Electronic principles - Chapter 3 ppt

Chương 3 Lý thuyết diode Từ Vựng (1) • 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 Từ Vựng (2) • maximum forward current = dòng thuận cực đại • nonlinear device = dụng cụ phi tuyến • Ohmic resistance = điện trở Ohm • power rating = định mức công suất • up-down analysis = phân tích tăng-giảm Nội dung chương 3 3-1 Các ý tưởng cơ bản 3-2 Diode lý tưởng 3-3 Xấp xỉ bậc 2 3-4 Xấp xỉ bậc 3 3-5 Trounleshooting 3-6 Phân tích mạch tăng-giảm 3-7 Đọc bảng dữ liệu 3-8 Cách tính điện trở khối 3-9 Điện trở DC của diode 3-10 Đường tải 3-11 Diode dán bề mặt Properties of Diodes Properties of Diodes Kristin Ackerson, Virginia Tech EE Kristin Ackerson, Virginia Tech EE Spring 2002 Spring 2002 Figure 1.10 – The Diode Transconductance Curve Figure 1.10 – The Diode Transconductance Curve 2 2 • V V D D = Bias Voltage = Bias Voltage • I I D D = Current through = Current through Diode. I Diode. I D D is Negative is Negative for Reverse Bias and for Reverse Bias and Positive for Forward Positive for Forward Bias Bias • I I S S = Saturation = Saturation Current Current • V V BR BR = Breakdown = Breakdown Voltage Voltage • V V φ φ = Barrier Potential = Barrier Potential Voltage Voltage V V D D I I D D (mA) (mA) (nA) (nA) V V BR BR ~V ~V φ φ I I S S Properties of Diodes Properties of Diodes The Shockley Equation The Shockley Equation Kristin Ackerson, Virginia Tech EE Kristin Ackerson, Virginia Tech EE Spring 2002 Spring 2002 • The transconductance curve on the previous slide is characterized by The transconductance curve on the previous slide is characterized by the following equation: the following equation: I I D D = I = I S S (e (e V V D D / / η η V V T T – 1) – 1) • As described in the last slide, I As described in the last slide, I D D is the current through the diode, I is the current through the diode, I S S is is the saturation current and V the saturation current and V D D is the applied biasing voltage. is the applied biasing voltage. • V V T T is the thermal equivalent voltage and is approximately 26 mV at room is the thermal equivalent voltage and is approximately 26 mV at room temperature. The equation to find V temperature. The equation to find V T T at various temperatures is: at various temperatures is: V V T T = = kT kT q q k = 1.38 x 10 k = 1.38 x 10 -23 -23 J/K T = temperature in Kelvin q = 1.6 x 10 J/K T = temperature in Kelvin q = 1.6 x 10 -19 -19 C C ∀ η η is the emission coefficient for the diode. It is determined by the way 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 the diode is constructed. It somewhat varies with diode current. For a silicon diode silicon diode η η is around 2 for low currents and goes down to about 1 is around 2 for low currents and goes down to about 1 at higher currents at higher currents Diode Circuit Models Diode Circuit Models Kristin Ackerson, Virginia Tech EE Kristin Ackerson, Virginia Tech EE Spring 2002 Spring 2002 The Ideal Diode The Ideal Diode Model Model The diode is designed to allow current to flow in The diode is designed to allow current to flow in only one direction. The perfect diode would be a only one direction. The perfect diode would be a perfect conductor in one direction (forward bias) perfect conductor in one direction (forward bias) and a perfect insulator in the other direction and a perfect insulator in the other direction (reverse bias). In many situations, using the ideal (reverse bias). In many situations, using the ideal diode approximation is acceptable. diode approximation is acceptable. Example: Assume the diode in the circuit below is ideal. Determine the Example: Assume the diode in the circuit below is ideal. Determine the value of I value of I D D if a) V if a) V A A = 5 volts (forward bias) and b) V = 5 volts (forward bias) and b) V A A = -5 volts (reverse = -5 volts (reverse bias) bias) + + _ _ V V A A I I D D R R S S = 50 = 50 Ω Ω a) With V a) With V A A > 0 the diode is in forward bias > 0 the diode is in forward bias and is acting like a perfect conductor so: and is acting like a perfect conductor so: I I D D = V = V A A /R /R S S = 5 V / 50 = 5 V / 50 Ω Ω = 100 mA = 100 mA b) With V b) With V A A < 0 the diode is in reverse bias < 0 the diode is in reverse bias and is acting like a perfect insulator, and is acting like a perfect insulator, therefore no current can flow and I therefore no current can flow and I D D = 0. = 0. Diode Circuit Models Diode Circuit Models Kristin Ackerson, Virginia Tech EE Kristin Ackerson, Virginia Tech EE Spring 2002 Spring 2002 The Ideal Diode with The Ideal Diode with Barrier Potential Barrier Potential This model is more accurate than the simple This model is more accurate than the simple ideal diode model because it includes the ideal diode model because it includes the approximate barrier potential voltage. approximate barrier potential voltage. Remember the barrier potential voltage is the Remember the barrier potential voltage is the voltage at which appreciable