Micron School of Materials Science and Engineering. Problem Set 10 Solutions

Size: px

Start display at page:

Download "Micron School of Materials Science and Engineering. Problem Set 10 Solutions"

Leslie Bruce
6 years ago
Views:

1 Problem Set 10 Solutions In-Exercises Using the p-n Diode Program and n-p-n Diode Program 1. Create the p-n junction (i.e., diode) for the following doping conditions: a. N A = 5x10 15 cm -3, N D = 5x10 15 cm -3 b. N A = 1x10 15 cm -3, N D = 5x10 18 cm -3 For each case, write down and briefly explain what happens (i.e., trends) to the following when changing the doping from part a to part b. Look up and write down the equation governing each effect to help you answer each question. i. Depletion widths, W p and W n. See the values below with the figures. The depletion width is the same distance on the n-side and the p-side when the two are evenly doped. For charge neutrality, Na*Wp=Nd*Wn [Kasap, p. 477, equation 6.1]. So in case-b, since the n-type side is much more heavily doped than the p-type side, the depletion width Wn must decrease to satisfy the charge neutrality equation. The total depletion width can also be calculated ( Na Nd ) V bi by: WDEP where Vbi is the Built-In Potential or the Flat Band Voltage enand [Kasap, p. 479, equation 6.7 & 6.6]. The Built-In Potential is discussed in the next problem. ii. Built-in voltage (i.e., flat band voltage or barrier height voltage). Note: Look up the equations for depletion width and built-in voltage/potential in chapter 6 of Kasap. Can you use the equations to make sense of i. & ii? kt N AN D The Built-in potential is defined as Vbi ln 2, therefore the product Na*Nd q ni determines which doping conditions will have a larger built-in potential (case-b). The sketches and relationships for charge density, electric field, and potential can be determined using Poisson s equation [Kasap, p. 479, equation 6.6]. In this case, a simplified de dv one dimension (x) version gives:, and E where E is electric field, ρ is dx dx r 0 charge density, and V is potential [Kasap, p. 477, equation 6.1.5]. See the values below with the figures. 1

2 iii. 1a. For each case, hand sketch or copy & paste the plots from the program of the energy band diagram, charge density, electric field, and potential (i.e., voltage) 2

3 1b. 3

Energy (ev) MSE 410-ECE 340 2. For case 1a, attempt to bias the junction at the flat band condition. a. The program will eventually fail (i.e., blow up), but how close (i.e., what voltage) did you come to the calculated built-in voltage (i.

4 Energy (ev) MSE 410-ECE For case 1a, attempt to bias the junction at the flat band condition. a. The program will eventually fail (i.e., blow up), but how close (i.e., what voltage) did you come to the calculated built-in voltage (i.e., flat band voltage or barrier height voltage)? Around V. b. Is the barrier increasing or decreasing? In order to bias at flat-band, a positive voltage is applied to the p-side of the junction. As the potential is increased, the bands flatten out which lowers the barrier height. c. Is this the forward or reverse bias condition? Briefly explain. This case is called forward bias. Since V bi is V, bias the p-side to V and the bands are nearly flat. 3. Now change your doping to N A = 1x10 19 cm -3 and N D = 2x10 19 cm -3 and bias to -10V. a. Is this a forward or reverse bias condition? Briefly explain. Reverse bias since applying a negative voltage to the p-side. b. Write down your depletion widths, W p and W n, and explain what happens using the equation for the depletion width to help you. See the values below with the figures. For charge neutrality, Na*Wp=Nd*Wn. So since the n-type side is more heavily doped than the p-type side, the depletion width Wn must decrease, and thus is less than Wp, to satisfy the charge neutrality equation c. Look up Zener Breakdown in chapter 6 of Kasap. Could this happen with the current doping conditions? Explain. See Kasap Ch. 6.5 page Since both the p-side and n-side are highly doped, the depletion widths are narrow. As a negative voltage is applied to the p-side with increasing magnitude, Zener breakdown will occur (tunneling) since the energy of the valence band on the p-side is higher than the energy of the conduction band on the n-side and the depletion region is very narrow. It appears (from simulation) that the threshold for where we might expect to see Zener breakdown would occur at around -4V. Energy Band Diagram Evac (ev) Ec (ev) Distance (nm) 4

