Solid State Electronics

Name: Solid State Electronics
Uploaded: 2017-10-18T03:10:28+00:00
Duration: PTM32S40
Description: Solid State Electronics

Solid State Electronics
Bipolar Junction Transistors Ronan Farrell Recommended Book: Streetman, Chapter 7 Solid State Electronic Devices BJTs

Solid State Electronics
Topics we’ll be covering Qualitative explanation of the operation of the BJT BJT equations and the Eber Molls model. Physical Construction of BJT Imperfections such as base narrowing, avalanche breakdown, thermal effects, emitter crowding Small signal model and parasitic correspondence. High Frequency Transistors BJTs

Bipolar Junction Transistors
The BJT consists of a sandwich of either an N type material with P type on either side, or visa-versa. The names used to identify the different type of bipolar junction transistor (BJT) is given by the structure of the device PNP P-type N-type P-type NPN N-type P-type N-type BJTs

Collector Emitter Base +V -V PNP P N P N P N NPN +V -V Historically, the left hand side region (the most positive side) is called the EMITTER (of holes). The right hand side (the most negative) the COLLECTOR (of holes). The centre is called the BASE. BJTs

This construction for the transistor gives rise to two PN junctions within the device. In comparison with the FET, the current in a BJT device travels across the junctions. The behaviour of the BJT is governed by the behaviour of charge carriers at a PN junction. These are the same equations that we worked out in the PN junction section of this course. BJTs

In normal operation, one PN junction is forward biased and the other is reversed biased. In a forward bias junction, the main mechanism for current flow is diffusion. Holes diffuse from the P-type material to the N-type material and recombine. There is also the reverse current of electrons diffusing from the N-type to the P-type material. If one side is much more heavily doped than the other, then the majority of the current is carried by the carriers from the more heavily doped side. Simply because there are more of them and the concentration differential is greater. BJTs

In reverse bias junctions the main current mechanism is drift current as the energy potential required for majority carriers to overcome the energy difference at the barrier is too great. We said that the reverse bias drift current was dependent on the number of minority carriers available on each side of the junction. “Near” the junction these fall across the barrier (being of a different sign to the majority carriers). As long as there is a potential difference all minority carriers “near” the junction will cross it. The only and major limit on the current that can flow is due to the limited number of minority carriers, an in a PN junction they are created by thermal generation near the junction. BJTs

Now if we could increase the number of minority carriers in the reverse bias PN junction then we could increase the current flow. So consider this... P N N P Holes Holes Holes Holes Majority Minority Minority Majority Majority carriers when they cross the PN junction become minority carriers in the other material. BJTs

So consider this... P N N P Holes Holes Holes Holes P N P So the left-hand side is positive bias and lots of holes are crossing into the N side, becoming minority carriers. If they come “near” the right-hand junction, reverse biased, then they’ll fall across the junction. The more carriers the left hand side provides, the greater the current across the right-hand side junction. BJTs

Now all is well and good with this idea, the principle of adding additional minority carriers into a substance is called carrier injection and the forward biased device is a minority carrier injection, or in or case a “hole injection device” (because we drew a PNP device). Two questions need to be asked to make this a practical device Will all the injected holes get to the reverse biased junction? Will the forward biased junction actually work as we’ve drawn it? BJTs

Now minority carriers in a material, outside of the depletion region, quickly meet a majority carrier and recombine, maintaining the long-term equilibrium. However this takes some time, the lifetime of the minority carrier, , and this lifetime can be extended by making the doping of the material less. Less dopant, therefore fewer majority carriers and hence a greater chance of a minority carrier lasting longer, thus the average lifetime of a minority carrier increases. BJTs

But while the minority carrier is avoiding recombination it is also moving physically, with some speed (it physically crossed a barrier to become a minority carrier) So as before, the speed of the carrier (electrons are faster than holes) and the average lifetime of the carrier gives us the average distance it will travel. So if the width of the centre region is a small fraction of the average distance a carrier will travel then it is safe to say that the majority of the carriers will actually be caught by the pull of the electric field of the reverse biased junction and fall across the junction. BJTs

So the other question was will it work as shown, well the answer is no. No matter how dominant one current carrier is over the other, there is still some current contribution from the other type. So if P-type is massively dominant, the majority of the current will be carried by holes, but some, though a small amount, will be carried by electrons. Also in the centre region, some recombination will occur, not all holes will get to the reverse biased junction (assuming a PNP device). Recombination removes a hole and an electron and these electrons have to be replenished from somewhere. BJTs

