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Trying to measure mA and don't understand what I'm seeing


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#1 John West

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Posted 28 December 2013 - 08:06 AM

Hi all.  To anyone more versed in electronics than me, this is probably going to be a simple question :)...

 

I am measuring mAs being used by my netduino mini.  Currently I'm just testing an led that blinks on and off every 3 seconds.

 

I'm measing mAs right off the power supply.  The results I'm seeing are 26mA when the led is off and 39mA when it's on.  This seems right.  The question I have is:

 

The results are exactly the same whether I plug in a 9v power supply or a 12v (which shows as a 16v on the voltmeter).  Shouldn't the mAs be lower for the 12v than for the 9v?  Am I totally not understanding this?

 

Oh, and I know there's a regulator (only from reading other threads here) that takes whatever voltage the mini is fed and converts it to 5v.  But would this be reflected straight off the power supply?

 

Anything in layman's terms would be great!  Thx!



#2 Paul Newton

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Posted 28 December 2013 - 09:51 AM

Hi John,

 

Consider a simple LED circuit (battery + resistor + LED).

When you use an LED you add a series resistor to limit the current and set the correct voltage at the LED terminals.

If you started with a 9V battery you would calculate the resistor value for 9 Volts.

If you then changed to a 12V battery, you would use a bigger resistor to maintain the same current through the LED.

Obviously you had to fit a new resistor to get the same current and voltage at the LED.

 

But, rather than calculating the value, you could have used a voltmeter and a variable resistor to acheive the same thing.

You would connect a voltmeter across the the LED, and then starting with the highest resistance, turn down the resistance until the voltage on the LED was correct.

When the voltage is correct the same current will be flowing through the circuit.

If you added another LED, the voltage would drop because not enough current is flowing to drive both LEDs, and so you would turn down the resistance until the correct voltage was present again. Now there should be double the current.

 

The Netduinos have linear voltage regulators.

These are basically a transistor and a voltage detector.

The transistor operates as a series resistor whose impedance is controlled by the voltage detector.

The voltage detector measures the output voltage to the circuit (5V or 3.3V), and sets the amount of current through the transistor so that the voltage is correct.

The more current the circuit takes, the lower the voltage - the more the transistor is turned on to compensate.

 

Ring any bells?

If the supply voltage is increased, the transistor compensates and the same current flows.

 

There are other types of voltage regulator (not used on the Netduino) - "switching regulators" use a transistor(s) as a switch rather than a resistor. The switch rapidly turns on and off causing current to flow into some combination of inductors and capacitors. Depending on the arangement, this sort of voltage regulator can reduce or increase the output voltage (some are able to do both depending on the input voltage).

Now the regulator is more like an AC transformer: power in = (power out + losses)

So if you have more input voltage, the regulator takes less input current.

 

Hope this helps - Paul






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