, (Norm Freedman)
wrote:

Any electronics experts out there? Would appreciate help with a simple
circuit:

to light an LED when current in a 12vdc circuit goes above
approximately 10 amps (adjustable from about 8 - 13 amps would be
nice) and drop out when current goes below that set point. Maximum
current is about 20 amps.

Any ideas?

Thanks for the help

On 24 Feb 2004 11:48:17 -0800, (Norm Freedman) wrote:

Any electronics experts out there? Would appreciate help with a simple
circuit:

to light an LED when current in a 12vdc circuit goes above
approximately 10 amps (adjustable from about 8 - 13 amps would be
nice) and drop out when current goes below that set point. Maximum
current is about 20 amps.

Any ideas?

Thanks for the help

I'll be retiring in a couple of months, and with any luck at all, it'll be on a boat.
I have a horror of not doing any more hardware or software projects. Although
I'll probably do some contract work for my current company, it'll be more like
*work* than fun. The project you suggest is interesting, and something I could
use on my own boat.

If you haven't found a solution in a few months, email me and I'd be glad to
send you a unit at cost. LOL..I never make *one* of anything.

Norm

On 24 Feb 2004 11:48:17 -0800, (Norm Freedman) wrote:

Any electronics experts out there? Would appreciate help with a simple
circuit:

to light an LED when current in a 12vdc circuit goes above
approximately 10 amps (adjustable from about 8 - 13 amps would be
nice) and drop out when current goes below that set point. Maximum
current is about 20 amps.

Any ideas?

Thanks for the help

I'll be retiring in a couple of months, and with any luck at all, it'll be on a boat.
I have a horror of not doing any more hardware or software projects. Although
I'll probably do some contract work for my current company, it'll be more like
*work* than fun. The project you suggest is interesting, and something I could
use on my own boat.

If you haven't found a solution in a few months, email me and I'd be glad to
send you a unit at cost. LOL..I never make *one* of anything.

Norm

Norm,

Several people have suggested a low resistance current shunt, and
then amplifying the voltage across the resistor. That idea is workable,
but somewhat problematic.

For example, if you use a .1 ohm resistor in series with your circuit,
it will produce a 1V differential voltage at 10 amps, and a 2V drop
at your max of 20 amps. You could then use a simple comparator with
a potentiometer circuit to adjust the "LED on/off" voltage. The problem
is that at 20 amps, you are dropping your 12V dc down to 10V, and your
sense resistor is dissipating 40 watts!

If you reduce the size of the series resistor to .01 ohms, you only get
..1V at your 10 amps, and .2V at 20 amps. This voltage is a little low
to drive the comparator directly, so you'll need to add an op amp to
amplify the voltage and then drive the amplified voltage into the comparator.
You can still set up the pot circuit to adjust the current at which the
LED is turned on. Unfortunately, your sense resistor is still dissipating
4 watts when the current gets to 20 amps. That is going to be one HOT
power resistor. (Hot enough to burn you if you put your finger on it).

A better solution is to use a Hall Effect current sensor such as a HAW 20-P,
or LA 20-PB. These sensors measure a DC current and put out a voltage
that is proportional to the current. You can purchase these from Digikey
(www.digikey.com) and although they are a little pricey ($22-$38) in
small quantities, they are a much more elegant solution to the problem
you are trying to solve. You can drive the voltage output of the sensor
directly into a comparator and then adjust the voltage at which the
comparator switches with the pot circuit.

For a discussion of hall sensors and battery current sensing, see
http://powerweb.grc.nasa.gov/elecsys/doc/hall.html.

Good luck with it,

Don W.

Norm Freedman wrote:

Any electronics experts out there? Would appreciate help with a simple
circuit:

to light an LED when current in a 12vdc circuit goes above
approximately 10 amps (adjustable from about 8 - 13 amps would be
nice) and drop out when current goes below that set point. Maximum
current is about 20 amps.

Any ideas?

Thanks for the help

Norm,

Several people have suggested a low resistance current shunt, and
then amplifying the voltage across the resistor. That idea is workable,
but somewhat problematic.