Another case where electrons play a role is that across even a reverse biased junction, there will be a current comprising of drift current. This will result in electrons entering into the base (assuming a PNP device). This will provide some of the electrons needed to replenished the electrons removed by recombination, but it will be a very small portion of the number required. The first two mechanisms far outweigh the third and ensure that a source of electrons for the base (assuming a PNP device) is required. Thus we need a base current to inject electrons. BJTs

This diagram indicates the three current flows for the holes and electrons. (1) holes lost to recombination (2) holes reaching reverse bias junction and crossing over (3) reverse bias junction drift current (4) electrons supplied by base to replace electrons lost through recombination (5) electrons injected across the forward biased junction. BJTs

Collector-Emitter Current Ratio
Now these currents can be related by simple equations In this equation B is called the Base Transport Factor, not to be confused with . The proportion of holes that leave the emitter and get to the reverse bias junction. The ratio of the emitter current which is due to holes (assuming a PNP device) as a fraction of the total emitter current is called the emitter injection efficiency  Ideally, we’d like both of these to be as close to 1 as possible but both will be slightly less than one. BJTs

Collector-Emitter Current Ratio
Now taking these two, B and  we can come up with a relationship for IE and IC In this equation a is called the current transfer ratio, and ideally we wish this to be as close to one as possible. BJTs

Base-Collector Current Ratio
The base current is to be found by investigating the current required to keep supplying electrons (assuming a PNP device) to the base to replace the net loss in electrons due to recombination, reverse bias junction drift and forward bias diffusion. Reverse biased junction drift current (very small, neglect) The current required to replenish the electrons lost through hole recombination. The current component of the forward biased PN junction (emitter base) that is carried by electrons from the base to the emitter. BJTs

So looking at this more closely BJTs

But we had a relationship for IE(p) and IC So Therefore b is the base-collector current amplification factor. This should be a large number as a is small. BJTs

Quantitative Analysis of BJT
What we shall be doing in this section is investigating the behaviour of the system so as to be able to quantify the values of a and b. Throughout this analysis we shall be assuming a PNP device. First we’ll make some simplifying assumptions: The current across the forward biased junction is purely diffusion, there is no drift current component. The emitter current is purely composed of holes. (IE(n)=0, g=1) The base collector reverse biased junction current is neglible and can be ignored. The cross-sectional area of the device is constant. All currents and voltages are stable and constant. BJTs

The key is to find the current that flows from the emitter to the collector. The difference is the base current. The current in a PN junction is based on the number of excess carriers in the base.. The excess hole concentration at the edge of a depletion region on the N side of a PN junction can be easily found from the diode equations. pn is equilibrium number of holes BJTs

The excess carriers in the base at the edge of the depletion regions caused by the collector and emitter regions CB is a reverse biased voltage. Assuming emitter and collector regions equally doped. BJTs

These can be simplified by assuming that VEB is much greater than 0.026V and that VCB is much less than V. Using the diffusion equation we can work out the excess hole concentration as a factor of distance away from the junction, into the base, dp(x). Let xn be the variable and it represents distance in the base from the edge of the EB depletion region. The effective width of the base, from depletion region edge to depletion region edge is WB BJTs

So using the diffusion equation, Diffusion length for a hole The solution for an equation of it type, in general is given by (it’s something you’d know or look up). You’d solve this by looking at boundary conditions, in this case at xn=0 and xn=WB BJTs

From this we would get the following values for C1 and C2 Note: Not worth remembering. However complicated this looks, lets make a simplifications. Assume that collector-base strongly reversed bias, then DpC = 0 The constants simplify to BJTs

And the excess distribution equation changes to Note: Not worth remembering. So now that we have the excess carrier distribution in the base we can work out the current due to holes in the base. The emitter-base current will be that at x=0 and the collector current with be that at x=WB. The difference is the base current. Note: other format. BJTs

The current in the base due to holes is given by So quickly looking at what the differential of the excess carriers is with respect to xn BJTs

So back to the current equation. The current at the emitter edge of the base is when xn=0 The current at the collector edge of the base is when xn=0 BJTs

If we substitute the values of C1 and C2, we get expressions that are best described in hyperbolic functions. co-sec-h co-tan-h Again not something really worth worrying about in detail, just understanding what we are trying to do is more important. BJTs

