EIE209 Basic Electronics Diode Circuits

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EIE209 Basic Electronics Diode Circuits

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What is a diode? Simplest view (no physics): — a unidirectional device that allows current to flow in one direction but not the other. Ideally, we regard a diode as short circuit when voltage applied to it in the forward manner is positive.

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  1. EIE209 Basic Electronics Diode circuits Contents • Diode models • Typical diode circuits & applications • Load line concepts for nonlinear load analysis Prof. C.K. Tse: Diode circuits 1
  2. What is a diode? Simplest view (no physics): — a unidirectional device that allows current to flow in one direction but not the other. Ideally, we regard a diode as short circuit when voltage applied to it in the forward manner is positive. Prof. C.K. Tse: Diode circuits 2
  3. Ideal characteristic Bias conditions Forward bias : vd > 0 — current can flow and id > 0. Reverse bias : vd < 0 — current cannot flow and id = 0. id id Ideal diode with finite forward drop (more realistic) vd vd 0.7V Prof. C.K. Tse: Diode circuits 3
  4. Real characteristic Take a closer look at the characteristic around the turning point. The i-v characteristic is an exponential function. id From physics, we have id = I ss (e q vd /kT - 1) Also, the diode can only stand the –VBD negative voltage up to a certain vd threshold VBD , beyond which breakdown 0.7V the diode conducts reverse current (breakdown). Prof. C.K. Tse: Diode circuits 4
  5. Which model to use? The choice depends on the external voltage magnitudes. Ideal model: Ideal model: Ideal model: id = 100/100 = 1 A id = 10/100 = 100 mA id = 2/100 = 20 mA With 0.7V drop: With 0.7V drop: With 0.7V drop: id = (100–0.7)/100 id = (10–0.7)/100 id = (2–0.7)/100 = 0.997 A = 93 mA = 13 mA Prof. C.K. Tse: Diode circuits 5
  6. Example: rectifier circuit The ideal model is valid if the external voltages are well above 0.7V. What is the magnitude of vr ? Crude — 50 V Better — 50 – 1.4 = 48.6 V Prof. C.K. Tse: Diode circuits 6
  7. Application examples clamper dc restorer half-wave doubler full-wave doubler Prof. C.K. Tse: Diode circuits 7
  8. A nonlinear circuit problem Suppose we wish to find vd , given that the external voltage Vs is not large enough to validate the use of the ideal diode model. ?? Prof. C.K. Tse: Diode circuits 8
  9. Step 1 : locating the operating points Recall: The characteristic curve/line for a device actually defines where the point (v, i) can lie. We know 1 the operating point (vd, 2 the operating point (vR, iR) id) is somewhere on the is somewhere on the diode characteristic curve resistor characteristic curve Prof. C.K. Tse: Diode circuits 9
  10. Step 2 : KVL & KCL constraints We also know from KCL : id = iR AND from KVL : vd + vR = Vs = Vs O+ = Prof. C.K. Tse: Diode circuits 10
  11. Step 3 : enforcing KVL & KCL Method: flip the resistor curve horizontally; and push the two curves together horizontally until the y-axes are Vs apart. push push add to Vs Prof. C.K. Tse: Diode circuits 11
  12. Solution : load line The flipped resistor line is called the LOAD LINE. Prof. C.K. Tse: Diode circuits 12
  13. General problem How to find vd and id ? + vR – iR id R id + + Vs _ :) vd – nonlinear device vd Prof. C.K. Tse: Diode circuits 13
  14. Basic load line construction + vR – iR i R id slope = –1/R + + Vs _ :) vd id device – characteristic nonlinear device v vd Vs Prof. C.K. Tse: Diode circuits 14
  15. Alternative construction + vR – iR i R id id when device is + + short-circuit Vs _ :) vd device – characteristic nonlinear device load line v vd when device is open- circuit Prof. C.K. Tse: Diode circuits 15
  16. Tutorial problem Find the operating point. Prof. C.K. Tse: Diode circuits 16
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