Question In the circuit below, assume the diodes are ideal. Draw the voltage-current characteristics at  the input side of the circuit, with voltage (v) in the x-axis and current (i) in the y-axis as the  voltage is varied from -∞ to +∞ 1. In the circuit below, assume the diodes are ideal. Draw the voltage-current characteristics at the input side of the circuit, with voltage \( (v) \) in the \( x \)-axis and current \( (i) \) in the \( y \)-axis as the voltage is varied from \( -\infty \) to \( +\infty \). ( 25 points)

I35AS8 The Asker · Electrical Engineering

In the circuit below, assume the diodes are ideal. Draw the voltage-current characteristics at 
the input side of the circuit, with voltage (v) in the x-axis and current (i) in the y-axis as the 
voltage is varied from -∞ to +∞

Transcribed Image Text: 1. In the circuit below, assume the diodes are ideal. Draw the voltage-current characteristics at the input side of the circuit, with voltage \( (v) \) in the \( x \)-axis and current \( (i) \) in the \( y \)-axis as the voltage is varied from \( -\infty \) to \( +\infty \). ( 25 points)
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Transcribed Image Text: 1. In the circuit below, assume the diodes are ideal. Draw the voltage-current characteristics at the input side of the circuit, with voltage \( (v) \) in the \( x \)-axis and current \( (i) \) in the \( y \)-axis as the voltage is varied from \( -\infty \) to \( +\infty \). ( 25 points)
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&#12304;General guidance&#12305;The answer provided below has been developed in a clear step by step manner.Step1/2From the questions, we have to draw a V-I characteristics of the circuit given in the question..We already knows the basic principle of the Diode working i.e.Diode is forward biased when the anode is connected to positive(+) terminal of the battery and cathode is connected to the negative(-) terminal of the battery.Diode is reverse biased when the cathode is connected to positive(+) terminal of the battery and anode is connected to the negative(-) terminal of the battery.Diode offers zero resistance while it is connected as forward bias. Practically, it is not working condition until voltage drop of 0.7V for silicon and 0.3V for germanium.Diode offers high resistance while it is connected as reverse bias.From the circuit given in the question, we have to take a values from -10V to +10V as the input voltage.1.5&#937;5VD2D12&#937;10V<path d=" M 299 222 L 180 222 " marker-end="url(#MTzaCefJ5IzN3G2HeZztk-end-arrowhead)" fill="#FFFFFF" stroke="rgb(51, 51, 51)" stroke-width="2" stroke-dasharray="0" stroke-linecap="round" stroke-linejoin="round" stroke-miterlimit="10" dominant-baseline="text-before-edge" vector-effect="non-scaling-stroke" data-id="MTzaCefJ5IzN3G2HeZztk"><path transform="" transform-origin="180 160" d=" M 180 160 L 301 160 " fill="#FFFFFF" stroke="rgb(51, 51, 51)" stroke-width="2" stroke-dasharray="0" stroke-linecap="round" stroke-linejoin="round" stroke-miterlimit="10" dominant-baselin ... See the full answer