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ES330 Laboratory Experiment No. 1 NPN Common-Emitter Amplifier [Reference: Section 7.5.2 of Sedra & Smith (pp. 470-471)] Objectives: 1. Design the amplifier for voltage gain AV to be at a minimum of -100 (V/V) and choose resistor values of RC and RE by calculation. 2. Measure the voltage gain of the amplifier to see how it compares with your calculated voltage gain. Display waveforms on an oscilloscope. 3. Measure the output resistance RO of the amplifier looking into the output port. 4. Replace NPN transistor with a different device and see what changes. [extra-credit] Materials: 1. Breadboard 2. One NPN transistor – type 2N3904 3. Three large 47 microfarad capacitors 4. Several resistors of various values (two resistors are 10 k in value) 5. Jumper wires for use on breadboard 6. Function generator, digital multimeter and oscilloscope Circuit Schematic: 1 Circuit Parameters: The parameters of the circuit are listed in the table below. Component Value CC1, CC2 and CE 47 F each NPN BJT 2N3904 V+ = (-V_ ) 10 volts (nominal) Resistors RB and RL 10 k each Signal Generator Rsig 50 or 600 Not specified in the table are values for the emitter resistor RE and collector resistor RC. These you will determine using specified design goals, namely, the collector current and the voltage gain of the amplifier. The minimum voltage gain AV of the amplifier is to be at least -100 (V/V) in magnitude – the minus sign indicates that the common-emitter BJT amplifier is inverting (i.e., produces a 180 degree phase shift in the output waveform). In addition, we want the DC collector current IC equal to 1 milliampere (1 mA). AV -100 (V/V) and IC = 1 mA Transistor Parameters: Nominal parameters for the 2N3904 NPN bipolar transistor can either be measured using the transistor you selected, or estimated from the 2N3904 Data Sheet. The Data Sheet is reproduced in the Appendix. The DC current gain of the transistor () is designated as hFE on the Data Sheet. You should use a nominal value for hFE rather than the minimum or maximum value for hFE for design purposes. We will ignore the transistor’s output resistance ro in the small-signal model of the transistor because it very large for the 2N3904. Determining RE and RC: The value of IC is set by choosing RE because the forward-biased base-emitter junction voltage is close to VBE = 0.7 volt at IC = 1 mA. The transconductance gm is gm IC I (mA) 1 mA C 0.0385 (A/V) . kT / q 26 mV 26 mV Remember the base node potential (VB) is close to ground potential (i.e., 0 volts) if the base current IB flowing in RB is quite small. You should check to see that this is true in your design. 2 In the small-signal hybrid-pi model the parameter r is given by r kT / q IB 26 mV I B (mA) where hFE IC IB The voltage gain AV is then AV gm RC RL The value of resistor RC controls the voltage gain because it is in parallel with resistor RL. However, there is another constraint on resistor RC because the collector-emitter voltage VCE must be large enough to keep the transistor in the forward-active region of operation while allowing for adequate voltage swing along the load line. In other words, be sure to pay attention to the Q-point location of the transistor. Note: There are effectively two load lines – a DC load line which excludes RL and an AC load line where the load line resistance is RC and RL in parallel. DC Operating Point Analysis: Sketch the DC schematic circuit. The three large capacitors are all open circuits for DC analysis. Also, the signal generator assumes no role in the DC analysis. Begin by determining the values of IB and IE where IC = IB + IE = 1 mA. The nominal power supply V_ = - 10 volts. A. What is the base node voltage VB? VB = ______volt B. Determine the value of RE required to set the collector current at IC = 1 mA. The resistor RE = _______ ohms C. Is the value of RE equal to commonly available resistor values? If not, what can you do to establish IC = 1 mA? Comment on this. AC Analysis: We next determine VCE and RC. Sketch the AC small-signal circuit. The three large capacitors now become short circuits (i.e., the capacitors have negligible impedance at higher frequencies). Also, the power supply connections are assumed to be AC grounded. A. What happens with the emitter resistor RE in the AC analysis? 3 B. Let vi be the small-signal voltage at the base node of the transistor and vsig is the internal voltage of the signal generator. You should recognize that the signal generator is shown in its Thévenin equivalent circuit. Be sure to check the value of resistance Rsig is for the signal generator (i.e., function generator) you use in this laboratory. Note that we use vi in the voltage gain expression for the amplifier. Estimate of vi/vsig = ____________ C. Derive an expression for AV = vo/vi, where vo is the output voltage as defined in the schematic circuit on page 1. Find a value for resistor RC giving a voltage gain equal to, or a little larger than, AV = -100 (V/V). Be sure that this value of RC also gives an acceptable value for VCE. What value of VCE do you calculate? VCE = _______ volts D. Now that you know the value of resistor RC, calculate the output resistance Ro at the output node as defined in the schematic circuit. Ro = _______ ohms Prototyping: Now you have reached the point where you can prototype the circuit you designed. It should look something like the one in the photograph. Be sure to keep wire lengths and lead lengths short to prevent unwanted oscillations. 4 Measurements: Make the following measurements: A. Using a digital multimeter, measure the DC voltages at the base node (VB), emitter node (VE) and collector node (VC) of the transistor. B. Using a function generator set its sinusoidal “peak-to-peak” amplitude to 10 mVpk-pk with a frequency of 1,000 Hz (i.e., 1 kHz). This is the small-signal voltage vi driving the amplifier. Now measure the amplifier’s midband voltage gain AV. C. Using an oscilloscope generate plots of vo and vi versus time t. D. Measure the output resistance Ro. You can do this by replacing the 10 k RL load resistor with say a 1 Meg resistor and again measuring the voltage gain. This gives a maximum voltage gain value. By adjusting (i.e., lowering) the value of RL we can find a value of RL such that the voltage gain is one-half the value you found with the 1 Meg resistor. That particular value of RL is equal to the output resistance Ro of the amplifier. E. Increase the sinusoidal “peak-to-peak” amplitude of the function generator to a level where the output sinusoidal waveform distorts. Describe what you see with respect to the distorted waveform’s shape. Post-Measurement Exercise: A. Calculate the values of VBE and VCE at the DC bias from the ones you measured in the lab. How do they compare with the calculated values? B. Compare the measured and the calculated voltage gain values. Explain the difference. C. Compare the measured and calculated output resistance values. Extra-Credit (NPN Bipolar Transistor Substitution) Obtain a different NPN bipolar transistor (not a 2N3904) from the instructor and plug it into the breadboard amplifier. Measure the voltage gain, bias point (VBE and VCE) values, and output resistance Ro. How does it differ from results you got with the 2N3904? APPENDIX (2 pages) 5 6 7