In the schematic above, there's a PN junction with a source \$V_d\$, the barrier potential \$V_J\$, contact potentials \$aV_{J0}\$ on the P-side and (1-a)\$V_{J0}\$ on the N-side. \$V_{J0}\$ indicates the barrier potential of the PN junction with no source connected, \$V_J\$ the barrier potential with this source present and 'a' is a fraction in \$aV_{J0}\$.
How can we interpret the whole circuit in the point of view of erergy-providing and energy-consuming?
1) There are voltage drops \$aV_{J0}\$, (1-a)\$V_{J0}\$. Does that mean a test positive unit charge will "loose" energy while going from the (+) side to the (-) side? If so, in what means does the test charge get it's energy? What kind of form of energy is it transferred to?
2) Across the depletion region from P to N-side, there is an electric field pointing to the P-side. I guess there must be a reverse electric field acting on a test positive charge for it to pass through the build-in field. Where does that reverse electric field come from?
3) If the external source \$V_d\$ > \$V_{J0}\$(the barrier potential of the PN junction with no source connected), will there still be a barrier voltage inside the PN junction?
Thank you!
Best Answer
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.
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.
See 2, and get back to me if it doesn't make sense.