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DC current alarm. What would be the most cost effective way to trip an audible alarm if current goes over 2 amps? Answered

I have a test fixture that I use.  I am testing a circuit board and I need to be able to trip an audible alarm if the current going from my 15 V DC power supply goes over 2 amps, or more precisely 1.5 amps. As of now I have to watch the amp meter to see if during a particular portion of the test the amps go over 1.5.  This only occurs for 2-3 seconds of a 60 second test.  If I happen to not be watching I have to start over.  I would like to be able to reset the alarm after it is tripped to test the next board.  Obviously I would like to do this a inexpensively as possible.  Thanks for any help.

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steveastrouk

3 years ago

Measure current with a low value resistor or Hall sensor, use a comparator to drive an alarm.

We tried using an arduino with a Hall sensor but it was not accurate enough. Will do some research on low value resistor and comparator.

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-max-brentatespicorpdotcom

Answer 3 years ago

Arduino only has like a 8 or 10 bit ADC. That means it has 256 or 1024 resolution. Like Steve pointed out, the arduino uses the 0-5V range for the ADC by default, so you will need to either amplify the hall effect sensor output, or reduce the voltage scale on the arduino. I believe the arduino MEGA can be configured to use a 1.1V reference.

You need to get the scale right. There's an option to make the ADC in an arduino work off its internal 1.6 V reference, and you may need some gain to make the signal strong enough to measure anyway

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-max-

3 years ago

A crude way to do this would be to use a low voltage relay with the coil in series with the supply, and use the contacts to activate a buzzer.

My linear power supply has a shunt resistor in it that is about 1 ohm, so that if 1.8-2A of current flows, 1.8-2V is generated across the shunt resistor, and a red LED + resistor are directly in parallel across it. Because LEDs have very nonlinear characteristics (where they do not work until a specific voltage is reached), it works reasonably well.

A slightly better method might be to use a low value resistor in series with the load (in between the +15Vdc and load), and a PNP transistor, preferably with high gain. The emitter will connect to the +15Vdc, and a shunt resistor will connect between +15Vdc and the load, and the base of the transmitter will go to a small resistor (1K or so) to the output. The collector will connect to a buzzer, the other end of which goes to ground. The value of the shunt resistor used will need to have a value such that when the maximum desirable current is flowing, at least 0.6V of voltage drop is generated across the shunt resistor. 0.4 ohms should work.

Probably the most precise and configurable way to do this would be to use an op amp or comparator based circuit, using a comparator or op amp, zener diode, and some resistors. The comparator is really just a differential amplifier that has insanely high gain. Even the tiniest difference in voltage between the + and - input gets amplified so high, that the output is basically either only at maximum positive voltage, or the minimum negative voltage. I'll leave it up to you to see if you can figure out how to use it to activate a buzzer when a large enough voltage difference across the shunt resistor (hence the maximum current) is reached.

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-max--max-

Answer 3 years ago

I should point out that (especially with the first 3 solutions) that there will be a significant voltage drop across the shunt resistor. The Resistors + LED option for instance will drop the voltage by 2ish V @ 1.5ish A!!! That means that there under full load, you will only see 13V. Ideally, you would have the shunt resistor as part of the feedback loop. (the feedback pin of the power supply should connect to the output after the shunt resistor) Or if you are using a 7815, you can actually stick the current shunt before the regulator, just keep in mind that the regulator will see that much less voltage to the input voltage terminal, and the regulator needs 2V more than the nominal output to regulate properly.

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rickharris

3 years ago

Is the amp meter digital or analogue?

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If your willing to break open your power supply you could monitor the b and e bar on the final digit. These are the only ones that refer to the number 6 being shown b will be off and e on if 6 is on the display This is unique to the number 6 (UNLESS you display Alpha characters) , you can then take this and use it to trigger a simple alarm (use a 555 for example).