Nội dung

COMSATS Institute of Information Technology Virtual campus Islamabad Dr. Nasim Zafar Electronics 1 - EEE 231 Fall Semester – 2012 DC Analysis of Transistor Circuits-II Lecture No: 17 Contents:  DC Current and Voltage Analysis.  Examples and Exercises. References:  Microelectronic Circuits: Adel S. Sedra and Kenneth C. Smith.  Electronic Devices : Thomas L. Floyd ( Prentice Hall ).  Integrated Electronics Jacob Millman and Christos Halkias (McGraw-Hill).  Electronic Devices and Circuit Theory: Robert Boylestad & Louis Nashelsky ( Prentice Hall ).  Introductory Electronic Devices and Circuits: Robert T. Paynter. Lecture No. 17 DC Analysis of Transistor Circuits-II Reference: Chapter 5.4 Microelectronic Circuits Adel S. Sedra and Kenneth C. Smith. DC Analysis of Transistor Circuits Basic Transistor Operation  Consider this circuit as two separate circuits:  The Base-Emitter Circuit  The Collector-Emitter Circuit  The amount of current flow in the baseemitter circuit controls the amount of current that flows in the collector circuit.  Small changes in base-emitter current yields a large change in collector-current. DC Analysis of Transistor Circuits Analysis of this transistor circuit to predict: DC Voltages and Currents requires use of : Ohm’s law,  Kirchhoff’s voltage law  and the ß for the transistor. DC Analysis of Transistor Circuits Kirchhoff’s voltage law:  In the Base Circuit: VBB is distributed across the base-emitter junction and RB In the collector circuit: We determine that VCC is distributed proportionally across RC and the transistor, VCE. Transistor Characteristics and Parameters  There are three dc voltages and three dc currents to be considered. IB: dc base current IE: dc emitter current IC: dc collector current VBE: dc voltage across baseemitter junction VCB: dc voltage across collector-base junction VCE: dc voltage from collector to emitter BJT-Current and Voltage Analysis  For all circuits: Assume the NPN transistor operates in the linear region:  write B-E voltage loop  write C-E voltage loop  When the base-emitter junction, in an NPN transistor is forward biased, it is like a forward biased diode and has a forward-voltage drop of: VBE = 0.7 V NPN BJT-Current and Voltage Analysis: Input Circuit: Forward Biased E-B Junction. Since the emitter is grounded, by Kirchhoff’s voltage law, the voltages in the input circuit are: VBB = VRS + VBE VRS = VBB -- VBE Using Ohm’s law: VRS = IB RS IB RS = VBB -- VBE IB = (VBB – VBE)/RS NPN BJT-Current and Voltage Analysis:  Output Circuit: Reverse Biased B-C Junction.  Using Kirchhoff’s voltage law, the voltages in the Output Circuit are: VCC = VCE + VRL VCE = VCC -- VRL The voltage drop across RL is: VRL = IC RL The collector voltage is: VCE = VCC -- IC RL DC Voltages for the Biased Transistor  Collector voltage: VCE = VCC - ICRC  Base voltage: VBE = VCE – VCB – IB = (VBB – VBE)/RB – for silicon transistors, VBE = 0.7 V – for germanium transistors, VBE = 0.3 V DC Analysis of Transistor Circuits  Transistor Currents: IE = IC + IB  alpha (DC): IC = DCIE  beta (DC): IC = DCIB – DC typically has a value between 20 and 200 Examples and Exercises: DC Analysis of Transistor Circuits  Q: What is IB, IC, IE and also VCE, VCB, VBE ??  A: I don’t know ! But, we can find out—IF we complete each of the four steps required for BJT DC analysis. DC Analysis of Transistor Circuits  Step 1 – Assume an operating mode. Let’s assume the BJT is in the linear region ! Remember, this is just a guess; we have no way of knowing for sure what mode the BJT is in at this point.  Step 2 - Enforce the conditions of the assumed mode.  Step 3 – ANALYZE the circuit.  Step 4 -“Write KVL equations for the base-emitter “leg”. Modes of BJT Operation: IC(mA) Saturation Region IB = 200 A 30 Active Region IB = 150 A 22.5 IB = 100 A 15 IB = 50 A 7.5 Cutoff Region IB = 0 0 VCE (V) 0 5 10 15 20  Active: BJT acts like an amplifier (most common use).  Saturation: BJT acts like a short circuit.  Cutoff: BJT acts like an open circuit. Nasim Zafar 18 Transistor Characteristics and Parameters The Cutoff Region With no IB the transistor is in the cutoff region and just as the name implies there is practically no current flow in the collector part of the circuit. With the transistor in a cutoff state the full VCC can be measured across the collector and emitter(VCE) Example 5.4 - Figure 5.34 Consider the circuit shown in Fig. 5.34(a), which is redrawn in Fig. 5.34(b) Example 5.4  Example: 5.4  We wish to analyze this circuit to determine:  all node voltages  and branch currents.  We will assume that β is specified to be 100. Example 5.4 - Figure 5.34 Solution - Example 5.4 Solution - Example 5.4 Input Circuit: Forward Biased E-B Junction: Step 1:  The circuit in Fig. 5.34(b) shows that the base is connected to +4 V and the emitter is connected to ground through a resistance RE.  The base–emitter junction will be forward biased.  Since the emitter is grounded, by Kirchhoff’s voltage law, the voltages in the input circuit are: VBB = VRE + VBE VRE = VBB -- VBE Solution - Example 5.4  Assuming that VBE is approximately 0.7 V, it follows that the emitter voltage will be:  Step 2:  We know the voltages at the two ends of RE and thus can determine the current IE through it, Solution - Example 5.4  Step 3:  We can evaluate the collector current from: Solution - Example 5.4  Step 4: We are now in a position to use Ohm’s law to determine the collector voltage :  Since the base is at +4 V, the collector–base junction is reverse biased by 1.3 V, and the transistor is indeed in the active mode as assumed.  Step 5: It remains only to determine the base current IB, as follows: Example 5.5 Example 5.5 - Figure 5.35 Example 5.5 - Figure 5.35 Solution: Example 5.5  Assuming active-mode operation, we have: Solution: Example 5.5  Since the collector voltage calculated, appears to be less than the base voltage by 3.52 V, it follows that our original assumption of active-mode operation is incorrect. In fact, the transistor has to be in the saturation mode. Assuming this to be the case, we have: Solution: Example 5.5  Also: Some More Examples Example17-1 Input Circuit: Forward Biased Junction. B-E Voltage Loop: VBB = VRB + VBE VBB = IBRB + VBE b = 100 Solve for IB, IC, VCC: IC IB IE IB RB = VBB -- VBE IB = (5 - VBE)/RB = (5-0.7)/100k = 0.043mA IC = bIB = (100)0.043mA = 4.3mA VCC = 10 - ICRC = 10 - 4.3(2) = 1.4V Example 17-2  The voltages in the input circuit are: VE = VB - VBE = 4V - 0.7V = 3.3V IE = (VE - 0)/RE = 3.3/3.3K = 1mA IC IE IC  IE = 1mA VC = 10 - ICRC = 10 - 1(4.7) = 5.3V Exercise Summary of DC Analysis  Bias the transistor so that it operates in the linear region  B-E junction forward biased, C-E junction reversed biased.  Use VBE = 0.7 (NPN), IC  IE, IC = bIB  Write B-E, and C-E voltage loops.  For DC analysis, solve for IC, and VCE.  For design, solve for the resistor values (IC and VCE specified).