DESIGN OF ELECTRONIC SYSTEMS Course Code 11 EC

DESIGN OF ELECTRONIC SYSTEMS Course Code : 11 -EC 201 DEPARTMENT OF ELECTRONICS & COMPUTER ENGINEERING

The two modes of operation of ideal diodes and the use of an external circuit to limit the forward current (a) and the reverse voltage (b).

A Simple Application: The Rectifier A fundamental application of the diode, one that makes use of its severely nonlinear i-v curve, is the rectifier circuit. The circuit consists of the series connection of a diode D and a resistor R. Figure 3 (a) Rectifier circuit. (b) Input waveform.

Another Application: Diode Logic Gates Diodes together with resistors can be used to implement digital logic functions. Diode logic gates: (a) OR gate; (b) AND gate (in a positive-logic system).

A silicon diode said to be a 1 -m. A device displays a forward voltage of 0. 7 V at a current of 1 m. A. Evaluate the junction scaling constant 7; in the event that n is either 1 or 2. What scaling constants would apply for a 1 -A diode of the same manufacture that conducts 1 A at 0. 7 V? example Solution Since then For the 1 -m. A diode: If n = 1: Is = 10 -3 e-700/25 = 6. 9 x 10 -16 A, or about 10 -15 A If n = 2: Is = 10 -3 e-700/50 = 8. 3 x 10 -10 A, or about 10 -9 A The diode conducting 1 A at 0. 7 V corresponds to one-thousand 1 -m. A diodes in parallel with a total junction area 1000 times greater. Thus IS is also 1000 times greater, being 1 p. A and 1µA, respectively for n=1 and n=2.

Example-1 For the circuits shown in Fig. below using ideal diodes, find the values of the voltages and currents indicated

Solution a) In this case the diode is conducting, such that the voltage at the top end of the diode equals that at the bottom end, so V = − 3 V. Then, I = 6 V/10 kΩ = 0. 6 m. A. b) The diode is not conducting, so I = 0, and V = 3 V. c) The diode is conducting so I = 0. 6 m. A as before, but now V = 3 V. d) The diode is not conducting, so I = 0 and V = − 3 V.

Ex: 2 : - For the circuits shown in Fig. below using ideal diodes, find the values of the voltages and currents indicated

Solution (a) For D 1: The diode is conducting so VD 1 = +1 V; For D 2: The diode is conducting so VD 2 = +3 V; I = (3 -(-3))/2 kΩ = 3 m. A; Two voltages (1 V, 3 V) try to appear at the point V, but at a node no more than one voltage can appear. Hence, V = 3 V and I = 3 m. A. (b) For D 1: VD 1 = +1 V; For D 2: VD 2 = +3 V; I = (3 -(1))/2 kΩ = 1 m. A; Hence V = 1 V; I = 1 m. A.

Ex: 3 : - In each of the ideal-diode circuits shown in Fig. P 3. 4, v 1 is a 1 k. Hz, 10 V peak sine wave. Sketch the waveform resulting at v 0. What are its positive and negative peak values? Cont…

Solution (a) Piecewise expression for V 0 is: Minimum value: 0 V Maximum value: 10 V (b) Piecewise expression for V 0 is: Minimum value: -10 V Maximum value: 0 V (c) Current never flows, so V 0 = 0 always. (e) Current always flows so V 0 = V 1 Minimum values: − 10 V. Maximum value: 10 V. Cont…

(g) Piecewise expression for V 0 is: Minimum value: -10 V Maximum value: 0 V (h) The output is always connected to ground, V 0 = 0 always. (i) When V 1 < 0 this acts as a voltage divider. Otherwise they two are equal. Piecewise expression is: Minimum value: − 5 V. Maximum value: 10 V. Cont…

(k) The current source causes 1 V voltage drop across the resistor, such that V 0 is always one volt higher than the voltage at the bottom of the resistor. When V 1 > 0 the voltage at the bottom of the resistor is ground. When V 1< 0 the voltage at the bottom of the resistor is V 1. The piecewise expression for V 0 is:

Ex: 4 : - The circuits shown in Fig. below can function as logic gates for input voltages are either high or low. Using “ 1" to denote the high value and “ 0” to denote the low value, prepare a table with four columns including all possible input combinations and the resulting values of X and Y. What logic function is X of A and S? What logic function is Y of A and B? For what values of A and 5 do X and F have the same value? For what values of A and B do X and Y have opposite values? Sol: X = A. B Sol: Y = A + B

Ex: 5(a) : - Assuming that the diodes in the circuits of Fig. below are ideal, find the values of the labeled voltages and currents.

