/elec/propeller-clock

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6 by edam
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Why can't I use a potentiometer to get a different voltage?
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===========================================================
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Suppose that you have 12V and you need 6V.  Why can't you use an
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arrangement like this to get your 6Vs?
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    +12V -------+---------
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                |
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               |¯|
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               |_| R
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                |
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                +--------o  6V, yes?
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                |
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               |¯|
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               |_| R
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                |
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     GND -------+---------
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You *do* get 6V, but it isn't actually practical to do this.  The
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problem is that, when R is suitably high, you've limited the current
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at the 6V pin so much that it isn't really useful.
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And when R is suitably low, there isn't enough resistance between the
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+12V and GND rails to prevent a lot of current flowing.  Lets work
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this out (lets suppose that R is 1 ohm).
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	V = IR, therefore I = V/R
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	P = VI = V(V/R) = (V^2)/R = ( 12 ^ 2 ) / 1 = 144W
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	P = VI, therefore I = P/V = 144/12 = 12A
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That's a lot of current and a lot of power, flowing continuously, just
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to provide 6V!
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Also, fluctuations in resistance and current drawn on the 6V pin would
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actually change the ratio of the potentiometer, so you wouldn't get 6V
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anyway.
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The solution is to use a voltage regulator.  The 7805 will give out 5V
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(so long as you can supply it with at least 7V).  But the arduinos
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already have something similar on-board and they can take an input
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voltage in the range 6-20V (although 7-12V is recommended).
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Wiring up multiple LEDs in series to a single arduino pin
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=========================================================
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First, lets think about a single LED.
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             |
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    Arduino  |     ___      ,,
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            o-----|___|-----►|---- GND
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             |      R       D
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             |
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The arduino pin, when raised high, is at 5V and no more than 20mA can
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be taken from it.  The LED will take about 10mA and wants about 1.5V.
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You can think of this as a potentiometer arrangement, with the
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resistance of R and D proportionally splitting the voltage between the
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two components.  We want 1.5V across the LED, so we need 3.5V (that's
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5 - 1.5) across the resistor.  Now, we don't know the resistance of
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the LED, but we don't need to.  If we know that we want 10mA through
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the whole series (the resistor and the diode), then we can use V=IR as
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follows...
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    V=IR, therefore R=V/I
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    R = 3.5 / .01 = 350Ω
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Which is why the standard resistor you'd use with one LED is orange,
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orange, brown (actually, 340 ohms).
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Now lets consider more than one LED in series.
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             |
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    Arduino  |     ___      ,,     ,,
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            o-----|___|-----►|-----►|---- GND
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             |      R       D1     D2
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             |
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Ok, so now you still need 10mA through the whole series, but you want
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1.5V across the first diode and 1.5V across the second as well.  That
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leaves 2V across R, the resistor (5 - 1.5 - 1.5).
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    R = V/I = 2 / .01 = 200Ω
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You can see that beyond three or perhaps, at a push, four LEDs you're
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not going to get the required 1.5V across each LED. So three (or four)
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is the limit to how many LEDs you can drive in series from one pin on
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the arduino.
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Wiring up multiple LEDs in parallel to a single arduino pin
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===================================--======================
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Imagine we have this
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             |
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    Arduino  |     ___        ,,
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            o-----|___|---+---►|---.
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             |       R    |   D1   |
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             |            |        |
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             |            |   ,,   |
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             |            '---►|---+--- GND
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             |                D2
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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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             |
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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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                          |¯|
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                          |_| R2
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                           |
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                           |
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                           ▼  D1
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                           ¯``
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                           |
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                           |
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             |             ▼  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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                     ===
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                      | C
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                      |
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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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    5V
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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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TRANSISTORS
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===========
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When using an NPN transistor as a switch, a typical set up might look
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something like this:
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    o--------------+---- 5V
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                   |
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                   []
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                  (  )  some load
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                   []
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                   |
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         ___     ,-|c
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    o---[___]--b(|↘ ) T (NPN)
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          R      `-|e
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                   |
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               ----+---- GND
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You would typically connect the pin to 5V to turn on the transistor.
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The current between the emitter and base turns on a larger current
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between the emitter and collector.  The resistor, R, limits the
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turn-on current and prevents a short (effectively) across the
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transistor (and whatever the pin is connected to, such as an
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arduino!).
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PNP transistors were used in the days when a -5V rail was typical
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instead of a 5V rail.  In this case, a typical set up would be exactly
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the same as above, but with -5V used for the top rail and a PNP
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transistor.  The thing to note here is that the direction of the
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current would also have changed. Now imagine flipping this diagram
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up-side-down and offsetting both rails by +5V (so that -5V becomes GND
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and GND becomes 5V, respectively).  Then you'd have this set up, which
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is a modern-day typical usage scenario for a PNP transistor.
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               ----+---- 5V
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                   |
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         ___     ,-|e
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    o---[___]--b(|↙ ) T (PNP)
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          R      `-|c
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                   |
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                   []
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                  (  )  some load
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                   []
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                   |
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    o--------------+---- GND
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In this set up, as before, a current is required between the base and
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emitter to turn on a larger current between the collector and emitter.
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But the difference this time is that the pin must be connected to
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ground to achieve this.
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CAPACITORS
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==========
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Two capacitors in parallel are equivelant to one capacitor whose
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value is the sum of the two.
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COMMON PARTS LIST AND USEFUL VALUES
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===================================
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NPN transistors
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	BC548/BC547, 5V switch
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PNP transistors
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	BC327/BC328, -5V switch
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LEDs
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	~1.5V, ~10mA
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	With 5V across them, a 560Ω resistor is required.