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Unformatted text preview: EE40 P10.7 Homework #8 Solution Assuming the simple piece‐wise model as shown in Fig. 10.23 for diodes. (a) When V is equal to ‐1.2V, the two diodes in series are both on, while the third diode is off; When V is equal to 0.6V, the two diodes in series are both off, while the third diode is on; When V is in between ‐1.2V and 0.6V, none of the diodes are on, the current is effectively zero. So the I‐V curve is Note that since we assume the simple piece‐wise model, the lines at V=‐1.2V and V=0.6V can be perfectly vertical and intersect the V‐axis at V=‐1.2V and V=0.6V, respectively. Also note that: When we connect two diodes in series, according to KCL, the current flowing through either diode should be exactly the same. That implies that the two diodes can only be both ON, or both OFF; it is generally impossible that one diode is ON while the other is OFF. This time we have one diode and one Zener diode connected in series for each branch. Notice that according to our diode model, the Zener diode can be conducting current when its voltage is either +0.6V or –Vx where Vx is reverse‐
Page 1 of 1 (b) breakdown voltage. In stark contrast, the normal diode conducts current only when its voltage is +0.6V. Upon the above observation, we conclude that if the applied voltage V is such direction that the normal diode would be reverse biased, the whole branch (with that normal diode in series with Zener diode) cannot conduct any current. So the I‐V curve is (c) (d) When V increases to 0.6V, the single diode (which lies alone in the branch) starts to conduct current. Notice that when this diode conducts current, according to our model, it effectively “clamps” its voltage right at 0.6V. Therefore, the voltage V cannot exceed 0.6V in any event, which means that, the couple of diodes (which lie together in one branch) are always OFF, since they require V=1.2V to turn them on. So the I‐V curve is Page 2 of 2 Again, due to our choice of the diode model, the I‐V curve above can be a vertical line without the curvature shown in the picture. Part b Why might generating I‐V curves be useful? P10.10 The generated I‐V curve can be used in load‐line analysis. Note: Let us assume Id = Is exp( Vd ) , in order for the diode current to increase nVT to 10X of its original value, the diode voltage only has to increase by Page 3 of 3 ΔVd = nVT ln10 = 120mV . That is, for n=2, for every 120mV increase in the diode voltage, the diode current will increase by 10X! P10.17 ix = Vs − Vx 20 − Vx = = 0.0002(20 − Vx) ...(1) Rs 5000 ix = 0.01(1 − vx / 5)3 ...(2) Equation (1) and (2) correspond to two curves on the ix‐Vx plane. The solution of (Vx, Ix) corresponds to the intersection point of the two lines. It is roughly Vx = 4.1V, Ix = 3.1mA. P10.19 Page 4 of 4 P10.28 Page 5 of 5 Firstly, the voltage regulator attemps to regulate VL to be roughly equal to 5V. Secondly, the fact that the Zener diode is ideal implies that, when the voltage across the Zener diode is 5V, the current Iz flowing through that diode can be zero or any POSITIVE value; in other words, when the VL is regulated to be 5V, Iz can be any value so long as it is flowing in the reference direction shown in the picture. Thirdly, based upon the “Iz flowing direction” constraint, we are pretty sure that the current flowing through Rs, when VL = 5V, should be equal to, or greater than 100mA. If the current flowing through Rs is less than 100mA, according to the problem statement, iL can be as large as 100mA; that means if iRS is less than 100mA and iL is 100mA, iz should be flowing in a direction opposite to its reference direction! Even an “ideal” Zener diode CANNOT tolerate such negative current when its voltage is equal to the reverse breakdown voltage. Building upon our forgoing argument, the choice of Rs should be such that “the minimum current flowing through Rs should be equal to, or greater than, 100mA”. Why do we say “the minimum current”? Because as Vs is changing, the current flowing through Rs also changes. Since IRS = (Vs‐5V)/Rs, the minimum current flowing through Rs is the current value when Vs takes its minimum value of 10V, which leads to the following solution: P10.36 Page 6 of 6 An ideal diode means we should apply the piece‐wise model to it assuming VDON=0V. (a) (b) (c) (d) The diode is on, V=0.6V. I=(10V‐0.6V)/2700ohm = 3.5mA The diode is off, I=0A, V=10V The diode is on, V=0.6V, I=(0.6V)/2700ohm = 0.2mA The diode is on. According to Vdon=0.6V I × 1kΩ − (10mA − I) × 1kΩ = 0.6V Solving gives I=5.3mA. V=5.3mA*1kohm=5.3V. P10.63 An ideal diode means we should apply the piece‐wise model to it assuming