TECHGURU CLASSES for ENGINEERS (Your Dedication + Our Guidance = Sure Success) CHAPTER -2 (pn - JUNCTION DIODE) : ELECTRONIC DEVICES EQUIVALENT CIRCUIT OF PN-JUNCTION DIODE Ideal diode : An ideal diode is a two terminal device which has the following properties : PERSONAL REMARK : Refer figure 1 for Q. 1 and Q. 2. Assume D1 and D2 to be ideal diodes. (i) conducts with zero resistance when forward-biased, and (ii) offers an infinite resistance when reverse-biased. –2V The ideal diode acts like an automatic switch. The switch is closed when the diode is forward bias and is open when it is reverse biased as shown in figure 1. (Anode) + (Cathode) – D1 V0 2V D2 Closed I Forward bias (Anode) – (Cathode) + Open –9V Reverse bias Figure 1 Figure 1 Which one of the following statements is true ? (a) Both D1 and D2 are ON (b) Both D1 and D2 are OFF (c) D1 is ON and D2 is OFF (d) D2 is ON and D1 is OFF Sol.(d)Here the diode D 2 is ON while D1 is OFF. Since first off all D2 turn ON which makes V0 = 2 V(because diodes are ideal) Now, the anode of the diode D1 is connected to – 2 V while cathode is connected to + 2 V which makes the diode D1 OFF. 1. Practical Diode : Practical diode is a two terminal device which has the following properties: (i) conducts with low resistance (of the order of few ) when forwardbiased, and (ii) offers an very high resistance (of the order of M) when reversebiased. The Equivalent Circuit of Practical Diode Shown in Figure 2. (Anode) + (Cathode) – Rf Forward bias (Anode) – (Cathode) + Reverse bias VB Rr Values of V 0 and I, respectively, are (a) 2 V and 1.1 mA (b) 0 V and 0 mA (c) 2 V and 0.7 mA (d) 4 V and 1.3 mA Sol.(a)The given circuit can be redrawn according to the question 2. Figure 2 Table given below illustrates the comparision between Ge and Si diode : Parameter Ge - diode Si - diode Material used Germanium Silion Cut-in-voltage (Vg) 0.2 V 0.6 V D1 O.C. – 2V Knee voltage 0.3 V 0.7 V Reverse saturation current Breakdown voltage in µA in nA Lower Applications Low voltage low 2V LUCKNOW GORAKHPUR 9919526958 0522-6563566 ALLAHABAD AGRA 9919751941 9451056682 + – V0 I – 9V + – 9V Higher Rectifiers, clippers, temperature applications LUCKNOW S.C. D2 PATNA 9919751941 clampers etc. or I = 1.1 mA and V0 = 2V NOIDA SUMMER CRASH COURSE ONLINE TEST SERIES 9919751941 WINTER CRASH COURSE OFF-LINE TEST SERIES 58 TECHGURU CLASSES for ENGINEERS (Your Dedication + Our Guidance = Sure Success) CHAPTER -2 (pn - JUNCTION DIODE) : ELECTRONIC DEVICES PROBLEMS BASED ON GATE/IES/PSUs PERSONAL REMARK : 1. Determine the current I for each of the configurations of figure 1 using the approximate equivalent model for the diode. 30 V 100 Si – I + Si – 10 I 12 V + 20 Ex. The values of I & V for the circuit shown is (Assume forward drop across each diode is 0.7 V) (a) 6 mA, 2.3 V (b) 0.3 mA, 1.3 V (c) 1.3 mA, 2.3 V (d) 2.3 mA, 2.3 V D3 +3V (a) D2 (b) I D1 Si + 10 V I 1 k +1V 10 – V +2V Si (c) Figure 1 Sol. (a) I = 0 mA , since diode is reverse-biased Ans. (b) V20 = 30 V – 0.7 V = 29.3 V (Kirchhoff's Voltage Law) 29.3 V = 1.465 A 20 Ans. 10 V = 1 A , since center branch is open 10 Ans. I= (c) I = Determine Vo and ID for the networks of figure 2 2. ID –5V + 12 V V0 V0 1.2 k Si 4.7 k 2.2 ID Sol.