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#1
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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 |
#2
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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 |
#3
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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 |
#4
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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 |
#5
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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 |
#6
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#7
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#8
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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 |
#9
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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 |
#10
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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 |
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