- ideal diode = diode lý t ưở ng. - Figure 1.10 – The Diode Transconductance Curve Figure 1.10 – The Diode Transconductance Curve 2 2. - V V φ φ = Barrier Potential = Barrier Potential Voltage. - The transconductance curve on the previous slide is characterized by The transconductance curve on the previous slide is characterized by the following equation:. - 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. - 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. - For a the diode is constructed. - 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. - The Ideal Diode The Ideal Diode. - 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 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. - In many situations, using the ideal (reverse bias). - In many situations, using the ideal 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). - 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:. - 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. - 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.. - 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. - 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. - 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. - I I D D = V = V A A - V - V φ φ = 4.7 V = 94 mA = 4.7 V = 94 mA. - with Barrier with Barrier Potential and Potential and Linear Forward Linear Forward. - 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. - this is usually not necessary since the R F F (forward (forward resistance) value is pretty constant. - 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. - 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. - Once again, write a KVL equation Once again, write a KVL equation for the circuit:. - for the circuit:. - Values of ID for the Three Different Diode Circuit Models Values of ID for the Three Different Diode Circuit Models. - Ideal Diode Model. - Ideal Diode Model with. - Barrier Potential and Linear Forward. - I D 100 mA 94 mA 85.5 mA. - These are the values found in the examples on previous These are the values found in the examples on previous slides where the applied voltage was 5 volts, the barrier slides where the applied voltage was 5 volts, the barrier potential was 0.3 volts and the linear forward resistance potential was 0.3 volts and the linear forward resistance. - The operating point or Q point of the diode is the quiescent or no- 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 condition. - The Q point is obtained graphically and is really only. - needed when the applied voltage is very close to the diode’s barrier needed when the applied voltage is very close to the diode’s barrier potential voltage. - potential voltage. - The example 3 3 below that is continued on the next below that is continued on the next slide, shows how the Q point is determined using the. - slide, shows how the Q point is determined using the transconductance curve and the load line.. - transconductance curve and the load line.. - different values of V φ φ into the equation for I into the equation for I D D using using the ideal diode with barrier potential model for the the ideal diode with barrier potential model for the. - Using V φ φ values of 0 volts and 1.4 volts we obtain values of 0 volts and 1.4 volts we obtain I I D D values of 6 mA and 4.6 mA respectively. - Next values of 6 mA and 4.6 mA respectively. - on the graph with the transconductance curve.. - This line is the load line.. - transconductance transconductance curve below is for a curve below is for a Silicon diode. - Q point in this Q point in this example is located example is located at 0.7 V and 5.3 mA.. - at 0.7 V and 5.3 mA.. - Q Point: The intersection of the The intersection of the load line and the. - transconductance curve.. - The dynamic resistance of the diode is mathematically determined The 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:. - across the diode in the situation where a voltage source is across the diode in the situation where a voltage source is. - The ac component of the diode voltage is found using the The ac component of the diode voltage is found using the. - The voltage drop through the diode is a combination of the ac and The voltage drop through the diode is a combination of the ac and. - dc components and is equal to:. - Example: Use the same circuit used for the Q point example but 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, v. - source voltage is now, v in in = 6 + sin(wt) Volts. - It is a silicon diode so the barrier potential voltage is still 0.7 volts.. - barrier potential voltage is still 0.7 volts.. - The DC component of the circuit is the The DC component of the circuit is the. - 6V – 0.7 V 6V – 0.7 V = 5.3 mA = 5.3 mA. - I I D D 5.3 mA 5.3 mA
Xem thử không khả dụng, vui lòng xem tại trang nguồn hoặc xem
Tóm tắt