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Transistors Lesson #8 Chapter 4

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Transistors Lesson #8 Chapter 4 Transistors Lesson #8 Chapter 4 BME 372 Electronics I – 225 J.Schesser Configurations of the BJT • Common Emitter and Emitter Follower Collector C o iC iB B + o vCE + - Base - vBE iE o E Emitter npn BME 372 Electronics I – 226 J.Schesser Configurations of the BJT • Common Collector Emitter E o i i E B vBE - - B + o vCE Base + ic o C Collector • Common Base npn BME 372 Electronics I – 227 J.Schesser Characteristics of the BJT npn Common Emitter Configuration iC iB 100μA 25 10mA iB 20 5mA 50μA 15 vBE 20V .6 .8 1 vCE Collector-emitter is a family of Base-emitter junction looks curves which are a function of like a forward biased diode base current BME 372 Electronics I – 228 J.Schesser BJT Equations iE iC iB Collector C i o I =β i =α i C i c B E B + i B o vCE E + - Base - iB (1)iE vBE iE o E Emitter iC iB iB 1 Since α is less than unity then β will be greater than unity and there is current gain from base 1 to collector. BME 372 Electronics I – 229 J.Schesser Example • Calculate the values of β and α from the transistor shown in the previous graphs. iC 100μA i β = ic / ib 10mA B = 5m/50µ = 100 5mA 50μA α = β /(β+1) = 100/101 20V =.99 vCE BME 372 Electronics I – 230 J.Schesser BJT Analysis RC 2k RB + 10V VCC + 50k Vin - - 1.6V + - VBB Here is a common emitter BJT amplifier: What are the steps? BME 372 Electronics I – 231 J.Schesser BJT Analysis – Inputs and Outputs RC 2k iC RB + + VCE 10V VCC ib + + 50k Vin - VBE - - 1.6V - + - VBB We would want to know the collector current (ic), collector-emitter voltage (VCE), and the voltage across RC. To get this we need to fine the base current (ib) and the base-emitter voltage (VBE). BME 372 Electronics I – 232 J.Schesser BJT Analysis – Input Equation RC 2k iC RB + + VCE 10V VCC iB + + 50k Vin - VBE - - 1.6V - + - VBB To start, let’s write KVL around the base circuit. Vin(t) + VBB = iB(t)RB + VBE(t) BME 372 Electronics I – 233 J.Schesser BJT Analysis – Output Equations RC 2k iC RB + + VCE 10V VCC iB + + 50k Vin - VBE - - 1.6V - + - VBB Likewise, we can write KVL around the collector circuit. VCC= iC(t)RC + VCE(t) BME 372 Electronics I – 234 J.Schesser BJT Analysis Use Superposition: DC & AC sources • Note that both equations are written so as to calculate the transistor parameters (i.e., base current, base-emitter voltage, collector current, and the collector-emitter 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 Q-point • 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 BME 372 Electronics I – 235 J.Schesser 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. BME 372 Electronics I – 236 J.Schesser BJT DC Analysis Base-Emitter Circuit Q point 39 i B 34 amps 29 VBE = 0; 24 iB = VBB / RB Q POINT = 1.6 /50k = 32A 19 IBQ = 14 20A iB = 0; VBE = VBB = 1.6 V 9 4 VBEQ = 0.6 V -1 -0. -0. -0. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 3 2 1 v BE volts First let’s set Vin(t) =0 to get the Q-point for the BJT. We start with the base circuit. VBB = iBRB + VBE And the intercepts occur at iB = 0; VBE = VBB = 1.6 V and at VBE = 0; iB = VBB / RB = 1.6 /50k = 32A The Load Line intersects the Base-emitter characteristics at VBEQ = 0.6 V and IBQ = 20A BME 372 Electronics I – 237 J.Schesser BJT DC Analysis Collector-Emitter Circuit Q point 7.E-03 i C amps VCC 5 ma 6.E-03 iB=50 μa RC 5.E-03 40 μa 4.E-03 30 μa 3.E-03 Q POINT 20 μa ICQ 2.5 ma 2.E-03 10 μa 1.E-03 V = 5.9 V CEQ 0 μa 0.E+00 0246810 v CE volts VCC 10 v Now that we have the Q-point for the base circuit, let’s proceed to the collector circuit. VCC= iCRC + VCE The intercepts occur at iC = 0; VCE = VCC = 10 V; and at VCE = 0; iC = VCC / RC = 10 /2k = 5mA The Load Line intersects the Collector-emitter characteristic, iB=20mA at VCEQ = 5.9 V and ICQ = 2.5mA β = 2.5m/20m = 125 BME 372 Electronics I – 238 J.Schesser BJT DC Analysis Summary • Calculating the Q-point 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 base-emitter circuit and using load line analysis on the base-emitter characteristics, we obtained the base current Q-point – We then applied KVL to the collector-emitter circuit and using load line analysis on the collector-emitter characteristics, we obtained the collector current and voltage Q-point • This process is also called DC Analysis • We now proceed to perform AC Analysis BME 372 Electronics I – 239 J.Schesser BJT AC Analysis • How do we handle the variable source Vin(t) ? • When the variations of Vin(t) are large we will use the base-emitter and collector-emitter characteristics using a similar graphical technique as we did for obtaining the Q-point • When the variations of Vin(t) are small we will shortly use a linear approach using the BJT small signal equivalent circuit. BME 372 Electronics I – 240 J.Schesser BJT AC Analysis • Let’s assume that Vin(t) = 0.2 sin(ωt). • Then the voltage sources at the base vary from a maximum of 1.6 + .2 = 1.8 V to a minimum of 1.6 -.2 = 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 BME 372 Electronics I – 241 J.Schesser BJT AC Analysis Base-Emitter Circuit i 39 B “load line” for minimum input voltage 34 amps 29 24 “load line” for maximum input voltage 19 IBQ = 14 20μA 9 4 VBEQ = 0.6 V -1 -0. -0. -0. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 3 2 1 v BE volts From this graph, we find: At Maximum Input Voltage: VBE = 0.63 V, iB = 24μa At Minimum Input Voltage: VBE = 0.59 V, iB = 15μa Recall: At Q-point: VBE = 0.6 V, iB = 20μa Note the asymmetry around the Q-point of the Max and Min Values for the base current and voltage which is due to the non-linearity of the base-emitter characteristics ΔiBmax=24-20=4μa; ΔiBmin=20-15=5μa BME 372 Electronics I – 242 J.Schesser BJT AC Analysis Base-Emitter Circuit i 39 B “load line” for minimum input voltage 34 amps 29 24 “load line” for maximum input voltage 19 IBQ = 14 20A 9 4 VBEQ = 0.6 V -1 -0. -0. -0. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 3 2 1 v BE volts 2 0.00004 1.8 0.000035 1.6 0.00003 1.4 1.2 0.000025 vin 1 0.00002 ib 0.8 0.000015 0.6 0.00001 0.4 0.2 0.000005 0 0 0102030 BME 372 Electronics I – 243 J.Schesser BJT Characteristics-Collector Circuit 7.E-03 i C amps 6.E-03 iB=50 μa 5.E-03 40 μa 4.E-03 30 μa 3.E-03 24 μa Q POINT 20 μa ICQ 2.5 ma 2.E-03 15 μa 10 μa 1.E-03 V = 5.9 V CEQ 0 μa 0.E+00 0246810 v CE volts Using these maximum and minimum values for the base current on the collect circuit load line, we find: At Maximum Input Voltage: VCE = 5 V, iC = 2.7ma At Minimum Input Voltage: VCE = 7 V, iC = 1.9ma Recall: At Q-point: VCE = 5.9 V, iC = 2.5ma Note that in addition to the asymmetry around the Q-point there is an inversion between the input voltage and the collector to emitter voltage BME 372 Electronics I – 244 J.Schesser BJT Characteristics-Collector Circuit 7.E-03 i C amps 6.E-03 iB=50 μa 5.E-03 40 μa 4.E-03 30 μa 3.E-03 24 μa Q POINT 20 μa ICQ 2.5 ma 2.E-03 15 μa 10 μa 1.E-03 V = 5.9 V CEQ 0 μa 0.E+00 0246810 v CE volts 0.025 0.00004 4 8 0.000035 0.02 3.5 7 0.00003 3 6 0.015 0.000025 ic 2.5 5 0.00002 vin ib 2 4 0.01 0.000015 vc e 1.5 3 0.00001 0.005 1 2 0.000005 0.5 1 0 0 0 0 0102030 0 102030 BME 372 Electronics I – 245 J.Schesser 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 = Δic / Δib = (2.7 – 1.9) m / (24 -15) μ = 0.8/9 ✕ 103 = 88.9 – Voltage gain = Vo / Vi = ΔVCE / ΔVBE = (5 – 7) / (.63 - .59) = -2/0.04 = - 50 – Voltage gain = Vo / Vs = ΔVCE / ΔVin = (5 – 7) / .4 = -2 / .4 = - 5 BME 372 Electronics I – 246 J.Schesser BJT AC Analysis Summary • Once we complete DC analysis, we analyze the circuit from an AC point of view.
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