Biasing Circuits and Transistor Amplifier Operation

Transcription

1 Biasing Circuits and Transistor Amplifier Operation Base Bias Circuit As was previously seen the circuit below turns a transistor on. The circuit now can be referred to as Base Bias. Operation Initially with no base current flowing the output or collector voltage V OUT = V C = V CC the supply voltage. The resistor R B is connected from V CC to the base of the transistor. Base current will flow in the transistor causing the transistor to turn on. The base-emitter junction in the transistor is just like a diode. Transistor On Base-Emitter voltage V BE = 0.7 V The base current causes a collector current to flow I C = β x I B Current flow through the collector resistor R C will cause a voltage drop on this resistor of V RC and the V OUT = V C will start to decrease from the value of V CC the supply voltage. This follows from KVL. V CC - V RC - V C = 0 so V C = V CC - V RC Q Point This term is used to describe the values of I C and V CE for a biased-on transistor and they are referred to as I CQ and V CEQ. Q Point Values I CQ and V CEQ

2 Example Calculate the values of I B, V BE, I C and V C. Β = 150 I B = (12 V V)/ 270 kω = 42 µa V RC = I C x R C = 6.28 ma x 1 kω = 6.28 V I C = β x I B = 150 x 42 µa = 6.28 ma V C = 12 V 6.28 V = 5.72 V The Q Point values for this circuit are I CQ = 6.28 ma and V CEQ = 5.72 V The analysis and example above show clearly that the Q Point values are dependent on the value of β. Recall from previous notes that the value of β is highly variable. The result is that if we were to construct a large number of the above Base Bias circuits (in manufacturing) the Q Point values I CQ and V CEQ for the circuit would vary widely. Transistor Amplifier Basics What does an amplifier do? An amplifier takes a very small AC input or signal voltage and makes it much larger at the output. Consider the case where you plug your ipod into an amplifier so you can play its music through external speakers at a party. The audio output signal from the ipod may only be a few hundred mv, not nearly

3 enough to drive a pair of speakers. The amplifiers job is raise the few hundred mv to 10 or 20 V or more without changing the shape of the waveform in any way. Most audio amplifiers are integrated circuit types these days, but we will look at a basic transistor amplifier (which is what you have inside an Integrated Circuit). Amplifier Distortion The other important job that an amplifier is concerned with (other than making the output signal larger than the input signal) is to be sure that the input and output signals look the same that there is no distortion in the output signal. Distortion in the output signal if it is music - can cause the music to sound badly and not as the artist intended it to. No Bias Circuit Operation Let s examine the case where the transistor is not turned on by a Bias circuit and then apply an AC waveform at the input. Notice that there is no bias resistor connected to a DC voltage source so the DC base current I B will be 0, the DC collector current I C will be 0, the DC output voltage V CE will be V CC. What happens if an AC signal voltage is applied at V IN as shown? The positive half cycle of the input voltage will cause base current to flow in RB and the Base-Emitter junction. This will cause collector current flow and the collector-emitter voltage will decrease from a value of V CC. On the negative half cycle of the input voltage the transistor already being in the Off state will not turn on and the collectoremitter voltage will stay at a value of V CC. Input and Output Amplifier Waveforms The most significant observation from these waveforms is that the Input and Output waveforms do not look the same the output is distorted because the transistor has not been turned on by a Bias circuit.

4 Amplifier Operation with a Bias Circuit Consider the circuit below that uses base bias and has an AC signal of 200 mv P-P connected at V IN. β = 100 Here is a very basic transistor amplifier using a Base Bias circuit that we discussed earlier. It has some serious limitations as a practical amplifier but will serve to illustrate how the transistor amplifies a small input signal. Recall that the base current I B was determined as: I B = (V CC - V BE )/R B = (10 V V)/186 kω = 50 µa And I C = β x I B = 100 x 50 µa = 5.0 ma And And V out = V C = V CE = V CC I C x R C = 10 V (5 ma)(1 kω) = 5V V BE = 0.7 V

