Lecture 13 Review - Whats in the Exam Lecture slides...

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1 ECTE290 Spring 2007 Lecture 13 2 What’s in the Exam ± Lecture slides ± Homework ± Tutorial questions ± Textbook for additional reading 3 Diodes and Applications ± Diodes ± The IDEAL Diode model ± Constant Voltage Drop Model ± Zener Diodes ± Applications of diodes: ± Half Wave Rectification ± Battery Charger ± Full Wave Rectification circuits ± Centre tapped transformer ± Bridge rectifier ± Clamping and Limiting circuits 4 Ideal Diode Model: I-V Characteristics In Reverse Bias Diode is Cutoff or Simply ‘Off’ In Forward Bias Diode is turned on or Simply ‘On’ Symbol for Ideal Diode
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2 5 Methodology: Solving The Ideal Diode Circuit 1. Assuming a diode conduction state. ± ON (V D =0, I D >0) ± OFF (V D <0, I D =0) 2. Substitute ideal circuit model into circuit, short circuit if ‘ON’, open circuit if ‘OFF’ 3. Solve the diode current and voltage using KVL, KCL… 4. If the solution is consistent with the assumption, then the initial assumption was correct. If not, the diode conduction state is opposite to that initially assumed. In an ideal diode model, the diode can be in one of two states: it is either OFF, or it is ON. In general, when we solve a circuit using this model, we do not know which state the diode is in. Therefore, we use the following approach. We guess, and then we test that guess. 6 Example 2: Ideal Diode Given V S =12 V; V B =11 V; R 1 =5 Ω; R 2 =4 Find the conduction state V S R 1 v D + - R 2 V B Answer : Assume: Diode is off 2 1 R R V V i B s + = Rearranging gives: 0 2 1 = + + + B s V iR iR V V S R 1 v D + - R 2 V B v 1 i i KVL around outer loop gives: (V D <0, I D =0) 7 Example 2: Ideal Diode cont Given V S =12 V; V B =11 V; R 1 =5 R 2 =4 Find the conduction state V S R 1 v D + - R 2 V B Answer : Assuming: Diode is off 0 1 = + D s iR V ν V S R 1 v D + - R 2 V B v 1 i i KVL around leftmost loop gives: 2 1 R R V V i B s + = Since: Rearranging gives for Diode voltage: 1 2 1 R R R V V V v B s s D + + = + - (V D <0, I D =0) 8 Example 2: Ideal Diode cont Given V S =12 V; V B =11 V; R 1 =5 R 2 =4 Find the conduction state V S R 1 v D + - R 2 V B Answer : Assuming: Diode is off V S R 1 v D + - R 2 V B v 1 i i Subbing in values we get: Volts R R R V V V v B s s D 44 . 11 1 2 1 = + + = + - This is consistent with the assumption So it is true that the diode is off Reverse biased (V D <0, I D =0)
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3 9 Diode Models Forward Bias: Constant Voltage Drop Model ± For Silicon diodes, very little current flows until V 0.5V ± At V 0.7V, the diode characteristics are nearly vertical ± In the vicinity of V 0.7V, a wide range of current may flow. ± The forward voltage drop of a diode is often assumed to be R D =0, V D = 0.7V (for silicon) ± Diodes made of different material have different voltage drops ± V 0.2V - 2.4V Also called: Offset Diode model 10 Example 2: Offset Diode model ± Let’s try out this on some simple circuits.
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Lecture 13 Review - Whats in the Exam Lecture slides...

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