(d) From the given circuit Three voltage (2.3 V, 1.3 V and 0.3 V) try to appear at the pant V, but at a node more than one voltage can appear. Hence one voltage will appear forcing the other two not to appear through diodes D1 and D2. Hence, V = 2.3 Volts and I 2.3V 2.3 mA 1 k Ex. Obtain the values of V and I for the circuit shown in figure (Assume forward drop across each diode is 0.7 V) (a) 1.3 mA (b) 1.3 m A (c) 6.3 mA (d) None of these +5V Si I (a) (b) Figure 2 Sol. (a) Diode forward-biased, Sol.(c)V = 1.3 V and 0 V0 0 (4.3 V) 1.95 mA R 2.2 k LUCKNOW GORAKHPUR 9919526958 0522-6563566 ALLAHABAD AGRA 9919751941 9451056682 V +1V V0 = – 4.3 V LUCKNOW +3V 2V Kirchhoff's Voltage Law : – 5 V + 0.7 V – V0 = 0 IR = ID = 1K PATNA 9919751941 I 5 1.3 6.3 mA 1 NOIDA SUMMER CRASH COURSE ONLINE TEST SERIES 9919751941 WINTER CRASH COURSE OFF-LINE TEST SERIES 59 TECHGURU CLASSES for ENGINEERS (Your Dedication + Our Guidance = Sure Success) CHAPTER -2 (pn - JUNCTION DIODE) : ELECTRONIC DEVICES (b) Diode forward-biased, PERSONAL REMARK : Ex. The value of V & I for the circuit shown assuming the drop across the diode of 0.7 V +5 V 12 V – 0.7 V = 1.915 mA .2 k 4.7 k ID = V 0 = V4.7k + VD = (1.915 mA) (4.7 k ) + 0.7 V = 9.7 V 3. Determine the level of Vo for each network figure 3 + 20 V Si Ge 2 k + 10 V 2.5 K Si 2 k V0 D1 4.7 k (a) (b) Figure 3 4 k (20 V – 0.7 V – 0.3 V) 2 = (20 V – 1 V) 4 k 2 k 3 I= 5 0.7 5 0.7 0.7 2.5K 5K 1.72 1 0.72 mA Ans. (10 V 4 V – 0.7 V) 13.3 V = 1.985 mA 2 k 4.7 k 6.7 k Applying KVL, 0.7 0.7 V = 0 .Gives,V = 0 V Hence alternative (b) is the correct choice. V' = IR = (1.985 mA) (4.7 k ) = 9.33 V V0 = V' – 4 V = 9.33 V – 4 V = 5.33 V 4. 5V I 2 or V0 = (19 V) = 12.67 V 3 (b) V D2 (a) 0 V, 0.72 mA (b) 0 V, 0.72 mA (c) 1.4 V, 0.72 mA (d) 1.4 V, 0.72 mA Sol. Assume that both the diode are 'ON' .From the given circuit –4 V V0 = I V0 4 k Sol. (a) Ans. Ex. The values of I and V assuming drop across the diode of 0.7 V D1 Determine V0 and ID for the networks of figure 4. ID V0 1V Si I 10 mA 2.2 k 2.2 k ID V0 + 20 V D2 –4V 6.8 k Si (a) (a) (b) (c) (d) (b) Figure 4 Sol.(a) Applying source transformation for the 10 mA source and 2.2 k resistor as shown in figure 5. V I 3V 2.7 mA, 2.3 V 2.3 m A, 2.3 V 7.3 mA, 2.3 V 7.3 mA, 1.7 V 1K 5V Sol.(c) V = 2.3 V and V = IR = (10 mA) (2.2 k ) = 22 V I R = 2.2 k 2.3 ( 5) 7.3 7.3 mA 1K Diode forward-biased ID = LUCKNOW GORAKHPUR 9919526958 0522-6563566 LUCKNOW 22 V – 0.7 V = 4.841 mA 2.2 k 2.2 k ALLAHABAD AGRA 9919751941 9451056682 PATNA 9919751941 Ans. NOIDA SUMMER CRASH COURSE ONLINE TEST SERIES 9919751941 WINTER CRASH COURSE OFF-LINE TEST SERIES 60 TECHGURU CLASSES for ENGINEERS (Your Dedication + Our Guidance = Sure Success) CHAPTER -2 (pn - JUNCTION DIODE) : ELECTRONIC DEVICES V0 = ID (2.2 k ) PERSONAL REMARK : = (4.841 mA) (2.2 k ) = 10.65 V Ans. (20 V – 0.7 V + 4 V) = 3.43 mA 6.8 kΩ (b) Diode forward-biased, ID = Ans. Ex. The states of the diodes D1 and D2 for the figure shown under extremely high negative Vs >> V is D1 + V0 – 0.7 V + 4 V = 0 or V0 V0 = – 3.3 V Ans. R1 5. Determine V01 and V02 for the network of figure 6. Si + 12 V 4.7 k – 10 V Ge Vs Si V02 V01 V02 V01 3.3 k (a) (b) Figure 6 (b) V01 = 12 V – 0.7 V = 11.3 V Ans. V02 = 0.3 V Ans. Ex. For the circuit shown below, using ideal diode, the values of voltage and current are [IES-EC-2011] + 3V 10 k 10 V – 0.7 V – 0.3 V 9V = 2 mA 1.2 k 3.3 k 4.5 k V=? V02 = – (2 mA) (3.3 k ) = – 6.6 V Ans. 6. Determine V0 and ID for the network of figure 7 ID ID Si Si V0 Si Si I=? – 3V (a) (b) (c) (d) 15 V + 15 V R3 (a) Both D1 and D2 OFF (b) D1 OFF and D2 ON (c) Both D1 and D2 ON (d) D1 ON and D2 ON Sol.