The problem asks to find the AC signal output $V_0$ of the given circuit. The circuit consists of a $5V$ DC source, a $3 \sin(\omega t) mV$ AC source, a $1k\Omega$ resistor, a Germanium diode (Gc), and a Silicon diode (Si) connected in series.

1. Problem Description

The problem asks to find the AC signal output $V_0$ of the given circuit. The circuit consists of a $5V$ DC source, a $3 \sin(\omega t) mV$ AC source, a $1k\Omega$ resistor, a Germanium diode (Gc), and a Silicon diode (Si) connected in series.

2. Solution Steps

The given circuit has a DC source of $5V$ and an AC source of $3\sin(\omega t) mV$. The diodes are connected in series with opposite polarities. We need to determine the output voltage $V_0$.

First, consider the DC voltage. Since the $5V$ source is connected in series with the diodes, we need to consider the forward voltage drop of the diodes. The Germanium diode (Gc) has a forward voltage drop of approximately $0.3V$, and the Silicon diode (Si) has a forward voltage drop of approximately $0.7V$.

The total forward voltage drop is $0.3V + 0.7V = 1V$. Since the DC source is $5V$, the DC voltage across the series combination of the resistor and two diodes is $5V$. The total voltage drop across the diodes is $1V$. Therefore, the DC current through the resistor is $\frac{5V - 1V}{1k\Omega} = \frac{4V}{1000\Omega} = 4mA$.

Now, consider the AC signal $3\sin(\omega t) mV$. Since the DC voltage establishes a forward bias across the diodes, the AC signal will be superimposed on the DC bias. The incremental resistance of the diodes depends on the DC current flowing through them. The incremental resistance $r_d$ of a diode is given by

$r_d = \frac{V_T}{I_D}$

where $V_T$ is the thermal voltage (approximately $26mV$ at room temperature) and $I_D$ is the DC current flowing through the diode.

For the Germanium diode, $r_{d,Ge} = \frac{26mV}{4mA} = 6.5\Omega$

For the Silicon diode, $r_{d,Si} = \frac{26mV}{4mA} = 6.5\Omega$

The total incremental resistance of the two diodes is $r_{d,total} = r_{d,Ge} + r_{d,Si} = 6.5\Omega + 6.5\Omega = 13\Omega$.

Now we can calculate the AC output voltage. The AC signal voltage is divided between the $1k\Omega$ resistor and the total dynamic resistance of the two diodes. The output voltage can be determined using the voltage divider rule:

$V_{0,AC} = 3\sin(\omega t)mV \times \frac{r_{d,total}}{1000\Omega + r_{d,total}} = 3\sin(\omega t)mV \times \frac{13\Omega}{1000\Omega + 13\Omega} = 3\sin(\omega t)mV \times \frac{13}{1013} \approx 3\sin(\omega t)mV \times 0.0128$

$V_{0,AC} \approx 0.0384\sin(\omega t)mV \approx 38.4\sin(\omega t)\mu V$

However, we are looking for the AC signal output at $V_0$. Since the diodes are connected with opposite polarities, the AC voltage will likely be significantly attenuated. A better estimate is that the diodes effectively clamp the signal to approximately $\pm (0.7 - 0.3)V = \pm 0.4V$. The small signal of $3mV$ will be significantly attenuated.

Consider the voltage divider formed by the $1 k\Omega$ resistor and the series diodes' dynamic resistances, as calculated above. The change in output voltage will be:

$V_{o,AC} \approx 3mV (\frac{r_{d,Ge}+r_{d,Si}}{R + r_{d,Ge}+r_{d,Si}}) = 3mV (\frac{13}{1000+13}) = 3mV (\frac{13}{1013}) = 3mV * 0.0128 = 0.0384 mV = 38.4 \mu V$.

Therefore, the AC signal output is approximately $38.4\sin(\omega t) \mu V$.

3. Final Answer

$38.4 \sin(\omega t) \mu V$