VDON=0V. Page 7 of 7 P10.67 An ideal diode means we should apply the piece‐wise model to it assuming VDON=0V. Label the joint point where the diode, the 2kohm resistor, and the 3kohm resistor connect with each other as node X, hence the nodal voltage Vx. i) First, assume the diode is OFF. Then 3kohm and 2kohm resistors are in series, since the current flowing through either of them is the same. Then Vx=2V. Vo=Vin, since the current flowing through the 1.2kohm resistor is zero. Of course, to satisfy our assumption that the diode is OFF, Vo should be smaller than Vx, which means Vin should be smaller than 2V. So now, for Vin < 2V, Vout = Vin. When Vin > 2V, the diode is on. Then Vx = Vout. According to KCL Vin − Vout Vout Vout − 5V = + , 1.2kΩ 2kΩ 3kΩ which yields ii) Vout = 0.5Vin + 1 Page 8 of 8 So the Vout‐Vin curve looks like: part (b) What is the physical meaning of slope changes? The diode is ON or OFF, depending on whether Vin is greater than, or smaller than 2V. In these two different regions, different resistors are active, so the slope changes. For example, when Vin>2V, we have a voltage divider between 1.2kohm and (2kohm3kohm) resistors, with the voltage division ratio of 1:1, hence the slope of 1/2. Problem 9 (a) According to the problem statement, assume R_1=0ohm. In the first 1/4 cycle, Vin(t) increases from 0 to 10V, V_2(t) will track this change, except that V_2(t) is lower than Vin(t) by Vdon=0.6V. At t=T/4, V_2(t)=10V‐ 0.6V=9.4V. Right after that, Vin(t) is lower than V_2(t), and the diode D1 is off. Since the time constant is much larger than 1/60 s, the voltage across the capacitor C remains roughly unchanged. That means V_2(2) will retain the 9.4V value for several cycles. At any time instant, if V_2(t) is smaller than VBD=5V, V_3(t) just tracks the value of V_2(t) since the Zener diode is off and R_L is much greater than R_2 (Recall the voltage divider relationship). If V_2(t) is higher than 5V, V_3(t) will be Page 9 of 9 clamped to 5V since the Zener diode is now conducting current and effectively clamps that voltage. (b) According to the problem statement, assume R_1=0ohm. In the first 1/4 cycle, Vin(t) increases from 0 to 10V, V_2(t) will track this change, except that V_2(t) is lower than Vin(t) by Vdon=0.6V, due to the presence of the diode D1. At t=T/4, V_2(t)=10V‐0.6V=9.4V. Right after that, Vin(t) is lower than V_2(t), and the diode D1 is off. Since the time constant is much larger than 1/60 s, the voltage across the capacitor C remains roughly unchanged. That means V_2(2) will retain the 9.4V value for several cycles. At any time instant, if V_3(t) is smaller than VBD=5V, V_3(t) is V_2(t)/11 since the Zener diode is off and R_L is equal to 0.1R_2 (Recall the voltage divider relationship). Since V_2(t) has a maximum value of 9.4V, which is smaller than 5V*11 = 55V, V_3(t) will be always smaller than 5V, and the Zener diode will be always off. According to the problem statement, assume R_2=0ohm. Then V_2(t) = V_3(t). If V_3(t) < 5V, the Zener diode is off. Then in the first cycle, when Vin(t) increases from 0.6V to 5.6V, V_3(t) will track Vin(t) except that V_3(t) is lower than Vin(t) by 0.6V. After Vin(t) passes 5.6V, V_3(t) will be clamped to 5V due to the conduction of the Zener diode, which is now turned on. After Vin(t) is smaller than 5.6V, the Zener diode and the diode D1 is off, the capacitor gets discharged through R_2 and R_L. Since the time constant is on the order of 1/60 s, V_3(t) will decrease from 5V. At some time instant, when Vin(t) is greater than V_3(t)+0.6V, the diode D1 is on again and then V_3(t) again tracks Vin(t) except with a 0.6V voltage drop. At some time, Vin(t) is higher than 5.6V, then V_3(t) will be clamped to 5V again… According to the problem statement, assume R_1=R_2=0ohm. Then V_2(t) = V_3(t). In the first 1/4 cycle, Vin(t) increases from 0 to 10V, V_2(t) will track this change, except that V_2(t) is lower than Vin(t) by Vdon=0.6V. At t=T/4, V_2(t)=10V‐ 0.6V=9.4V. Right after that, Vin(t) is lower than V_2(t), and the diode D1 is off. Since the time constant is on the order of 1/60 s, the capacitor will be discharged through R_L, and V_2(t) drops with time. At some time instant, when Vin(t) is (c) (d) Page 10 of 10 greater than V_3(t)+0.6V, the diode D1 is on again and then V_2(t) again tracks Vin(t) except with a 0.6V voltage drop… So the solution curves should look like the following. Note: rough sketch only. Also note that the 3 row corresponds to curves for part d) while the last row corresponds to curves for part c). Page 11 of 11 ...
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 Fall '07
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