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A designer requires a shunt regulator of approximately 20 V. Two kinds of Zener diodes are available: 6.8-V devices with rz of 10 Ω and 5.1-V devices with rz of 30 Ω. For the two major choices possible, find the load regulation. In this calculation neglect the effect of the regulator resistance R.
- a)-30mV/mA and 120mV/mA respectively
- b)30mV/mA and 60mV/mA respectively
- c)-60mV/mA and +60mV/mA respectively
- d)-30mV/mA and -120mV/mA respectively
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A designer requires a shunt regulator of approximately 20 V. Two kinds...
Three 6.8v zeners provide 3*6.8 = 20.4v with 3 * 10 =30Ω Resistance, neglecting R, we have
load Regulation = -30mV/mA.
For 5.1 Zeners we need 4 diodes to provide 20.4v with 4 * 30 =120Ω Resistance.
load Regulation = -120mV/mA .
load Regulation = -30mV/mA.
For 5.1 Zeners we need 4 diodes to provide 20.4v with 4 * 30 =120Ω Resistance.
load Regulation = -120mV/mA .
Most Upvoted Answer
A designer requires a shunt regulator of approximately 20 V. Two kinds...
Ω and 5.1-V devices with rzof 5 Ω. Determine which type of Zener diode should be used in the shunt regulator and calculate the necessary resistance to achieve the desired output voltage.
To determine which type of Zener diode to use, we need to calculate the minimum input voltage required for each type of diode to regulate at 20 V. The formula for the minimum input voltage is:
Vin = Vz + Vr
where Vin is the input voltage, Vz is the Zener voltage, and Vr is the voltage across the resistor. We can rearrange this formula to solve for Vr:
Vr = Vin - Vz
For the 6.8-V Zener diode, the minimum input voltage required to regulate at 20 V is:
Vr = Vin - Vz = 20 V - 6.8 V = 13.2 V
Using Ohm's Law, we can calculate the necessary resistance:
Vr = IR
13.2 V = I * 10 Ω
I = 1.32 A
Therefore, the resistance required is:
R = Vr / I = 13.2 V / 1.32 A = 10 Ω
For the 5.1-V Zener diode, the minimum input voltage required to regulate at 20 V is:
Vr = Vin - Vz = 20 V - 5.1 V = 14.9 V
Using Ohm's Law, we can calculate the necessary resistance:
Vr = IR
14.9 V = I * 5 Ω
I = 2.98 A
Therefore, the resistance required is:
R = Vr / I = 14.9 V / 2.98 A = 5 Ω
Comparing the two results, we see that the shunt regulator using the 5.1-V Zener diode requires a lower resistance and therefore would be a better choice. However, we should also consider the power dissipation in the Zener diode:
Pz = Vz * Iz
where Pz is the power dissipation, Iz is the Zener current, and Vz is the Zener voltage. We can calculate the Zener current using Ohm's Law:
Iz = Vr / R
For the 5.1-V Zener diode:
Iz = Vr / R = 14.9 V / 5 Ω = 2.98 A
Pz = Vz * Iz = 5.1 V * 2.98 A = 15.2 W
For the 6.8-V Zener diode:
Iz = Vr / R = 13.2 V / 10 Ω = 1.32 A
Pz = Vz * Iz = 6.8 V * 1.32 A = 8.98 W
We can see that the power dissipation in the 5.1-V Zener diode is much higher than in the 6.8-V Zener diode. Therefore, we should use the 6.8-V Zener diode with a 10-Ω resistor to achieve the desired output voltage of 20 V.
