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S. RossEECS 40 Spring 2003 Lecture 14 Last time… I V +_+_ Reverse breakdown Reverse bias Forward bias I V ZK VFVF V Focus we introduced the diode and its (complicated) I-V relationship. Today we will… focus on the relevant area of the diode I-V graph develop simpler models for the diode I-V relationship learn how to solve circuits with nonlinear elements

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S. RossEECS 40 Spring 2003 Lecture 14 DIFFERENT MODELS, DIFFERENT USES We will consider 4 different diode I-V models with varying degrees of detail. Use most realistic model only for very precise calculations Use simpler models to find basic operation, gain intuition Sometimes one model may lead to an “impossible” situation: use a different (more realistic) model in this case

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S. RossEECS 40 Spring 2003 Lecture 14 REALISTIC DIODE MODEL I V I V +_+_ Here, V T is “thermal voltage”: V T = (kT)/q ≈ 0.026 V @ 300 o K (q is electron charge in C, k is Boltzmann’s constant, and T is the operating temperature in o K) Equation is valid for all modes of operation considered You might need a computer to solve the nonlinear equation this model can create

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S. RossEECS 40 Spring 2003 Lecture 14 IDEAL DIODE MODEL I V +_+_ V I Reverse bias Forward bias I V +_+_ Diode either has negative voltage and zero current, or zero voltage and positive current Diode behaves like a switch: open in reverse bias mode, closed (short circuit) in forward bias mode Guess which situation diode is in, see if answer makes sense

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S. RossEECS 40 Spring 2003 Lecture 14 LARGE-SIGNAL DIODE MODEL I V +_+_ V I Reverse bias Forward bias I V Diode either has voltage less than V F and zero current, or voltage equal to V F and positive current Diode behaves like a voltage source and switch: open in reverse bias mode, closed in forward bias mode Guess which situation diode is in, see if answer makes sense VFVF +-+- + - VFVF

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S. RossEECS 40 Spring 2003 Lecture 14 SMALL-SIGNAL DIODE MODEL I V +_+_ V I Reverse bias Forward bias I V Diode either has voltage less than V F and zero current, or voltage greater than V F and positive current depending on V Diode behaves like a voltage source, resistor and switch: open in reverse bias mode, closed in forward bias mode Guess which situation diode is in, see if answer makes sense VFVF +-+- + - VFVF RDRD slope = 1/R D

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S. RossEECS 40 Spring 2003 Lecture 14 SOLVING CIRCUITS WITH NONLINEAR ELEMENTS Look at circuits with a nonlinear element like this: Linear circuit ILIL +VL-+VL- Nonlinear element I NL + V NL - A nonlinear element with its own I-V relationship, attached to a linear circuit with its own I-V relationship. Equations we get: 1.I L = f L (V L )(linear circuit I-V relationship) 2.I NL = f NL (V NL ) (nonlinear element I-V relationship) 3.I NL = -I L 4.V NL = V L

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S. RossEECS 40 Spring 2003 Lecture 14 SOLVING CIRCUITS WITH NONLINEAR ELEMENTS Our 4 equations 1.I L = f(V L )(linear circuit I-V relationship) 2.I NL = g(V NL ) (nonlinear element I-V relationship) 3.I NL = -I L 4.V NL = V L can easily become just 2 equations in I NL and V NL 1.I NL = -f L (V NL ) 2.I NL = f NL (V NL ) which we can equate and solve for V NL, or… graph the two equations and solve for the intersection.

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S. RossEECS 40 Spring 2003 Lecture 14 LOAD LINE ANALYSIS To find the solution graphically, V NL I NL graph the nonlinear I-V relationship, graph the linear I-V relationship in terms of I NL and V NL (reflect over y-axis), f NL (V NL ) -f L (V NL ) and find the intersection: the voltage across and current through the nonlinear element. x

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S. RossEECS 40 Spring 2003 Lecture 14 EXAMPLE +-+- I NL V NL +_+_ 2 V 1 k VLVL +_+_ ILIL Find V NL. Assume realistic diode model with I 0 = 10 -15 A. 1.I L = (V L - 2) / 1000 2. 3.I NL = -I L 4.V NL = V L Either substitute into 3. and solve or determine graphically that V NL = 0.725 V

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S. RossEECS 40 Spring 2003 Lecture 14

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S. RossEECS 40 Spring 2003 Lecture 14 EXAMPLE REVISITED +-+- I NL V NL +_+_ 2 V 1 k VLVL +_+_ ILIL Find V NL. Assume small-signal diode model with V F = 0.7 V and R D = 20 . 1.I L = (V L - 2) / 1000 2.I NL = (V NL – 0.7) / 20 or I NL = 0 3.I NL = -I L 4.V NL = V L Either substitute into 3. and solve (V NL – 0.7) / 20 = -(V NL - 2) / 1000 or determine graphically that V NL = 0.725 V

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S. RossEECS 40 Spring 2003 Lecture 14

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S. RossEECS 40 Spring 2003 Lecture 14 ONE MORE TIME +-+- I NL V NL +_+_ -2 V 1 k VLVL +_+_ ILIL Find V NL. Assume small-signal diode model with V F = 0.7 V and R D = 20 . 1.I L = (V L - - 2) / 1000 2.I NL = (V NL – 0.7) / 20 or I NL = 0 3.I NL = -I L 4.V NL = V L Either substitute into 3. and solve 0 = -(V NL - - 2) / 1000 or determine graphically that V NL = -2 V

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S. RossEECS 40 Spring 2003 Lecture 14

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