Question

Four ideal diodes are connected to form a circuit as shown in the figure. An $AC$ signals $V_{in}=10\sin \left(100\pi t\right)$ volt is applied across points $1$ and $2$ and the output is measured across points $5$ and $6$. The output voltage across points $5$ and $6$. The output voltage $\left(V_{out}\right)$ is:

• A. $10\sin \left(100\pi t\right)V$
• B. $10V$
• C. $7.07V$
• D. $3.2V$

Answer 1

Answer: (C)

$7.07V$

The Output of this configuration be slightly more than $7.1$ volts DC when measured as an open circuit at the output terminals.

This due to the discharge rate of the capacitor versus the $100$ Hz fullwave rectification waveform.

The capacitor will hold the DC output voltage slightly higher than zero, decaying from about $3$ volts to $1.8$ volts before the increasing input voltage from the next half sinwave pulse pushes it back to $10$ volts peak.
If a load current of $0.7$ milliamps or more is drawn from this circuit, the capacitor will have no effect on the output voltage (See the black curve below) which will then be $\frac{10}{\sqrt{2}}=7.07$VoltsDC.