Figure 1. Simple Transistor Model with Resistive Loads and Bias Circuit.
The circuit shown in Error! Reference source not found. is a simplified model of a bipolar junction transistor (BJT). The portion of the circuit inside the dotted rectangle is a Thévenin equivalent circuit of the biasing circuit. The base, emitter, and collector terminals are denoted by , , and respectively. The base-emitter diode junction is modeled by a constant voltage drop, the source , which for most silicon transistors is about 0.6 Volts.
A typical v-i curve for a BJT is shown in Error! Reference source not found.. The transistor model used in Error! Reference source not found. applies for the transistor operating between cutoff and saturation. The purpose of the bias circuit is to provide an operating point for which the model is valid. The operating point must be far enough from saturation and cutoff so that signals of magnitude for which the circuit is designed will not cause the transistor to be in saturation or cutoff during part of the waveform.
Figure 2. Idealized v-i Curves and Load Line for a BJT Transistor.
The nodes in the circuit of Error! Reference source not found.are the base, emitter, and collector nodes. These are denoted on the schematic by , and , respectively. Note that the emitter and base are a supernode when the constant voltage drop model is used for the base-emitter diode. The node equations are
Ohm's law equations for the base, emitter, and collector currents are
Substituting the Ohm's law forms for the currents into the KCL equations
Putting these equations into matrix form gives us
Note that the top row has no dependence on the collector voltage . This is because the current source presents infinite impedance between the collector model and the emitter and base portion of this simple transistor model. As a result, we can solve for the emitter voltage directly and use it in the equation for the collector circuit to find the collector voltage .
Now we solve for the emitter voltage
and the collector voltage
The base current is found by Ohm's law as
We substitute from into and obtain
This circuit is similar to those considered in Example 1.3, pages 10-12, and homework problem 1.65 on page 33. The principal difference is that the transistor model in Figure 1Error! Reference source not found. models the base-emitter junction as a voltage drop rather than a resistance. The effective dynamic resistance of the base-emitter junction is modeled here as part of the Thévenin resistance in the bias circuit.
Looking at the components connected to the emitter using the KCL, the emitter current is the sum of the base current and the collector current . Since, in this idealized transistor model, the collector current is the current gain times the base current, we have
With the base current we have the collector current as
The v-i curves of the transistor are for the collector-emitter voltage, which we find from and to be
The load line is a straight line on the v=i curve that reflects the Thévenin equivalent circuit made up of the supply voltage and the collector circuit resistance . The intercepts on the v-i plot are the supply voltage on the voltage axis and on the current axis.
The circuit can operate as a linear amplifier only when the operating point is on the load line in a region between saturation and cutoff.
An important addition to the load line is a curve of allowable power dissipation for the transistor. This is a hyperbola with asymptotes of the v and i axes, and the region between this curve and the axes represents the allowable operating region for the transistor without exceeding the power dissipation specified. The equation for the allowed region is
and is an important part of the design process. The load line can cross this curve and part of the operation of the transistor may be outside the allowable region, but the quiescent (zero signal) operating point must be in the allowed region.
We have the Thévenin equivalent circuit for the power supply voltage source and collector resistor. There is a Norton equivalent circuit for the transistor side of the collector circuit: the controlled current source. The Norton equivalent resistance is the dynamic collector resistance of the transistor, which we neglect in the simple transistor model that we are using here. In an actual transistor, the horizontal lines in the i-v curves for the transistor at different base currents will have a slight upward slope, corresponding to a high but finite dynamic collector resistance. Typical values of this parameter for BJTs are to a few megohms.
The circuit of Figure 1can be used as a model of an emitter follower by omitting and taking its value as zero in the circuit solution. The open circuit voltage at the emitter is given by and the short circuit current is given by taking as zero in ,
The Thévenin resistance is the ratio of the open circuit voltage as given by and the short circuit current as given by ,
An interesting way that the Thévenin resistance can be viewed is as the parallel combination of the emitter resistance and the base resistance divided by ,
Note that the reciprocals of both Thévenin equivalent resistances are seen as the diagonal terms of the matrix in .
The inverting amplifier voltage gain can be obtained from by noting that a signal added to the bias voltage will result in a signal component in the collector voltage, mapped linearly through the dependence of the collector voltage on the bias voltage. Thus the inverting amplifier voltage gain is
Note that the emitter resistor can limit the signal gain, which can never reach . This gain limit can be incorporated into the design as a way to define gain independently of or it may be eliminated, at least partially, by bypassing the emitter resistor with a large capacitor.
