The reverse breakdown voltage for a silicon d | vedclass

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(B) The circuit consists of two parallel branches connected to a $10 \ V$ battery.Branch $1$ (Top): The diode is forward-biased. The voltage drop acro

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$1.3$

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(B) The circuit consists of two parallel branches connected to a $10 \ V$ battery.Branch $1$ (Top): The diode is forward-biased. The voltage drop across the diode is $0.3 \ V$. The voltage across the $10 \ \Omega$ resistor is $V_1 = 10 \ V - 0.3 \ V = 9.7 \ V$. The current through this branch is $i_1 = \frac{9.7 \ V}{10 \ \Omega} = 0.97 \ A$.Branch $2$ (Bottom): The diode is reverse-biased. Since the battery voltage $(10 \ V)$ is greater than the reverse breakdown voltage $(6 \ V)$,the diode operates in the breakdown region. The voltage across the $10 \ \Omega$ resistor is $V_2 = 10 \ V - 6 \ V = 4 \ V$. The current through this branch is $i_2 = \frac{4 \ V}{10 \ \Omega} = 0.4 \ A$.The total current through the battery is $I = i_1 + i_2 = 0.97 \ A + 0.4 \ A = 1.37 \ A$. The closest approximate value is $1.3 \ A$.

The reverse breakdown voltage for a silicon diode is $6 \ V$ for a particular doping concentration and the forward bias voltage drop is $0.3 \ V$. Find the approximate current through the battery in the given circuit in $Amp$.

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