(d) Since Vs is extremely high negative therefore diode D1 will 'ON' and D2 will 'OFF'. V01 = – 10 V + 0.3 V + 0.7 V = – 9 V I= R2 D2 V 1.2 k Ge Sol. (a) – 3 V and 0.6 mA 3 V and 0.0 mA 3 V and 0.6 mA –3 V and 0.0 mA 61(a) Let the diode is ON + 3V Si V0 4.7 k 10k 2.2 k V –5V (a) I (b) – 3V Figure 7 Sol.(a) Both diodes forward-biased , IR = 15 V – 1.4 V = 2.894 mA 4.7 k I 2.894 mA Assuming identical diodes , ID = R = 1.45 mA 2 V0 = 15 V – 0.7 V – 0.7 V = 13.6 V LUCKNOW GORAKHPUR 9919526958 0522-6563566 LUCKNOW ALLAHABAD AGRA 9919751941 9451056682 I= 3 – (–3) = 0.6 mA 10 K Direction of current is from p to n- Ans. side, So the diode is ON. Clearly V = –3 V PATNA 9919751941 NOIDA SUMMER CRASH COURSE ONLINE TEST SERIES 9919751941 WINTER CRASH COURSE OFF-LINE TEST SERIES 61 TECHGURU CLASSES for ENGINEERS (Your Dedication + Our Guidance = Sure Success) CHAPTER -2 (pn - JUNCTION DIODE) : ELECTRONIC DEVICES (b) Right diode forward-biased : ID = PERSONAL REMARK : 15 V 5 V – 0.7 V = 8.77 mA 2.2 k V0 = 15 V – 0.7 V = 14.3 V Ans. Ex. Two Ge diodes are connected in series opposition across a 5 V battery as shown in figure. Determine Vo and I for the networks shown in figure 8. 7. + 10 V V1 + 16 V I Ge Si + D2 I + Si Si V2 + D1 I Si 5V V0 V0 1 k 4.7 k +8V (a) (b) Figure 8 Sol. (a) Ge diode "on" preventing Si diode from turning "on": 10 V – 0.3 V 9.7 V = 9.7 mA 1 k 1 k Ans. V0 = I × 1 k = 9.7 mA × 1 k = 9.7 V Ans. I= (b) I = 16 V – 0.7 V – 0.7 V – 8 V 6.6 V = 1.404 mA 4.7 k 4.7 k V0 = 8 V + (1.404 mA) (4.7 k ) = 14.6 V Ans. Ans. Determine V01, V02, and I for the network of figure 9. Both diodes forward-biased : Vg (si) = 0.7 V and Vg (Ge) = 0.3 V 8. Sol. V – 0.7 V I2 k = k I (Si diode) = I2 k V01 = 0.7 V V02 = 0.3 V (a) Find voltage across each diode assuming breakdown voltage of diodes is greater than 5 V. (b) Find current if Vz = 4.9 V and I0 = 5 A Sol. Since diodes are in series, same current I flows in D1 and D2 . From the given circuit we observe that diode D 2 in forward direction while diode D 1 in reverese direction ,therefore current in the circuit is limited by diode D1. Since D2 is forward biased, V2 will be very small and therefore, V1 (i.e. 5 V2) will be much larger that VT (i.e. 0.026 V). So ,the current will be Consider .7 V – 0.3 V diode D2 : I I 0 (e V/VT 1) Since, I = I0 & V = V2 = 9.65 mA , I0.47 k = = 0.851 mA . k – I0.47 k = 9.65 mA – 0.851 mA = 8.799 mA Ans. Ans. Ans. We get , I0 I0 (eV/VT 1) or V2 VT In2 0.026 0.693 0.018 V V1 5 V2 5 0.018 4.892 V 9. Determine V0 and ID for the network of figure 10. Vo1 (b) 2 k 0.47 k Vo2 ID I Vo +10 V 20 V Si Si Ge 2 k – Figure 9 LUCKNOW GORAKHPUR 9919526958 0522-6563566 LUCKNOW 2 k Si + e V /VT 1 1 If V z is 4.9 V, then D 1 will breakdwon.So,V1 = 4.9 V.Therefore , V2 = 5 V1 = 5 4.9 = 0.1 V Putting , I0 = 5 A, V2 = 0.1 V We get, I 5(e 0.1/ 0.026 1) 229 A 2 k Figure 10 ALLAHABAD AGRA 9919751941 9451056682 PATNA 9919751941 NOIDA SUMMER CRASH COURSE ONLINE TEST SERIES 9919751941 WINTER CRASH COURSE OFF-LINE TEST SERIES 62 TECHGURU CLASSES for ENGINEERS (Your Dedication + Our Guidance = Sure Success) CHAPTER -2 (pn - JUNCTION DIODE) : ELECTRONIC DEVICES Sol. For the parallel Si – 2 k branches a Thevenin equivalent will result (for "on" diodes) in a single series branch or 0.7 V and 1 k resistor as shown in figure 11. 