Biasing of BJT Inverting Amplifiers and Emitter Followers

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*      The Base Circuit

*      Solving the Circuit

*      Kirchhoff's Current Law with Node Voltage Notation

*      The Circuit Voltages and Currents

*      The Load Line

*      The Thévenin or Norton Equivalent Circuits for the Emitter and Collector

*      The Norton Equivalent for the Collector of a BJT Inverting Amplifier

*      The Thévenin Equivalent for the Emitter of a BJT Emitter Follower

*      The Voltage Gain for Signals

*      Inverting Amplifier Voltage Gain

*      Emitter Follower Voltage Gain

*      Robustness of the Design for Varying Current Gain for Different Bias Circuits

*      Designing an Inverting Amplifier Stage

*      Experiment

*      The Circuit

*      Lab Report Requirements

The Base Circuit

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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.

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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.

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Figure 2.  Idealized v-i Curves and Load Line for a BJT Transistor.

Figure

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Solving the Circuit

Kirchhoff's Current Law with Node Voltage Notation

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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

                                             

Equation.DSMT4

Equation.DSMT4

Ohm's law equations for the base, emitter, and collector currents are

                                                      

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Equation.DSMT4

Equation.DSMT4

Substituting the Ohm's law forms for the currents into the KCL equations

                                

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Putting these equations into matrix form gives us

                              .                     

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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 .

The Circuit Voltages and Currents

Now we solve for the emitter voltage

                                                                               

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and the collector voltage

                                       .                              

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The base current is found by Ohm's law as

                                                                                                    

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We substitute  from into and obtain

                                                                                                     

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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.

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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

                                                      

Equation.DSMT4

Equation.DSMT4

With the base current  we have the collector current  as

                                                                                         

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The v-i curves of the transistor are for the collector-emitter voltage, which we find from and to be

                                    .                           

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The Load Line

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.

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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.

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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

                                                                                                                      

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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.

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The Thévenin or Norton Equivalent Circuits for the Emitter and Collector

The Norton Equivalent for the Collector of a BJT Inverting Amplifier

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.

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The Thévenin Equivalent for the Emitter of a BJT Emitter Follower

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 ,

                                                   .                                         

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The Thévenin resistance is the ratio of the open circuit voltage as given by and the short circuit current as given by ,

                                                                                            

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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 ,

                                                       .

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Equation.DSMT4

Note that the reciprocals of both Thévenin equivalent resistances are seen as the diagonal terms of the matrix in .

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The Voltage Gain for Signals

Inverting Amplifier Voltage Gain

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

                                                                              

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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.

Emitter Follower Voltage Gain

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

                                              .                                  

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Robustness of the Design for Varying Current Gain for Different Bias Circuits

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 .

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A simple indicator of how sensitive the design is to variation in current gain  is the sensitivity of  to , which is

                                    

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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

                                                    

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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.

Equation.DSMT4

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Designing an Inverting Amplifier Stage

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 .

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Experiment

The Circuit

See Figure 3 below for reference.  Use a 2N3904 transistor with the TO-92 case.  Find a data sheet for this transistor on the Internet.  Use the design process given above.  The design constraints are:

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.

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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.

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Figure 3.  Circuit for BJT Biasing Experiment.

Figure

Lab Report Requirements

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.

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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.

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Measure the frequency response of your amplifier by finding the 3 dB rolloff points for the high frequency limit.

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