Bài giảng Mạch điện I - Chương 3: Lý thuyết Diode

Chương 3Lý thuyết diodeTừ Vựng (1)anodebulk resistance = điện trở khốicathodediodeideal diode = diode lý tưởngknee voltage = điện áp gốilinear device = dụng cụ tuyến tínhload line = đường tảiTừ Vựng (2)maximum forward current = dòng thuận cực đạinonlinear device = dụng cụ phi tuyếnOhmic resistance = điện trở Ohmpower rating = định mức công suấtup-down analysis = phân tích tăng-giảmNội dung chương 33-1 Các ý tưởng cơ bản3-2 Diode lý tưởng3-3 Xấp xỉ bậc 23-4 Xấp xỉ bậc 33-5 Trounleshooting3-6 Phân tích mạch tăng-giảm3-7 Đọc bảng dữ liệu3-8 Cách tính điện trở khối 3-9 Điện trở DC của diode3-10 Đường tải3-11 Diode dán bề mặtProperties of DiodesKristin Ackerson, Virginia Tech EESpring 2002Figure 1.10 – The Diode Transconductance Curve2VD = Bias VoltageID = Current through Diode. ID is Negative for Reverse Bias and Positive for Forward BiasIS = Saturation CurrentVBR = Breakdown VoltageV = Barrier Potential VoltageVDID(mA)(nA)VBR~VISProperties of DiodesThe Shockley EquationKristin Ackerson, Virginia Tech EESpring 2002The transconductance curve on the previous slide is characterized by the following equation:ID = IS(eVD/VT – 1)As described in the last slide, ID is the current through the diode, IS is the saturation current and VD is the applied biasing voltage.VT is the thermal equivalent voltage and is approximately 26 mV at room temperature. The equation to find VT at various temperatures is:VT = kT	 q k = 1.38 x 10-23 J/K T = temperature in Kelvin q = 1.6 x 10-19 C is the emission coefficient for the diode. It is determined by the way the diode is constructed. It somewhat varies with diode current. For a silicon diode  is around 2 for low currents and goes down to about 1 at higher currentsDiode Circuit ModelsKristin Ackerson, Virginia Tech EESpring 2002The Ideal Diode ModelThe diode is designed to allow current to flow in only one direction. The perfect diode would be a perfect conductor in one direction (forward bias) and a perfect insulator in the other direction (reverse bias). In many situations, using the ideal diode approximation is acceptable.Example: Assume the diode in the circuit below is ideal. Determine the value of ID if a) VA = 5 volts (forward bias) and b) VA = -5 volts (reverse bias)+_VAIDRS = 50 a) With VA > 0 the diode is in forward bias and is acting like a perfect conductor so: ID = VA/RS = 5 V / 50  = 100 mAb) With VA 0 the diode is in forward bias and is acting like a perfect conductor so write a KVL equation to find ID:	0 = VA – IDRS - V 	ID = VA - V = 4.7 V = 94 mA 	 RS 50  V+V+Diode Circuit ModelsThe Ideal Diode with Barrier Potential and Linear Forward Resistance This model is the most accurate of the three. It includes a linear forward resistance that is calculated from the slope of the linear portion of the transconductance curve. However, this is usually not necessary since the RF (forward resistance) value is pretty constant. For low-power germanium and silicon diodes the RF value is usually in the 2 to 5 ohms range, while higher power diodes have a RF value closer to 1 ohm.Linear Portion of transconductance curveVDIDVDIDRF = VD IDKristin Ackerson, Virginia Tech EESpring 2002+VRFDiode Circuit ModelsThe Ideal Diode with Barrier Potential and Linear Forward Resistance Kristin Ackerson, Virginia Tech EESpring 2002Example: Assume the diode is a low-power diode with a forward resistance value of 5 ohms. The barrier potential voltage is still: V = 0.3 volts (typical for a germanium diode) Determine the value of ID if VA = 5 volts.+_VAIDRS = 50 V+RFOnce again, write a KVL equation for the circuit:0 = VA – IDRS - V - IDRFID = VA - V = 5 – 0.3 = 85.5 mA RS + RF 50 + 5Diode Circuit ModelsKristin Ackerson, Virginia Tech EESpring 2002Values of ID for the Three Different Diode Circuit ModelsIdeal Diode ModelIdeal Diode Model with Barrier Potential VoltageIdeal Diode Model with Barrier Potential and Linear Forward ResistanceID100 mA94 mA85.5 mAThese are the values found in the examples on previous slides where the applied voltage was 5 volts, the barrier potential was 0.3 volts and the linear forward resistance value was assumed to be 5 ohms.Các mô hình diodeGraphical PN-Junction Diode V-I CharacteristicReverse breakdownForward Bias RegionReverse Bias RegionMathematical ApproximationIdeal Diode ModelAn ideal diode is a two-terminal device defined by the following non-linear (currentvoltage) iv-characteristic:Constant-Voltage-Drop ModelMô hình với điện trở thuậnMô hình tín hiệu nhỏDiode Spice ModelRs is inevitable series resistance of a real device structure. Current controlled current source represents ideal exponential behavior of diode. Capacitor specification includes depletion-layer capacitance for reverse-bias region as well as diffusion capacitance associated with junction under forward bias.Typical default values: Saturation current = 