So with IC and IE all we need to do now is subtract IC from IE and we’ll get the number of holes that recombined, and hence IB This formula for IB however is more interesting as it gives us a formula in terms of parameters we actually have. BJTs

If we make an assumption that the reverse bias junction is very reverse bias (as we have already assumed elsewhere) then we could say that the minority hole concentration close to the junction on the N side is pretty close to zero, ie Hence we can get the following simplifications. BJTs

Now we can make an even further simplification by approximating tanh(y) with a first-order approximation of y BJTs

So working on that basis, we get This accounts for hole recombination current but not the current injected back from the base to the emitter across the forward biased PN junction. However this can be kept small and neglected. BJTs

We can see from this that the base current, which we want to be small as it reduces a and b, is inversely proportional to tp. So if we can increase tp we could decrease IB and corresponding increase b. tp is the time between a hole finding an electron, so if there are less free electrons then tp will go up. Thus reducing the doping in the N-type material increases tp and reduces IB. Making the N in the middle of the sandwich very lightly doped reduces recombination and improves the transistor performance. BJTs

A term which we’ll be investigating further is WB, the effective width of the base, ignoring the depletion regions. Increasing the length increases IB as it clearly would increase the recombination rate. Normally the base is made very narrow compared to the diffusion length, LP, to keep the recombination rate and thus IB low. Another important parameter is A. Modifying the area of the junctions will decrease the base current. In practice the cross-sectional area of the channel is different across the device so as to optimise performance. BJTs

To get b and a, we need just to divide the two relevant currents. Now these equations were defined for the normal operational mode of the transistor but they are cumbersome and where a range of assumptions have been made that are not true in real devices. This analysis is important in that we can see how different parameters can effect the overall performance of the system, in particular, t, the carrier lifetime. BJTs

Coupled Diode Equations (Ebers Moll)
This all works fine assuming that the collector junction is reversed biased. However this does not have to be the case though it is the normal mode of operation. When it is not the case, we cannot discount the excess hole concentration pc in the base if this junction is forward biased. If this is the case then we need to consider more generalised expressions for the currents across the junctions. This approach uses the idea of coupled diodes and was first used by J.J. Ebers and J.L. Moll. BJTs

Consider in general that the current flowing across each junction can be split into a superposition of currents. As before we only need consider the hole current as it’s a PNP device. We call the currents in the expected directions, normal currents. IEN: the current from emitter to base. ICN: the current from base to collector. On a similar basis, current components flowing in the opposite direction are called inverted currents. IEI: the current from base to emitter. ICI: the current from collector to base. BJTs

These current components can be found from: providing that the cross-sectional area of the device is constant. Let us also assume that the doping in the collector and the emitter is the same. BJTs

From these equations we can define some constants: Using these, the current components can now be expressed as Minus sign because they are flowing in the opposite direction to our expected current flows BJTs

Combining the different current components. Looking at IE a little more closely, we get that BJTs

Similarly we can express IC in a similar form. Now these equations are perfectly fine for symmetric devices (constant area, equal doping) but to factor in asymmetric devices. Consider IC when pE=0, This would be the collector current when VEB=0. ICS would be a number representing the collector current at that time. Similarly for IE BJTs

Now if we define a new term  for the proportion of each current contribution that gets across to the other region, ie the number of carriers from the emitter to the collector, and visa-versa. BJTs

Finally adding all these together BJTs

These are the basic Eber Moll equations and are valid for all configurations and bias voltages. It defines the collector and emitter currents in terms of only four parameters which could be measured. N, I, IES and ICS It can be shown, but we won’t prove it, that BJTs

It is possible to simplify the equations by noting that IEN and ICI are defined by diode equations. This is why they are called the coupled-diode equations. If we use the relationship we can get similarly The Ebers Moll Equations are often expressed in this form as it leads to an easy circuit representation of the result. BJTs

This has a circuit representation as E C IC IE IB B This model is the basis of all the expanded models used in computer SPICE simulations. BJTs

Non-Idealities - Base Narrowing
Now in all our discussions so far we made an assumption that recombination of holes in the base region was small (assuming a PNP device). This could be ensured by lightly doping the base so that there are less free electrons and hence less chance of recombination Keeping the width of the base small compared to the diffusion length. An effect of the light doping is that if the emitter or the collector are heavily doped (which they normally are) then most of the depletion region will be within the base. BJTs

