I am noob to arduino and to most electronics. I have ordered a arduino uno for a testing base for a specific project. I want to learn on a basic setup then apply it on a larger scale.
I have not yet received the board but am trying to do the math first and have better understanding on the interactions of the components first.
I understand that components have a specific working range for Volt and Amps and I need to control these with a resistor. I know that for led to preform properly i need to appreciate the forward voltage and and amps that are needed and apply the correct resistor to not burn out led.
I want to figure out what resistor i would need for an IR circuit and want to make sure i am understanding correctly and doing the math properly.
These are the specs i was going off of for a Gikfun 5mm 940nm LEDs Infrared Emitter and IR Receiver Diode for Arduino (Pack of 20pcs) EK8443
Product description
Diameter : 5MM, Wave length:940nm
Forward Voltage: 1.2-1.3V, Power: 0.15 W
Receiving Angle: 40 degrees
Voltage (VF) : 1.1-1.4 V (0.1 V each file)
Receiving tube (limit parameters (Absolute Maximum Ratings) (Ta = 25 ).
Storage temperature: - 25 ~ + 80
Transmitting and receiving distance: about 7 - 8 M.
Maximum power: 70 mw;
Maximum forward current: 30 ma;
Maximum reverse voltage: 5 v;
Maximum pulse current peak: 75 ma;
Welding temperature/time: 260 / acuities 5 S
Working environment: - 25 ~ + 70
If I am powering off a 3.3V pin, 3.3V-1.3V=2V, 2V/.03A=66.6Ohms. So i would use a minimum of 66 Ohms resistor?
Also i don't see a different set of specs for emitter/Receiver so I assume that these specs apply to both?
I honestly can't remember the last time a noob actually did their homework.
Well Done except for ONE detail:
WHAT does THIS say (EXACTLY) ? Maximum forward current: 30 ma;
You should back off the maximum by at least 10-15% , or at the very least 5%.
No , I would NOT use 66 ohms, although technically you're correct that is the MINIMUM, but
it is not recommended to operate a semiconductor device at the ragged edge of it's rating.
I would go with a 75 or 82.
75 ohms would give you (3.3V-1.3V)/75 = 0.0266 A (26.6mA)
Ok great, I kinda figured that the 30mA was the max and believe that the working range of the pin outputs was 20mA and was thinking that a 100 Ohms resistor was more inline with what i should be using, but i will use the recommended resistor.
Thank you so much, I know its a simple answer for you but really helps me.
So assuming using your suggested resistance of 75 or 82. If i was to use 22 AWG wire at a length of 50 feet, would the resistance of the wire replace the need for a resistor?
22 AWG 0.0254 (Diameter in inches) 16.14 (in Ohms per 1000ft)
No it doesn't. I will do some more homework before I ask any more questions, to try and make them less noobish.
You have provided me with alot of food for thought.
Re: resistors, affording or getting them isn't the issue. I am trying to appreciate and figure out the project and just ask for verification on what I think to be true. I am a mechanical guy with some electrical background and am trying to create something and want to understand what i have to do, without being spoon fed the answers.
Basically i want to run sensor to detect and falling object in a enclosed tube that will be falling at a rate of approx 6 times per second. This is will be taking place approx 25 feet away from the arduino. I want to output a signal to let me know when this happens or when it stops happening.
raschemmel:
The 75 ohm resistor was the current limiting resistor.
2V/75 ohms = 0.0266A (26.6mA)
If you used the only the wire ,
2V/0.807 ohms = 2.47A !!!!
Perhaps more to the point if you add the 0.807 Ohms to 75 Ohm resistor you'll get
2V / 75.807 Ohms = 26.4mA
So 50ft of wire reduces the current by 0.2mA
And that's a lot less than the variation caused by the fact that you'll be using a resistor with maybe 5% tolerance so 75 is really somewhere between 71.25 and 78.75).
Don't try for too much spurious precision with these sort of calculations. Everything has tolerances and if you try to take them all into account you'll drown in useless numbers.
I will do some more homework before I ask any more questions, to try and make them less noobish.
Your losing points at an alarming rate. Don't make me take back your Gold Star .
Here's the homework you requested :
Write V = I*R on the blackboard 100 times
then write R = V/I 100 times
and then I = V/R 100 times
Post a photo of the blackboard.
I don't know where you've been but the OP was not trying for precision, he was asking if he could replace
the 75 ohm resistor with the 0.807 ohm 50 foot wire.
I think Larry's idea of using 5000 feet of wire to replace the 75 ohm resistor was the best option.
Im a toolmaker and am use to being very precise, idk how fussy i have to be with electronics.
Ok, let's not try to put a positive spin on a mistake.
You get credit for good choices and corrected for poor choices.
There's no way replacing the 75 ohm resistor with a 0.807 ohm wire is going to look like a good choice.
Just for the record, this is what you said:
So assuming using your suggested resistance of 75 or 82. If i was to use 22 AWG wire at a length of 50 feet, would the resistance of the wire replace the need for a resistor?
22 AWG 0.0254 (Diameter in inches) 16.14 (in Ohms per 1000ft)
16.14/1000 = .01614Ohms/foot x 50ft = .807 Ohms
So you correctly calculated the resistance of the 50 foot length of wire BEFORE you asked if you could
replace the resistor with the wire.
FYI, if you had put a LEADING 0 in front of the DECIMAL POINT (Like I ALWAYS DO), I seriously doubt
you would have made that mistake, but .807 still doesn't look like 80 ohms.
Does replacing a 75 ohm resistor with a 0.807 ohm length of wire sound like an attempt at precision to you ?
The better question would have been :
"Can i replace the 75 ohm resistor with 5000 ft of wire ?" (because that would be mathematically and electronically correct.)
