Bipolar Junction Transistors

Bipolar Junction Transistors Conotinuing the study of semiconductor devices, we now turn to a more complicated structure of three alternating doped semiconductor regions, better known as the bipolar junction transistor (BJT). The BJT, like all other transistors, 1 is capable of being used as a large-signal (DC) switch or as a small-signal (AC) amplifiers. In this chapter, we will build on our analysis of the diode to characterize the BJT as an ON/OFF DC switch and as a physical amplifier that produces a gain greater than one. In later chapters, we will consider the fundamentals of logic operations using these DC switches. Just as an introduction to the lingo, we will be considering DC bias calculations to obtain the bulk characteristics of the BJT as well as a relatively small superimposed small-signal or AC component that contains information. Physical Structure of the BJT The semiconductor structure of a BJT is that of three alternating semiconductor regions, either npn or pnp. We will perform our analyses for the npn-bjt shown in Figure X, which can be modified easily for the pnp. Figure 1. Bipolar Junction Transistor The three regions of the BJT are labeled base (B), collector (C), and emitter (E) to reflect what their basic operation is in terms of physical charge carriers. You should immediately notice that the BJT s structure is approximately the same as two reversed diodes shown in Figure X; the analysis will use this observation as a starting point. 1 Besides the BJT, electrical engineers also use (Metal-Oxide-Semiconductor) Field Effect Transistors (MOSFETs or just FETs) and Junction Field Effect Transistors (JFETs), each of which may be used as DC switches or AC amplifiers. The physics and DC bias equations are more complicated, so we will focus on the BJT as the prototypical transistor. 1

2 Figure 2 For diodes, we generally considered two possible cases of the diode voltage: V D < 0 Volts (0.7 Volts for the more practical diode) and V D > 0 Volts (again, 0.7 Volts for practical diodes) corresponding to the diode either being ON or OFF. With three semiconductor regions (or two diodes), there are now four possible combinations of voltages between the terminals. We will only consider the forward-active mode, which is when the collector-base voltage, V CB, and the base-emitter voltage, V BE, are both positive (consequently, V CE will be positive as well). Figure 3. BJT Operating in Forward-Active Biasing Region. We will most often treat the base terminal as the input terminal and the collector as the output terminal. Delving into the internal operation of the BJT, we do indeed see diode characteristics from base to emitter (with the same polarity as V BE ) and another diode from base to collector (this time opposite to the polarity of a positive V CB ). The relative magic of the BJT is caused by a physical process called avalanche multiplication across the p-type semiconductor region. When operated with DC voltages in the forward-active mode, electrons are excited along the edge of the emitter region with enough energy to overcome the base-emitter electric field, 2 and hop into the base region. Once in the base region, the electrons are quickly propelled through the base due to the reverse-biased collector base junction; the overall effect is a flow of electrons being emitted by the Emitter and collected by the Collector. If you are still having trouble picturing these descriptions, consider the quasimechanical diagram in Figure X. 2 Recall that the position of an individual electron is a question for probability rather than deterministic equations.

3 Figure 4 We draw loosely-held marbles in an otherwise flat board to represent the electrons in a the emitter region, a flat base region filled with holes where the marbles can fall into and become trapped, and then a strongly sloping collector region (due to the reverse-biased base-collector junction). When enough energy is given to the electrons in the emitter, they will be propelled towards the base region with enough momentum to pass straight through the base (although a small number will become trapped in the holes) and fly through to the collector. Also, you an imagine these mobile marbles (electrons) colliding with other electrons, creating an avalanche of carriers over the junction between regions. You should notice that the voltage polarities are seemingly labeled backwards: gravity makes objects fall down, while an electron in an electric field actually falls up. A more careful analysis 3 will show that the number of electrons crossing into the base region and then directly into the collector is a constant factor β, known as the BJT s performance parameter, of the number of electrons exiting through the base, measured as I B. Typical β s of good BJTs range from about 50 to 200, so the positive current from collector to emitter (flow of electrons from emitter to collector) will largely dominate the base current. The nearly-completed model for the BJT is shown in Figure X. Figure 5 The currents in each of the diodes and the controlled current source may be written using the reverse saturation currents of each of the diodes. 3 Refer to Pierret s Semiconductor Device Fundamentals

