Will need to use provided data sheets for the components
tran4.ps Project 2 Audio Amplifier September 4, 2018 1 Introduction In the first project we used transistors as switches. In digital circuits, such as computers, that is the way transistors are used. When you use transistors that way, you either have them turned fully on (saturated) or fully off (cut-off). ON-OFF-ON-OFF-ON-OFF-ON-OFF-ON-OFF-ON-OFF-ON-OFF. Gets pretty boring after a while. However there is also a whole region between saturation and cut-off in which transistors can be used to do other things. This region is called the active region. (Cue spooky music from the old TV series ”The Twilight Zone”). In the active region the transistor is neither fully on nor fully off. It is half on and half off at the same time. The collector-emitter voltage VCE and the collector current IC are both not zero at the same time. That means the transistor is dissipating power P = VCEIC just like a resistor. That sounds a bit boring because we could probably just use an ordinary resistor. Except that the value of this ”transistor-resistor” can be adjusted by changing the base current IB . Now we have a variable resistor that we can control electronically, instead of turning a knob. (By the way, the name ”transistor” is a short form of the original name ”transfer-resistor”. How is that useful? Well, it means that we can use voltages to drive the base current to vary the resistance. If we put another resistor in the collector circuit (in series with the transistor ”resistance”) that forms a voltage divider. If there is a fixed voltage across the voltage divider then the voltage at the middle (on the collector) will vary up and down in the same way that the (signal) voltage driving the current into the base varies up and down. Now we already know that a transistor has gain, so that more current flows through the collector than we drive into the base. That means the signal (up and down) voltage we get on the collector (at the centre of the voltage divider) is bigger than the signal voltage that we use to drive the base circuit. So we can use transistors to amplify signals (such as the signal from a microphone when you speak into it) from micro-volts or milli-volts up to volts, giving enough signal to drive a speaker or do something else useful. You voice is an analogue signal, not a digital signal. When we use transistors in the active region we are doing analogue signal processing. Transistors can be used to do both digital signal processing and analogue signal processing. There are not many electronic devices today that don’t use transistors. (Before transistors existed we used valves. But that is another story). 2 Amplifying Analogue Signals 2.1 The Simplest Circuit Figure 1 shows one of the simplest transistor circuits for amplifying analogue signals. The resistor RL plays the role of the top resistor in the voltage divider, and the part of the circuit from collector to emitter through the transistor plays the role of the bottom part of the voltage divider. Changing the base current IB changes the effective resistance of the circuit from collector to emitter. When this happens the current through the collector changes, and the voltage on the collector (at the centre of the voltage divider) also changes. If the (fixed) voltage supply is VS then we can used Kirchhoff’s voltage law to add up all of the voltage dropped as current flows from the supply (VS) through the resistor (minus ICRL) and through the transistor (minus VCE) to ground (zero). If we write that as an equation we get: VS − ICRL − VCE = 0 (1) 1 Q1 2N3904 RL 100 power out in ground Figure 1: Simple transistor amplifier circuit. This gives the collector current in terms of the other variables as: IC = VS − VCE RL (2) If we are designing the circuit, we can choose the voltage supply VS and the collector resistor RL. But we can’t calculate the collector current unless we know VCE . 2.2 The Load Line Transistors are non-linear devices, and there is no convenient mathematical equation to describe how they behave. It is much easier to represent their behaviour in terms of a diagram which relates the voltages and currents on the base, emitter and collector to one another. Figure 2 shows the relationship between collector current IC , base current IB and collector-emitter voltage VCE . Diagrams like figure 2 can be found in transistor datasheets from some manufacturers. (The diagram I am showing here is not for any particular transistor, but the shapes of the curves are more or less correct. You need to use the diagram for whatever transistor you are using in your circuit). In order to solve for the collector current in our simple amplifier circuit we need to combine equation 2 and figure 2. We do this graphically, by converting equation 2 into another line drawn onto figure 2. Looking at equation 2 we might realise this represents a straight line y = mx+ c where y = IC , m = −1/RL, X = VCE and c = VS/RL, going through point #1: IC = 0 when VS = VCE , and going through point #2: IC = VS/RL when VCE = 0. You can check that those points are on the line be substituting VS = VCE and VCE = 0 into equation 2. Plotting the two points #1 and #2 on figure 2 and drawing a straight line between them produces the extra line sloping down to the right in figure 3. This new straight line, which represents Kirchhoff’s voltage law, is called the ”load line”. 