A p-n junction germanium diode is shown here, The current is excess of 10 mA through the diode produces the large joule-heating which damages (burns) the diode. If we want to use a 1.5 V battery to forward bias the diode, want should be the diode, that the maximum current does not exceed 6 mA? (V_forward for Ge is 0.3V)

The given circuit has two ideal diodes connected as shown in the figure below. The current flowing through the resistance $R_1$ will be

Identify the semiconductor devices whose characteristics are given below, in the order $\left(\mathrm{a}\right),\mathrm{ }\left(\mathrm{b}\right),\mathrm{ }\left(\mathrm{c}\right),\mathrm{ }\left(\mathrm{d}\right)$:

The non-zero potential difference across diode $D_1$ and that across diode $D_2$ are equal in the circuit shown in the figure (both the diodes are identical in characteristics)

The $V-I$ characteristic of a diode is shown in the figure. The ratio of forward to reverse bias resistance is:

In the following circuit diagram, the current through the battery is

$Ge$ and $Si$ diodes start conducting at $0.3 V$ and $0.7 V$ respectively. In the following figure if $Ge$ diode connection are reversed, the value of $V_0$ changes by: (assume that the Ge diode has large breakdown voltage)

In the following circuit, the equivalent resistance between $A$ and $B$ is

Determine the current passing through the diode $D_2$ in the circuit given. Assume these are Silicon diodes.