5 d. Do you think that Zener breakdown could be exploited in an application? Explain. Look up Esaki diode and find resulting energy band diagrams for the Esaki diode to help you with your explanation. What most notable award was given to Leo Esaki for his invention? Provide a one-sentence moral of the story. Solution: Zener breakdown is described on pages and 573 in the course textbook by Kasap. 5

6 Energy (ev) Energy (ev) MSE 410-ECE Construct an n-p-n junction with N A = 1x10 16 cm -3, N D = 5x10 15 cm -3 (for both n-type regions). Bias each junction in the following condition: a. Vpn = -1V and Vpn = -1V. Examining the barrier of the np junction and the barrier of the pn junction, will electrons flow from the left n-type side to the right n-type side? Use the barriers to aid you in explaining why. The barrier has increased, hence electrons would not flow from the left n-type side to the right n- type side. b. Vpn = -1V and Vpn = 0.1V. Examining the barrier of the np junction and the barrier of the pn junction, will electrons flow from the left n-type side to the right n-type side? Use the barriers to aid you in explaining why. The barrier has increased, hence electrons would not flow from the left n-type side to the right n- type side. c. Vpn = 0.5V and Vpn = -1V. Examining the barrier of the np junction and the barrier of the pn junction, will electrons flow from the left n-type side to the right n-type side? Use the barriers to aid you in explaining why. The barrier has decreased on the left side, hence electrons would flow from the left n-type side to the right n-type side which signifies an on-state for the bipolar transistor. d. Vpn = 0.5V and Vpn = 0.5. Examining the barrier of the np junction and the barrier of the pn junction, will electrons flow from the left n-type side to the right n-type side? Use the barriers to aid you in explaining why. The barriers have been reduced on both sides by the same amount, hence electrons would flow from the left n-type side to the right n-type side but they would flow an equal amount the opposite direction which would result in a zero net flow. a) b) Energy Band Diagram Energy Band Diagram Distance (nm) Evac (ev) Ec (ev) Ev (ev) Efi (ev) Efp (ev) Efn (ev) Distance (nm) Evac (ev) Ec (ev) Ev (ev) Efi (ev) Efp (ev) Efn (ev) 6

7 Energy (ev) Energy (ev) MSE 410-ECE 340 c) d) Energy Band Diagram Energy Band Diagram Distance (nm) Evac (ev) Distance (nm) Evac (ev) Ec (ev) Ec (ev) Ev (ev) Ev (ev) Efi (ev) Efi (ev) Efp (ev) Efp (ev) Efn (ev) Efn (ev) e. Of the 4 conditions you just examined, which is the best condition for current to flow? For the condition you chose, is the n-p junction in forward or reverse bias conditions and is the p-n junctions in forward or reverse bias condition? Typically, the emitter is doped n + and the collector is doped n. For the forward bias condition, forward bias the emitter-base junction and reverse bias the collector-base junction. In this case, both n-regions are doped the same, and image-c depicts the best condition for current to flow with the np junction forward biased, and the pn junction reverse biased. (From left to right: n/p/n, emitter/base/collector) 5. Is a bipolar transistor a majority or minority carrier device? Explain. Do not have to do this problem. On next problem set. 7

8 6. Given a Si p-n junction (i.e., diode), start with Poisson s equation and derive the equation for the electric field using the charge density and the equation voltage using the electric field. Additionally, hand sketch the charge density, electric field and voltage potential (see 1a c). Show all work. Label all known values. Assume that NA > ND (e.g., say NA = 5x10 15 cm -3, ND = 2x10 15 cm -3 ). Hint: For electric field, integrate from boundaries towards zero. a. Charge density b. Electric field 8

9 9

10 c. Potential (i.e., voltage) 10

11 11

Leaving Cert Physics Long Questions Semiconductors

Leaving Cert Physics Long Questions 2017-2002 10. Semiconductors Please remember to photocopy 4 pages onto one sheet by going A3 A4 and using back to back on the photocopier Contents Ordinary level questions...