diodes
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What exactly does a diode do?
What happens in a diode to make current only flow one way around a circuit? My physics teacher said it was something to do with the metals in it and the impurities in them. Can anyone give me a detailed explanation. Any help much appreciated.
Connect a diode from the 12V side of the aux lights to the 12V side of the headlight. That way the headlight will still be powered when the aux lights are on. Actually they will be powered just a little less. A diode drops around a Volt in this case, so they will get a little dimmer, but not a lot. This should be more than offset by the aux lights coming on.
You need a diode that is rated for whatever current your headlight draws. I am not familiar enough with motorcycles to know what typical headlight current is, but I expect a few Amps at least. You can figure this out from the light wattage rating. Divide that by 12 to get a rough idea of current in Amps. For example, if the headlight is "100 Watt", then this tells you the current is about 8.3 Amps. There is some slop in this, so in this case you want a diode rated for at least 10 A.
Diodes also have a voltage rating, but in this case that shouldn't be much of a issue because 12 V is rather low. A "20 V" diode should be OK, but I'd feel better about anything rated for 30 V or more. If you can find something called a "Schottky" diode, that will have a little better characteristics but may be hard to find and more expensive if you do. A ordinary diode is good enough.
The diode has to be installed with the right orientation. You want the "anode" end at the aux light and the "cathode" end at the headlight. These are usually not spelled out explicitly like that. Sometimes the cathode end is marked with a band if the diode is the shape of a cylinder. Otherwise, there might be a sort of arrow symbol:
It is there to keep the transistor's current less susceptible to temperature changes.
In the case of Q1:
Suppose that instead of having R1 and D1, Q1base was connected directly to ground.
Emitter current would be:
$$ I_{e} = \frac{20V - V_{be}}{R_{2}}$$
You can see Ie is susceptible to variations in Vbe, which has a known dependency on temperature (T), so you might as well express it as: $$ I_{e}(T) = \frac{20V - V_{be}(T)}{R_{2}}$$
But with the diode, if they are matched and thermally bonded: $$ V_{diode}(T) = V_{be}(T) $$
So now: $$ I_{e} = \frac{20V+V_{diode}(T)-V_{be}(T)}{R_{2}} $$ Which simplifies to: $$ I_{e} = \frac{20V}{R_{2}} $$ Independent of Vbe, and its variations with temperature.
The diode is effectively providing the little voltage offset that would be needed to compensate for Vbe changes with T, in order to maintain a constant current.
Best Answer
This is probably better off asked on the physics stack, but essentially:
You have two differently doped semiconductors, one P-type and one N-type. The N-type has excess electrons, and the P-type has excess "holes" (which means the absence of an electron in an electron shell)
When you bring the P and N type materials together, you form a P-N junction. At the junction, some of the excess electrons cross over and fill the holes in the P-type material, forming a depletion zone (essentially a built in potential):
This depletion zone is (almost) non-conductive, as it has no free charge carriers. Now if we apply a potential (voltage) across the junction, depending on the polarity, the depletion zone either widens or narrows. If the region widens, the non-conductive region increases and no current can flow through the diode.
If it narrows, at a certain potential (~0.7V for a silicon diode) the diode will begin to conduct. This (hopefully) explains why the diode acts in the way it does.
Above images from this introduction to diodes.
Also, this image (from the Wikipedia P-N Junction link above) shows the P-N junction with a little more information than the above images: