Physics 623Transistor Characteristicsand Single Transistor AmplifierSept. 13, 20061 Purpose• To measure and understand the common emitter transistor characteristic curves.• To use the base current gain (β), and load line analysis to predict and experimentallyto verify the DC operating point (often called the “Q point”) for your transistor in thecommon emitter configuration.• To predict, using the transresistance model, the AC properties of your amplifier circuitand to verify them experimentally.• To distinguish between current and voltage driven base signals.• To understand the effect of emitter resistor by-pass (degenerative f eedback) through atransresistance analysis.2 Procedure1. Characteristic Curves (common emitter)The first step in this experiment is to measure and record the common emitter char-acteristic curves for a silicon NPN 2N1480 transistor. These are diffused junctiontransistors with a fairly small base current gain.(a) Your lab instructor will illustrate the basic principles of operation of the curvetracer. The curve tracer will be used to measure the characteristics of your tran-sistor and to make hard copies for your notebook.(b) Be sure to record the appropriate scale readings for your transistor for later anal-ysis. Note also the wide variation between your transistor and those of yourneighbors. In the Appendix we give the specifications for this transistor as listedby the manufacturer.(c) Compare your measured β = IC/IBto the range given on the spec sheet. Exceptfor Q-point calculations, the generally more useful quantity is the small signalparameter hfe= dIC/dIB. The difference in value is usually not important.2. The “Q point” (or DC quiescent point)So far we have been concerned only with the properties of the transistor itself. Inpractical amplifier applications, it is desirable to design a circuit whose properties arepredictable, i.e. which don’t depend strongly on your accidental choice of a specifictransistor. In Fig. 1 we s how a very common configuration of base and emitter1VinVoutVcc = +15 VFigure 1: Basic Single Transistor Amplifierbiasing for the common emitter amplifier configuration. This circuit is reasonablystable against variations in transistor properties (for β = hferanging from 14 to 40,the Q point stays within 20% of VCE' 6.5 V ).3. Before you connect your circuit, predict what the DC operating point (Q point) willbe. This may be done in the following way:(a) Construct the The venin equivalent circuit for the base bias circuit.(b) The base-emitter voltage drop for a forward-biased Si transistor is 0.6 V. ThereforeVE= VB− 0.6 V.(c) The emitter current IE, will be IE=VEREfrom Ohm’s law. The base currentIB=IEβ. You could load your Thevenin equivalent of the base bias circuit withthe current IB, calculate a new VBand solve these equations simultaneously toget the exact operating point, but this is seldom necessary. If the bias networkdesign is good (meaning not too sensitive to β, a single iteration will yield a veryaccurate approximation to the operating point.(d) Using Ohm’s law, calculate IE, and, thus IC, since they are about the same.(e) From the equation for the collector-emitter circuit:VCC= ICRC+ VCE+ IEREcalculate VCE. You have now determined the DC operating point (the Q point).The equation for the load line can be written (assuming IC= IE),IC=(VCC− VCE)(RE+ RC)2vinvoutrinroutAv vinFigure 2: Thevenin Equivalent Circuit for an Amplifier.ibCollectorEmitterBasertrrbFigure 3: Tranresistance Model Equivalent Circuit.4. Draw the load line on your characteristic curves and mark your calculated Q point.5. Next connect the circuit as shown schematically in Fig. 1. Be sure to check thetransistor pin-out diagram at the end of the transistor data sheet. (Transistors arenormally drawn as if you were looking at the transistor leads from the lead sidewhile integrated circuits are normally drawn as if you were looking at the pins fromabove the circuit board, ie from the side away from the IC pins.) Measure VCE,VE,VB,and VBEand compare with your calculated values. (They should agree within ∼ 10%).6. AC properties. Three important parameters, defined by the coupled Theveninequivalent circuits of the amplifier input and output shown in Fig. 2, determine theAC behavior of the amplifier. They are:(a) The input impedance (ri),(b) The output impedance (ro), and(c) The voltage gain (Av= vo/vi).Before actually measuring these quantities, it is instructive to estimate them. Usethe transresistance model, in which the transistor is replaced by an equivalent circuitcontaining a current generator ic= βibin series with the base emitter resistor rtr(called3the transresistance). The equivalent circuit is shown in Fig. 3. By differentiatingthe basic diode equation (for a p-n junction) we find that the dynamic base emitterresistance or “transresistance” is inversely proportional to the current IE.rtr=∂V∂I=0.026 ohm/AIE.Thus, the transresistance is:rtr= 0.026 ohm/AIC+ 2 ohm!where we have assumed IC' IEand the “2 ohm” is a rough average value for the ohmicresistance of the internal connection to the emitter junction. Note that an AC signalintroduced on the base lead can pass to AC ground through three paths: R1, R2, andthrough the transistor. So, ri(common emitter input impedance) is calculated as:ri= R1k R2k (rb= β(rtr+ RE))(with “k” meaning “in parallel” so you add reciprocals to get the reciprocal of thecombination).The common emitter AC voltage gain is then:Av=vovi= −icRCic(rtr+ RE)= −RC(rtr+ RE).In this approximation. the output impedance ro= RC.7. (a) Now introduce a AC signal (a sine wave from the waveform generator) of frequencyf = 5 KHz into the base through a coupling capacitor Cc≥ 1.0 µF and measureri, Avand ro. To measure ri, introduce a series resistor (R3), and calculate riusing the voltage divider equations.(b) Vary the magnitude of the input signal and note any distortion in the outputsignal which results.Interpret the distortion using the output characteristics and the load line.(c) Measure the frequency response (or band width) of your amplifier (Av(ω)) and tryto identify the lower corner frequency in terms of the component values.8. Emitter Resistor Bypass ... Current and Voltage Dr ive. To have a truecommon emitter configuration for an amplifier, the emitter resistor RE, is bypassed bya large capacitor (∼ 1 to 2 µF). The result is that the emitter is at AC ground, while