4 I BE = I 0BE (e V BE V T 1) I BC = I 0BC (e V BC V T 1) 0 I CE = β I 0BE The final modification is to notice that the semiconductor junction (diode) between base and collector is reverse-biased and replace it by an open circuit (diode turned OFF). The final model of the BJT operated in forward-active mode is shown in Figure X. Figure 6. Block equivalent of a BJT. DC Model of the BJT The first task we undertake whenever introducing any new electrical device is to obtain the IV characteristics; for a BJT, we will be interested in how the collector current, I C, changes with respect to the voltages V BE and V CE. The voltage V BE sets the current into the base, I B, while the voltage V CE controls the width of the semiconductor junctions and as a result contributes to the current from collector to emitter. The classic IV curve for the BJT appears in Figure X (the multiple curves are for different values of V BE ). Figure 7

5 The characteristics to notice are the ratio between the collector current and the base current, which we earlier termed the performance parameter β, and the small degree of inclination in the forward active region (V CE > V BE 0.7V ). This slope actually represents a non-ideality of the resistor corresponding to the AC output resistance between collector and emitter. We quantify this slope by the imaginary voltage intercept created by extrapolating the line far to the left. Calculation of this Early voltage will later allow us to calculate AC characteristics of the device. Figure 8. Calculation of the Early Voltage, V A Example: Calculate the performance parameter and the Early voltage of a BJT with the IV curve shown in Figure X. Figure 9 The first and easiest to determine is the performance parameter, β. By estimating the average value of I C to be roughly 2.2 ma, we calculate β as the ratio of I C to I B. β = I C 2.2 ma I B 35 µa = 63 To calculate the early voltage, we first determine the slope of I C versus V CE

6 0.2 ma slope = 0.02 ma 10 V V and then extrapolate the line to the left to find the imaginary intercept. I C 2.2 ma = V A A Simple DC BJT Circuit (2.2 ma) (0.02 ma/v ) = 110 V In the last chapter, we saw that the diode could be treated as a type of voltageinduced switch; the BJT may also be used as a DC switch, but we will have the added flexibility of a control voltage that is not in the same series voltage loop as the switch. The simplest circuit in which we can see the BJTs response as a switch is the BJT inverter in Figure X. Figure 10. BJT Inverter Do not be fooled by the way that the circuit is drawn in Figure X: the bottom node has a ground symbol signifying 0 volts, while V DD at the top represents a fixed DC potential (must be positive if in forward-active mode). The equivalent method of drawing the BJT with obvious loops is shown in Figure X; you should quickly realize how much easier Figure X is to draw than Figure X. Figure 11. BJT Inverter Testing the extreme values of input voltages will allow us to see that the current flow in the BJT can be turned ON or OFF depending on the base voltage. When V B is set to zero, the diode voltage from base to emitter is zero, which corresponds to a flow of zero current. The controlled source between collector and emitter will also produce zero current, forcing the current through, and therefore the voltage dropped across, the resistor to be zero. As a result, V out = V DD. When V B is set to V DD, the BJT will try to transmit as much current as the rest of the system will allow the maximum current that may flow from collector to emitter (and consequently through the resistor) is VDD R. Such a large current flow forces the entire voltage V DD to be dropped across the resistor, leaving an output voltage of zero.

7 V out = { V DD V in = 0 0 V in = V DD The output voltage (a real, physical quantity) must be continuous, so we require that there be some input voltage between 0 and V DD where the output transitions between V DD and 0, or simply high and low. These input/output relations also show an interesting new tool for circuit design: a voltage-controlled switch. The full input/output characteristics of the BJT inverter with V DD = 15 V appear in Figure X. Figure 12.. You should notice that the calculated extreme points coincide well with the plot and previous approximations: the output transition from high to low occurs at an input voltage in the neighborhood of 0.7 volts (where the diode-like base-emitter junction turns ON). Further, calculation of the base current will show that I B never exceeds the ratio VDD βr which is typically on the order of a few µa. Taking these results as a general set of assumptions for all BJTs, we will always use the approximations V BE = 0.7 Volts and I B = 0 Amps (a BJT with 0 < V BE 0.7 V would not conduct much current and would therefore be considered as OFF the proper name is cutoff mode rather than forward-active).

8 Figure 13.. AC Model of the BJT Using the block-level model of the BJT in Figure XX, we may characterize an AC response whenever the device is ON (operating in forward active) corresponding to the addition of a small signal voltage to the DC base voltage. Take V B to be the superposition of a DC voltage ( 0.7 Volts to turn the diode ON ) and a small sinusoidal signal, V B (magnitude < 5 mv). The current into a diode, and consequently the base of the transistor, can be written using the linear Taylor approximation, I B = I B (V BDC )(1 + V B ) = I B (V BDC ) + I B (V BDC ) V B V t V t from which we separate the AC portion of the base current I B = I B (V BDC ) V B V t The corresponding collector current in the controlled source will be the same AC base current multiplied by the factor β. By transformming these characterizations into a linear circuit, we obtain the circuit shown in Figure X as the AC model of the transistor (comomnly known as the hybrid-π model). Figure 14.. The various parts of this AC model can be explained easily (all quantities X are the small signal perturbations of the DC quantities, X): g m is the transconductance parameter that ultimately converts between changes in base voltage and changes in collector current. Mathematically, g m = I C = β I B = βi B(V ) BDC = I C V B V B V t V t r π is the input resistance that relates changes in base voltage to base current.

