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Recall Lecture 10 Introduction to BJT 3 modes of operation Cut-off Active Saturation Active mode operation of NPN NPN PNP IE = IS [ e VBE / VT ] IE = IS [ e VEB / VT] IC = IB IC = IE IE = IB( + 1) =[ /+1] = [ / 1 - ] Based on KCL: IE = IC + IB DC Analysis of BJT Circuit and Load Lines DC analysis of BJT BE Loop (EB Loop) – VBE for npn and VEB for pnp CE Loop (EC Loop) - VCE for npn and VEC for pnp When node voltages are known, branch current equations can be used. Common-Emitter Circuit The figures below is showing a common-emitter circuit with an npn transistor and the DC equivalent circuit. Assume that the B-E junction is forward biased, so the voltage drop across that junction is the cut-in or turnon voltage VBE (on). Unless given , always assume VBE = 0.7V KVL at B-E loop You can calculate IC using IC = βIB Calculate IE using IE = IB + IC Input Load Line IB versus VBE Input Load Line – IB versus VBE Derived using B-E loop The input load line is obtained from Kirchhoff’s voltage law equation around the B-E loop, written as follows: VBE – VBB + IBRB = 0 y = mx + c Both the load line and the quiescent base current change as either or both VBB and RB change. For example; = 200 The input load line is essentially the same as the load line characteristics for diode circuits. VBE – 4 + IB(220k) = 0 IB = -VBE + 4 220k 220k y = mx + c IBQ = 15 μA KVL at C-E loop of the circuit You have calculated the value of IC from the value of IB Output Load Line IC versus VCE Output Load Line – IC versus VCE Derived using C-E loop For the C-E portion of the circuit, the load line is found by writing Kirchhoff’s voltage law around the C-E loop. We obtain: y = mx + c For example; IC = -VCE + 10 2k 2k y = mx + c = 200 Q-point is the intersection of the load line with the iC vs vCE curve, corresponding to the appropriate base current To find the intersection points setting IC = 0, VCE = VCC = 10 V setting VCE = 0 IC = VCC / RC = 5 mA Examples BJT DC Analysis Common-Emitter Circuit Example 1 Calculate the base, collector and emitter currents and the C-E voltage for a common-emitter circuit by considering VBB = 4 V, RB = 220kΩ, RC = 2 kΩ, VCC = 10 V, VBE (on) = 0.7 V and β = 200. Example 2 = 0.99 KVL at BE loop: Hence, 0.7 + IERE – 4 = 0 IE = 3.3 / 3.3 = 1 mA IC = IE = 0.99 mA IB = IE – IC = 0.01 mA KVL at CE loop: ICRC + VCE + IERE – 10 = 0 VCE = 10 – 3.3 – 4.653 = 2.047 V Common-Emitter Circuit - PNP Example 3 Find IB, IC, IE and RC such that VEC = 2.5 V for a common-emitter circuit . Consider: VBB = 1.5 V, RB = 580 kΩ, VCC = 5 V, VEB (on) = 0.6 V, = 100. Example 4: DC Analysis and Load Line Calculate the characteristics of a circuit containing an emitter resistor and plot the output load line. For the circuit, let VBE (on) = 0.7 V and β = 75. Output Load Line Use KVL at B-E loop to find the value of IB Use KVL at C-E loop – to obtain the linear equation IC RC + VCE + IE RE – 12 = 0 IC RC + VCE + (IC / )RE – 12 = 0 VCE = 12 – IC (1.01) IC = - VCE + 12 1.01 IB = 75.1 A VCE = 12 – 5.63 (1.01) VCE = 6.31 V Example: DC Analysis – Common Base (npn) E C B C B E = 100 5k BE Loop 0.7 + IERE – 5 = 0 15V IE = 0.43 mA CE Loop 10 k VCE + ICRC + IERE – 5 – 15 = 0 VCE = 20 – 4.3 – 2.13 5V VCE = 13.57 V