2 k (10 V – 0.7 V) 2 V0 = (9.3 V) = 6.2 V k 2 k 3 I2 k = 6.2 V = 3.1 mA 2 k ID = PERSONAL REMARK : V0 10V 0.7V 1 k 2K Ans. Eth , R th I 2k 3.1 mA = 1.55 mA Ans. 2 2 Figure 11 10. Determine V0 for the network of figure 12. 5V Si V0 0V Si 2.2 Figure 12 5V Sol. The Si diode with –5 V at the cathode is "on" while the other is "off". The result is , V0 = 0 – 0.7 V = – 0.7 V 11. Determine V0 for the network of figure 13. –5V Si 0V Si V0 Figure 13 Sol. 0 V at one terminal is "more positive" than – 5 V at the other input terminal. Therefore assume lower diode "OFF" and upper diode "ON".The result : V0 = – 5 V + 0. 7 V = – 4.3 V 12. Determine the level of V0 for the gate of figure 14. 10 V Si 10 V V0 Si Figure 14 Sol. Since all the system terminals are at 10 V the required difference of 0.7 V across either diode cannot be established. Therefore, both diodes are "OFF" and V0 = + 10 V as established by 10 V supply connected to 1 k resistor.. LUCKNOW GORAKHPUR 9919526958 0522-6563566 LUCKNOW ALLAHABAD AGRA 9919751941 9451056682 PATNA 9919751941 NOIDA SUMMER CRASH COURSE ONLINE TEST SERIES 9919751941 WINTER CRASH COURSE OFF-LINE TEST SERIES 63 TECHGURU CLASSES for ENGINEERS (Your Dedication + Our Guidance = Sure Success) CHAPTER -2 (pn - JUNCTION DIODE) : ELECTRONIC DEVICES 13. Assuming that the diodes are ideal in figure, the current in diode D1 is (GATE-EE-2004) 1 k 1 k PERSONAL REMARK : (a) 8 mA D1 (b) 5 mA (c) 0 mA D2 I 5V 8V (d) –3 m A Sol.(c) Let Both the diodes are ‘ON’ and replaced by short circuit 1 k VA 1 k I 5V 8V From the diagram VA = 0 V and Apply nodal at VA 5 – VA V 8 I A or 5 – VA = I + VA + 8 or 5 = I + 8 or I = –3 mA 1 1 The direction of current flow is against the diode convention. The diode is reverse bised I = 0 mA. 14. Assume that D1 and D2 in figure are ideal diodes. The value of current I is (GATE-EE-2005) (a) 0 mA D1 2k (b) 0.5 mA 1mA (DC) (c) 1 mA I 2k (d) 2 mA D2 Sol.(a) The diode D1 is forward biased. The diode D2 is reverse biased as its convention is opposite to the direction of current. D2 is reverse biased and is replaced by open-circuit. Therefore, I = 0 mA 15. What are the states of the three ideal diodes of the circuit shown in figure ? (GATE-EE-2006) 1 1 + 10V D2 D1 1 5A D3 (a) D1 ON, D2 OFF, D3 OFF (b) D1 OFF, D2 ON, D3 OFF (c) D1 ON, D2 OFF, D3 ON (d) D1 OFF, D2 ON, D3 ON LUCKNOW GORAKHPUR 9919526958 0522-6563566 LUCKNOW ALLAHABAD AGRA 9919751941 9451056682 PATNA 9919751941 NOIDA SUMMER CRASH COURSE ONLINE TEST SERIES 9919751941 WINTER CRASH COURSE OFF-LINE TEST SERIES 64 TECHGURU CLASSES for ENGINEERS (Your Dedication + Our Guidance = Sure Success) CHAPTER -2 (pn - JUNCTION DIODE) : ELECTRONIC DEVICES Sol.(a) Let all the diodes be reverse biased. PERSONAL REMARK : 1 Aply KCL at node V1 1 V 1 V1 – 10 V1 – V2 0 1 1 or 2V1 – V2 = 10 ...