LÝ THUYẾT DIODE (MẠCH điện tử SLIDE)

        anode bulk resistance = điện trở khối cathode diode ideal diode = diode lý tưởng knee voltage = điện áp gối linear device = dụng cụ tuyến tính load line = đường tải      maximum forward current = dòng thuận cực đại nonlinear device = dụng cụ phi tuyến Ohmic resistance = điện trở Ohm power rating = định mức cơng suất up-down analysis = phân tích tăng-giảm 3-1 Các ý tưởng 3-2 Diode lý tưởng 3-3 Xấp xỉ bậc 3-4 Xấp xỉ bậc 3-5 Trounleshooting 3-6 Phân tích mạch tăng-giảm 3-7 Đọc bảng liệu 3-8 Cách tính điện trở khối 3-9 Điện trở DC diode 3-10 Đường tải 3-11 Diode dán bề mặt Properties of Diodes Figure 1.10 – The Diode Transconductance Curve ID (mA) • VD = Bias Voltage • ID = Current through Diode ID is Negative for Reverse Bias and Positive for Forward Bias IS VBR ~V VD • IS = Saturation Current • VBR = Breakdown Voltage • V = Barrier Potential Voltage (nA) Kristin Ackerson, Virginia Tech EE Spring 2002 Properties of Diodes The Shockley Equation • The 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: k = 1.38 x 10-23 J/K VT = kT q 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 for low currents and goes down to about at higher currents Kristin Ackerson, Virginia Tech EE Spring 2002 Diode Circuit Models The Ideal Diode Model The 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 = volts (forward bias) and b) VA = -5 volts (reverse bias) a) With VA > the diode is in forward bias and is acting like a perfect conductor so: RS = 50  ID VA + _ ID = VA/RS = V / 50  = 100 mA b) With VA < the diode is in reverse bias and is acting like a perfect insulator, therefore no current can flow and ID = Kristin Ackerson, Virginia Tech EE Spring 2002 Diode Circuit Models The Ideal Diode with Barrier Potential This model is more accurate than the simple ideal diode model because it includes the approximate barrier potential voltage Remember the barrier potential voltage is the + V voltage at which appreciable current starts to flow Example: To be more accurate than just using the ideal diode model include the barrier potential Assume V = 0.3 volts (typical for a germanium diode) Determine the value of ID if VA = volts (forward bias) RS = 50  ID VA + _ With VA > the diode is in forward bias and is acting like a perfect conductor so write a KVL equation to find ID: = VA – IDRS - V V + ID = VA - V = 4.7 V = 94 mA RS 50  Kristin Ackerson, Virginia Tech EE Spring 2002 Diode Circuit Models The Ideal Diode with Barrier Potential and Linear Forward Resistance + V RF RF =  VD 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 to ohms range, while higher power diodes have a RF value closer to ohm ID Linear Portion of transconductance curve  ID  ID VD  VD Kristin Ackerson, Virginia Tech EE Spring 2002 Diode Circuit Models The Ideal Diode with Barrier Potential and Linear Forward Resistance Example: Assume the diode is a low-power diode with a forward resistance value of ohms The barrier potential voltage is still: V = 0.3 volts (typical for a germanium diode) Determine the value of ID if VA = volts RS = 50  Once again, write a KVL equation for the circuit: ID VA + _ V = VA – IDRS - V - IDRF + ID = VA - V = – 0.3 = 85.5 mA RS + RF 50 + RF Kristin Ackerson, Virginia Tech EE Spring 2002 Rs 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 = 0, Transit time = seconds Loop equation for given circuit is: V  I D R  VD V and R may represent Thévenin equivalent of a more complex 2terminal 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 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 Problem: Find Q-point Given data: V=10 V, R=10k Analysis: 10  I D 104  VD To define the load line we use, VD= I D (10V / 10k) 1mA VD= V, ID =0.5 mA These 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 V Analysis: By 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) 10  I D 104  VD or significant digits for VD is   VD     1 10 13  exp 40VD   1 usually plenty since IS, n, VT, and R I D  I S exp   nVT   are rarely know to better precision  10 10410 13  exp 40VD   1  VD This equation is transcendental, so it has no closed form solution Typically, we use SPICE if we want to use full math model If 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 < Thus diode is assumed to be either on or off Analysis is conducted in following steps: • Guess diode’s region of operation from circuit • Analyze circuit using diode model appropriate for assumed operation region • Check results to check consistency with assumptions Assume diode is on (since it looks like anode will be at high voltage than cathode) I  (10  0)V 1mA D 10k I D 0 (our assumption is right) Q-point is(1 mA, 0V) Assume diode is off Hence ID =0 Loop equation is: 10  VD  104 I D 0  VD  10V our assumption is right Q-point is (0, -10 V) Analysis: Assume diode is on (10  Von )V 10k (10  0.6) V  0.94 mA 10k ID  vD = Von for iD >0 and vD = for vD < Von Ideal diode model is CVD model with Von = 0V Analysis: 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, I1  (15  0)V 1.5 mA 10k I1 I D1  I D2 I D2 0  ( 10)V 2 mA 5k I D1 1.5   0.5 mA Q-points are (-0.5 mA, V) and (2.0 mA, V) But, ID1