Consider the ideal diodes $D_1$ and $D_2$ as shown in the figure with cut-in voltage $V_\gamma=0$ Volt and $v_i(t)$ is in Volt. The maximum voltage (Volt) of the output $v_o(t)$ is (rounded off to two decimal places).

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Step 1: Understand the circuit. The source $v_i(t)=6\sin(\omega t)$ drives a $10\text{ k}\Omega$ resistor $R_1$ to a node $A$, which is where the output $v_o(t)$ is measured with respect to the ground rail. From node $A$, a second $10\text{ k}\Omega$ resistor $R_2$ leads to a node $N$. Node $N$ connects to ground through two parallel ideal-diode branches: $D_1$ in series with a $2$ V battery, and $D_2$ in series with a $4$ V battery. Since $V_\gamma=0$, each diode turns on the instant the voltage across it tries to go positive. Step 2: Find when the diodes are off. If both diodes are off, no current flows anywhere in the circuit, so there is no drop across $R_1$ or $R_2$, and node $A$ and node $N$ both sit at $v_i(t)$. This holds as long as $v_i(t)<2$ V, since that keeps $D_1$, the lower clamp, reverse biased. In this range, $$ v_o(t)=v_i(t) $$ Step 3: Find what happens once D1 turns on. As $v_i(t)$ rises past $2$ V, node $N$ tries to rise above $2$ V too, which turns $D_1$ on. Since $D_1$ is ideal, it pins node $N$ at exactly $2$ V, and current now flows from the source through $R_1$ and $R_2$ into the $2$ V battery. Because $2$ V is smaller than $4$ V, node $N$ never gets a chance to reach $4$ V, so $D_2$ stays off the whole time. Only $D_1$ is active. Step 4: Write the current when D1 conducts. With $D_1$ on, the same current $I$ flows through the series pair $R_1+R_2$ between the source and the clamped node: $$ I=\frac{v_i(t)-2}{R_1+R_2}=\frac{v_i(t)-2}{20\text{ k}\Omega} $$ Step 5: Find the output voltage in this range. The output is measured at node $A$, before $R_2$, so $$ v_o(t)=v_i(t)-IR_1=v_i(t)-\frac{R_1}{R_1+R_2}\big(v_i(t)-2\big) $$ With $R_1=R_2$, $\dfrac{R_1}{R_1+R_2}=\dfrac{1}{2}$, so $$ v_o(t)=v_i(t)-\frac{1}{2}\big(v_i(t)-2\big)=\frac{1}{2}v_i(t)+1 $$ Step 6: Find the maximum of v_o(t). Since $v_o(t)=\dfrac{1}{2}v_i(t)+1$ grows with $v_i(t)$ in this region, and $v_i(t)=6\sin(\omega t)$ reaches its largest value of $6$ V at its peak, the maximum output occurs there: $$ v_{o,\max}=\frac{1}{2}(6)+1=3+1=4\ \text{V} $$ Step 7: Final conclusion. For the lower part of the swing, $v_i(t)<2$ V, including all negative values down to $-6$ V, $v_o(t)=v_i(t)$ stays below $2$ V, so it never beats the clamped region. The overall maximum output voltage is $$ \boxed{4.00\ \text{V}} $$

Consider the ideal diodes \(D_1\) and \(D_2\) as shown in the figure with cut-in voltage \(V_\gamma=0\) Volt and \(v_i(t)\) is in Volt.

The maximum voltage (Volt) of the output \(v_o(t)\) is (rounded off to two decimal places).

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Correct Answer: 4

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Top GATE EC Analog Circuits Questions

Top GATE EC Questions

Consider the ideal diodes \(D_1\) and \(D_2\) as shown in the figure with cut-in voltage \(V_\gamma=0\) Volt and \(v_i(t)\) is in Volt. The maximum voltage (Volt) of the output \(v_o(t)\) is (rounded off to two decimal places).