In the circuit below the maximum value of \(v_{out}\) is ________ V. (rounded off to the nearest integer)
Show Hint
Treat the diode as open below its 3 V knee (so vout=vin) and as a fixed 3 V drop once conducting, with the remaining voltage dividing across the two resistors.
Step 1: Identify the circuit and label the elements.
The source \(15\cos(\omega t)\) drives a series loop made up of a \(4\ \Omega\) resistor, a diode with a \(3\) V breakdown/knee marking, and an \(8\ \Omega\) resistor, all connected in a single loop back to the source. The output \(v_{out}\) is measured across the series combination of the diode and the \(8\ \Omega\) resistor, taken from the node between the \(4\ \Omega\) resistor and the diode down to the return rail.
Step 2: Understand how the diode behaves.
The diode is marked with \(3\) V, which is its breakdown/turn-on threshold in the direction it is connected. As long as the voltage across the diode stays below \(3\) V in that direction, the diode does not conduct and behaves like an open circuit. Once the voltage across the diode tries to exceed \(3\) V, the diode conducts and holds the voltage across itself fixed at \(3\) V, similar to a Zener diode in breakdown.
Step 3: Analyze the circuit while the diode is OFF.
While the diode is not conducting, no current can flow anywhere in this single-loop circuit, so the loop current \(I = 0\). With zero current, there is no voltage drop across either resistor, so the entire source voltage appears directly across the diode:
\[
v_{diode} = v_{in}
\]
Since \(v_{out}\) is measured across the diode plus the \(8\ \Omega\) resistor, and the resistor carries no current at this stage,
\[
v_{out} = v_{diode} = v_{in}
\]
This continues as \(v_{in}\) rises, until \(v_{out}\) (which equals \(v_{diode}\)) reaches the \(3\) V breakdown point, that is, when \(v_{in} = 3\) V.
Step 4: Analyze the circuit once the diode turns ON.
Once \(v_{in}\) exceeds \(3\) V, the diode conducts and clamps its own voltage at
\[
v_{diode} = 3 \text{ V}
\]
Current now flows around the loop. Applying Kirchhoff's voltage law, the source voltage splits between the \(4\ \Omega\) resistor, the diode, and the \(8\ \Omega\) resistor:
\[
v_{in} = I(4) + 3 + I(8)
\]
Solving for the current,
\[
I = \frac{v_{in} - 3}{4+8} = \frac{v_{in}-3}{12}
\]
Step 5: Write vout in terms of vin for this stage.
\(v_{out}\) is the drop across the diode plus the \(8\ \Omega\) resistor, so
\[
v_{out} = v_{diode} + I(8) = 3 + 8\cdot\frac{v_{in}-3}{12} = 3 + \frac{2}{3}(v_{in}-3)
\]
This shows that once the diode turns on, \(v_{out}\) still rises with \(v_{in}\), but at a reduced rate of \(\tfrac{2}{3}\) instead of rising one-for-one, because part of the extra voltage above \(3\) V is now dropped across the \(4\ \Omega\) resistor.
Step 6: Find the maximum value of vout.
The source \(15\cos(\omega t)\) has a peak value of \(15\) V, so the largest possible value of \(v_{in}\) is \(15\) V, which is well above the \(3\) V threshold, meaning the diode is conducting at this peak. Substituting \(v_{in} = 15\) V into the Step 5 expression:
\[
v_{out,\max} = 3 + \frac{2}{3}(15-3) = 3 + \frac{2}{3}(12)
\]
\[
v_{out,\max} = 3 + 8 = 11 \text{ V}
\]
Final Answer:
The maximum value of \(v_{out}\) is
\[ \boxed{11 \text{ V}} \]
Find dV/dt