5 The Q Point values for this circuit are I CQ = 5.0 ma and V CEQ = 5.0 V These values are the DC Bias conditions for the circuit. Biasing is required so that the transistor can react to small input voltages which are not large enough to turn on the transistor (recall that the transistor starts to conduct at approximately 0.5 V - the voltage at the base emitter - and the normal base emitter operating voltage is about 0.7 V). An approximation of the I B vs V BE response curve looks something like this for the transistor input I B 50µa 0.7V V BE If a 200 mv P-P sine wave is applied to the base emitter junction and when the sine wave is at its positive peak of 100 mv V BE will now be 0.8 V (0.7 V V) The input curve shows that base current will rise. Let us assume it rises to 80 µa. The collector current I C will rise to β x I B = 100 x 80 µa = 8 ma. The output voltage V CE will decrease to V CE = V CC (I C x R C ) = 10V (8 ma)(1kω) = 2V By a similar procedure, when V IN goes to -100 mv, V BE drops to 0.6 V so that I B falls to about 20 µa. Therefore I C will decrease to 2 ma and V CE will increase to 8 V. To put this together we have for the AC voltages and currents: V IN,P-P = 200 mv I B,P-P = 60 µa I C,P-P = 6 ma V CE,P-P = 6 V

6 Amplifier Waveforms Current, Voltage and Power Gain Gain represents how much bigger (or smaller) and output quantity is than the corresponding input quantity. Gains can be greater than 1 or less than 1 (a loss). Gain is represented by the letter A and is subscripted according to type current, voltage or power The AC Current gain β AC (also called A I ) is found as A I = β AC = I C /I B = 6 ma/60 µa = 100 The Voltage Gain A V is found as A V = V CE,P-P /V IN,P-P = 6 V/200 mv = 30 The Power Gain A P = Current Gain x Voltage Gain A P = 100 x 30 = 3000 There are a few things you should notice about this basic amplifier: The output voltage is 180 out of phase with the input voltage (not a problem) The input and output waveforms have the same shape no distortion. This is a good thing

7 The transistor is a current controlled device, we apply an input voltage, but it is the resulting change in base current that causes the collector current to change. The output voltage is changed as a result of the changed collector current flowing through collector resistor R C The output voltage is a mixture of AC and DC since the average V CE is the bias voltage of 5V. V OUT is a 6 V P-P sine wave added to 5 V DC. The output sine wave changes from 2 V to 8 V, not from -3 V to +3 V (this IS a problem) What the transistor is actually doing is controlling the flow of current from the DC supply as the input signal varies. The maximum peak to peak output voltage possible is 10 V P-P (or in other words it is set by V CC ). We set the bias point at 5 V in the middle of the output range so that the transistor output can swing an equal amount in both directions (from 5V up to 10V, and from 5V down to 0V) Importance of the Location of Q Point on DC Load Line Recall from before the drawing of the DC load line. This graph of V CE vs I C shows the Cut-Off and Saturation points. V CE Cut-Off (V CE =V CC, I C =0) Q Point for smaller β Q Point (centered) Q Point for larger β I C Saturation (V CE 0, I c,max ) Also shown on the graph is the location of the Q Point for a bias circuit which represents a particular set of V CEQ and I Cq values. This location of the Q Point is not unique if the value of β changes the Q Point will move up or down the load line. The ideal location for the Q Point is centered on the DC load line. When an AC signal is applied to the bias circuit - to use the circuit as an amplifier - the instantaneous value of V OUT (V CE ) moves up and down the DC load line from the location of the Q Point. The output voltage is a symmetrical signal (to avoid

8 clipping distortion) and must move equally up or down the load line. A maximum output voltage will be achieved when the position of the Q Point is centered on the load line. Base Bias Dependence on β The bias circuit we used is called a fixed Base Bias circuit because the base current is I B = (V CC - V BE )/R B All of these are constant so if a replacement transistor with a different value of β is plugged in the base current is still 50 µa but the collector current I C and V CE Q point bias values will be different because of the variability in β. If, for example a different transistor where β = 180 is used the new I C will be 180 x 50 µa = 9 ma and the new V CE will be 10 V (9 ma)(1 kω) = 1V. This means that V OUT can only fall 1 V and because we must have symmetry in output for most waveforms, we are limited to 1V P-P as our maximum output. We need a bias circuit that is not affected by variations in β. Temperature Sensitivity The other big problem is that transistors are quite sensitive to variations in temperature and if temperature rises, for example, the collector current will increase as more electrons in the transistor structure will pick up enough energy to break away from atoms and add to the current flow. The effect is the same as an increase in β and the bias values will drift upward. A bias circuit is needed which is immune to temperature change as well. Note: This discussion about Q Point, the DC load line and the maximum output voltage is not taking into account the concept of an AC load line.