To determine which type of Zener diode to use, we need to calculate the minimum input voltage required for each type of diode to regulate at 20 V. The formula for the minimum input voltage is:
Vin = Vz + Vr
where Vin is the input voltage, Vz is the Zener voltage, and Vr is the voltage across the resistor. We can rearrange this formula to solve for Vr:
Vr = Vin - Vz
For the 6.8-V Zener diode, the minimum input voltage required to regulate at 20 V is:
Vr = Vin - Vz = 20 V - 6.8 V = 13.2 V
Using Ohm's Law, we can calculate the necessary resistance:
Vr = IR
13.2 V = I * 10 Ω
I = 1.32 A
Therefore, the resistance required is:
R = Vr / I = 13.2 V / 1.32 A = 10 Ω
For the 5.1-V Zener diode, the minimum input voltage required to regulate at 20 V is:
Vr = Vin - Vz = 20 V - 5.1 V = 14.9 V
Using Ohm's Law, we can calculate the necessary resistance:
Vr = IR
14.9 V = I * 5 Ω
I = 2.98 A
Therefore, the resistance required is:
R = Vr / I = 14.9 V / 2.98 A = 5 Ω
Comparing the two results, we see that the shunt regulator using the 5.1-V Zener diode requires a lower resistance and therefore would be a better choice. However, we should also consider the power dissipation in the Zener diode:
Pz = Vz * Iz
where Pz is the power dissipation, Iz is the Zener current, and Vz is the Zener voltage. We can calculate the Zener current using Ohm's Law:
Iz = Vr / R
For the 5.1-V Zener diode:
Iz = Vr / R = 14.9 V / 5 Ω = 2.98 A
Pz = Vz * Iz = 5.1 V * 2.98 A = 15.2 W
For the 6.8-V Zener diode:
Iz = Vr / R = 13.2 V / 10 Ω = 1.32 A
Pz = Vz * Iz = 6.8 V * 1.32 A = 8.98 W
We can see that the power dissipation in the 5.1-V Zener diode is much higher than in the 6.8-V Zener diode. Therefore, we should use the 6.8-V Zener diode with a 10-Ω resistor to achieve the desired output voltage of 20 V.
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A designer requires a shunt regulator of approximately 20 V. Two kinds of Zener diodes are available: 6.8-V devices with rzof 10 Ω and 5.1-V devices with rzof 30 Ω. For the two major choices possible, find the load regulation. In this calculation neglect the effect of the regulator resistance R.a)-30mV/mA and 120mV/mA respectivelyb)30mV/mA and 60mV/mA respectivelyc)-60mV/mA and +60mV/mA respectivelyd)-30mV/mA and -120mV/mA respectivelyCorrect answer is option 'D'. Can you explain this answer?
Question Description
A designer requires a shunt regulator of approximately 20 V. Two kinds of Zener diodes are available: 6.8-V devices with rzof 10 Ω and 5.1-V devices with rzof 30 Ω. For the two major choices possible, find the load regulation. In this calculation neglect the effect of the regulator resistance R.a)-30mV/mA and 120mV/mA respectivelyb)30mV/mA and 60mV/mA respectivelyc)-60mV/mA and +60mV/mA respectivelyd)-30mV/mA and -120mV/mA respectivelyCorrect answer is option 'D'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A designer requires a shunt regulator of approximately 20 V. Two kinds of Zener diodes are available: 6.8-V devices with rzof 10 Ω and 5.1-V devices with rzof 30 Ω. For the two major choices possible, find the load regulation. In this calculation neglect the effect of the regulator resistance R.a)-30mV/mA and 120mV/mA respectivelyb)30mV/mA and 60mV/mA respectivelyc)-60mV/mA and +60mV/mA respectivelyd)-30mV/mA and -120mV/mA respectivelyCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A designer requires a shunt regulator of approximately 20 V. Two kinds of Zener diodes are available: 6.8-V devices with rzof 10 Ω and 5.1-V devices with rzof 30 Ω. For the two major choices possible, find the load regulation. In this calculation neglect the effect of the regulator resistance R.a)-30mV/mA and 120mV/mA respectivelyb)30mV/mA and 60mV/mA respectivelyc)-60mV/mA and +60mV/mA respectivelyd)-30mV/mA and -120mV/mA respectivelyCorrect answer is option 'D'. Can you explain this answer?.
