/elec/propeller-clock

To get this branch, use:
bzr branch http://bzr.ed.am/elec/propeller-clock
6 by edam
added electronics info and (unfinished) scematic and changed Notes to
1
Why can't I use a potentiometer to get a different voltage?
2
===========================================================
3
7 by edam
finished writing up electronics notes
4
Suppose that you have 12V and you need 6V.  Why can't you use an
6 by edam
added electronics info and (unfinished) scematic and changed Notes to
5
arrangement like this to get your 6Vs?
6
7
    +12V -------+---------
8
                |
7 by edam
finished writing up electronics notes
9
               |¯|
10
               |_| R
6 by edam
added electronics info and (unfinished) scematic and changed Notes to
11
                |
12
                +--------o  6V, yes?
13
                |
7 by edam
finished writing up electronics notes
14
               |¯|
15
               |_| R
6 by edam
added electronics info and (unfinished) scematic and changed Notes to
16
                |
7 by edam
finished writing up electronics notes
17
     GND -------+---------
6 by edam
added electronics info and (unfinished) scematic and changed Notes to
18
7 by edam
finished writing up electronics notes
19
You *do* get 6V, but it isn't actually practical to do this.  The
6 by edam
added electronics info and (unfinished) scematic and changed Notes to
20
problem is that, when R is suitably high, you've limited the current
7 by edam
finished writing up electronics notes
21
at the 6V pin so much that it isn't really useful.
6 by edam
added electronics info and (unfinished) scematic and changed Notes to
22
23
And when R is suitably low, there isn't enough resistance between the
7 by edam
finished writing up electronics notes
24
+12V and GND rails to prevent a lot of current flowing.  Lets work
25
this out (lets suppose that R is 1 ohm).
6 by edam
added electronics info and (unfinished) scematic and changed Notes to
26
8 by edam
cleaned-up info
27
	V = IR, therefore I = V/R
7 by edam
finished writing up electronics notes
28
	P = VI = V(V/R) = (V^2)/R = ( 12 ^ 2 ) / 1 = 144W
8 by edam
cleaned-up info
29
	P = VI, therefore I = P/V = 144/12 = 12A
6 by edam
added electronics info and (unfinished) scematic and changed Notes to
30
31
That's a lot of current and a lot of power, flowing continuously, just
32
to provide 6V!
33
34
Also, fluctuations in resistance and current drawn on the 6V pin would
35
actually change the ratio of the potentiometer, so you wouldn't get 6V
36
anyway.
37
7 by edam
finished writing up electronics notes
38
The solution is to use a voltage regulator.  The 7805 will give out 5V
39
(so long as you can supply it with at least 7V).  But the arduinos
40
already have something similar on-board and they can take an input
6 by edam
added electronics info and (unfinished) scematic and changed Notes to
41
voltage in the range 6-20V (although 7-12V is recommended).
42
7 by edam
finished writing up electronics notes
43
44
Wiring up multiple LEDs in series to a single arduino pin
45
=========================================================
46
47
First, lets think about a single LED.
48
49
             |
50
    Arduino  |     ___      ,,
51
            o-----|___|-----►|---- GND
52
             |      R       D
53
             |
54
55
The arduino pin, when raised high, is at 5V and no more than 20mA can
56
be taken from it.  The LED will take about 10mA and wants about 1.5V.
57
58
You can think of this as a potentiometer arrangement, with the
59
resistance of R and D proportionally splitting the voltage between the
60
two components.  We want 1.5V across the LED, so we need 3.5V (that's
61
5 - 1.5) across the resistor.  Now, we don't know the resistance of
62
the LED, but we don't need to.  If we know that we want 10mA through
63
the whole series (the resistor and the diode), then we can use V=IR as
64
follows...
65
66
    V=IR, therefore R=V/I
67
    R = 3.5 / .01 = 350Ω
68
69
Which is why the standard resistor you'd use with one LED is orange,
70
orange, brown (actually, 340 ohms).
71
72
Now lets consider more than one LED in series.