10 fA, Rs = 0W, Transit time = 0 secondsGiải tích mạch diodeDiode Circuit Analysis: BasicsV and R may represent Thévenin equivalent of a more complex 2-terminal network.Objective of diode circuit analysis is to find quiescent operating point for diode, consisting of dc current and voltage that define diode’s i-v characteristic.Loop equation for given circuit is:This is also called the load line for the diode. Solution to this equation can be found by: Graphical analysis using load-line method. Analysis with diode’s mathematical model. Simplified analysis with ideal diode model. Simplified analysis using constant voltage drop model.Load-Line Analysis (Example)Problem: Find Q-pointGiven data: V=10 V, R=10kW.Analysis:To define the load line we use,VD= 0VD= 5 V, ID =0.5 mAThese points and the resulting load line are plotted.Q-point is given by intersection of load line and diode characteristic:Q-point = (0.95 mA, 0.6 V)Problem: Find Q-point for given diode characteristic.Given data: IS = 10-13 A, n = 1, VT = 0.025 VAnalysis: This equation is transcendental, so it has no closed form solution.Analysis using Mathematical Model for DiodeBy iteratively guessing values for VD and increasing or decreasing VD until right side of equation equals 10:Q-point = ( 0.943 mA, 0.574 V)2 or 3 significant digits for VD is usually plenty since IS, n, VT, and R are rarely know to better precision.Typically, we use SPICE if we want to use full math model.Analysis using Ideal Model for DiodeIf diode is forward-biased, voltage across diode is zero. If diode is reverse-biased, current through diode is zero.vD =0 for iD >0 and iD =0 for vD 0 and vD = 0 for vD < Von.Ideal diode model is CVD model with Von = 0V.Two-Diode Circuit AnalysisAnalysis: Ideal diode model is chosen. Since 15V source is forcing positive current through D1 and D2 and -10V source is forcing positive current through D2, assume both diodes are on.Since voltage at node D is zero due to short circuit of ideal diode D1,Q-points are (-0.5 mA, 0 V) and (2.0 mA, 0 V) But, ID1 <0 is not allowed by diode, so try again.Two-Diode Circuit Analysis (contd.)Since current in D1 is zero, ID2 = I1,Q-points are D1 : (0 mA, -1.67 V):off D2 : (1.67 mA, 0 V) :onAnalysis: Since current in D2 but that in D1 is invalid, the second guess is D1 off and D2 on.Điểm tĩnh QThe Q PointKristin Ackerson, Virginia Tech EESpring 2002The operating point or Q point of the diode is the quiescent or no-signal condition. The Q point is obtained graphically and is really only needed when the applied voltage is very close to the diode’s barrier potential voltage. The example 3 below that is continued on the next slide, shows how the Q point is determined using the transconductance curve and the load line.+_VA= 6VIDRS = 1000 V+First the load line is found by substituting in different values of V into the equation for ID using the ideal diode with barrier potential model for the diode. With RS at 1000 ohms the value of RF wouldn’t have much impact on the results.ID = VA – V  RSUsing V  values of 0 volts and 1.4 volts we obtain ID values of 6 mA and 4.6 mA respectively. Next we will draw the line connecting these two points on the graph with the transconductance curve. This line is the load line.The Q PointID (mA)VD (Volts)246810120.20.40.60.81.01.21.4The transconductance curve below is for a Silicon diode. The Q point in this example is located at 0.7 V and 5.3 mA.4.6Kristin Ackerson, Virginia Tech EESpring 20020.75.3Q Point: The intersection of the load line and the transconductance curve.Dynamic ResistanceKristin Ackerson, Virginia Tech EESpring 2002The dynamic resistance of the diode is mathematically determined as the inverse of the slope of the transconductance curve. Therefore, the equation for dynamic resistance is:rF = VT IDThe dynamic resistance is used in determining the voltage drop across the diode in the situation where a voltage source is supplying a sinusoidal signal with a dc offset.The ac component of the diode voltage is found using the following equation:vF = vac rF 	rF + RSThe voltage drop through the diode is a combination of the ac and dc components and is equal to:VD = V + vFDynamic ResistanceKristin Ackerson, Virginia Tech EESpring 2002Example: Use the same circuit used for the Q point example but change the voltage source so it is an ac source with a dc offset. The source voltage is now, vin = 6 + sin(wt) Volts. It is a silicon diode so the barrier potential voltage is still 0.7 volts.+vinIDRS = 1000 V+The DC component of the circuit is the same as the previous example and therefore ID = 6V – 0.7 V = 5.2 mA 1000  rF = VT = 1 * 26 mV = 4.9 	 ID 5.3 mA = 1 is a good approximation if the dc current is greater than 1 mA as it is in this example.vF = vac rF = sin(wt) V 4.9  = 4.88 sin(wt) mV rF + RS	 4.9  + 1000 Therefore, VD = 700 + 4.9 sin (wt) mV (the voltage drop across the 	 diode)