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Committer: Dan
Date: 2011-11-03 00:54:13 UTC
Revision ID: dan@waxworlds.org-20111103005413-miheyjsrpm3a0r1e

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files removed:
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electronics-information

pcb.psd

project

project/propeller-clock.pro

project/propeller-clock.sch

src/propeller-clock.pde

src/propeller-clock.sch

src/utility

src/utility/Bounce.cpp

src/utility/Bounce.h

test

test/fan-test.pde

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Notes => Notes.rtf

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electronics-information

Why can't I use a potentiometer to get a different voltage?

===========================================================

Suppose that you have 12V and you need 6V. Why can't you use an

arrangement like this to get your 6Vs?

+12V -------+---------

|¯|

|_| R

+--------o 6V, yes?

|¯|

|_| R

GND -------+---------

You *do* get 6V, but it isn't actually practical to do this. The

problem is that, when R is suitably high, you've limited the current

at the 6V pin so much that it isn't really useful.

And when R is suitably low, there isn't enough resistance between the

+12V and GND rails to prevent a lot of current flowing. Lets work

this out (lets suppose that R is 1 ohm).

V = IR, therefore I = V/R

P = VI = V(V/R) = (V^2)/R = ( 12 ^ 2 ) / 1 = 144W

P = VI, therefore I = P/V = 144/12 = 12A

That's a lot of current and a lot of power, flowing continuously, just

to provide 6V!

Also, fluctuations in resistance and current drawn on the 6V pin would

actually change the ratio of the potentiometer, so you wouldn't get 6V

anyway.

The solution is to use a voltage regulator. The 7805 will give out 5V

(so long as you can supply it with at least 7V). But the arduinos

already have something similar on-board and they can take an input

voltage in the range 6-20V (although 7-12V is recommended).

Wiring up multiple LEDs in series to a single arduino pin

=========================================================

First, lets think about a single LED.

Arduino | ___ ,,

o-----|___|-----►|---- GND

| R D

The arduino pin, when raised high, is at 5V and no more than 20mA can

be taken from it. The LED will take about 10mA and wants about 1.5V.

You can think of this as a potentiometer arrangement, with the

resistance of R and D proportionally splitting the voltage between the

two components. We want 1.5V across the LED, so we need 3.5V (that's

5 - 1.5) across the resistor. Now, we don't know the resistance of

the LED, but we don't need to. If we know that we want 10mA through

the whole series (the resistor and the diode), then we can use V=IR as

follows...

V=IR, therefore R=V/I

R = 3.5 / .01 = 350Ω

Which is why the standard resistor you'd use with one LED is orange,

orange, brown (actually, 340 ohms).

Now lets consider more than one LED in series.

Arduino | ___ ,, ,,

o-----|___|-----►|-----►|---- GND

| R D1 D2

Ok, so now you still need 10mA through the whole series, but you want

1.5V across the first diode and 1.5V across the second as well. That

leaves 2V across R, the resistor (5 - 1.5 - 1.5).

R = V/I = 2 / .01 = 200Ω

You can see that beyond three or perhaps, at a push, four LEDs you're

not going to get the required 1.5V across each LED. So three (or four)

is the limit to how many LEDs you can drive in series from one pin on

the arduino.

Wiring up multiple LEDs in parallel to a single arduino pin

===================================--======================

Imagine we have this

Arduino | ___ ,,

o-----|___|---+---►|---.

100

| R | D1 |

101

| | |

102

| | ,, |

103

| '---►|---+--- GND

104

| D2

105

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This wouldn't work. We can't use one resistor in series with the two

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LEDs because the resistance of the two LEDs wont actually be exactly

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the same. You'll end up running one brighter than the other and burn

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one out quicker.

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So, imagine we have this instead

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Arduino | ___ ,,

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o----+---|___|----►|---.

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| | R1 D1 |

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| | |

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| | ___ ,, |

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| '---|___|----►|---+--- GND

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| R2 D2

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This is ok. R1 and R2 are just the usual 340 ohms. But we have to

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bare in mind that each LED requires 10mA. So the total current drawn

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from the arduino pin will be 20mA, which is the most you're allowed to

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draw. So two LEDs is the most that we can drive, in parallel, directly

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from a pin.

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Using a transistor to drive multiple LEDs

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=========================================

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In both of the following diagrams, the resistor on the arduino pin,

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R1, just needs to be something suitably high to provide a small

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current on the base of the transistor. So, R1 could be 1kΩ.

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Here's an "in series" set up

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.----- 12V

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|¯|

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|_| R2

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144

▼ D1

145

¯``

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148

| ▼ D2

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| ¯``

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| |

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| |

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| ▼ D3

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| ¯``

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| |

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Arduino | ___ ,-|

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o----|___|--(|< ) T

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| R1 `-|

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| |

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| '----- GND

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This is fine. D1, D2 and D3 will require a total of 4.5V across them

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(1.5V each), leaving plenty of headroom (you've got 12V to play with).

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R2 would be

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R2 = V/I = ( 12 - 1.5 - 1.5 - 1.5 ) / 0.01 = 750Ω

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So the limit here is about 8 LEDs.

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Something else to consider here is that the transistor actually also

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requires 0.7V across it. So in that calculation, the desired voltage

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across R2 should actually be 12 - 1.5 - 1.5 - 1.5 - 0.7, but we can

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safely ignore it in this example.

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Here's an "in parallel" set up

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.-------+-------+----- 12V

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| | |

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|¯| |¯| |¯|

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|_| R3 |_| R4 |_| R5

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| | | |

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| | | |

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| ▼ D1 ▼ D2 ▼ D3

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| ¯`` ¯`` ¯``

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| | | |

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| +-------+-------'

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| |

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Arduino | ___ ,-|

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o----|___|--(|< ) T

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| R1 `-|

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| |

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| '----- GND

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Here, R3 = R4 = R5 = 340Ω, as usual. The numbher of LEDs is limited

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only by the current that can be drawn from the power supply.

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RC CIRCUITS

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===========

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An RC (resistor capacitor) curcuit is a basic low-pass filter. Here

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we're talking about giving it a pulse wave signal (a voltage that

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oscillates between 0V and approx. 5V).

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___

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Vin o----[___]----+-----o Vc

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R |

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===

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| C

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0V o-------------+-----o

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If the resistor weren't there, the capacitor on it's own would act

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like an open circuit to a pulse wave (or to an AC power supply). This

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is because the capacitor would charge to the voltage supplied across

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it and discharge almost immediately. The purpose of the resistor is to

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limit the current that is available to charge the capacitor, so it

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charges slowly and it takes some time before Vc becomes almost equal

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to Vin.

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With a pulse wave on Vin, Vc would look something like this:

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_ _ _ _

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,'' \ ,'' \ ,'' \ ,''

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/ \ / \ / \ /

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0V / ',,_/ ',,_/ ',,_/

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If you were to draw the tangent to the curve at time=0 (so, the

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initial rate of change of charging), and you note the time that this

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line crossed 5V, then this time is T. That is to say, time T is the

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time it would take the capacitor to charge to the target voltage if it

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charged constantly at the rate it actually initially charges at. Then

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this equation applies

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T = RC

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