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MSE 310-ECE 340 Dept. of Materials Science and Engineering Electrical Properties Of Materials Fall 2014/Bill Knowlton

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Problem Set 10 Solutions

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. NA = 5x1015 cm-3, ND = 5x1015 cm-3 b. NA = 1x1015 cm-3, ND = 5x1018 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, Wp and Wn .

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 by: 122 ( ) Va d bi

DEPa d

N NW

eN N

ε +=

where Vbi is the Built-In Potential or the Flat

Band Voltage [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?

The Built-in potential is defined as 2

ln A Dbi

i

kT N NV

q n

=

, therefore the product Na*Nd

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

one dimension (x) version gives: 0

, and r

dE dVE

dx dx

ρε ε

= = − where E is electric field, ρ is

charge density, and V is potential [Kasap, p. 477, equation 6.1.5]. See the values below with the figures.


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iii. 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)

1a.


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1b.


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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.e., flat band voltage or barrier height voltage)? Around 0.660 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 Vbi is 0.661 V, bias the p-side to 0.660 V and the bands are nearly flat.

3. Now change your doping to NA = 1x1019 cm-3 and ND = 2x1019 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, Wp and Wn, 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 502-503. 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.


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4. Construct an n-p-n junction with NA = 1x1016 cm-3, ND = 5x1015 cm-3 (for both n-type regions). Bias each junction in the following condition: a. Vnp = -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. Vnp = -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. Vnp = 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. Vnp = 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)


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c) d)

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. A bipolar transistor is a minority carrier device because the base is the control part of the device and is opposite in type from the emitter and collector. The emitter emits majority carriers in the emitter into the base where they become minority carriers.


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6. Given a Si p-n junction (i.e., diode), start with Poisson’s equation, derive and hand sketch (see aa – cc)). Show all work. Label all known values. Assume that NA > ND (e.g., say NA = 5x1015 cm-3, ND = 2x1015 cm-3). Hint: For electric field, integrate from boundaries towards zero.

a. Charge density

b. Electric field


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c. Potential (i.e., voltage)


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7. Draw the band diagrams of a GaAs p-n-p bipolar transistor in the following conditions. Label your plots

thoroughly including the emitter, base and collector: a. Flat band condition.

b. Equilibrium condition.

c. Equilibrium condition and add Ohmic contacts.


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d. Operating condition (i.e., when the transistor is in the on state). Show the flow of carriers in your drawing.

problem set 3 solutions - college of...

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