Solution (a) Assume that both the diodes are conducting. ID 2= (10 -0)/10 = 1 m. A; At node B: I+1 = (0 -(-10))/5 = 2 m. A; Results in I = 1 m. A. Thus D 1 is conducting as originally assumed, and the final result is I = 1 m. A and V = 0 V. (b) For the circuit in Fig. (b). If we assume that both diodes are conducting, then VB=0 and V=0. The current in D 2 is obtained from ID 2= (10 -0)/5 = 2 m. A; At node B: I+2 = (0 -(-10))/10 = 1 m. A; Which yields I = -1 m. A. Since this is not possible, our original assumption is not correct. We start again, assuming that D 1 is off and D 2 is on. The current ID 2 is given by ID 2= (10 -(-10))/15 = 1. 33 m. A;

Ex: 5(b) : - Assuming that the diodes in the circuits of Fig. below are ideal, find the values of the labeled voltages and currents. I 1 2 2 I 3 D 1& D 2 Conducting I 1=1 m. A; I 3=0. 5 m. A; I 2=0. 5 m. A V= 0 V I 3 D 1=off, D 2=On I 1= I 3=0. 66 m. A V = -1. 7 V

Ex: 6 : - Assuming that the diodes in the circuits of Fig. below are ideal, utilize Thevenin's theorem to simplify the circuits and thus find the values of the labeled currents and voltages.

Solution (a) D conducting I=0. 225 m. A V=4. 5 V (b) D is not conducting I=0 A V=-2 V

Ex: 7 : - The circuit of Fig. below can be used in a signaling system using one wire plus a common ground return. At any moment, the input has one of three values: +3 V, 0 V, - 3 V. What is the status of the lamps for each input value? (Note that the lamps can be located apart from each other and that there may be several of each type of connection, all on one wire!). Solution: V 3 V 0 V -3 V Red On Off Green Off On D 1 conducts D 2 conducts

Ex: 8 : - Listed below are the results of measurements taken on several different junction diodes. For each diode, the data provided are the diode current /, the corresponding diode voltage V, and the diode voltage at a current 1/10. In each case, estimate Is, n, and the diode voltage at 10/. (a) 10. 0 m. A, 700 m. V, 600 m. V (b) 1. 0 m. A, 700 m. V, 600 m. V (c) 10 A, 800 m. V, 700 m. V (d) 1 m. A, 700 m. V, 580 m. V (e) 10 µA, 700 m. V, 640 m. V

Ex: 9 : - The circuit in Fig. below utilizes three identical diodes having n = 1 and Is = 10 -14 A. Find the value of the current I required to obtain an output voltage V 0 = 2 V Assume n=1. If a current of 1 m. A is drawn away from the output terminal by a load, what is the change in output voltage? Assume n=1

Solution The circuit shown utilizes three identical diodes having n=1 and Is= 10 -14 A. To find the value of the current I required to obtain an output voltage Vo=2 V : we know that The voltage across each diode is

Solution If a current of 1 m. A is drawn away from the output terminal by a load, what if the change in the output voltage. Load current = 1 m. A, therefore IDγ = 2. 81 m. A

Ex: 10 : - For the circuit shown in Fig. below, both diodes are identical, conducting 10 m. A at 0. 7 V and 100 m. A at 0. 8 V. Find the value of R for which V = 80 m. V.

Solution Both the diodes are identical, therefore Is, η, VT are same For diode 1 : VD 1 = 0. 7 V; ID 1 = 10 m. A. For diode 1 : VD 2 = 0. 8 V; ID 2 = 100 m. A. To find η: To find R if V=80 m. V:

Ex: 11 : - A designer of an instrument that must operate over a wide supply-voltage range, noting that a diode's junction voltage drop is relatively independent of junction current, considers the use of a large diode to establish a small relatively constant voltage. A power diode, for which the nominal current at 0. 8 V is 10 A, is available. Furthermore, the designer has reason to believe that n = 2. For the available current source, which varies from 0. 5 m. A to 1. 5 m. A, what junction voltage might be expected? What additional voltage change might be expected for a temperature variation of +25°C?

As an alternative to the idea suggested in Problem (Ex: 10), the designer considers a second approach to producing a relatively constant small voltage from a variable current supply: It relies on the ability to make quite accurate copies of any small current that is available (using a process called current mirroring). The designer proposes to use this idea to supply two diodes of different junction areas with the same current and to measure their junction-voltage difference. Two types of diodes are available; for a forward voltage of 700 m. V, one conducts 0. 1 m. A while the other conducts 1 A. Now, for identical currents in the range of 0. 5 m. A to 1. 5 m. A supplied to each, what range of difference voltages result? What is the effect of a temperature change of ― 25°C on this arrangement? Assume n=1.

Ex: 12 : - A “ 1−m. A” diode (i. e. one that has v. D = 0. 7 V at i. D = 1 m. A is connected in series with a 200 resistor to a 1. 0 − V supply. 1. Provide a rough estimate of the diode current you would expect. Solution: 1. For the rough estimate we assume the voltage drop across the diode is v. D = 0. 7 V. In that case the voltage drop across the resistor is v. R = 1. 0 − 0. 7 = 0. 3 V. The curent is then i. D = VR /R = 0. 3 200 = 1. 5 m. A.