KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU  Control and Automation Dept. 1 7 DC BIASING FETS (CONT D)


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1 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 1 7 DC BIASING FETS (CONT D) Most of the content is from the textbook: Electronic devices and circuit theory, Robert L. Boylestad, Louis Nashelsky, 11 th ed, 013
2 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 7.7 Enhancementtype MOSFETS The transfer characteristics of the enhancementtype MOSFET are quite different from those encountered for the JFET and depletiontype MOSFETs, resulting in a graphical solution quite different from those of the preceding sections. First and foremost, recall that for the n channel enhancementtype MOSFET, the drain current is zero for levels of gatetosource voltage less than the threshold level V T. For levels of V GS greater than V T the drain current is defined by I D = k V GS V T Since specification sheets typically provide the threshold voltage and a level of drain current (I D (on)) and its corresponding level of V GS on, two points are defined immediately. To complete the curve, the constant k of must be determined from the specification sheet data I D = k V GS V T I D (on) = k V GS on V T
3 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 3 Example 7.9: (Feedback Biasing) V DD = 1 V I D on = 6 ma V GS on = 8 V V T = 3 V R G = 10 M R D = k C 1 = 1 μf C = 1 μf V DD R G for the enhancementtype MOSFET determine V GS, I D, V DS R D C Using I D on and V GS on k = 0.4 x 10 3 A/V I D = k V GS V T V GS = V DD I D R D I D =.75 ma V GS = 6.4 V V DS V GS = 6.4 V Intersection of two equations is the Q point V i C 1 V GS V DS V o
4 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 4 Example 7.10: (Voltage Divider Configuration) V DD = 40 V I D on = 3 ma V GS on = 10 V V T = 5 V R 1 = 0 M R = 18 M R D = 3 k R S = 80 For the n channel depletiontype MOSFET Determine V GS, I D, V DS R 1 R I G V DD V GS R D I D I S R S V DS Using I D on and V GS on k = 0.1 x 10 3 A/V V DD V G = R = 18 V R 1 R I D = k V GS V T V GS = V G I D R S I D = 6.7 ma V GS = 1.5 V V DS = 14.4 V Intersection of two equations is the Q point
5 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept Combination Networks Now that the dc analysis of a variety of BJT and FET configurations is established, the opportunity to analyze networks with both types of devices presents itself. Fundamentally, the analysis simply requires that we first approach the device that will provide a terminal voltage or current level. Example 7.11: Determine V G, V C for the network V DD = 16 V I DSS = 1 ma V P = 6 V β = 180 R 1 = 8 k R = 4 k R G = 1 M R D =.7 k R E = 1.6 k R 1 R R G V DD R D R E V D V C 5 R V TH = V DD = 3.63 V R R 1 R TH =18.57 k V TH = R TH I B V BE R E I E I B = 18.57x10 3 = μa x103 I E =1.717mA I C =1.707mA I D = I S = I C = 1.707mA I D = I DSS 1 V GS V P V GS = V V G = V BE R E I E =3.447 V V S = V C = V G V GS = V V DD =R D I D V D V D = V V DS = 4.07 V
6 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 6 Example 7.1: Determine V D for the network V DD = 16 V I DSS = 8 ma V P = 4 V β = 80 R B = 470 k R C = 3.6 k R S =.4 k R B V CC R C V C I D = I DSS V GS = I D R S V GS =.4 V I D = 1 ma 1 V GS V P I E = 1mA I C = ma I B = μa V D V CC =R C I C V C V C =1.443 V V CC =R B I B V BE V D V D =9.497 V R S
7 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept Design Example 7.13: To have network with V D = 14 V, I D = ma determine values of R D and R S V DD = 0 V I DSS = 6 ma V P = 3 V V DD R D I D I G V GS I S V DS R S V DD = R D I D V D 6 R D = = 3k x10 3 I D = I DSS I D I DSS V GS V P = 1 V GS 3 V GS = { 1.68, 4.73} V 1.68 = R S I D R S = 634
8 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 8 Example 7.14: To have network with V D = 1 V, V GS = V determine value R S V DD = 16 V R 1 = 91 k R = 47 k R D = 1.8 k R 1 V DD R D V DD V G = R = 5.44 V R 1 R V DD = R D I D V D 6 I D = =. ma 1.8x103 V GS = V G I D R S I G I D R S = 3.35k R V GS I S V DS R S
9 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 9 8 OPERATIONAL AMPLIFIERS Most of the content is from the textbook: Electronic devices and circuit theory, Robert L. Boylestad, Louis Nashelsky, 11 th ed, 013
10 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept Introduction An operational amplifier, or opamp, is a very high gain differential amplifier with high input impedance and low output impedance. Typical uses of the operational amplifier are to provide voltage amplitude changes (amplitude and polarity), oscillators, filter circuits, and many types of instrumentation circuits. An opamp contains a number of differential amplifier stages to achieve a very high voltage gain. SingleEnded Input Singleended input operation results when the input signal is connected to one input with the other input connected to ground.
11 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept OPAMP BASICS An operational amplifier is a very high gain amplifier having very high input impedance (typically a few mega ohms) and low output impedance (less than 100 ). The basic circuit is made using a difference amplifier having two inputs (plus and minus) and at least one output. The plus () input produces an output that is in phase with the signal applied, whereas an input to the minus () input results in an oppositepolarity output.
12 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 1 The basic circuit connection using an opamp is shown in the figure. The circuit provides operation as a constantgain multiplier. An input signal V 1 is applied through resistor R 1 to the minus input. The output is then connected back to the same minus input through resistor R f. The plus input is connected to ground. Since the signal V 1 is essentially applied to the minus input, the resulting output is opposite in phase to the input signal.
13 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 13
14 KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 14 V i = R f V R 1 R 1 R 1 A f R 1 R V V i f R f V i = V R f (1 A V )R 1 1 if A V 1 and A V R 1 R f V i = R f A V R V 1 1 V o = A VV i V i V i V o = R f V 1 R 1 = A V V i R f A V R 1 V 1 = R f R 1 V i V 1
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