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### Topic: How do I determine how much current my circuit is using? (Read 1 time)previous topic - next topic

#### brucemac

##### Jul 30, 2012, 09:31 pm
If I use a 2.2K Ohm resistor on an 5 Volt Arduino output pin, according to Ohm's law, the amperage calculates to about 2.7 milliamps.  But knowing that
a resistor reduces voltage and amperage, how much current is actually being used in this circuit.  I am using the 2.2K Ohm resistor on some darlington
arrays and optocouplers and they work fine.  I tested the voltage after the resistor and it reads 2.9 volts but I don't how many milliamps the Arduino
is actually using.  Is it actually the 2.7 or because the voltage is reduced, is it something else and if so how is it calculated?

Thanks.
Bruce

#1
##### Jul 30, 2012, 10:04 pm
3.141592653 5 ...
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

#### James C4S

#2
##### Jul 30, 2012, 10:32 pm

If I use a 2.2K Ohm resistor on an 5 Volt Arduino output pin, according to Ohm's law, the amperage calculates to about 2.7 milliamps.

Yes, Ohm's Law says 5V / 2.2Kohms = 2.7mA

This is assuming the only thing in the "circuit" is the resistor.  If there are other devices, then their voltage drops need to be considered.

But knowing thata resistor reduces voltage and amperage,

Instead of using the word "amperage" use the word "current."  Resistors reduce (or restrict) current flow; they, by definition, resist.  They do not reduce voltage.  Because they restrict current, this creates a voltage drop across them.  This is defined by.... Ohm's Law.

If you have a 2.2K Ohm resistor with 2.7mA flowing through it, its voltage drop will be 5V.  (See the math works in reverse.)

I tested the voltage after the resistor and it reads 2.9 volts but I don't how many milliamps the Arduino

Are you talking about the Arduino board as a whole or just on a single I/O pin?
Capacitor Expert By Day, Enginerd by night.  ||  Personal Blog: www.baldengineer.com  || Electronics Tutorials for Beginners:  www.addohms.com

#### brucemac

#3
##### Jul 31, 2012, 05:01 am
Just 1 pin.  I am driving some 817 optocouplers, several tip120 darlington, and a ULN2802 array.  Because I have about 12 different things going on
in my Arduino at one time I don't want to overload the board.  I am "offloading" all the output devices (led's, LCD display, DC motor, etc.) with the
darlingtons using a 2amp, 9 volt wall wart.  I am trying to see how many things I can get the Arduino to do in one program.  12 so far.
Bruce

#### brucemac

#4
##### Jul 31, 2012, 05:02 am
This is the first time I have used this forum to get help.  I am in awe at the responses.

Many thanks.
Bruce

#### Docedison

#5
##### Jul 31, 2012, 11:13 am
You Ain't Seen Nuthin Yet Sir... It gets crezzy sometimes, I've seen two or three different topics in one thread (If you are real unlucky).

Doc
--> WA7EMS <--
"The solution of every problem is another problem." -Johann Wolfgang von Goethe
I do answer technical questions PM'd to me with whatever is in my clipboard

#### brucemac

#6
##### Jul 31, 2012, 05:34 pm
I'm having trouble wraping my head around the concept that the resistor also reduces the voltage in a circuit in terms of what I really
wind up with in a circuit.   In my example with the 2.2K Ohm resistor, I've reduced the current to less that 3 milliamps.  That is great.
My goal is to limit, as much as possible, the current used by each pin on my Arduino by offloading each circuit's output device with a darlington.
I think, at this point, that even if I were to connect 50 pins to darlingtons I would still only accumulate less than a total of 150 milliamps load on my Arduino.
I believe that is way under the limit for my Mega.  But because of the (unintended consequence) additional voltage drop that occurs with
a resistor, I need to check to be sure there is enough voltage and current to "trigger" the darlingtons.  Am I making sense and do I have this correct?

Bruce
Bruce

#### James C4S

#7
##### Aug 01, 2012, 04:37 am
As I said before, resistors do not reduce voltage.  They drop voltage.  They drop the difference of whatever voltages are present on either side of them.

Diodes are different and will only drop whatever their forward voltage is.  Most diodes are around 0.7V.  (Light Emitting Diodes, LEDs, can be between 2 and 4V depending on the color.)  Regardless of what voltage you apply to the diode, it will only drop its forward voltage.  They will conduct whatever current is available, which is why you need a current limiting resistor in series with them (depending on the application.)

This is important to understand because...

My goal is to limit, as much as possible, the current used by each pin on my Arduino by offloading each circuit's output device with a darlington.

A darlington is a pair of BJT transistors.  In terms of voltage drops, you should consider each BJT's Base to Emitter a diode's worth of voltage drop, which is 0.7V.  So for a Darlington pair (DP), you are around 1.4V.

That means if you put a resistor between the pin of your arduino and the input of your DP, the DP will drop 1.4V leaving a 5V-1.4V=3.6V to drop across the resistor.  Using Ohm's Law, you can see that 3.6V/2.2Kohms = 1.63mA.  Since the resistor is in series with the pin an the DP, only 1.63mA will flow.

BJTs are not really voltage devices.  They are current devices.  They amplify the amount of current flowing through them.  The more current flowing from base to emitter, means more current is allowed to flow from collector to emitter.
Capacitor Expert By Day, Enginerd by night.  ||  Personal Blog: www.baldengineer.com  || Electronics Tutorials for Beginners:  www.addohms.com

#### brucemac

#8
##### Aug 01, 2012, 05:01 pm
Thanks James!

That makes complete sense to me now.  I just wasn't getting the voltage drop thing.  And I forgot about the drop from the darlington itself.
Based on what you just said, I completed one of my circuits, as a test, with a 817 optocoupler driving a 5mm red led with 5 volts.  I used the 2.2K Ohm
resistor between the Arduino pin and the optocoupler and 150 Ohm resistor between the led and +5 Volts.  It works just fine.  Actually, I consider
the additional voltage drop a bonus because it further reduces the amount of current the Arduino uses on each pin.
I tried some higher value resistors on the opto's input side and it did, in fact, reduce the brightness level on the led. So I view it more as a variable
resistor than just an on and off switch.  One circuit down and lots more to go!

Again, thanks so much for your help. I am completely impressed with all my experiences with the Arduino community.  It is just unbelievable
how much support and information is available.  But it is the people like you that contribute so willingly that makes this such a positive experience.
Thank you.
Bruce
Bruce

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