Electronic principles - Chapter 3

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Nội dung Text: Electronic principles - Chapter 3

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 Figure 1.10 – The Diode Transconductance Curve2 ID (mA) • VD = Bias Voltage • ID = Current through Diode. ID is Negative Diode. for Reverse Bias and Positive for Forward IS Bias Bias VBR • IS = Saturation Current Current VD ~Vφ • VBR = Breakdown Voltage Voltage • Vφ = Barrier Potential Voltage Voltage (nA) Kristin Ackerson, Virginia Tech EE Spring 2002
Properties of Diodes Properties The Shockley Equation • The transconductance curve on the previous slide is characterized by The the following equation: the ID = IS(eVD/η 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. the • VT is the thermal equivalent voltage and is approximately 26 mV at room temperature. The equation to find VT at various temperatures is: temperature. VT = kT kT q k = 1.38 x 10-23 J/K T = temperature in Kelvin q = 1.6 x 10-19 C 1.38 ∀ η 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 η is around 2 for low currents and goes down to about 1 at higher currents at Kristin Ackerson, Virginia Tech EE Spring 2002
Diode Circuit Models Diode The Ideal Diode The The diode is designed to allow current to flow in The Model only one direction. The perfect diode would be a Model 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. diode Example: Assume the diode in the circuit below is ideal. Determine the Example: value of ID if a) VA = 5 volts (forward bias) and b) VA = -5 volts (reverse value bias) bias) a) With VA > 0 the diode is in forward bias RS = 50 Ω and is acting like a perfect conductor so: and ID = VA/RS = 5 V / 50 Ω = 100 mA ID + b) With VA < 0 the diode is in reverse bias VA _ and is acting like a perfect insulator, therefore no current can flow and ID = 0. therefore Kristin Ackerson, Virginia Tech EE Spring 2002
Diode Circuit Models Diode The Ideal Diode with The This model is more accurate than the simple This Barrier Potential ideal diode model because it includes the Barrier approximate barrier potential voltage. Remember the barrier potential voltage is the + Vφ voltage at which appreciable current starts to flow. flow. Example: To be more accurate than just using the ideal diode model Example: include the barrier potential. Assume Vφ = 0.3 volts (typical for a include germanium diode) Determine the value of ID if VA = 5 volts (forward bias). germanium RS = 50 Ω With VA > 0 the diode is in forward bias and is acting like a perfect conductor ID so write a KVL equation to find ID: so + 0 = VA – IDRS - Vφ VA _ + ID = VA - Vφ = 4.7 V = 94 mA Vφ 50 Ω RS Kristin Ackerson, Virginia Tech EE Spring 2002
Diode Circuit Models Diode The Ideal Diode The This model is the most accurate of the three. It includes a linear forward resistance that is calculated from the slope of with Barrier the linear portion of the transconductance curve. However, Potential and this is usually not necessary since the RF (forward this Linear Forward resistance) value is pretty constant. For low-power Resistance germanium and silicon diodes the RF value is usually in the germanium 2 to 5 ohms range, while higher power diodes have a RF to value closer to 1 ohm. value ID + Linear Portion of Vφ RF Linear transconductance curve curve RF =  VD  ID  ID VD  VD Kristin Ackerson, Virginia Tech EE Spring 2002
Diode Circuit Models Diode The Ideal Diode The Example: Assume the diode is a low-power diode Example: with Barrier with a forward resistance value of 5 ohms. The Potential and barrier potential voltage is still: Vφ = 0.3 volts barrier Linear Forward (typical for a germanium diode) Determine the value of ID if VA = 5 volts. of Resistance RS = 50 Ω Once again, write a KVL equation Once ID for the circuit: for + VA 0 = VA – IDRS - Vφ - IDRF _ + Vφ ID = VA - Vφ = 5 – 0.3 = 85.5 mA RS + RF 50 + 5 RF Kristin Ackerson, Virginia Tech EE Kristin Spring 2002
Diode Circuit Models Diode Values of ID for the Three Different Diode Circuit Models Ideal Diode Ideal Diode Model with Model with Ideal Diode Barrier Barrier Model Potential and Potential Linear Forward Voltage Resistance ID 100 mA 94 mA 85.5 mA These are the values found in the examples on previous These 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. value Kristin Ackerson, Virginia Tech EE Spring 2002
The Q Point The The 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 signal 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. transconductance First the load line is found by substituting in First different values of Vφ into the equation for ID using different RS = 1000 Ω the ideal diode with barrier potential model for the diode. With RS at 1000 ohms the value of RF diode. ID wouldn’t have much impact on the results. wouldn’t + ID = V A – V φ VA _ RS + = 6V Vφ Using V φ values of 0 volts and 1.4 volts we obtain Using 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. This Kristin Ackerson, Virginia Tech EE Spring 2002
The Q Point The The The ID (mA) transconductance (mA) curve below is for a 12 Silicon diode. The Q point in this example is located 10 at 0.7 V and 5.3 mA. at 8 Q Point: The intersection of the Point: load line and the 6 transconductance curve. transconductance 5.3 4.6 4 2 VD (Volts) (Volts) 0.2 0.4 0.6 0.8 1.0 1.2 1.4 0.7 Kristin Ackerson, Virginia Tech EE Spring 2002
Dynamic Resistance Dynamic The dynamic resistance of the diode is mathematically determined The as the inverse of the slope of the transconductance curve. Therefore, the equation for dynamic resistance is: Therefore, rF = η VT ID The dynamic resistance is used in determining the voltage drop The across the diode in the situation where a voltage source is supplying a sinusoidal signal with a dc offset. supplying The ac component of the diode voltage is found using the The following equation: following vF = vac rF ac rF + RS The voltage drop through the diode is a combination of the ac and The dc components and is equal to: dc Kristin Ackerson, Virginia Tech EE Spring 2002
Dynamic Resistance Dynamic Example: 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 source barrier potential voltage is still 0.7 volts. barrier The DC component of the circuit is the The RS = 1000 Ω same as the previous example and therefore ID = 6V – 0.7 V = 5.3 mA therefore 6V ID 1000  + rF = η VT = 1 * 26 mV = 4.9  26 vin + ID 5.3 mA 5.3 Vφ η = 1 is a good approximation if the dc current is greater than 1 mA as it is in this example. 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 Kristin Ackerson, Virginia Tech EE across the Spring 2002