Now the width of the depletion region of the reverse biased collector junction is going to be mainly in the base. As the base is lightly doped it may extend quite a significant degree into the base. The width of the region will vary according to the collector voltage, ie it will vary with VCE. But it needs to be remembered that the length of the base is an important factor in determining the number of holes that recombine. The smaller the distance, the less holes. BJTs

The less recombination, the more efficient the transistor,  goes up, more collector current flows for the same base current. So simply, as VCE increases, for the same base current, the collector current increases. Thus in saturation we have a slope to the curves, and this is nearly a linear relationship. If these linear curves were extrapolated back they would intersect at the same point on the IC=0 axis, at a point which we called -VA, where VA is the Early Voltage BJTs

BJTs

If the depletion region were to grow such that it filled the base, then we’d have a phenomenon called Punch-through. All emitter holes would be collected in the collector. There would be no recombination and hence no transistor effect. This normally doesn’t occur in practice as avalanche failure normally occurs first. It can be avoided by minimising the base-collector reverse bias voltage. BJTs

Non-Idealities - Avalanche Failure
This occurs when the voltage across the reverse bias junction is so great that avalanche failure commences. This produces additional current that feeds into the circuit. The common-emitter configuration is more prone to avalanche failure than common base BJTs

Non-Idealities - Thermal Runaway
In BJT’s the parameters can vary with temperature. In particular  can vary substantially. In silicon or germanium, the lifetime of an excess carrier, , increases with temperature. However mobility decreases by approximately T-3/2. However the net effect is normally that the increased lifetime dominates and  increases. If  increases then for the same IB the collector current increases. BJTs

Non-Idealities - Thermal Runaway
So consider this case. The transistor has been biased such that it has a large IC. This current by the voltage dropped across VCE determines the power dissipated in the BJT transistor. Power equals heat, if the heat cannot dissipate quickly enough then the silicon warms up. The silicon warms up,  increases, so the collector current increases. This means that more power is produced and hence more heat. This increases  until so much heat is produced that the device fails. Graphically.. It smokes… This is a BJT effect and has to be designed against in the circuit using the BJT. Emitter degenerated Common Emitter circuits are immune to this. BJTs

Large Signal Parasitics
N Emitter P Base N Collector Before we look at small signal, there are some large signal non-idealities that should be considered. Any current flowing through a piece of semiconductor experiences a resistance. Any PN junction also includes a capacitor. So what we will do on the next page is indicate where these parasitics are. BJTs

Large Signal Parasitics
N Emitter P Base N Collector RC Parasitic junction capacitances. Less and smaller parasitics than in MOSFETs. CjC C Rb B CjE E RE BJTs

Small Signal Model N Emitter P Base N Collector + vBE - vCE iB iC hi
The small signal models of the BJT that are most commonly used are the hybrid  or the H-parameter models. These are very similar to each other. The simplified ideal H-parameter model is shown below. + vBE - vCE iB iC hi ho hf iB B E C BJTs

Small Signal Model N Emitter P Base N Collector
rin/hie the “equivalent” resistance across the forward biased junction and also the base and emitter bulk resistances. Very low. ho/ro the equivalent” resistance across the reverse biased junction and also the collector, base and emitter bulk resistances. Very high due to reverse bias junction. hfe/ is a value that is heavily dependent on the emitter efficiency and the base transport factor. Ideally high. BJTs

Parasitics in the Small Signal Model
N Emitter P Base N Collector The main parasitics are the capacitors and the emitter resistance. The emitter resistance can be ignored as well in many analyses. iB iC Cjc C B Cje + vBE - + vCE - hi hf iB ho E BJTs

Example Assuming that the PNP device is biased as per normal with an emitter base voltage of +1 Volt, calculate the current transport factor, , and the current gain term, b. Calculate the change in b for an emitter base voltage of +2 Volt. Assume the Base transport factor B is ideal and equal to one. Device cross-sectional Area = 10-4 /cm2 Effective width of base region = 10-4 cm ni2 = /cm3 BJTs

Example BJTs

Example Explain the process of base narrowing. Assuming that the forward bias region has a negligible depletion region within the base, determine the voltage at which punch-through occurs in the following NPN device at degC (a hot but still functioning transistor). Device cross-sectional Area = 10-4 /cm2 Width of base region = 10-3 cm ni2 = /cm3 BJTs

Example BJTs

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