current starts to voltage at which appreciable current starts to flow. flow. Example: To be more accurate than just using the ideal diode model Example: To be more accurate than just using the ideal diode model include the barrier potential. Assume V include the barrier potential. Assume V φ φ = 0.3 volts (typical for a = 0.3 volts (typical for a germanium diode) Determine the value of I germanium diode) Determine the value of I D D if V if V A A = 5 volts (forward bias). = 5 volts (forward bias). + + _ _ V V A A I I D D R R S S = 50 = 50 Ω Ω With V With V A A > 0 the diode is in forward bias > 0 the diode is in forward bias and is acting like a perfect conductor and is acting like a perfect conductor so write a KVL equation to find I so write a KVL equation to find I D D : : 0 = V 0 = V A A – I – I D D R R S S - V - V φ φ I I D D = V = V A A - V - V φ φ = 4.7 V = 94 mA = 4.7 V = 94 mA R R S S 50 50 Ω Ω V V φ φ + + V V φ φ + + Diode Circuit Models Diode Circuit Models The Ideal Diode The Ideal Diode with Barrier with Barrier Potential and Potential and Linear Forward Linear Forward Resistance Resistance This model is the most accurate of the three. It includes a This model is the most accurate of the three. It includes a linear forward resistance that is calculated from the slope of linear forward resistance that is calculated from the slope of the linear portion of the transconductance curve. However, the linear portion of the transconductance curve. However, this is usually not necessary since the R this is usually not necessary since the R F F (forward (forward resistance) value is pretty constant. For low-power resistance) value is pretty constant. For low-power germanium and silicon diodes the R germanium and silicon diodes the R F F value is usually in the value is usually in the 2 to 5 ohms range, while higher power diodes have a R 2 to 5 ohms range, while higher power diodes have a R F F value closer to 1 ohm. value closer to 1 ohm. Linear Portion of Linear Portion of transconductance transconductance curve curve V V D D I I D D   V V D D   I I D D R R F F = =   V V D D   I I D D Kristin Ackerson, Virginia Tech EE Kristin Ackerson, Virginia Tech EE Spring 2002 Spring 2002 + + V V φ φ R R F F Diode Circuit Models Diode Circuit Models The Ideal Diode The Ideal Diode with Barrier with Barrier Potential and Potential and Linear Forward Linear Forward Resistance Resistance Kristin Ackerson, Virginia Tech EE Kristin Ackerson, Virginia Tech EE Spring 2002 Spring 2002 Example: Assume the diode is a low-power diode Example: Assume the diode is a low-power diode with a forward resistance value of 5 ohms. The with a forward resistance value of 5 ohms. The barrier potential voltage is still: V barrier potential voltage is still: V φ φ = 0.3 volts = 0.3 volts (typical for a germanium diode) Determine the value (typical for a germanium diode) Determine the value of I of I D D if V if V A A = 5 volts. = 5 volts. + + _ _ V V A A I I D D R R S S = 50 = 50 Ω Ω V V φ φ + + R R F F Once again, write a KVL equation Once again, write a KVL equation for the circuit: for the circuit: 0 = V 0 = V A A – I – I D D R R S S - - V V φ φ - I - I D D R R F F I I D D = V = V A A - V - V φ φ = 5 – 0.3 = 85.5 mA = 5 – 0.3 = 85.5 mA R R S S + R + R F F 50 + 5 50 + 5 [...]... potential was 0 .3 volts and the linear forward resistance value was assumed to be 5 ohms Kristin Ackerson, Virginia Tech EE Spring 2002 The Q Point The operating point or Q point of the diode is the quiescent or nosignal 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... graph with the transconductance curve This line is the load line Kristin Ackerson, Virginia Tech EE Spring 2002 The Q Point ID (mA) 12 10 8 5 .3 4.6 The transconductance curve below is for a Silicon diode The Q point in this example is located at 0.7 V and 5 .3 mA Q Point: The intersection of the load line and the transconductance curve 6 4 2 VD (Volts) 0.2 0.4 0.6 0.7 0.8 1.0 1.2 1.4 Kristin Ackerson,... diode so the barrier potential voltage is still 0.7 volts The DC component of the circuit is the RS = 1000 Ω same as the previous example and therefore ID = 6V – 0.7 V = 5 .3 mA ID 1000  + rF = η VT = 1 * 26 mV = 4.9  vin + ID 5 .3 mA Vφ vF = vac η = 1 is a good approximation if the dc current is greater than 1 mA as it is in this example rF = sin(wt) V 4.9  = 4.88 sin(wt) mV rF + RS 4.9  + 1000  . suất • up-down analysis = phân tích tăng-giảm Nội dung chương 3 3- 1 Các ý tưởng cơ bản 3- 2 Diode lý tưởng 3- 3 Xấp xỉ bậc 2 3- 4 Xấp xỉ bậc 3 3- 5 Trounleshooting 3- 6 Phân tích mạch tăng-giảm 3- 7 Đọc. Trounleshooting 3- 6 Phân tích mạch tăng-giảm 3- 7 Đọc bảng dữ liệu 3- 8 Cách tính điện trở khối 3- 9 Điện trở DC của diode 3- 1 0 Đường tải 3- 1 1 Diode dán bề mặt Properties of Diodes Properties of Diodes Kristin. is: V V T T = = kT kT q q k = 1 .38 x 10 k = 1 .38 x 10 -2 3 -2 3 J/K T = temperature in Kelvin q = 1.6 x 10 J/K T = temperature in Kelvin q = 1.6 x 10 -1 9 -1 9 C C ∀ η η is the emission