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EE 321 Analog Electronics Problems

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EE 321 Analog Electronics, Fall 2013 Homework #5 solution 3.26. For the circuit shown in Fig. P3.26, both diodes are identical, conducting 10 mA at 0.7 V, and 100 mA at 0.8 V. Find the value of R for which V = 80 mV. V1 is the voltage across diode 1, V2 across diode 2, I1 is current through diode 1, and I2 through diode 2. Then we have (using exponential model) I = I1 + I2 I2 = Is exp  V2 nVT  I1 = Is exp  V2 − V nVT  = Is exp or alternatively, V nVT  I = I1 + I2 = I1 1 + exp  I2 = I1 exp and thus   and V nVT  I I1 = 1 + exp and V V = R= I1 I   V nVT 1 + exp  Now we only need to determine nVT . Note that   ID V log = IS nVT 1  V nVT   V2 nVT    V exp − nVT and log  IDa IS  log − log  IDa IDb nVT =   IDb IS =  = Va − Vb nVT Va − Vb nVT Va − Vb   log IIDa Db Now inserting Va = 0.8 V, Ia = 100 mA, Vb = 0.7 V, and Ib = 10 mA, we get nVT = 0.1 = 43.4 mV log 10 Inserting that in the expression for R, we get    80 80 1 + exp = 58.5 Ω R= 10 43.4 3.66. A shunt regulator utilizing a zener diode with an incremental resistance of 5 Ω is fed through an 82 − Ω resistor. If the raw supply changes by 1.3 V, what is the corresponding change in the regulated output voltage? The circuit looks like this, with RL = ∞. The output is across the Zener diode, so across the supply and rz . The relationship between the input and output voltages is rz R + rz To get the regulation, the rati ...
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