ECE 340 Lecture 31 : Narrow Base Diode Class Outline:

Transcription

1 ECE 340 Lecture 31 : Narrow Base Diode Class Outline: Narrow-Base Diodes

2 Things you should know when you leave Key Questions What is a narrow-base diode? How does current flow in a narrow-base diode?

3 Quick Introduction Bardeen changes everything! UIUC Professor ( ) The modern age is ushered in with the invention of the transistor Discovered in 1947 Consisted of a germanium base with gold contacts Bardeen realized that electrons did not all travel through semiconductors the same way. Surfaces are different Leads to more complex radios and computers M.J. Gilbert ECE 340 Lecture 31 11/04/11

4 Quick Introduction We can make things with discrete transistors Transistor radio from 1954 which uses 4 discrete transistors. Let s try using many INTEGRATED transistors Integrated circuits fabricate all transistors and metal interconnects on the same piece of semiconductor. Jack Kilby UIUC 47, patent TI 1959, Nobel prize Used germanium. Robert Noyce 1961, co-founder of Fairchild, then Intel. Used silicon

5 Quick Introduction Put many of the transistors together and we get The first microprocessor Intel 4004 (1971) 2250 transistors 740 khz operation Comperable computation with ENIAC Built on 2 wafers We use 12 today 10 µm line widths We use 45 nm today 4-bit bus width We use > 800 kb today Used first in the Busicom Calculator

6 In the final few weeks of the class, we will discuss the operation of the bipolar junction transistor Our plan will be to build up our knowledge about related devices and then discuss the operation of the bipolar junction transistor. To this end, we begin with the narrow-base diode. The narrow-base diode is comprised of: p + -n diode. Heavily doped n + contact. Typically, the neutral portion of the lightly doped region is much less than the average hole diffusion length away.

7 What do we expect to happen with this device? Neutral portion of lightly doped n-region is much less than L p. Hole diffusion current is still dominant, as we expect, but the boundary conditions are altered. Minority holes cross x n = l are assumed to recombine when they enter the n + contact as the minority lifetime will be extremely small. New boundary conditions:

8 Clearly to describe how current flows through the narrow-base diode, we will need to solve the diffusion equation But we can simplify it Most of the minority injected holes will diffuse across the n-type base. I(x = 0) ~ I(x = x n ). Make the straight line approximation for the variation of the excess carrier profile Identical current requires constant gradient. We may also determine how much charge is injected.

9 Assuming low-level injection, minority holes may only contribute through the diffusion process. Let s write down the hole current density in the n-region: But we ve used the straight line approximation so we can simplify this equation But the current here is much larger than in a normal p+-n diode at the same voltage. This is because l replaces L p in the denominator of the diode equation since l << L p. Theoretically speaking, all excess holes recombine immediately at x n = l. Since the excess concentration is ~ 0 this is far below the normally expected concentration in a normal p + -n diode.

10 The much larger change in hole concentration across the neutral n-region which sets up a large diffusion. We already know that only a few holes will recombine within the n-region. Each time this happens another electron must flow in from the n+ contact to preserve the charge neutrality. Or mathematically This sets up an electron current flowing into the n-region at x n = l compensates for the small decrease in hole diffusion current due to recombination The difference in the brackets is small as l << Lp so the hole diffusion current remains roughly constant up to the n + contact.

11 Let s examine this visually x > l inside the n+ contact the entire current is carried by majority electrons.

12 Using the straight line approximation, we can get some additional insight into the system behavior Assuming a constant cross-section, we can solve the diffusion equation exactly using linear combinations of exponentials (as before) So, the hole diffusion current in the n-region is given by And we already know the boundary conditions. And the current injected at the boundary becomes: Use small angle expansion:

13 What is the hole current at x n = l at the n + contact Which reduces to The difference between these two hole currents will represent the electron current flowing into the base from the n + contact.

14 So, let s summarize Hole concentration gradient: Stored minority charge: Electron Current injected into the p + contact: In the three terminal p+- n p device there exists a small electron base current corresponding to the sum of the electron injection and recombination currents which we can adjust to control the much large hole current. Therefore, we get current multiplication.