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### Transcript of Field Effect Transistor (FET) 1. Introduction Field Effect Transistor (FET) Junction Field Effect...

• Slide 1
• Field Effect Transistor (FET) 1
• Slide 2
• Introduction Field Effect Transistor (FET) Junction Field Effect Transistor (JFET) Metal Oxide Semiconductor FET (MOSFET) Metal Oxide Semiconductor FET (MOSFET) Depletion Type MOSFET Depletion Type MOSFET Enhancement Type MOSFET Enhancement Type MOSFET 2
• Slide 3
• Junction Field Effect Transistor (JFET) n-channel JFET p-channel JFET 3
• Slide 4
• JFET Introduction JFET is always operated with the gate source p-n junction reversed biased. 4
• Slide 5
• JFET Introduction Channel width and thus the channel resistance can be controlled by varying the gate voltage. JFET biased for constructionGreater V GG narrows the channel Less V GG widens the channel Water analogy for the JFET control 5
• Slide 6
• JFET Characteristics and Parameters For V GS = 0 v, the value of V DS at which I D becomes essentially constant is the pinch-off voltage (V p ) and is denoted as I DSS. Breakdown occurs at point C when I D begins to increase very rapidly with any further increase in V DS. 6
• Slide 7
• V GS controls I D. The value of V GS that makes I D approximately zero is the cutoff voltage V GS(off). The JFET must operate between V GS = 0 and V GS(off). 7
• Slide 8
• Transfer Characteristics William Bradford Shockley derived a relationship between I D and V GS which is known as Shockleys equation and is given by The above equation suggests that when V GS = 0, I D = I DSS. When V GS = V p, I D = 0 8
• Slide 9
• Transfer curve from the drain characteristics 9
• Slide 10
• Example The following parameters are obtained from a certain JFET datasheet: V P = -8 v and I DSS = 5 mA. Determine the values of I D for each value of V GS ranging from 0 v to -8 v in 1 v steps. Plot the transfer characteristic curve from these data. Solution: 10
• Slide 11
• 11
• Slide 12
• V GS IDID 12
• Slide 13
• FET Biasing The following relations can be applied to the dc analysis of most of the FET amplifiers: 13
• Slide 14
• JFET Biasing: Fixed Bias Circuit 14
• Slide 15
• JFET Biasing: Fixed Bias Circuit 15 Circuit for dc analysis
• Slide 16
• Fixed Bias Circuit GS Loop: Apply KVL 16 Apply the Shockleys Equation: Plot Shockleys equation:
• Slide 17
• Fixed Bias Circuit Q-Point: 17
• Slide 18
• Fixed Bias Circuit DS Loop 18 Also note that In addition
• Slide 19
• Example: Determine the following for the given Fig. (a)V GSQ (b) I DQ (c) V DS (d) V D (e) V G (f) V S. Solution: (a) V GSQ = -V GG = -2 V 19 (b) (c) (d) V D = V DS = 4.75 V (e) V G = V GS = -2 V (f) V S = 0 V
• Slide 20
• JFET Biasing: Self Bias Configuration 20
• Slide 21
• Self Bias Circuit: DC Analysis 21 Self-bias Circuit for dc analysis
• Slide 22
• JFET Self Bias Circuit I G = 0 I S = I D From GS Loop: -V GS = V RS or V GS = -I S R S Substituting I S = I D V GS = -I D R S. 22
• Slide 23
• JFET Self Bias Circuit Shockley Equation: 23
• Slide 24
• JFET Self Bias Circuit: Q-Point Self-Bias Line: Since V GS = -I D R S. If I D = 0 then V GS = 0 and I D = I DDS /2 (say), then V GS = -I DDS R S /2 Superimposing this straight line on the transfer curve, we get Q-point as shown in the Fig. 24 Self Bias line Transfer Curve (Shockley equation)
• Slide 25
• JFET Self Bias Circuit DS Loop: Using KVL Substituting I S = I D, or In addition 25
• Slide 26
• JFET Self Bias Circuit: Example 1 Determine the following: V GSQ, I DQ, V DS, V S, V G, and V D. Solution: Step 1: Draw the self bias line: V GS = - I D R S, When I D = 0, V GS = 0. Choosing I D = 4 mA, V GS = -4mA1 k = -4 v The line is drawn below: 26
• Slide 27
• V GS (volts) JFET Self Bias Circuit: Example 1 Step 2: Plot the Shockley equation: (I DSS = 8mA, V P = -6v) 27 V GS 0-3 -4-5-6 I D (mA) 85.5520.880.220 I D (mA)
• Slide 28
• JFET Self Bias Circuit: Example 1 Step 3: Show the Shockley curve and the self bias line on the same graph paper From the graph, V GSQ = -2.6 v, I DQ = 2.6 mA 28 I D (mA) V GS (volts) Self bias line Shockley Curve Q-Point
• Slide 29
• JFET Self Bias Circuit: Example 1 Step 4: Find the remaining quantities: V DS = V DD I D (R S + R D ) = 20 2.6mA( 1 k + 3.3 k ) = 8.82 v V S = I D R S = (2.6mA)(1k ) = 2.6 v V G = 0 v V D = V DS + V S = 8.82 + 2.6 = 11.42 v (or V D = V DD I D R D = 11.42 v) 29
• Slide 30
• JFET Biasing: Voltage Divider Circuit 30
• Slide 31
• JFET Biasing: Voltage Divider Circuit dc analysis 31 V G Applying KVL, or But V RS = I S R S = I D R S Therefore
• Slide 32
• Voltage Divider Circuit: Q-Point Bias Line: (i) When I D = 0 V GS = V G I D R S = V G (0)(R S ) V GS = V G (ii) When V GS = 0 32 Plot this line along with the Shockley Curve, as shown in the Figure.