9 r π = V B V t = I B I B (V BDC ) = β V t = β I C g m r o is an output resistance that accounts for imperfections of the model in Figure XX and for the physical limitations of semiconductor manufacturing. The formula for this output resistance is r o = V A + V CE I C, where V A is the Early voltage and V CE = V C V E is the difference in DC voltages between collector to emitter. The intuition underlying r o is grounded in the fundamental device physics of the BJT, 4 affected largely by the control of silicon and doping impurities. The numerical values of V A can be anywhere from 50-200 Volts, corresponding to r o values on the order of 100 kω. The non-idealities of a transistor can be explored far beyond those presented above, but do not generally have a large effect. Hand analysis using the hybrid-π model is the typical example of an engineering trade-off: simpler analyses are used to obtain rough, but quick, solutions to hard problems. More precise answers are then obtained by using a computer. To summarize the basic AC parameters in terms of the DC bias points: g m = I C V t r π = β g m r o = V A + V CE I C Single-Stage BJT Amplifier Now that we have characterized the transistor for both DC and AC voltages, we are ready to build a single-stage BJT audio amplifier. The schematic in Figure XX shows such an amplifier. 4 A good reference is Grey & Meyer

10 Figure 15.. We will assume that the values of the DC source voltage, V DD, Early voltage, V A, the BJTs performance parameter, β, and all the resistors (typically measured in kωs) are all given. Further, we will generally disregard the specific values of the capacitors C 1, C 2, and C E, since they are generally large (perhaps a few hundred µf ) and used only to block the flow of DC currents while freely admitting the flow of AC currents (within a certain range of frequencies). We can verify this property mathematically by calculating the magnitudes of the capacitor impedances for the extreme values of frequencies in the audio range (20 Hz to 20 khz) and comparing them to the impedances for DC signals (corresponding to ω = 0). Example: Show that each of the capacitors in Figure X may be approximated as DC open circuits and AC short circuits. Assume 330 µf capacitors. To get an idea of the magnitude of the capacitor impedances, we simply plug in a frequency of 0 Hz for DC signals and a range of values from 20 Hz to 20 khz for AC signals. When ω = 0, 1 Z C = jωc = 1 0 = When ω = 20 2π, 1 Z C = jωc = 1 0.0415 = 24 Ω When ω = 20000 2π, 1 Z C = jωc = 1 41.469 = 0.024 Ω The approximation that the capacitors are DC open circuits is extremely good, while the largest resistance value within the audio range (20 Hz - 20 khz) is no more than 25Ω; compared to resistor values in the kω range, this capacitor impedance is near enough to being a short circuit.

11 Figure 16 The other important trick that will come into play is that DC sources are AC zeros; that is, the AC part, or time-varying component, of a DC source (constant) is always zero. This observation suggests that the DC source voltage, V DD, will be a ground node for the AC analysis. DC Analysis The massive amplifier circuit can be broken into three separate circuits (separated by C 1 and C 3 ) for DC analysis. Figure 17. DC Equivalent of the Single-Stage BJT Amplifier. We will calculate: 1. The DC input voltage 2. The DC output voltage 3. The DC bias points of the transistor (all the voltages and currents at the transistor terminals) DC Input voltage The capacitor C 1 blocks DC current from flowing out of the input source V in into the amplifier. Therefore, the DC portion of the input voltage is negligible to other parts of the circuit and will often be taken as zero. If we consider multiple-stage amplifiers, the DC input voltage may not be zero, but will not directly affect the

12 gain. DC Output Voltage The DC output voltage will not affect the actual amplifier block due to C 3, but it may be useful in its own right. We will calculate V outdc as a voltage divider between R out1 and R out2. Figure 18 V outdc = R out2 R out1 + R out2 V DD BJT Bias points We will use the approximations learned from the BJT inverter, namely that I B = 0 and V BE = 0.7, to solve for the DC values of voltages and currents in the amplifier circuit, called the DC bias. The order of solution will always be V B, V E, I E, I C, and V C. We can then re-approximate I B using the preceding solutions. Base Voltage: Based on the initial assumption that I B = 0 and the fact that the DC current left through C 1 is also zero, any current flowing downward in R 1 must also flow through R 2, which means that they are in series for the DC source voltage V DD. The voltage V B is thus obtained by a routine voltage divider between R 1 and R 2.

