Electronic – Simple pn junction / diode question

Voltage is always relative. When you just hold one probe of your DMM to the + of a battery, for instance it will display 0 V, like it's saying "this is nothing, what do I have to compare it with?".

Any circuit should have a ground (yours hasn't). Ground is your reference everything else is compared to. Usually that's the negative terminal of the power supply. It's only by calling ground 0 V that the supply's +12 V makes sense: it's 12 V higher than ground. If you measure 2.7 V in your circuit the "referenced to ground" is implied, unless indicated otherwise. That's the case when you say "relative to X", then you don't use ground as a reference, but the voltage at X.

For DC normal arithmetic applies: if A is +5 V relative to ground and B = +12 V relative to ground then A is -7 V relative to B.

1. Yes, a test positive unit charge will lose energy when it goes across the diode. The actual process of the transfer of energy is a somewhat more statistical thing, though. Thinking about the test charge as a single charge carrier like an electron or hole isn't a very precise thing, although it is sufficient to understand the operation of the diode for most practical uses. The actual transfer of energy is due to a change in the energy level of the unit charge (well, actually, net nechange in energy level of many charge carriers). In steady state, which is where all practical uses of diodes takes place (the physics definition of it - really short time scales), this energy loss comes in two ways. One is resistive, and ends up as heat. This is often described as being due to collisions. The other is the effect of a charge carrier moving from area A to B with different local density of states. There, you'd say the energy goes into the lattice, which ultimately turns up as heat (in steady state). This energy is actually liberated from the lattice as an EM wave with a frequency that works out to contain that much energy. In the case of LEDs, you can actually take the forward drop and calculate the frequencies, which would give you what is essentially the frequency of light emitted. Since diodes are better efficient than not, the frequencies are infrared or less and absorbed by the lattice or casing which ultimately becomes heat. Note that even high speed electronics are usually slow enough to qualify as steady state in this context.

2. The reverse electric field has to be applied from outside. This is where the forward drop voltage of a diode comes in (note that the forward resistance of the diode is more a part of point 1 than 2). The voltage drop that is generated across a diode is basically the energy you need for your test charge to cross the barrier. You may think of it as the toll you pay to get on the highway, and the resistance as the fuel you use to drive your car across it.

3. See 2, and get back to me if it doesn't make sense.