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The main contents of the chapter consist of the following: Base-biased (fixed bias) transistor circuits, voltage-divider-bias transistor circuits, examples and exercises.
COMSATS Institute of Information Technology Virtual campus Islamabad Dr. Nasim Zafar Electronics 1 EEE 231 Fall Semester – 2012 PotentialDividerBiasing Circuits: Lecture No: 19 Examples and Exercises. Nasim Zafar References: Ø Microelectronic Circuits: Adel S. Sedra and Kenneth C. Smith. Ø Integrated Electronics : Jacob Millman and Christos Halkias (McGrawHill) Ø Introductory Electronic Devices and Circuits Robert T. Paynter Ø Electronic Devices : Nasim Zafar Thomas L. Floyd ( Prentice Hall ). Basic Circuits of BJT: NPN Transistor IE = IC + IB Nasim Zafar Transistor Output Characteristics: Nasim Zafar Transistor Output Characteristics: Load Line – Biasing and Stability: Ø Active region: – – Ø Ø Ø BJT acts as a signal amplifier BE junction is forward biased and CB junction is reverse biased Graphical construction for determining the dc collector current IC and the collectortoemitter voltage VCE . The requirement is to set the Qpoint such that that it does not go into the saturation or cutoff regions when an a ac signal is applied. Maximum signal swing depends on the bias voltage Nasim Zafar The DC Operating Point: Biasing and Stability v Active region Amplifier: BJT acts as a Signal Amplifier 1. BE Junction Forward Biased C C IC VBE ≈ 0.7 V for Si 2. BC Junction Reverse Biased IC B B IB IB E IE E 3. KCL: IE = IC + IB Nasim Zafar IE The DC Operating Point: Biasing and Stability Slope of the Load Line: VCC = VCE + VRC VCE = VCC VRC VCE = VCC IC RC Ic ( )VCE Rc VCC RC Nasim Zafar Current Equations in a BJT: NPN Transistor Ø Ø Ø Collector Current iC iB Base Current Emitter Current iE Nasim Zafar In I se iC Is iC Is v BE e v BE e VT VT v BE VT 1. FixedBiased Transistor Circuits. BaseBiased (Fixed Bias) Transistor Circuit: Single Power Supply Nasim Zafar 10 Circuit19.3; Problem 19.6 (e) Determination of VCE Output Loop VCC=VRC+VCE+VRE VCE = VCCVRCVRE VCE = 12V 5.2V 1.3V VCE = 5.5V Nasim Zafar 35 Results of Problem 19.6 IE = IC = 1.3mA VRC = 5.2V VCE = 5.5V VRE = 1.3V VBB = 2V βdc was never used in a calculation. Hence, voltage divider biased circuits are immune to changes in βdc. Nasim Zafar A single voltage source supplies 36 Review of equations: In Review VRE = VBB – VBE IE ≈ IC Nasim Zafar 37 Summary: Ø Ø Voltagedivider biased circuits are immune to changes in βdc. A single voltage source supplies both voltages, VCC and VBB Ø The circuit Qpoint values are stable against changes in hFE Ø Use of the Thevenin equivalent circuit for the base makes the circuit simpler Ø Make the current in the voltage divider about 10 times IB, to simplify the analysis. Ø For design, solve for the resistor values (IC and VC specified) Nasim Zafar 38 Nasim Zafar 39 Circuit 19.4; Problem 19.7 (a) Given: VB = 3V and I = 0.2mA (a) RB1 and RB2 form a voltage divider Assume I >> IB I I = VCC/(RB1 + RB2) IB 0.2mA = 9 /(RB1 + RB2) Nasim Zafar 40 Circuit 19.4; Problem 19.7 (b) Given: VB = 3V and I = 0.2mA. RB1 = 30KΩ , and RB2 = 15KΩ (b) Determination of the Thevenin voltage: I IB VB = VCC[RB2/(RB1 + RB2)] 3 = 9 [RB2/(RB1 + RB2)], Solve for RB1 and RB2. Nasim Zafar 41 Prob. 19.7 (c) Find the operating point The use of Thevenin equivalent circuit for the base makes the circuit simpler VBB = VB = 3V • RBB = RB1|| RB2 = 30KΩ | | 15KΩ = 10KΩ Nasim Zafar 42 Problem 19.7 (d) Write BE loop and CE loop BE Voltage Loop: VBB = VRBB + VBE + VRE VBB = IBRBB + VBE + IERE CE loop IE =2.09 mA CE Voltage Loop: VCC = ICRC + VCE + IERE BE loop VCE =4.8 V This is how all DC circuits are analyzed and designed! Nasim Zafar 43 Nasim Zafar 44 Example 19.7 Stage 2 • CE loop IE2 VCC = IE2RE2 + VEC2 +IC2RC2 15 = 2.8(2) + VEC2 + 2.8 (2.7) solve for VEC2 IC2 Nasim Zafar VCE2 = 1.84V 45 Example 19.7 CE loop neglect IB2 because it is IB2