/elec/propeller-clock : revision 9

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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.