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Analogue Electronics I EAB 2014 Jan 2009 Lecture 8 - MOSFETs DC Biasing 1 Topics To Cover • DC Biasing For Depletion-Type and Enhancement-Type MOSFETs 2 MOSFETs Biasing… why? We are interested in finding the quiescent point (Q-point) of the response in the circuit (current and/or voltage) Steps to obtain the Q-point 1. Plot the transfer characteristics curve (ID vs VGS) 2. Draw the DC load line (network characteristics) superimposed on the curve 3. Obtain the Q-point by finding the intersection point of the two plots. 3 Depletion type MOSFETs Use Shockley’s Equation for simpler calculation @ shorthand method VGS versus ID 2 ⎛ V GS ⎞ VGS ID I D = I DSS ⎜1 − ⎜ ⎟ ⎟ 0 IDSS ⎝ VP ⎠ 0.3 VP IDSS / 2 0.5 VP IDSS / 4 VP 0 4 Example 1 (voltage-divider) For the circuit given, determine: a) IDQ and VGSQ b) VDS 5 Example 1 … ID(mA) 11 Step 1: Plot the Transfer Characteristics Curve. 10 9 8 Using Shockley’s Equation: 7 IDSS 6 When VGS = 0, ID = IDSS 5 ID = 0, VGS = VP 4 3 When VGS = VP/2 = -1.5 V, 2 ID = IDSS/4 = 1.5 mA VGS(V) 1 -3 -2 -1 1 VP 6 Example 1 … ID(mA) 11 Consider when VGS = +1 V, 10 9 2 ⎛ V GS ⎞ 8 I D = I DSS ⎜1 − ⎜ ⎟ ⎟ 7 ⎝ VP ⎠ IDSS 6 2 ⎛ 1 ⎞ = (6 mA ) ⎜ 1 − 5 ⎟ ⎝ − 3 ⎠ 4 3 = 10 . 67 mA 2 VGS(V) 1 -3 -2 -1 1 VP 7 Example 1 … Step 2: Draw the DC load line. Find equation for VGS using voltage divider relationship: Characteristics curve ⎛ 10 M ⎞ VG = ⎜ ⎟ ×18V = 1.5V ⎝ 10 M + 110 M ⎠ DC Load Line VGS = VG − I D RS = 1.5V − I D (750 Ω ) Intersection Set I D = 0 mA (Q-point) VGS = VG = 1.5V Set VGS = 0V VG ID = = 2 mA RS 8 Example 1 … Step 3: Obtain the Q-point values a) From the plot: IDQ = 3.1 mA and VGSQ = – 0.8 V b) VDS = VDD – ID (RD + RS) = 18 – (3.1m)(1.8k + 750) = 10.1 V 9 Example 2 (self-bias) For the network given, determine: a) IDQ and VGSQ b) VD 10 Example 2 (self-bias) Simplify the self-bias network: 20 V 20 V 6.2kΩ 6.2kΩ Vo IDSS=8mA IDSS=8mA Vi V P= -8V VP= -8V 1M Ω 2.4kΩ 2.4kΩ 11 Example 2 (self-bias) Step 1: Plot the Transfer Characteristics Curve. Using Shockley’s Equation: ID(mA) When VGS = 0, ID = IDSS 11 10 ID = 0, VGS = VP 9 IDSS When VGS = VP/2 = -4 V, 8 ID = IDSS/4 = 2 mA 7 6 Consider when VGS = +1 V, 5 2 ⎛ V ⎞ 4 I D = I DSS ⎜ 1 − GS ⎟ ⎜ ⎝ VP ⎟ ⎠ 3 2 2 ⎛ 1 ⎞ = (6 m ) ⎜ 1 − ⎟ 1 ⎝ −8⎠ VGS(V) = 10 . 