2.3 The Operating Point Q The solution for the collector current IC must lie somewhere along the load line (otherwise we are breaking Kirchhoff’s voltage law). The exact solution depends on the value of the base current. Choose any base current and find where it intersects the load line. That is your solution. That point is called the ”operating point” or ”quiescent point” Q. Notice that you don’t have to use any nasty equations. If we choose a low value of base current, the operating point is somewhere to the bottom right on the load line. If we increase the base current, the operating point moves further left and up the load line. Notice that as we change the base current by a small amount, the collector current changes by a much greater amount. That means we can amplify signals. 2 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0 1 2 3 4 5 IB=0.1mA 0.2 0.3 0.4 0.5 0.6 0.7 co lle ct or c ur re nt I C ( am ps ) collector-emitter voltage VCE (volts) base current Figure 2: Transistor characteristics curves. 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0 1 2 3 4 5 IB=0.1mA 0.2 0.3 0.4 0.5 0.6 0.7 co lle ct or c ur re nt I C ( am ps ) collector-emitter voltage VCE (volts) base current static load line Figure 3: Load line with operating point Q for chosen base current. When we drive a signal into the base, causing the base current to swing up and down, we want to stay in the active region of the transistor. If we swing the base current too high we might run up the load line and smack into the highest or leftmost edge where all of the base current lines join together. The transistor is then saturated and the collector current can’t increase any more. We don’t want that. Also, if we swing the base current too low we might run down the load line and smack into the x-axis where the collector current is zero. The transistor is then turned off and the collector current can’t decrease any further. We don’t want that either. So the best place to put the operating point, when there is no signal or when we are in the middle of a signal swinging from one side to the other, is about half way between saturation and cutoff. So all we have to do to keep the transistor operating in the active region is choose the base current from the diagram in the datasheet to put our middle signal value in the middle of the piece of the load line running through the active region. You can read the value of the base current required to do this from the datasheet. 3 2.4 Adding Some Practical Bits Adding a base resistor RB to control the base current, and capacitors to get the (alternating) signal in and out without disturbing the steady state currents flowing in the base and collector circuits, leads to the amplifier circuit shown in figure 4. You can read the collector current at the operating point from the diagram in the datasheet. You don’t need to use the current gain hFE to get the base current IB. You can read the base current at the operating point from the datasheet. You can also read the base-emitter voltage VBE at the quiescent collector current from one of the other figures in the datasheet. Have a look. Then calculate the voltage across the base resistor taking the base-emitter voltage VBE into account, and use Ohm’s law to get the value of RB. You have done all of that before. This is just the same. Well easier really because you have replaced some of the calculations by looking up values on datasheets. C.in 1 µF RB 100 Q1 2N3904 RC 100 C.out 1 µF power out in ground Figure 4: Circuit with base resistor and input and output capacitors added. 2.5 Dynamic Load Line We can get more amplification out of the circuit if we can make the load line steeper. To do that we need to make the collector resistor RL smaller, because the slope of the load line is −1/RL. But if we do that, it allows more current to flow though the collector. That wastes power, and we could overheat and destroy the transistor. However there is a neat trick. Let’s modify the circuit by splitting the load resistor into two parts: RL = RC + RE , leaving RC in the collector circuit, and putting RE in the emitter circuit. Really this doesn’t change anything too much, because the current in the emitter is almost the same as in the collector. (Why? Because the current gain is high so the base current is small and the emitter current IE = IC + IB ≈ IC). The load line is hardly affected at all. So what is the advantage? Well there isn’t one yet. But now let’s put a capacitor in parallel with the (new) emitter