ECE 3455: Electronics

1. ECE 3455: Electronics Diode Physics: A Brief Tour

2. Current-Voltage Characteristics Figure 3.7 The i–vcharacteristic of a silicon junction diode. We’ll try to justify this… …but we’ll leave this alone until Solid State Devices. Figure 3.8 The diode i–v relationship with some scales expanded and others compressed in order to reveal details.

3. Semiconductors of practical interest are generally crystalline, meaning their atoms are arranged in periodic arrays. This figure shows the “diamond” lattice: if the atoms are carbon, this is diamond; if they are Si or Ge we get the corresponding semiconductors (silicon or germanium). If the atoms alternate between Ga and As, we get GaAs (gallium arsenide), one of the important optically sensitive semiconductors (LEDs, photo-detectors).

4. Conduction (i.e., current flow) can take place via movement of either electrons or holes when a voltage (i.e., and electric field) is applied. A hole is the absence of an electron and “moves” when successive electrons take its place. Often, either electrons or holes dominate the current; the other current component is then very small.

5. We can add either electrons or holes to increase the conductivity. This process is called “doping” and involves intentional addition of impurities to the semiconductor. Electrons are added to Si by doping with a Group V (periodic table) element, typically P but also Sb. These have a fifth valence electron which is very loosely bound to the lattice and can move if a voltage is applied. The P or Sb atoms are donors. Holes are added to Si by doping with a Group III element, typically B but also Al. The B or Al atoms are acceptors.

6. p-type n-type Let’s make a diode!! Ingredients: mostly holes Ingredients: mostly electrons p-type n-type Electrons have been added via doping with donors. Holes have been added via doping with acceptors. Here is a pn junction diode… …and the corresponding circuit symbol.

7. Why do electrons (or holes, for that matter) move in a semiconductor in the first place? There are two possibilities… Diffusion: if there are more electrons (or holes) in one place than another, they will tend to diffuse to “even out” the concentration. Due to the concentration gradient, electrons will tend to move toward the right. Drift: if an electric field is applied (by applying a voltage), electrons will move toward more positive potential. e - +

8. p-type n-type p-type - n-type + So when p and n meet… …holes here diffuse to n-side …electrons here diffuse to p-side But that sets up a positively charged region on the n-type side and a negatively charged region on the p-type side... (but see note below) eo …which generates an electric field that opposes further diffusion. The charge shown in the lower figure is not due to electrons and holes. Once the electrons get to the p-type side, they disappear when they find holes there (the electrons and holes recombine); the same goes for holes on the n-type side. The charge is due to donors (on the n-side), which become positively charged when they lose their electrons (because they move to the other side!!), and to acceptors (p-side), which become negatively charged when they lose their holes. The donors and acceptors do not move; they are bound to the lattice, and are responsible for the charged regions shown.

9. p-type - n-type + In fact, electrons and holes never stop moving to the other side via diffusion, and the electric field never stops pulling them back to where they started. But eventually we reach a balance between diffusion and drift; this is equilibrium. (We are assuming there is no applied voltage here.) eo We’re here Electron drift Electron diffusion Hole diffusion Hole drift

10. e- e- Another way to look at this is the following: at equilibrium, there is an energy barrier keeping more (net) electrons from getting to the p-type side. But we could get another electron to diffuse over if we add energy somehow. Electron energy (Joules) diffusion If we add energy, this electron can diffuse to the p-side. extra energy weak electron p n Instead of adding energy, could we reduce the energy barrier? Now it takes less energy to get the electron over the barrier. The number of electrons that get over the hill increases exponentially with voltage. Electron energy (Joules) Similar arguments can be made for holes.

11. e- ef Va How can we reduce the barrier? We can apply a voltage!!! The increase in the number of electrons that can get over the barrier is exponential: the current goes as eVa/VT, where VT is the ‘thermal voltage” Note that Va generates an electric field that is opposite in sign to eo. So the net field reduces to ef, and the barrier to diffusion drops. What if we reverse the sign of Va?

12. e- er Va Now Va generates an electric field that is in the same direction as eo. So the net field increases to er, and the barrier gets big. There’s no diffusion now. But what’s been happening to the drift current?

13. ACME Red Arrows (Box of 25) Wile E. Coyote Theorem small field ef big field er The drift current doesn’t change very much with field, because of the Wile E. Coyote Theorem: regardless of how big the cliff is (no matter what voltage is applied), Wile E. will either fall, or not. Electrons that do fall contribute to the drift current just the same, whether the field is big or small.

14. e- Now we put this all together: e- er ef v > 0 v < 0 Forward bias: the diffusion current, which is positive, increases exponentially with voltage. The drift current is negligible. Reverse bias: Diffusion is “off”: the only current is the drift current, which is small and negative.

15. p-type - n-type + What about that threshold voltage thing?* eo It turns out that the equilibrium electric field corresponds to a voltage of about 0.8 V or so. When the applied voltage is about 0.6 – 0.7 V, we start to see a significant current**. So we can approximate a threshold voltage Vth as being about 0.7 V in many cases. *Don’t confuse the threshold voltage (Vth or Vf in your textbook) with the thermal voltage VT; these are not the same thing!! **In fact, the applied voltage does not all go to changing eo. Some of it is dropped across the equivalent resistance of the n- and p-type regions. So at Va = 0.7 V, something less than that is appearing across the barrier and reducing it enough to generate a noticeable current.