measure the current

See...

for a circuit that was designed to show current-flow direction, with zero as the turn-over point.

Should take only minor modifications to do what you want.

In your particular case, only guessing, something much simpler would probably do.

...Jim Thompson

I have a commercial device that goes to sleep when it's not needed and draws a few uA and when is running draws around 4mA. I want to figure out with my microcontroller when this device wakes up. Reading the current that passes through is the only way to figure out if the device is awake or not.

I thought about using the analog comparator in the microcontroller but the only problem is that I can measure with it Volts and not Amps.

Regards!

I'm having problems understanding this circuit as low level electronics is not (yet) my thing. If you have time, please tell me what should I modify to get the minimal circuit (as the space on my pcb is very small) ... If not, I'll do my best to figure this out.

Thank you. Regards!

You can do this with two resisitors and one PNP transistor if the power source is the same for both. Emitter to power source, Base to device drawing the current, Collector outputs voltage when current is drawn, 1k resistor from collector to ground.

Second resistor connected between Base and Emitter. For 1 ma and .6 volts to turn on transistor, R=E/I = 0.6/.001 = 600 ohms.

VCC

| | -------| | | | | | .-. | | | 600 Ohms | | | | '-' | | >| | |- ----|----------- Sense 1ma /| | | Logic Out -----| | .-. | | | | 1k '-' | | | | | GND

(created by AACircuit v1.28.5 beta 02/06/05

Hello Yew,

Need more data here. Such as required tolerance, whether it's a mass product or a one-up lab thing etc.

For an experiment a Reed contact with lots of windings around it comes to mind. Then there are comparators etc.

Regards, Joerg

Thank you very much. I'm going to test this today. You are right, the VCC is the same from both.

I see that you used .6volts in the ohm law. Is this the tension on the VCC? If yes, I guess that I should use 3.3volts (as my power source) so the 2nd resistor should be 3.3kohms. Right?

Regards!

Hello Yew,

As Jim mentioned you can do it electronically. But if you aren't familiar with chips like the TL431 (there is now a lower power variant, the TLV431) then it needs to be simpler.

You can do it with a comparator, even one in a uC, but you have to understand its range limits and tolerance. You could use it to sense a voltage drop across a resistor if the voltages on both sides are in the "compliance range" of that built-in comparator. Safety is another matter. I don't know whether the commercial device is connected to mains but if it does then safety rules become an issue.

Regards, Joerg

Do you have a spare output on the micro?

Just write code similar to a "watchdog" - when it's running, periodically output something. When that pulse isn't there, your micro is asleep.

Added circuitry: Zero. Parts cost: Zero. Firmware Development: TBD. ;-)

Good Luck! Rich

I can't write anything for the device I want to monitor because I don't have access to it's software. I can only write code for my microcontroller.

Regards!

There may be another otion in that special case.

Most micros turn off their oscillators when they go to sleep mode. You can simply monitor if the oscillator is stopped (steady hi or low signal) or running to derive a sleep/non-sleep signal easily.

HTH Wolfgang

See....

Newsgroups: alt.binaries.schematics.electronic Subject: Measure the Current - from S.E.D - CurrentSense1.pdf Message-ID:

for a simple-minded way to do it.

If precision were desired I'd probably do it with a regulator with current sensing, to keep the load current from changing the output voltage.

...Jim Thompson

Nope, that would be, I guess, the voltage drop accross the transistor

Hello!

I tested the circuit in the scheme and it works.

The problems is that the voltage drops with 0.6volts for my device. Is there any way to fix this?

And whose circuit did you try?

...Jim Thompson

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