Homework 4 B

ECE 4339 Spring 2014
Han Q. Le (copyrighted) U. of Houston

Part B

4.  (50 pts)

This problem involves the same structure as in Part A, prob. 3. The Base length is 50 μm. We now apply 3 V between Collector and Emitter. (C is higher voltage than E). No voltage is applied to the Base, i. e. it is floated (No voltage does not mean zero voltage).

4.1  (10 pts)

What do you think the current and band diagram are like. First, hand sketch (or draw with computer programs such as ppt or paint or any graphics package) what you think the band diagram should look like (use the result in Prob. 3). Then discuss in plain English what the current should be (for simplicity, just assume the structure has uniform cross section, hence, the same current flows through 3 regions: E-B junction, along the Base, and B-C junction. (this part is about analytical thinking).

4.2  (30 pts)

We can solve the question in 4.1 more quantitatively by doing the follow: Let the current be (Subscript[I, 0]). we can plot I-V curve for each segment:

4.2.a (10 pts) EB junction

Its voltage is (Subscript[V, EB])[(Subscript[I, 0])] (obtained by the inverse of I-V) (we just plot the current density and not worry about the actual cross section)

4.2.b (10 pts) Along the length of the Base

The Base is long, hence, we should not assume that there is no voltage drop from one end to another, although as we will findout, it is small - but we must show it). The voltage is (Subscript[V, B])~R (Subscript[I, 0]) where R is the resistance.

4.2.c (10 pts)  BC junction

Its voltage is  (Subscript[V, BC])[(Subscript[I, 0])]

If we plot all three, we have to find a common value (Subscript[I, 0]) such that:
           (Subscript[V, EB])[(Subscript[I, 0])]+(Subscript[V, B]) +(Subscript[V, BC])[(Subscript[I, 0])]=3  Volt

We can use graphical technique to solve this equation. Approximation is acceptable, because you will see that the current is really limited by just one of the three above (you have to find out which one), and most of the voltage drop is across that part. Hence, you obtain (Subscript[V, EB]),(Subscript[V, B]) ,(Subscript[V, BC]).

4.3  (10 pts)

Plot the band diagram of the entire structure using the value of (Subscript[V, EB]),(Subscript[V, BC]) obtained above. For simplicity and since (Subscript[V, B])  is small we can just let it be 0. Plot the excess carrier density (on linear scale).

Below is the example for GaAs, not Si which is what you have to do. It is shown here only to give you a rough idea. Delete it and do NOT include in your work. It is ONLY an example.

5.  (50 pts)

This is about the same structure in problem 4. (You must finish Prob. 4 first)

5.1  (20 pts)

Apply -0.25 V between E and B, while maintaining -3 V between E and C (what is the voltage between B and C?). Plot the band diagram of the whole structure, the carrier density, and the diffusion current between E and B on linear scale in a vertical column with properly aligned horizontal axis (see example below). Use the following value of diffusion lengths (these values are not correct for Si, but we use it just to illustrate the idea): (Subscript[L, n])=10 μm, (Subscript[L, p])=10 μm.

Below is the example for GaAs, not Si which is what you have to do. It is shown here only to give you a rough idea. Delete it and do NOT include in your work. It is ONLY an example.

5.2  (10 pts)

Analyze results you obtained in 5.1. By applying a forward bias voltage 0.25 V between E and B, it is equivalent with “lowering the dam” to let the electrons (water) from the Emitter to flow through the Base (water flows over the lowered dam) and into the Collector. However, are there really a lot of diffusing electrons (this is a qualitative statement) from Emitter reaching the Base-Collector junction? would you estimate it at less than 1%? (calculate and explain your answer).

If all things are the same, you cannot change anything except for the dimension of the structure, what would you change to allow more Emitter electrons to reach the Collector?

5.3  (20 pts)

Do the same as in 5.1, except that the Base length now is 5 μm. Compare with your analysis of 5.2, are there alot of Emitter electrons reaching the Collector here? How much?