• Slide 33
• JFET Biasing: Voltage Divider Circuit dc analysis 33 V G From DS Loop:
• Slide 34
• Voltage Divider Circuit: Example Determine the following: (a)I DQ and V GSQ. (b)V D (c)V S (d)V DS (e)V DG. 34
• Slide 35
• Voltage Divider Circuit: Example 1 Solution: I DSS = 8 mA, V p = -4 v. Shockley Equation: Bias Line: 35 V GS -4-20 I D mA024.58 When I D = 0, V GS = 1.82 v For V GS = 0, I D = 1.82/1.5k = 1.21 mA
• Slide 36
• 36 From the Figure, I DQ = 2.4 mA, V GSQ = -1.8 v (b) V D = V DD - I D R D = 16 (2.4mA)(2.4k ) = 10.24 v (c) V S = I D R S = 16 (2.4mA)(2.4k ) = 10.24 v (d) V DS = V DD I D (R D + R S ) = 16 (2.4mA)(2.4k + 1.5k ) = 6.64 v (e) V DG = V D - V G = 10.24 1.82 = 8.42 v
• Slide 37
• Voltage Divider Circuit: Example 2 For the given network, Detrmine (a)V G. (b) I DQ and V GSQ. (c)V D and V S. (d)V DSQ. Solution: (a) (b) I DSS = 10mA, V p = -3.5 v 37
• Slide 38
• 38 V GS (volts)-3.5-20 I D (mA)01.85.110 Bias Line: V GS = V G I D R S = 2.16 I D (1.1k ) When I D = 0, V GS = 2.6 v When V GS = 0, I = 2.16/1.1k = 2mA From the graph, we see that I DQ = 3.3 mA, V GSQ = -1.5 v (c) V D = V DD I DQ R D = 20 - (3.3mA)(2.2k ) = 12.74 v V S = I D R S = 3.63 v (d) V DSQ = V DD I DQ (R D +R S ) = 9.11 v
• Slide 39
• Metal-Oxide -Semiconductor Field Effect Transistor (MOSFET) 39 MOSFET Depletion Type MOSFET Enhancement- Type MOSFET
• Slide 40
• N- Channel Depletion-Type MOSFET 40 Construction of D-MOSFET (n-Channel) The foundation of this type of FET is the substrate (p-type material). The source and drain terminals are connected through metallic contacts to n doped regions linked by an n channel. The gate is also connected to a metal contact surface but remains insulated from the n-channel by a SiO 2 layer.
• Slide 41
• Basic Operation and Charactersitics of N Channel D-MOSFET 41 I D = I S = I DSS V GS = 0 When V GS = 0 and V DS is applied, the drain current I D = I DSS flows through the circuit due to the free electrons of the n-channel.
• Slide 42
• Basic Operation and Characterstics of N-Channel D-MOSFET 42 When V GS < 0, recombination between electrons and holes occurs. The more negative the bias, the higher the rate of combination. The resulting level of I D is reduced and becomes zero at pinch-off voltage. Electrons repelled by negative Potential at gate.
• Slide 43
• Basic Operation and Charactersitics of D- MOSFET 43 When V GS > 0, the gate will draw additional electrons from the p- substrate due to the reverse leakage current and the drain current increases at a rapid rate.
• Slide 44
• Example: Sketch the transfer characteristics for an n- channel depletion type MOSFET with I DSS = 10 mA and V p = -4 v. Solution: 44 V GS -4-20+1 I D (mA)02.55.61015.6 The curve is plotted on the next slide.
• Slide 45
• 45 Gate Source Voltage Drain Current (A)
• Slide 46
• P-Channel depletion type MOSFET 46
• Slide 47
• Symbols 47 N-Channel P-Channel
• Slide 48
• Example1: For the n-channel depletion type MOSFET of the Fig., determine (a) I DQ and V GSQ. (b) V DS. Solution: Shockley Equation: 48 18 v V GS -3-201 I D (mA)00.72.7610.7
• Slide 49
• Bias Line: 49 When I D = 0, V GS = 1.5, When V GS = 0, I D = V G /R S = 1.5/750 = 2 mA From the graph, I DQ = 3.1 mA, V GSQ = -0.8 v V DS = V DD I D (R D + R S ) = 10.1 v
• Slide 50
• Example2: Determine the following for the given network. (a) I DQ and V GSQ (b) V D. Solution: (a) Shockley Equation: 50 V GS -8-6-5-4-2012 I D (mA) 00.51.1252.004.58.0010.12512.5
• Slide 51
• Bias Line: V GS = -I D R S. When V GS = 0, I D = 0. When I D = 2.5 mA (say) V GS = -2.5 10 -3 2.4 1000 = -6V 51 I D (mA) V GS (volts) From the graph paper V GSQ = 4.3 V, I D = 1.7mA (b) V D = V DD I D R D = 20 (1.7mA)(6.2k ) = 9.46 V Shockley Equation Bias Line Q-Point
• Slide 52