13 Figure 19.. Emitter Voltage: V B = R 2 R 1 + R 2 V DD To find V E, we need only apply the condition that V BE, the potential difference between the base and emitter, is a constant 700 mv. Figure 20.. V BE = V B V E = 700 mv = V E = V B 0.7 V Emitter Current: I E depends only on the resistor R E and the voltage across it, V E. We see this because any current flowing out of the emitter must flow downwards through

14 R E (since no DC current may flow through the capacitor, C E ) creating a voltage V E = I E R E. Figure 21 Collector Current: I E = V E R E We obtain I C by considering the BJT as a generic three-terminal device and applying KCL. The assumption that I B = 0 forces the collector and emitter currents to be identical. Figure 22 Collector Voltage: I E = I B + I C = I C The entire current entering the collector must go through R C in a similar fashion to the relationship between I E and R E ; this current creates a voltage of I C R C across

15 the resistor R C. The collector voltage is then equal to the source voltage, V DD, minus the voltage drop across R C. Figure 23 V C = V DD I C R C Base Current revisited: The initial approximation for the base current was zero since it is a (large) factor of β less than the emitter and collector currents. The error introduced is only a few percent, which is certainly acceptable for hand calculations. We may improve the initial approximation by setting I B = I C β AC Analysis Now that we have calculated all of the DC voltages and currents in the circuit, we will move forward to calculating the AC gain this is where the real magic, or at least practical concern, of the BJT as an amplifier becomes evident! We begin the AC analysis by substituting short circuits for the capacitors and redrawing DC voltage sources as AC zeroes (ground nodes). The resulting circuit is shown in Figure X.

16 Figure 24.. We can simplify this circuit slightly by redrawing it with all resistors collected in recognizable forms as shown in Figure X. Figure 25 You should notice that the emitter is now directly connected to ground (R E goes away since it is in parallel with a short circuit). We then make the next substitution, that of the AC model of the BJT, making sure to match up the correct nodes at base, collector, and emitter. Figure 26.. We then collect the resistors outside the BJT and create a single circuit. Notice that R 1 and R 2 are in parallel since they each go from base to emitter (ground node) for the AC circuit, as are R out1 and R out2. Also notice that the output voltage, V out is measured from collector to ground (emitter). The resulting overall AC circuit is shown in Figure X.

17 Figure 27.. Solving this circuit is much easier than it may appear. We break up the total gain, A = Vout V in, into a pair of voltage ratios that represent the effect of each side of the circuit. A = V out V in = V π V in Vout V π V π Starting with the first voltage ratio, V in represents a voltage divider between the parallel combination of R 1, R 2, and r π and the input resistor, R s. V π V in = R 1 R 2 r π R 1 R 2 r π + R s The second voltage ratio works out to just be an Ohm s law calculation between the controlled current source of magnitude g m V π and the parallel set of resistors R C, r o, R out1, and R out2. V out V π = g m (r o R C R out1 R out2 ) The negative sign in the last term shows up because the flow of current in the resistors is opposite the polarity of the output voltage. (Many students mistake a negative gain as the signal getting smaller, thus defeating the purpose of an amplifier. The actual effect is to boost the signal by a factor of A and induce a 180 phase shift.) Combining the two terms, we obtain the overall AC gain of the BJT amplifier. R 1 R 2 r π A = g m (r o R C R out1 R out2 ) R 1 R 2 r π + R s Example: Plot the output voltage of the single-stage BJT amplifier shown in Figure X. You may assume that the device is operating in the forward active region, and that the BJT s parameters are: V A = 50 V and β = 200. Figure 28 Step 1: Apply assumptions V BE = 0.7 V, short circuits. I B = 0 A, while the capacitors act as DC open circuits and AC Step 2: Solve for the DC bias points. We will repeat the same order as before: V B, V E, I E, I C, V C, andi B.

