Chapter 3. Diodes, Problem Solutions. 3.1 Problem Solution

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1 Chapter 3 Diodes, Problem Solutions 3.1 Problem 3.13 A square wave of 10 V peak-to-peak amplitude and zero average is applied to a circuit resembling that in Figure (3.1) and employing a 100 Ω resistor. Assuming an ideal diode what is the peak output voltage that results? What is the peak diode current? What is the average diode current? What is the maximum reverse voltage across the diode? vi vo vi R vo t -5 t Figure 3.1: Solution The peak output voltage ˆv o is: ˆv o = 5 V

2 2 CHAPTER 3. DIODES, PROBLEM SOLUTIONS However v o (t) is given by: v o (t) = 5 0 t T/2 v o (t) = 0 T/2 < t T Let the period of the input voltage be T, then the average out output voltage v oavg : v oavg = 1 T = 1 T = 1 T = 5 T T o [ T/2 0 v(t) dt v o (t) dt + [ T/2 5 dt + T 0 T/2 T/2 0 dt T T/2 ] v o (t) dt 0 dt ] = 5 T T 2 = 2.5 V The peak and average currents î and i avg are given by: Maximum reverse voltage is 5 V. î = ˆv o R = = 50 ma i avg = v oavg R = = 25 ma

3 3.2. PROBLEM Problem 3.27 The circuit shown in Figure (3.2) uses identical diodes for which I D = 1 ma at V D = 0.7 V with n = 1. At 20 C, voltage V is measured by a very high resistance meter to be 0.1 V. By what factor does the reverse leakage current of these diodes exceed I s? Estimate the value of V when the temperature is raised by 50 C. +10 V D1 V D2 Figure 3.2: Solution The diode reverse leakage current I D = 1 ma is defined by: I D = I s e V D/V T 10 3 = I s e 0.7/0.025 I s = 10 3 e 0.7/0.025 = 10 3 e 28 = A

4 4 CHAPTER 3. DIODES, PROBLEM SOLUTIONS At V = 0.1 V, I D is: I D = I s e 0.1/0.025 = I s e 4 = I s 54.6 I D I s = 54.6 The reverse leakage current doubles for every 10 C rise, so for a 50 C rise the current increases by a factor of 2 5. I S doubles for every 5 C rise, so for a 50 C rise I s increases by a factor of we then have: I D = I s e V/V T 2 5 I D = 2 10 I s e V/V T [ ] 2 5 I D V = V T ln 2 10 I [ s ] 54.6 = ln 2 5 = ln(1.706) = 13.4 mv

5 3.3. PROBLEM Problem 3.44 Calculate the built-in voltage of a junction in which the p and n regions are doped equally with atoms/cm 3. Assume the free electron concentration in intrinsic silicon n i 10 5 /cm 3. With no external voltage applied, what is the width of the depletion region, and how far does it extend into the p and n regions? If the the cross sectional area of the junction is 100 µm 2, find the magnitude of the charge stored on either side of the junction, and calculate the junction capacitance C j. Solution The built-in voltage of a p n junction is given by: [ ] NA N D V = V T ln n 2 i [ ] = ln (10 5 ) 2 = = 1.27 V Let W, x n, x p and ɛ s be the total width, the width in the n region, the width in the p region of the depletion region, and the electric permittivity of silicon respectively. W is given by: W = x n + x p = 2ɛ s q = ( 1 N A + 1 N D ) V = 0.57 µm ( ) Where ɛ s = κɛ, κ = 11.7 is the dielectric constant of silicon and ɛ = F/cm is the permittivity of free space. The ratio of the widths of the depletion region in the n and p regions is given by: x n x p = N A N D

6 6 CHAPTER 3. DIODES, PROBLEM SOLUTIONS Since N A = N D, then x n = x p = W/2 = 0.28 µm. Let A = 100 µm 2 be the area of the junction, then the charge on the junction C j = C p = C n is given by: q j = q N AN D N A + N D A W = = C The capacitance C j of the depletion region is given by: C j = ɛ sa W = = F = 1.82 pf

7 3.4. PROBLEM Problem 3.65 For the circuit shown in Figure (3.3), utilize the constant-voltage-drop model (0.7 V) for each conduction diode and show that the transfer characteristic can be described by: for v I 4.65 V v o = v I for v I V v o = V for v I V v o = V +10 V i1 A D1 D2 vi D3 i D1 i D4 i D2 D4 i L v o B i2-10 V Figure 3.3: Solution When V I is small ( close to zero) all four diodes conduct, v A = v I + v D1

