/elec/propeller-clock : revision 5

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

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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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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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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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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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----+---- 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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___ ,-|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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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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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.