bzr branch
http://bzr.ed.am/elec/propeller-clock
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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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|_| R |
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+--------o 6V, yes? |
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|¯| |
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|_| R |
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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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First, lets think about a single LED. |
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Arduino | ___ ,, |
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o-----|___|-----►|---- GND |
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| R D |
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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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Arduino | ___ ,, ,, |
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o-----|___|-----►|-----►|---- GND |
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| R D1 D2 |
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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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Imagine we have this |
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Arduino | ___ ,, |
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o-----|___|---+---►|---. |
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| R | D1 | |
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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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| '---|___|----►|---+--- 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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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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|_| R2 |
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▼ D1 |
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¯`` |
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| ▼ D2 |
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| ¯`` |
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| ▼ D3 |
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| ¯`` |
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Arduino | ___ ,-| |
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o----|___|--(|< ) T |
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| R1 `-| |
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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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|_| R3 |_| R4 |_| R5 |
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| ▼ D1 ▼ D2 ▼ 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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| '----- 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. |