8 8 CHAPTER 3. DIODES, PROBLEM SOLUTIONS and v o = v A v D2 = (v I + v D1 ) v D2 Since all diodes are identical, and v D1 = v D2 = v D3 = v D4 = 0.7 V, then we also have: The currents i 1, and i L can be calculated as: v o = v I v B = v I v D3 i 1 = 10 v A = 10 v I 0.7 = 9.3 v I ma (3.1) 10 v o i L = = v I ma (3.2) 10 The current i 1 splits at A into i D1 and i D2, so i D1 < i 1, similarly i D2 splits into i L and i D4, so i L < i D2. So, if D 1, D 2, and D 3 conducting, the following inequality must be satisfied: i L < i D2 < i V Now, as v I increases in the positive direction i 1 decreases and i L increases, this means that there will be a valu for v I at which the above inequality is not satisfied. Under this condition D 1 and D 4 will cut off, i D1 = i D4 = 0 and i L = i D2 = i 1. Using Equation (3.1) and Equation (3.2) we can find the value of v I that makes the three current equal. v I 10 = 9.3 v I 10 v I = 9.3 v I = = 4.65 V vi D3 i1 i2 A D2 i D2 D4 B i L v o -10 V Figure 3.4:

9 3.4. PROBLEM V A D1 v o vi D4 i2 B -10 V Figure 3.5: At v I = 4.65 V D 1 and D 4 are cut off while D 3 and D 2 conduct. This situation will continue when v I 4.65 V and v o remains constant at V as the circuit behave like the one shown in Figure (3.4). The symmetery of the circuit indicates that a similar limiting value occurs at negative values of v I spcifically when v I 4.65 V when D 1 and D 4 conduct and D 2 and D 3 cut off and the circuit reduces to that shown in Figure (3.5). In conclusion the circuit provides: for v I 4.65 V v o = v I for v I V v o = V for v I V v o = V

10 10 CHAPTER 3. DIODES, PROBLEM SOLUTIONS 3.5 Problem 3.70 In the circuit shown in Figure (3.6), I is a dc current and v s is a sinusoidal signal. Capacitor C is very large; its function is to couple the signal to the diode but block the dc current from flowing into the signal source. Use the diode small-signal model to show that the signal component of the output voltage is: nv T v o = v s nv T + IR s If v s = 10 mv, find v o for I = 1 ma, and 1 µa. Let R s = 1 kω and n = 2. At what value of I does v o become one-half of v s? Note that this circuit function as a signal attenuator with the attenuation factor controlled by the value of the dc current I. I Rs C vo vs Figure 3.6: Solution A large capacitor has a very small reactance to AC signals. The equivalent circuit is shown Figure (3.7), where r d is the diode resistance. The two resistors in the equivalent circuit

11 3.5. PROBLEM Rs vo vs rd Figure 3.7: form a voltage divider for the input signal voltage v s. The output voltage v o is then given by: r d v o = v s R s + r d since: r d = nv T I then v o becomes: v o = v s I R s + nv T I nv T = v s nv T IR s + nv T For v s = 10 mv, R s = 1kΩ, n = 2 and V T = 25 mv, where the current I is in ma. v o = I mv = I mv = 0.5 mv for I = 1 ma = 9.8 mv for I = 1 µa v o 0.05 = v s I = 0.05 I I = 0.05 ma = 50 µa

12 12 CHAPTER 3. DIODES, PROBLEM SOLUTIONS 3.6 Problem 3.91 The circuit in Figure (3.8) implements a complementary-output rectifier. Sketch and clearly label the waveforms of v o + and vo. Assume a 0.7 V drop across each conducting diode. If the magnitude of the average of each output is to be 15 V, find the required amplitude of the sine wave across the entire secondary winding. What is the PIV of each diode? + vs D4 D1 110 V(rms 60 Hz - + v + o - vs D2 D3 R R v - o Figure 3.8: Solution The voltage at the positive output terminal v + o is given by: v + o = v s sin(ωt) 0.7 for φ ωt π φ = 0 otherwise where φ is given by: The dependence of v + o φ = sin v s on ωt is shown in Figure (3.9) for φ = 10 and v s = 15 V. The

13 3.6. PROBLEM vo wt Figure 3.9: dependence of the voltage of negative output terminal vo on ωt is exactly the same as that of v o + (shown in Figure (3.9)) except that vo is always negative. The average of v + o, is: π φ v o = 1 (v s sin θ 0.7)dθ π φ = 1 π φ {v s sin θ dθ 0.7 π φ φ = 1 { } v s [ cos θ] π φ φ 0.7 [θ] π φ φ π π φ } dθ = 1 π {v s [ cos(π φ) + cos(φ)] 0.7 [π φ φ]} = 1 π {2v s cos φ 0.7π 1.4φ} If v s 0.7, then φ 1, cos φ 1 and 1.4φ 1 and v s becomes: v o = 2 π v s 0.7

14 14 CHAPTER 3. DIODES, PROBLEM SOLUTIONS We can also have: for v o = 15 V, then v s = V. windings is V. v s = π 2 (v o + 0.7) The required amplitude across the entire secondary The maximum reverse bias across each diode is 2v s 0.7 = 48.6 V. To be in the safe side one then need to use diodes with PIV of say = 73 V.

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