Transistors. Lesson #8 Chapter 4. BME 372 Electronics I J.Schesser

Transcription

1 Transistors Lesson #8 Chapter 4 225

2 Configurations of the BJT Common Emitter and Emitter Follower B Base ib o Collector C o vce v BE o E Emitter npn i C i E 226

3 Configurations of the BJT Common Collector Common Base B o Base ib v BE Emitter E o o C vce Collector npn i E i c 227

4 Characteristics of the BJT npn Common Emitter Configuration i C 25 i B 10mA 100μA i B mA 50μA v BE V v CE Baseemitter junction looks like a forward biased diode Collectoremitter is a family of curves which are a function of base current 228

5 i i i E B C i i i C C E i B (1 ) i i 1 1 B E BJT Equations i B B o Base ib Collector C o vce v BE o E Emitter I c =β i B =α i E Since α is less than unity then β will be greater than unity and there is current gain from base to collector. i E 229

6 Example Calculate the values of β and α from the transistor shown in the previous graphs. i C β = i c / i b = 5m/50µ = 100 5mA α = β /(β1) = 100/101 =.99 10mA 100μA i B 50μA 20V v CE 230

7 BJT Analysis R C 2k R B V in 50k 1.6V 10V V CC V BB Here is a common emitter BJT amplifier: What are the steps? 231

8 BJT Analysis Inputs and Outputs R C 2k i C R B V in 50k 1.6V i b V BE V CE 10V V CC V BB We would want to know the collector current (i c ), collectoremitter voltage (V CE ), and the voltage across R C. To get this we need to fine the base current (i b ) and the baseemitter voltage (V BE ). 232

9 BJT Analysis Input Equation R C 2k i C R B V in 50k 1.6V i B V BE V CE 10V V CC V BB To start, let s write KVL around the base circuit. V in (t) V BB = i B (t)r B V BE (t) 233

10 BJT Analysis Output Equations R C 2k i C R B V in 50k 1.6V i B V BE V CE 10V V CC V BB Likewise, we can write KVL around the collector circuit. V CC = i C (t)r C V CE (t) 234

11 BJT Analysis Use Superposition: DC & AC sources Note that both equations are written so as to calculate the transistor parameters (i.e., base current, baseemitter voltage, collector current, and the collectoremitter voltage) for both the DC signal and the AC signal sources. Let s use superposition, calculate the parameters for each separately, and add up the results First, the DC analysis to calculate the DC Qpoint Short Circuit any AC voltage sources Open Circuit any AC current sources Next, the AC analysis to calculate gains of the amplifier. Depends on how we perform AC analysis Graphical Method Equivalent circuit method for small AC signals 235

12 BJT DC Analysis Using KVL for the input and output circuits and the transistor characteristics, the following steps apply: 1. Draw the load lines on the transistor characteristics 2. For the input characteristics determine the Q point for the input circuit from the intersection of the load line and the characteristic curve (Note that some transistor do not need an input characteristic curve.) 3. From the output characteristics, find the intersection of the load line and characteristic curve determined from the Q point found in step 2, determine the Q point for the output circuit. 236

13 V BE = 0; i B = V BB / R B = 1.6 /50k = 32 A BJT DC Analysis BaseEmitter Circuit Q point i B amps I BQ = 20 A Q POINT V BEQ = 0.6 V v BE volts i B = 0; V BE = V BB = 1.6 V First let s set V in (t) =0 to get the Qpoint for the BJT. We start with the base circuit. V BB = i B R B V BE And the intercepts occur at i B = 0; V BE = V BB = 1.6 V and at V BE = 0; i B = V BB / R B = 1.6 /50k = 32 A The Load Line intersects the Baseemitter characteristics at V BEQ = 0.6 V and I BQ = 20 A 237

14 BJT DC Analysis CollectorEmitter Circuit Q point i C amps 7.E03 6.E03 5.E03 4.E03 VCC RC 5 ma i B =50 μa 40 μa 30 μa I CQ 2. 5 ma 3.E03 2.E03 1.E03 0.E00 V CEQ = 5.9 V v CE volts Q POINT 0 μa 20 μa 10 μa VCC 10 Now that we have the Qpoint for the base circuit, let s proceed to the collector circuit. v V CC = i C R C V CE The intercepts occur at i C = 0; V CE = V CC = 10 V; and at V CE = 0; i C = V CC / R C = 10 /2k = 5mA The Load Line intersects the Collectoremitter characteristic, i B =20mA at V CEQ = 5.9 V and I CQ = 2.5mA β = 2.5m/20m =

15 BJT DC Analysis Summary Calculating the Qpoint for BJT is the first step in analyzing the circuit To summarize: We ignored the AC (variable) source Short circuit the voltage sources Open Circuit the current sources We applied KVL to the baseemitter circuit and using load line analysis on the baseemitter characteristics, we obtained the base current Qpoint We then applied KVL to the collectoremitter circuit and using load line analysis on the collectoremitter characteristics, we obtained the collector current and voltage Qpoint This process is also called DC Analysis We now proceed to perform AC Analysis 239