18 V B = 20kΩ 20kΩ + 80kΩ (20V ) = 4 V I E = 3.3V 3.3kΩ = 1 ma V E = 4V 0.7V = 3.3 V I C = 1 ma V C = 20V 8kΩ 1mA = 12 V I B = 1mA 200 = 5 µa Other DC terms: Along with the DC bias of the transistors, we are also interested in the DC output voltage and the difference of collector and emitter voltages (to calculate r o. 20 kω V CE = 12 V 3.3 V = 8.7 V V outdc = (20 V ) = 10 V 20 kω + 20 kω Step 3: We calculate the BJT s AC parameters using the DC bias points. g m = 1 ma 25.86 mv = 38.7 200 mω 1 r π = 38.7 mω 1 = 5.17 kω r 100 V + 8.7 V o = 1 ma Step 4: AC Model = 108.7 kω We begin by replacing the capacitors with short circuits, and all DC sources by AC zeros (ground nodes). The second step is to substitute the AC model of the transistor, collecting resistors where possible. Solving for the intermediate gains, V out V π Figure 29 V π = 20k 80k 5.17k V in 20k 80k 5.17k + 50 = 0.987 = ( 38.7 m ) (8k 20k 20k 108.7k) = 165 we can obtain the overall AC gain, A. A = V out = V out Vπ = 163 V in V π V in The most common error in any of these calculations is to neglect the 10 3 or 10 3 implied in kω or ma. Step 5: Plot V out V out = V outdc + V outac = V outdc + A V in = 10 0.489 cos(1000t)v Figure 30

19 Exercise: Determine the values of R E and β such that the single-stage amplifier pictured in Figure X has a transconductance of 50 m and a total AC gain of 150. Assume that the Early voltage is infinite, V A =. Figure 31 Comments on the Analysis The hardest part of analyzing BJT amplifiers is to gain an intuition for whether or not the various calculations result in reasonable values. As rules of thumb, you should calculate all DC voltages to be between the upper and lower supply voltages, usually zero and V DD, values of g m measured in m, r π measured in kω, and r o in hundreds of kω, resulting in gains of roughly 50-200 (remember that gains are negative). These values are intentionally vague, but relatively consistent for practical single-stage audio amplifiers. You should also notice that the value of DC bias voltages, and consequently AC parameters and gain, can change by varying component values within the circuit, which is where we turn next. Effects of Varying Component Values Scenario: Your company has completed the manufacture of a major system, but one portion fails because a single-stage BJT amplifier has a gain that is 10% too low. The design was contracted out to another company who is unavailable on the weekend, so nobody can help. To make things worse, the amplifier is on a single chip, so you cannot change any of the component values (resistors or capacitors). What do you do? The only parameter that you still control is the power supply, so you pull a piece of scrap paper and begin playing with equations: when you increase the source voltage, the base voltage goes up, with a corresponding increase in the emitter voltage. That causes the emitter current (and collector current) to also go up, resulting in a larger transconductance. You know that gain is linearly related to g m (and that DC sources are AC zeros independent of value), so you suggest adding a couple of AA batteries out of your walkman in series with the source voltage. Bingo: the system works, and your two AA batteries earn you a nice bonus. Yes, it s a little farfetched. But it is by no means inconceivable: as an engineer of any form, people will expect you to have the answer to everything mechanical, electrical, industrial, etc, and will not always choose to be civil in their requests! The rest of the chapter on BJTs is a fairly simple analysis of what if; that is, what happens to the (gain/power/etc.) when we vary (R 1, R C, etc.). This is more common than it might seem since most practical resistors are only accurate within 5% of their nominal values, and other components may vary widely. Following the example, we saw that an increase in V DD caused increases in V B, V E, I E, I C, and g m. Had we checked, there would have been a decrease in r π and r o, as well as an increase in V outdc. Rather than explain these in words, it is much

20 easier to tabulate them as a positive, negative, or no change. The table below shows the effects of increasing V DD and that of reducing R 1 on the other circuit values. table example You will notice that the gain, A, depends on many terms, some certainly having more significance than others. The table above shows that the overall gain (magnitude) increased despite the fact that r π decreases, causing a marginal decrease in the voltage ratio Vπ V in. The overall gain equation depends more heavily on the increase in the transconductance than the decrease in input resistance (test small changes to prove this to yourself). The process above works well for incremental changes in component values, but what if we drastically change a value? Let s consider R C. Plugging and chugging through equations will work until you obtain a collector voltage of. Clearly, DC voltages should be between the source and ground, so something is wrong! R C means that the collector resistor is being replaced by an open circuit, which precludes current from entering the collector, effectively turning the transistor off as an amplifier. While we will not explore all the possible cases of extremes in component values, you should realize that the equations derived only hold when the BJT is biased in the forward active mode. STUFF TO ADD: One more exercise of one-stage amp, variation on R 2 exercise, variation on R out2 example with percentages, variation on beta exercise, complete table for changes, Questions: 1. For the BJT inverter shown in Figure X, calculate how much power is dissipated in the circuit when the input voltage is 0 V and when it is V DD. 2. Power dissipation is a major concern in battery operated electronics, many of which use an Inverter similar to that of the BJT inverter. What condition should be placed on the series resistance, R, to minimize the power dissipation?