Question

The figure shows a circuit containing two diodes \(D_1\) and \(D_2\) with threshold voltages \(V_{TH}\) of 0.7 V and 0.3 V, respectively. Considering the simplified diode model, which assumes diode I–V characteristics as shown in the plot on the right, the current through the resistor \(R\) is ......... µA. 

Hint:

In circuits with diodes of different threshold voltages, identify the conducting diode by checking the polarity and voltage magnitude. Use simplified diode model \(V = V_{TH}\) when conducting.

Answer 1

Correct Answer: 97

Step 1: Circuit Analysis

The circuit consists of a $\mathbf{10\ V}$ source connected to two diodes, $\mathbf{D_1}$ and $\mathbf{D_2}$, in parallel, which are then connected in series with a resistor $\mathbf{R = 100\ k\Omega}$ to ground.

Diode $D_1$ threshold voltage: $V_{TH1} = 0.7\ V$

Diode $D_2$ threshold voltage: $V_{TH2} = 0.3\ V$

Resistor: $R = 100\ k\Omega = 100 \times 10^3\ \Omega$

The simplified diode model states that a diode conducts current only when the forward voltage across it is equal to or greater than its threshold voltage $V_{TH}$, and when conducting, the voltage across it remains at $V_{TH}$.

Step 2: Determining the Conducting Diode(s)

For a diode to conduct, it must be forward-biased (which they are) and the $\mathbf{10\ V}$ source voltage must be greater than the diode's threshold voltage $V_{TH}$. Since $10\ V > V_{TH1} = 0.7\ V$ and $10\ V > V_{TH2} = 0.3\ V$, both diodes could potentially conduct.

However, since the two diodes are in parallel, the voltage across the combination will be the lower of the two threshold voltages.

If $D_1$ attempts to conduct, the voltage across the parallel branch is $V_{D1} = 0.7\ V$.

If $D_2$ attempts to conduct, the voltage across the parallel branch is $V_{D2} = 0.3\ V$.

Since the diodes are in parallel, the branch with the lower threshold voltage, $D_2$ ($0.3\ V$), will conduct first. Once $D_2$ conducts, the voltage across the parallel combination is fixed at $V_{D2} = 0.3\ V$.

At this fixed voltage of $0.3\ V$:

Diode $D_2$ is conducting, as the forward voltage ($0.3\ V$) is equal to its threshold ($0.3\ V$).

Diode $D_1$ remains OFF, as the forward voltage across it ($0.3\ V$) is less than its threshold ($0.7\ V$).

Therefore, only diode $\mathbf{D_2}$ is conducting and the equivalent voltage of the parallel combination is $V_{D} = V_{TH2} = 0.3\ V$.

Step 3: Calculating the Current through $R$

The circuit simplifies to a $10\ V$ source connected in series with the conducting diode $D_2$ (modeled as a $0.3\ V$ voltage source) and the resistor $R$.

By applying Kirchhoff's Voltage Law (KVL) to the loop:

$$V_{source} - V_{D} - V_{R} = 0$$

$$10\ V - 0.3\ V - I_R R = 0$$

Where $I_R$ is the current through the resistor $R$. The current $I_R$ is the total current flowing from the source.

The voltage across the resistor $V_R$ is:

$$V_R = V_{source} - V_{D}$$

$$V_R = 10\ V - 0.3\ V$$

$$V_R = 9.7\ V$$

The current $I_R$ is calculated using Ohm's Law:

$$I_R = \frac{V_R}{R}$$

$$I_R = \frac{9.7\ V}{100\ k\Omega}$$

$$I_R = \frac{9.7}{100 \times 10^3}\ A$$

$$I_R = 0.097 \times 10^{-3}\ A$$

$$I_R = 97 \times 10^{-6}\ A$$

Final Answer

The current $I_R$ in microamperes ($\mu A$) is:

$$I_R = 97\ \mu A$$