Im seriously new to arduino. I come from a programming background and have very limited electronics experience.
I bought
This kit to help start me out and as an introduction to programming with arduino its great. However I just copy the circuit exactly and dont understand anything about the creation process of a circuit.
I then look for information on ideas i have to do and immediately get swamped in terminology and equations i dont understand.
is there a spot where i can start from the beginning on circuit design. I seriously have no clue on the "whys" behind things
ie the book explains to use a 330Ma(i think) resistor with an led to make sure the led doesnt burn out. but why 330Ma and not the other one it came with?
Im just concerned i dont want to fry the ardunio when i start playing around with other ideas.
Mainly my goal would be to make an arduino that is capable of tone detection and sense how close to true it is (ie it knows the wavelength of C and compares that to the sound coming into the arduino)
Im not looking for the answer i just want to get to the point where i have a basis for understanding what im looking for and build up from there.
Sounds like you want a chromatic tuner, like a guitar tuner.
Do a search for those, I recall seeing a couple posted here in the forum.
You will need some kind of sound pickup - typically a microphone. That outputs a low level analog signal that you will have to amplify some and possibly manipulate a little to get into the 0-5V range the arduino can work work.
As far as the LEDs, most LEDs have a maximum current of 20mA.
The Arduino output is typically near 5V, and the voltage across an LED when it turns on is around 2.2V for a Red LED. The other 2.8V has to be dissipated somewhere, usually in a resistor. The current from 5V thru the LED and thru the resistor is constant. If the LED has 2.2V across, the resistor has 2.8. Now Ohms Law kicks in, Voltage = Current * Resistance. V=IR, V/R = I, V/I = R. Same thing, said three ways.
You know V, 2.8V. You have a current that you want, 0.02A (20mA).
So: 2.8V = 0.02 * R. A little math, 2.8V/.02 = R = 140 ohm.
If a higher value resistor is used, the current will be lower:
V/R = I 2.8V/330 ohm = 11.5mA, which or may not look dimmer to the eye.
For experimenting, the current/brightness is not all that critical. There are other cases where it may be.