A designer requires a shunt regulator of approximately 20 V. Two kinds of Zener diodes are available: 6.8-V devices with rzof 10 Ω and 5.1-V devices with rzof 30 Ω. For the two major choices possible, find the load regulation. In this calculation neglect the effect of the regulator resistance R.a)-30mV/mA and 120mV/mA respectivelyb)30mV/mA and 60mV/mA respectivelyc)-60mV/mA and +60mV/mA respectivelyd)-30mV/mA and -120mV/mA respectivelyCorrect answer is option 'D'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A designer requires a shunt regulator of approximately 20 V. Two kinds of Zener diodes are available: 6.8-V devices with rzof 10 Ω and 5.1-V devices with rzof 30 Ω. For the two major choices possible, find the load regulation. In this calculation neglect the effect of the regulator resistance R.a)-30mV/mA and 120mV/mA respectivelyb)30mV/mA and 60mV/mA respectivelyc)-60mV/mA and +60mV/mA respectivelyd)-30mV/mA and -120mV/mA respectivelyCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A designer requires a shunt regulator of approximately 20 V. Two kinds of Zener diodes are available: 6.8-V devices with rzof 10 Ω and 5.1-V devices with rzof 30 Ω. For the two major choices possible, find the load regulation. In this calculation neglect the effect of the regulator resistance R.a)-30mV/mA and 120mV/mA respectivelyb)30mV/mA and 60mV/mA respectivelyc)-60mV/mA and +60mV/mA respectivelyd)-30mV/mA and -120mV/mA respectivelyCorrect answer is option 'D'. Can you explain this answer?.
Solutions for A designer requires a shunt regulator of approximately 20 V. Two kinds of Zener diodes are available: 6.8-V devices with rzof 10 Ω and 5.1-V devices with rzof 30 Ω. For the two major choices possible, find the load regulation. In this calculation neglect the effect of the regulator resistance R.a)-30mV/mA and 120mV/mA respectivelyb)30mV/mA and 60mV/mA respectivelyc)-60mV/mA and +60mV/mA respectivelyd)-30mV/mA and -120mV/mA respectivelyCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE).
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A designer requires a shunt regulator of approximately 20 V. Two kinds of Zener diodes are available: 6.8-V devices with rzof 10 Ω and 5.1-V devices with rzof 30 Ω. For the two major choices possible, find the load regulation. In this calculation neglect the effect of the regulator resistance R.a)-30mV/mA and 120mV/mA respectivelyb)30mV/mA and 60mV/mA respectivelyc)-60mV/mA and +60mV/mA respectivelyd)-30mV/mA and -120mV/mA respectivelyCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for A designer requires a shunt regulator of approximately 20 V. Two kinds of Zener diodes are available: 6.8-V devices with rzof 10 Ω and 5.1-V devices with rzof 30 Ω. For the two major choices possible, find the load regulation. In this calculation neglect the effect of the regulator resistance R.a)-30mV/mA and 120mV/mA respectivelyb)30mV/mA and 60mV/mA respectivelyc)-60mV/mA and +60mV/mA respectivelyd)-30mV/mA and -120mV/mA respectivelyCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of A designer requires a shunt regulator of approximately 20 V. Two kinds of Zener diodes are available: 6.8-V devices with rzof 10 Ω and 5.1-V devices with rzof 30 Ω. For the two major choices possible, find the load regulation. In this calculation neglect the effect of the regulator resistance R.a)-30mV/mA and 120mV/mA respectivelyb)30mV/mA and 60mV/mA respectivelyc)-60mV/mA and +60mV/mA respectivelyd)-30mV/mA and -120mV/mA respectivelyCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A designer requires a shunt regulator of approximately 20 V. Two kinds of Zener diodes are available: 6.8-V devices with rzof 10 Ω and 5.1-V devices with rzof 30 Ω. For the two major choices possible, find the load regulation. In this calculation neglect the effect of the regulator resistance R.a)-30mV/mA and 120mV/mA respectivelyb)30mV/mA and 60mV/mA respectivelyc)-60mV/mA and +60mV/mA respectivelyd)-30mV/mA and -120mV/mA respectivelyCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice Electrical Engineering (EE) tests.
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