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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 number of LEDs is limited
only by the current that can be drawn from the power supply.
RC CIRCUITS
===========
An RC (resistor capacitor) circuit 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 equivalent 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
Electrolytic Capacitor (Radial, 4700uF 16V)
Maplin part no. VH57M
Arduino-powerable relay (DPDT, gold contacts, 5V, 27mA)
Maplin part no. N05AW
|