Things you should know when you leave ECE 340 Lecture 39 : Introduction to the BJT-II Fabrication of BJTs Class Outline: Key Questions What elements make up the base current? What do the carrier distributions in a BJT look like? How do I draw the band diagrams under bias and in equilibrium? How do I describe the current flow quantitatively? Now that we understand the basics of the BJT operation, let s investigate how we can fabricate a BJT Our goal: Fabricate a double polysilicon self-aligned n-p-n BJT for use in an integrated circuit. With the windows defined, it is time to implant dopants into the substrate Why n-p-n? Because electrons have a higher mobility in Si than that of holes. Photoresist and oxide will serve as a mask for the implant process. Implant dopants using isotopes which have a very small diffusivity such as As and Sb to form the highly conductive n-layer. We start with a p-type substrate and oxidize the surface. Apply photoresist and expose to define the etch area and etch windows in the oxide. Remove the oxide and photoresist and grow and epitaxially grow a lightly doped n- layer. We have now formed a buried collector layer which is guaranteed to have a small contact resistance. 1
When we perform the epitaxial growth, there is a step or notch which forms on the surface. We use this as an alignment mark for subsequent processes Use photolithography in conjunction with the the base/emitter to etch a hole in the polysilicon/oxide stack We will need many BJTs, so we need isolation Different devices cannot talk to one another. Therefore, we use LOCOS to isolate them We can also use channel stop implants to increase the isolation. While these schemes are most prevalent for isolation, other processes exist. One of these involves using RIE to etch a shallow trench which can then be filled with oxide by using the LPCVD process. Finish this step by depositing a p+ polysilicon layer, P+ dopant layer, and oxide layer using LPCVD. Diffuse in boron from the p+ polysilicon layer in order to provide a lowresistance, high-speed ohmic base contact. Deposit another oxide layer to close up the base window then implant boron into this window to form an intrinsic base where the majority of the current will flow Finish the fabrication by performing the final steps Deposit another oxide layer to close the base window. Etch the oxide to provide spacers on the sidewalls. Now the final steps The more heavily doped p regions around the intrinsic base will serve to lower the base series resistance. Failure to enclose the base allows for shorting to the p substrate. Deposit and pattern polysilicon emiter and collector contacts. Diffuse in As from the polysilicon to finish the formation of the contacts. Deposit another layer of oxide and pattern it to form the regions for the metal contacts. Sputter on the metal contacts. 2
We will now begin to take a closer look at the operation of a BJT All of the techniques that we will use will be very similar to those of the narrowbase diode. We assume that holes are injected from the forward biased emitter and then diffuse to the collector junction. What solution steps will we take: 1. Solve for the excess hole distribution in the base. 2. Evaluate the emitter and collector currents from the gradient of the distribution. 3. The base current can be found by current summation or from charge control analysis of recombination in the base. But, as with everything, we must make some assumptions to make the analysis tractable What are the critical assumptions that we must make? 1. The holes diffuse from the emitter to the collector and the drift current is negligible in the base region. 2. The emitter current is made up entirely of holes and the emitter has an injection efficiency of γ = 1. 3. The collector saturation current is negligible. 4. The active part of the base and the two junctions are of uniform cross-sectional area and the current flow in the base is essentially one-dimensional from the emitter to the collector. 5. All of the currents and voltages are in steady state. The holes we intent to inject will flow from the emitter to the collector by diffusion, we already know how to solve for the representative currents Assume that there is no recombination. We can solve for the distribution of excess holes in the base region to fine the currents. But remember that this simplified band diagram is not going to be correct. So what do the bands look like under bias? We want to consider normal operation mode: Forward bias on the emitter terminal which will reduce the barrier at the emitter-base junction. Reverse bias on the collector terminal which increases the collector-base junction barrier. We can now determine the excess hole concentration at the edge of the emitter,δp E, depletion region and the concentration on the collector side of the base, Δp C We are under bias so the Fermi level must split into quasi-fermi levels 3
With the excess hole concentrations peaks, we may proceed with a solution to the concentrations as a function of distance Since we now know the appropriate boundary conditions, we may now solve for the constants in the diffusion equation Start with the diffusion equation: To which we already know the general form of the solution: Where L P is the diffusion length of holes in the base region. But this diode is not necessarily semi-infinite in the n-region so we cannot eliminate one of the constants by assuming that all of the holes will disappear for large x n. By combining these with the general solution, we can determine the excess hole concentration We can simplify these expressions some by assuming that the collector junction is strongly reverse biased which would make p n small compared to the injected concentration, Δp E W b << L P in a good transistor so most of the holes reach the collector. Apply boundary conditions: Let s examine the implications of this equation graphically Almost a linear variation between emitter and collector. We can also gain some insight into the electron concentrations in the different regions Shown here for normal mode operation. Decay assumes a long diode. Deviations result from small changes to I B caused by recombination. 4
We now understand the band diagram and the carrier distributions, now we can understand the current flow All that we need are the gradients of the concentrations at each depletion region edge. But we must replace the constants in the current equations and this will complicate things as in the narrow-base diode The generic form of the hole diffusion current is: Now find the emitter current by applying the above equation at x n = 0 to the general solution Assume that the electrons crossing from the collector to the base are negligible and the reverse saturation current comes solely from holes entering from the base, and we can find the collector current We can use current summation, as we did in the case of the narrow-base diode, to obtain the base current, I B The sum of the base and collector currents must be equal to the emitter current that has been injected into the BJT Nevertheless, these equations are still very complicated and cumbersome to use. Can we make simplifications? If the collector is reverse biased: If the equilibrium hole concentration, p n, is small we can neglect the terms involving Δp C. Currents become Assuming that: 5
We can make further approximations by expanding the hyperbolic terms For small values of W b /L p we can neglect terms above the first order. If recombination dominates, we can obtain the base current from the charge control model. If we consider that this stored charge must be replenished every τ p seconds and relate the recombination rate. The rate at which electrons are supplied then becomes the base current. We can arrive at the same answer by looking at the difference between the first order approximations for the emitter and collector current This gives a clear demonstration that the base current is reduced for small W b and large τ p. We can increase τ p by using light doping in the base region which also improves the emitter injection efficiency. 6