(1) Apply KCL at node V2 V1 – V2 + 5 = V2 1 V2 D2 1 10V 5A or V1 – 2V2 = –5 ... (2) 20 25 V or V1 = V 3 3 Clearly, D1 is forward biased and D3 is reverse biased Solve (1) and (2) , V2 = 16. Assuming that the diodes in the given circuit are ideal, the voltage V0 is D1 (GATE-EE-2010) 10k D 2 (a) 4 V 10k (b) 5 V V0 10V 15V (c) 7.5 V 10k (d) 12.12 V Sol.(b) Diode D1 is always ‘ON’ and D2 is ‘OFF’ V0 = 10k V0 10V 10k 10 10 R 2 = =5V 20 R1 R 2 +6V 10k The voltage at V1 and V2 of the arrangement D2 shown in the figure will be respectively (IES-EE-1998) +6V ( Given the cut-in voltage of diode is 0.7 V ) D1 (a) 6 V and 5.4 V (b) 5.4 V and 6 V 10k (c) 3 V and 5.4 V (d) 6 V and 3 V +3V Sol.(a)Clearly D1 is forward biased and D2 is reverse biased. The n-side of D2 is connected to more positive terminal ,V1 = 6 V Taking cut-in voltage of diode 0.6 V we get ,V2 = 6 0.6 = 5.4 V 17. V1 V2 19. For the circuit shown in figure the diode D is ideal. The power dissipated by the 300 resistor is [GATE-IN- 2004] 100, 0.5W (a) 0.25 W 200 (b) 0.50 W 0.5W (c) 0.75 W + (d) 1.00 W 60 cos314t – Sol.(c) From the circuit, I rms or I rms Im (Since the given circuit represents HWR) 2 Vm 60 1 2(100 200 300) 2 600 20 LUCKNOW GORAKHPUR 9919526958 0522-6563566 LUCKNOW 300 1W ALLAHABAD AGRA 9919751941 9451056682 PATNA 9919751941 NOIDA SUMMER CRASH COURSE ONLINE TEST SERIES 9919751941 WINTER CRASH COURSE OFF-LINE TEST SERIES 65 TECHGURU CLASSES for ENGINEERS (Your Dedication + Our Guidance = Sure Success) CHAPTER -2 (pn - JUNCTION DIODE) : ELECTRONIC DEVICES Now, the power dissipated by the 300 resistor is PERSONAL REMARK : 2 2 P300 I rms R 300 300 1 0.75 W 300 or P300 20 400 20. A voltage 1000 sin t Volts is applied across YZ. Assuming ideal diodes, the voltage measured across WX in Volts is [GATE-EC-EE- 2013] 1 k (a) sin t (b) (c) (d) W ( sin t + | sin t |)/ 2 ( sin t | sin t |)/2 0 for all t Y X Z 1 k Sol.(d) The given circuit 1 k W Z Y X 1 k During positive half cycle, the equivalent circuit is shown 1 k Y X W Z + From the equivalent circuit ,VWZ = 0 During negative half cycle , the equivalent circuit is shown 1 k W Y X Z + 22. Which of the following is NOT associated with a p-n junction? (GATE-EC-2008) From the equivalent circuit ,VWZ = 0 21. In a forward biased pn junction diode the sequence of events that best describes the mechanism of current flow is [GATE-EC-2013] (a) injection and subsequenct diffusion and recombination of minority carriers (b) injection and subsequenct diffusion and generation of minority carriers (c) extraction, and subsequent diffusion and generation of minority carriers (d) extraction and subsequent drift and recombination of minority carriers LUCKNOW GORAKHPUR 9919526958 0522-6563566 LUCKNOW ALLAHABAD AGRA 9919751941 9451056682 Sol.(a) In a forward biased pn junction diode the sequence of events that best describes the mechanism of current flow is injection and subsequenct diffusion and recombination of minority carriers PATNA 9919751941 (a) Junction Capacitance (b) Charge Storage Capacitance (c) Depletion Capacitance (d) Channel Length Modulation Sol.(d)Channel length modulation takes place in MOSFET NOIDA SUMMER CRASH COURSE ONLINE TEST SERIES 9919751941 WINTER CRASH COURSE OFF-LINE TEST SERIES 66