13 mA -8 -7 -6 -5 -4 -3 -2 -1 VP 1 2 12 Example 2 (self-bias) Step 2: Draw the DC load line. From the input loop: VGS = − I D RS = − I D (2.4 kΩ ) Characteristics curve Set I D = 0 mA VGS = 0V DC Load Line Intersection (Q-point) Set VGS = −6V VG ID = − = 2.5 mA RS 13 Example 2 (self-bias) Step 3: Obtain the Q-point values a) From the plot: IDQ = 1.7 mA and VGSQ = – 4.3 V b) VD = VDD – IDRD = 20 – (1.7m)(6.2k) = 9.46 V 14 Enhancement-Type MOSFETs Different equation is used: I D = k (V GS − V T )2 I D ( on ) k = (V GS ( on ) − VT )2 VGS versus ID VGS ID VGS(on) ID(on) VT 0 15 Example 3 (feedback-biasing) For the circuit given, determine: a) IDQ and VDSQ 16 Example 3 … Simplify the self-bias network: 12 V 12 V 2kΩ 2kΩ Vo 10M Ω I D(on)= 6mA I D(on)= 6mA VGS(on)= 8V VGS(on)= 8V Vi VGS(th)= 3V VGS(th)= 3V 17 Example 3 … Step 1: Plot the Transfer Characteristics Curve. I D (on ) Find value of k: k = (V GS (on ) − V T ) 2 6m = (8 − 3 )2 = 0 . 24 × 10 − 3 A / V 2 Select a few points: For VGS = 6 V, For VGS = 10 V, ID = k(VGS – VGS(T))2 ID = (0.24m) x (10 – 3)2 = (0.24m) x (6 – 3)2 = 11.76 mA = 2.61 mA 18 Example 3 … 19 Example 3 … Step 2: Draw the DC load line. Characteristics curve From output circuit: VDS = VGS = VDD – IDRD DC Load Line Intersection ID = 0 (Q-point) VGS = VDD = 12 V VGS = 0 ID = VDD / RD = 6 mA 20 Example 3 … Step 3: Obtain the Q-point values a) From the plot: IDQ = 2.75 mA and VDSQ = 6.4 V 21 Example 4 (voltage-divider) For the circuit given, determine: a) IDQ b) VGSQ c) VDSQ 22 Example 4 (voltage divider) Step 1: Plot the Transfer Characteristics Curve. I D (on ) Find value of k: k = (V GS (on ) − V GS (T ) )2 3m = (10 − 5 )2 = 0 . 12 × 10 − 3 A / V 2 Select a few points: For VGS = 15 V, For VGS = 20 V, ID = k(VGS – VGS(T))2 ID = (0.12m) x (20 – 5)2 = (0.12m) x (15 – 5)2 = 27 mA = 12 mA 23 Example 4 (voltage divider) ID(mA) 40 30 20 10 ID(on) V GS(V) 5 10 15 20 25 30 VGS(th) VGS(on) 24 Example 4 (voltage divider) Step 2: Draw the DC load line. ⎛ R2 ⎞ VG =⎜⎜R +R ⎟ V DD ⎟ ⎝ 1 2 ⎠ = 18 V From the input circuit: At ID = 0, VGS = VG – IDRS VGS = 18 V = 18 – ID(0.82k) At VGS = 0, ID = 21.95 mA 25 Example 4 (voltage divider) Characteristics curve DC Load Line Intersection (Q-point) 26 Example 4 (voltage divider) Step 3: Obtain the Q-point values a) From the plot: IDQ = 6.7 mA and VGSQ = 12.5 V b) From the output circuit: VDS = VDD – ID(RS + RD) = 40 – (6.7m)(0.82k + 3k) = 14.4 V 27 Question & Answer 28

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mosfet, gate, drain, source, bias, dc bias, voltage divider, self bias, feedback bias

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posted: | 4/15/2012 |

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MOSFET DC Biasing for Depletion and Enhancement. Cover Voltage Divider, Self-Bias, and Feedback Bias.

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