The emitter follower voltage gain can be obtained from by noting that a signal added to the bias voltage will result in a signal component in the emitter voltage, mapped linearly through the dependence of the emitter voltage on the bias voltage. Thus the emitter follower voltage gain is
The principal variation between circuits in production is the current gain of the transistor, which will vary in a sample population far more than the 5% tolerance used for resistors. We have noted that the operating point of the transistor must be between saturation and cutoff, and maximum voltage swing for the output is achieved when the operating point is halfway between the minimum collector voltage at which the transistor enters the saturation region, and cutoff, when the collector voltage reaches .
A simple indicator of how sensitive the design is to variation in current gain is the sensitivity of to , which is
A good measure of the stability of the circuit is the ratio of the sensitivity of the output voltage to undesired variation in the current gain and desired change in the output voltage to input signal. This ratio is
Inspection of and shows the rationale for several design principles:
Even a small emitter resistor can have a dramatic stabilizing effect on the transistor operating point, particularly for very high current gain . Since high current gain is the most common parameter variation between transistors in a lot, this is a very important design feature.
Decreasing will improve stability in proportion exceeding circuit gain. This is the reason that most transistors are biased through a voltage divider from instead of making equal to .
Keeping the base resistor small improves stability. This is another reason to use a voltage divider from to reduce the value of and thus keep the required value of resistance for small.
Since does not appear in , the choice of collector resistor does not affect stability relative to gain.
Here we will take these parameters as given:
The supply voltage . Here we will use +12 Volts.
The mean transistor current gain and its expected range. Here we will use 300, with tolerances of -50% to +100%.
The remainder of the design is directed toward meeting the requirements of the design. The required performance of the inverting amplifier will vary with the application. Here we will ask for maximum voltage swing on output, and low output impedance. The steps in the design are
Draw a hyperbola on the v-i plot that shows the maximum power dissipation of the transistor that you will allow in your design. This will be below the maximum power dissipation given in the data sheet, and will reflect heat sinks and any dissipation limitations in your expected layout, and your design safety factor.
Find a first design load line, and thus the collector circuit load resistance . The minimum resistance, and thus the minimum output impedance, will be tangent to the parabola that you drew as the locus of maximum allowable power dissipation for the transistor.
Select an operating point on the load line. For maximum voltage swing, this operating point will be midway between saturation and cutoff. The operating point defines a base current from the i-v plots of the transistor, and, with the current gain , a base current. This operating point must satisfy .
The supply voltage is 12 Volts.
The design center current gain is 120.
The maximum allowable transistor power is 50 milliwatts, far below the 2N3905 limit of 625 milliwatts.
The gain is to be set to 50, so the ratio of is set to 50.
We use a forward voltage drop of the base-emitter junction of 0.7 Volts.
We find that a resistance of 720 Ohms will limit transistor power to 50 milliwatts with a 12 Volt power supply. We will use 1000 Ohms for and 50 Ohms for to get our ratio of 50 to 1 to set the gain. We set the operating point so that the emitter voltage is set properly. With the expected collector current and value of resistance for the emitter resistor, we find that a bias current of about 60 microamperes is needed for the nominal current gain of the 2N3904. For a base voltage of about 1 volt, a value of of about is needed for a of 6 Volts. We begin with a voltage divider of two resistors, which offers a Thévenin equivalent voltage of 6 Volts and an equivalent resistance of . We will vary the lower resistor as shown in Figure 3 to obtain a collector voltage of about 5 Volts.
Figure 3. Circuit for BJT Biasing Experiment.
Your lab report will describe the process used to find the circuit that you used. Show why a value of of is used, what the maximum possible transistor power dissipation is, give the v-i curve for the transistor using the simple transistor model presented here, and the load line for . Show the hyperbola that bounds the permitted region for an allowed maximum transistor dissipation of 50 milliwatts. Describe the process you used to find the final values for the voltage divider in the bias circuit. Provide measurements of the voltages on the base and collector.
Measure the gain of the completed inverting amplifier by adding coupling capacitors as shown in Figure 3 and using your signal generator and oscilloscope. Use a voltage divider on your signal generator output if necessary to provide a small enough signal to provide an undistorted output.
Measure the frequency response of your amplifier by finding the 3 dB rolloff points for the high frequency limit.