For example, if you use a .1 ohm resistor in series with your circuit,
it will produce a 1V differential voltage at 10 amps, and a 2V drop
at your max of 20 amps. You could then use a simple comparator with
a potentiometer circuit to adjust the "LED on/off" voltage. The problem
is that at 20 amps, you are dropping your 12V dc down to 10V, and your
sense resistor is dissipating 40 watts!

If you reduce the size of the series resistor to .01 ohms, you only get
..1V at your 10 amps, and .2V at 20 amps. This voltage is a little low
to drive the comparator directly, so you'll need to add an op amp to
amplify the voltage and then drive the amplified voltage into the comparator.
You can still set up the pot circuit to adjust the current at which the
LED is turned on. Unfortunately, your sense resistor is still dissipating
4 watts when the current gets to 20 amps. That is going to be one HOT
power resistor. (Hot enough to burn you if you put your finger on it).

A better solution is to use a Hall Effect current sensor such as a HAW 20-P,
or LA 20-PB. These sensors measure a DC current and put out a voltage
that is proportional to the current. You can purchase these from Digikey
(www.digikey.com) and although they are a little pricey ($22-$38) in
small quantities, they are a much more elegant solution to the problem
you are trying to solve. You can drive the voltage output of the sensor
directly into a comparator and then adjust the voltage at which the
comparator switches with the pot circuit.

For a discussion of hall sensors and battery current sensing, see
http://powerweb.grc.nasa.gov/elecsys/doc/hall.html.

Good luck with it,

Don W.

Norm Freedman wrote:

Any electronics experts out there? Would appreciate help with a simple
circuit:

to light an LED when current in a 12vdc circuit goes above
approximately 10 amps (adjustable from about 8 - 13 amps would be
nice) and drop out when current goes below that set point. Maximum
current is about 20 amps.

Any ideas?

Thanks for the help

Thanks for all the suggestions.

This is really a simple application. I have no trim tab indicators.
All I really need to know is if they are fully retracted or extended.
When that happens, the motor current goes from about 8 amps to about
15 amps. Most of the time, the circuit is off. I just thought this
would be a simple way to tell, using a waterproof LED.



Don W wrote in message igy.com...
Norm,

Several people have suggested a low resistance current shunt, and
then amplifying the voltage across the resistor. That idea is workable,
but somewhat problematic.

For example, if you use a .1 ohm resistor in series with your circuit,
it will produce a 1V differential voltage at 10 amps, and a 2V drop
at your max of 20 amps. You could then use a simple comparator with
a potentiometer circuit to adjust the "LED on/off" voltage. The problem
is that at 20 amps, you are dropping your 12V dc down to 10V, and your
sense resistor is dissipating 40 watts!

If you reduce the size of the series resistor to .01 ohms, you only get
.1V at your 10 amps, and .2V at 20 amps. This voltage is a little low
to drive the comparator directly, so you'll need to add an op amp to
amplify the voltage and then drive the amplified voltage into the comparator.
You can still set up the pot circuit to adjust the current at which the
LED is turned on. Unfortunately, your sense resistor is still dissipating
4 watts when the current gets to 20 amps. That is going to be one HOT
power resistor. (Hot enough to burn you if you put your finger on it).

A better solution is to use a Hall Effect current sensor such as a HAW 20-P,
or LA 20-PB. These sensors measure a DC current and put out a voltage
that is proportional to the current. You can purchase these from Digikey
(www.digikey.com) and although they are a little pricey ($22-$38) in
small quantities, they are a much more elegant solution to the problem
you are trying to solve. You can drive the voltage output of the sensor
directly into a comparator and then adjust the voltage at which the
comparator switches with the pot circuit.

For a discussion of hall sensors and battery current sensing, see
http://powerweb.grc.nasa.gov/elecsys/doc/hall.html.

Good luck with it,

Don W.

Norm Freedman wrote:

Any electronics experts out there? Would appreciate help with a simple
circuit:

to light an LED when current in a 12vdc circuit goes above
approximately 10 amps (adjustable from about 8 - 13 amps would be
nice) and drop out when current goes below that set point. Maximum
current is about 20 amps.

Any ideas?

Thanks for the help