16 BJT AC Analysis How do we handle the variable source V in (t)? When the variations of V in (t) are large we will use the baseemitter and collectoremitter characteristics using a similar graphical technique as we did for obtaining the Qpoint When the variations of V in (t) are small we will shortly use a linear approach using the BJT small signal equivalent circuit. 240

17 BJT AC Analysis Let s assume that V in (t) = 0.2 sin(ωt). Then the voltage sources at the base vary from a maximum of = 1.8 V to a minimum of = 1.4 V We can then draw two load lines corresponding the maximum and minimum values of the input sources The current intercepts then become for the: Maximum value: 1.8 / 50k = 36 µa Minimum value: 1.4 / 50k = 28 µa 241

18 BJT AC Analysis BaseEmitter Circuit i B amps I BQ = 20μA V BEQ = 0.6 V v BE volts From this graph, we find: At Maximum Input Voltage: V BE = 0.63 V, i B = 24μa At Minimum Input Voltage: V BE = 0.59 V, i B = 15μa Recall: At Qpoint: V BE = 0.6 V, i B = 20μa Note the asymmetry around the Qpoint of the Max and Min Values for the base current and voltage which is due to the nonlinearity of the baseemitter characteristics Δi Bmax =2420=4μa; Δi Bmin =2015=5μa load line for minimum input voltage load line for maximum input voltage 242

19 BJT AC Analysis BaseEmitter Circuit i B amps load line for minimum input voltage I BQ = 20 A V BEQ = 0.6 V load line for maximum input voltage v BE volts vin ib 243

20 BJT CharacteristicsCollector Circuit i C amps 7.E03 6.E03 i B =50 μa 5.E03 4.E03 40 μa 30 μa I CQ E03 ma 2.E03 1.E03 0.E00 V CEQ = 5.9 V Q POINT v CE volts Using these maximum and minimum values for the base current on the collect circuit load line, we find: Note that in addition to the asymmetry around the Qpoint there is an inversion between the input voltage and the collector to emitter voltage μa At Maximum Input Voltage: V CE = 5 V, i C = 2.7ma At Minimum Input Voltage: V CE = 7 V, i C = 1.9ma Recall: At Qpoint: V CE = 5.9 V, i C = 2.5ma 24 μa 20 μa 15 μa 10 μa

21 BJT CharacteristicsCollector Circuit i C amps 7.E03 6.E03 i B =50 μa I CQ E03 4.E03 3.E03 ma 2.E03 1.E03 0.E00 V CEQ = 5.9 V v CE volts 40 μa 30 μa 24 μa Q POINT 20 μa 15 μa 10 μa 0 μa ic ib vin vce 245

22 BJT AC Analysis Amplifier Gains From the values calculated from the base and collector circuits we can calculate the amplifier gains: β = 125 Current gain = Δi c / Δi b = ( ) m / (24 15) μ = 0.8/ = 88.9 Voltage gain = V o / V i = ΔV CE / ΔV BE = (5 7) / (.63.59) = 2/0.04 = 50 Voltage gain = V o / V s = ΔV CE / ΔV in = (5 7) /.4 = 2 /.4 = 5 246

23 BJT AC Analysis Summary Once we complete DC analysis, we analyze the circuit from an AC point of view. AC analysis can be performed via a graphical processes Find the maximum and minimum values of the input parameters (e.g., base current for a BJT) Use the transistor characteristics to calculate the output parameters (e.g., collector current for a BJT). Calculate the gains for the amplifier 247

24 BJT Characteristics Distortion i C amps 7.E03 6.E03 5.E03 i B =50 μa 40 μa I CQ 2. 5 ma 4.E03 3.E03 2.E03 1.E03 0.E00 V CEQ = 5.9 V Note that even though the input (base current) is symmetrical about its Qpoint, the output voltage is uneven v CE volts Q POINT 0 μa 30 μa 20 μa 10 μa 24 μa 15 μa This is due to the nonlinear spacing of the characteristic curves and leads to signal distortion 248

25 The pnp Transistor Basically, the pnp transistor is similar to the npn except the parameters have the opposite sign. The collector and base currents flows out of the transistor; while the emitter current flows into the transistor The baseemitter and collectoremitter voltages are negative Otherwise the analysis is identical to the npn 249

26 PNP Bipolar Junction Transistors Base Collector ptype ntype ptype Two junctions CollectorBase and EmitterBase Biasing v BE Forward Biased v CB Reverse Biased Collector B o Base ib v BE C o o E v CE Emitter i C i E V in R B 50k 1.6V R C 2k 10V V CC pnp Emitter 250

27 Homework Probs. 4.4, 4.5, 4.8, 4.10, 4.14, 4.15, 4.19, 4.20, 4.21, 4.22, 251