73
74
             |
75
    Arduino  |     ___      ,,     ,,
76
            o-----|___|-----►|-----►|---- GND
77
             |      R       D1     D2
78
             |
79
80
Ok, so now you still need 10mA through the whole series, but you want
81
1.5V across the first diode and 1.5V across the second as well.  That
82
leaves 2V across R, the resistor (5 - 1.5 - 1.5).
83
84
    R = V/I = 2 / .01 = 200Ω
85
86
You can see that beyond three or perhaps, at a push, four LEDs you're
87
not going to get the required 1.5V across each LED. So three (or four)
88
is the limit to how many LEDs you can drive in series from one pin on
89
the arduino.
90
91
92
Wiring up multiple LEDs in parallel to a single arduino pin
93
===================================--======================
94
8 by edam
cleaned-up info
95
Imagine we have this
7 by edam
finished writing up electronics notes
96
97
             |
98
    Arduino  |     ___        ,,
99
            o-----|___|---+---►|---.
8 by edam
cleaned-up info
100
             |       R    |   D1   |
7 by edam
finished writing up electronics notes
101
             |            |        |
102
             |            |   ,,   |
103
             |            '---►|---+--- GND
104
             |                D2
105
106
This wouldn't work. We can't use one resistor in series with the two
107
LEDs because the resistance of the two LEDs wont actually be exactly
108
the same.  You'll end up running one brighter than the other and burn
109
one out quicker.
110
111
So, imagine we have this instead
112
113
             |
8 by edam
cleaned-up info
114
    Arduino  |        ___     ,,
7 by edam
finished writing up electronics notes
115
            o----+---|___|----►|---.
116
             |   |     R1     D1   |
117
             |   |                 |
118
             |   |    ___     ,,   |
119
             |   '---|___|----►|---+--- GND
120
             |         R2     D2
121
122
This is ok.  R1 and R2 are just the usual 340 ohms.  But we have to
123
bare in mind that each LED requires 10mA.  So the total current drawn
124
from the arduino pin will be 20mA, which is the most you're allowed to
125
draw.  So two LEDs is the most that we can drive, in parallel, directly
126
from a pin.
127
128
129
Using a transistor to drive multiple LEDs
130
=========================================
131
132
In both of the following diagrams, the resistor on the arduino pin,
133
R1, just needs to be something suitably high to provide a small
134
current on the base of the transistor. So, R1 could be 1kΩ.
135
136
Here's an "in series" set up
137
138
                           .----- 12V
139
                           |
140
                          |¯|
141
                          |_| R2
142
                           |
143
                           |
144
                           ▼  D1
145
                           ¯``
146
                           |
8 by edam
cleaned-up info
147
                           |
7 by edam
finished writing up electronics notes
148
             |             ▼  D2
149
             |             ¯``
150
             |             |
151
             |             |
152
             |             ▼  D3
153
             |             ¯``
154
             |             |
155
    Arduino  |    ___    ,-|
33 by edam
updated schematic (switched to PNP) and added info about transistors
156
            o----|___|--(|↘ ) T
7 by edam
finished writing up electronics notes
157
             |     R1    `-|
158
             |             |
159
             |             '----- GND
160
161
This is fine. D1, D2 and D3 will require a total of 4.5V across them
162
(1.5V each), leaving plenty of headroom (you've got 12V to play with).
163
R2 would be
164
165
    R2 = V/I = ( 12 - 1.5 - 1.5 - 1.5 ) / 0.01 = 750Ω
166
167
So the limit here is about 8 LEDs.
168
169
Something else to consider here is that the transistor actually also
170
requires 0.7V across it.  So in that calculation, the desired voltage
171
across R2 should actually be 12 - 1.5 - 1.5 - 1.5 - 0.7, but we can
172
safely ignore it in this example.
173
174
Here's an "in parallel" set up
175
176
                           .-------+-------+----- 12V
177
                           |       |       |
178
                          |¯|     |¯|     |¯|
179
                          |_| R3  |_| R4  |_| R5
180
             |             |       |       |  
181
             |             |       |       |
182
             |             ▼  D1   ▼  D2   ▼  D3
183
             |             ¯``     ¯``     ¯``
184
             |             |       |       |
185
             |             +-------+-------'
186
             |             |
187
    Arduino  |    ___    ,-|
33 by edam
updated schematic (switched to PNP) and added info about transistors
188
            o----|___|--(|↘ ) T
7 by edam
finished writing up electronics notes
189
             |     R1    `-|
190
             |             |
191
             |             '----- GND
192
193
Here, R3 = R4 = R5 = 340Ω, as usual.  The numbher of LEDs is limited
194
only by the current that can be drawn from the power supply.
