/elec/propeller-clock

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-----|___|---+---►|---.

             |       R    |   D1   |

             |            |        |

             |            |   ,,   |

             |            '---►|---+--- GND

             |                D2

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

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

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

one out quicker.

So, imagine we have this instead

             |

    Arduino  |        ___     ,,

            o----+---|___|----►|---.

             |   |     R1     D1   |

             |   |                 |

             |   |    ___     ,,   |

             |   '---|___|----►|---+--- GND

             |         R2     D2

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

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

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

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

from a pin.

Using a transistor to drive multiple LEDs

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

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

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

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

Here's an "in series" set up

                           .----- 12V

                           |

                          |¯|

                          |_| R2

                           |

                           |

                           ▼  D1

                           ¯``

                           |

                           |

             |             ▼  D2

             |             ¯``

             |             |

             |             |

             |             ▼  D3

             |             ¯``

             |             |

    Arduino  |    ___    ,-|

            o----|___|--(|↘ ) T

             |     R1    `-|

             |             |

             |             '----- GND

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

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

R2 would be

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

So the limit here is about 8 LEDs.

Something else to consider here is that the transistor actually also

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

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

safely ignore it in this example.

Here's an "in parallel" set up

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

                           |       |       |

                          |¯|     |¯|     |¯|

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

             |             |       |       |  

             |             |       |       |

             |             ▼  D1   ▼  D2   ▼  D3

             |             ¯``     ¯``     ¯``

             |             |       |       |

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

             |             |

    Arduino  |    ___    ,-|

            o----|___|--(|↘ ) T

             |     R1    `-|

             |             |

             |             '----- GND

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

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

RC CIRCUITS

===========

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

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

oscillates between 0V and approx. 5V).

              ___

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

                R     |

                      |

                     ===

                      | C

                      |

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

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

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

is because the capacitor would charge to the voltage supplied across

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

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

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

to Vin.

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

    5V

             _           _           _           _

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

         /     \     /     \     /     \     /

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

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

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

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

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

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

this equation applies

    T = RC

TRANSISTORS

===========

When using an NPN transistor as a switch, a typical set up might look

something like this:

    o--------------+---- 5V

                   |

                   []

                  (  )  some load

                   []

                   |

         ___     ,-|c

    o---[___]--b(|↘ ) T (NPN)

          R      `-|e

                   |

               ----+---- GND

You would typically connect the pin to 5V to turn on the transistor.

The current between the emitter and base turns on a larger current

between the emitter and collector.  The resistor, R, limits the

turn-on current and prevents a short (effectively) across the

transistor (and whatever the pin is connected to, such as an

arduino!).

PNP transistors were used in the days when a -5V rail was typical

instead of a 5V rail.  In this case, a typical set up would be exactly

the same as above, but with -5V used for the top rail and a PNP

transistor.  The thing to note here is that the direction of the

current would also have changed. Now imagine flipping this diagram

up-side-down and offsetting both rails by +5V (so that -5V becomes GND

and GND becomes 5V, respectively).  Then you'd have this set up, which

is a modern-day typical usage scenario for a PNP transistor.

               ----+---- 5V

                   |

         ___     ,-|e

    o---[___]--b(|↙ ) T (PNP)

          R      `-|c

                   |

                   []

                  (  )  some load

                   []

                   |

    o--------------+---- GND

In this set up, as before, a current is required between the base and

emitter to turn on a larger current between the collector and emitter.

But the difference this time is that the pin must be connected to

ground to achieve this.

CAPACITORS

==========

Two capacitors in parallel are equivelant to one capacitor whose

value is the sum of the two.

COMMON PARTS LIST AND USEFUL VALUES

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

NPN transistors

	BC548/BC547, 5V switch

PNP transistors

	BC327/BC328, -5V switch

LEDs

	~1.5V, ~10mA

	With 5V across them, a 560Ω resistor is required.

DIODE

	1N4001