12 by edam
added info about RC circuits
195
196
197
RC CIRCUITS
198
===========
199
200
An RC (resistor capacitor) curcuit is a basic low-pass filter. Here
201
we're talking about giving it a pulse wave signal (a voltage that
202
oscillates between 0V and approx. 5V).
203
              ___
204
    Vin o----[___]----+-----o Vc
205
                R     |
206
                      |
207
                     ===
208
                      | C
209
                      |
210
     0V o-------------+-----o
211
212
If the resistor weren't there, the capacitor on it's own would act
213
like an open circuit to a pulse wave (or to an AC power supply). This
214
is because the capacitor would charge to the voltage supplied across
215
it and discharge almost immediately. The purpose of the resistor is to
216
limit the current that is available to charge the capacitor, so it
217
charges slowly and it takes some time before Vc becomes almost equal
218
to Vin.
219
220
With a pulse wave on Vin, Vc would look something like this:
221
222
    5V
223
             _           _           _           _
224
          ,'' \       ,'' \       ,'' \       ,''
225
         /     \     /     \     /     \     /
226
    0V  /       ',,_/       ',,_/       ',,_/
227
228
If you were to draw the tangent to the curve at time=0 (so, the
229
initial rate of change of charging), and you note the time that this
230
line crossed 5V, then this time is T. That is to say, time T is the
231
time it would take the capacitor to charge to the target voltage if it
232
charged constantly at the rate it actually initially charges at. Then
233
this equation applies
234
235
    T = RC
236
33 by edam
updated schematic (switched to PNP) and added info about transistors
237
238
TRANSISTORS
239
===========
240
241
When using an NPN transistor as a switch, a typical set up might look
242
something like this:
243
244
    o--------------+---- 5V
245
                   |
246
                   []
247
                  (  )  some load
248
                   []
249
                   |
250
         ___     ,-|c
251
    o---[___]--b(|↘ ) T (NPN)
252
          R      `-|e
253
                   |
254
               ----+---- GND
255
256
You would typically connect the pin to 5V to turn on the transistor.
257
The current between the emitter and base turns on a larger current
258
between the emitter and collector.  The resistor, R, limits the
259
turn-on current and prevents a short (effectively) across the
260
transistor (and whatever the pin is connected to, such as an
261
arduino!).
262
263
PNP transistors were used in the days when a -5V rail was typical
264
instead of a 5V rail.  In this case, a typical set up would be exactly
265
the same as above, but with -5V used for the top rail and a PNP
266
transistor.  The thing to note here is that the direction of the
267
current would also have changed. Now imagine flipping this diagram
268
up-side-down and offsetting both rails by +5V (so that -5V becomes GND
269
and GND becomes 5V, respectively).  Then you'd have this set up, which
270
is a modern-day typical usage scenario for a PNP transistor.
271
272
               ----+---- 5V
273
                   |
274
         ___     ,-|e
275
    o---[___]--b(|↙ ) T (PNP)
276
          R      `-|c
277
                   |
278
                   []
279
                  (  )  some load
280
                   []
281
                   |
282
    o--------------+---- GND
283
284
In this set up, as before, a current is required between the base and
285
emitter to turn on a larger current between the collector and emitter.
286
But the difference this time is that the pin must be connected to
287
ground to achieve this.
288
289
290
CAPACITORS
291
==========
292
293
Two capacitors in parallel are equivelant to one capacitor whose
294
value is the sum of the two.
295
296
297
COMMON PARTS LIST AND USEFUL VALUES
298
===================================
299
300
NPN transistors
301
	BC548/BC547, 5V switch
302
303
PNP transistors
304
	BC327/BC328, -5V switch
305
306
LEDs
307
	~1.5V, ~10mA
308
	With 5V across them, a 560Ω resistor is required.
41 by edam
modified schematic and updated notes
309
310
DIODE
311
	1N4001