At 0.3 V and 0.7 V, the diodes Ge and Si become conductor respectively. In given figure, if ends of

Q: Why doesn't current flow through both diodes initially?

Because the Germanium diode has a lower knee voltage ($0.3\text{ V}$), it turns on first and clamps the voltage across the parallel combination at $0.3\text{ V}$. This voltage is insufficient to turn on the Silicon diode, which requires $0.7\text{ V}$.

Q: What does 'overturned' mean in this context?

It means the diode's physical connections are reversed in the circuit, changing it from a forward-biased state to a reverse-biased state.

Q: Why do we subtract the diode voltage from the battery voltage?

According to Kirchhoff's Voltage Law, the sum of voltage drops in a closed loop equals the supplied voltage. The battery supplies the voltage, and the diode and resistor share it.

Q: What if the battery voltage was less than 0.3 V?

Neither diode would receive enough voltage to overcome their potential barriers. They would both act as open circuits, and the output voltage $V_0$ would be zero.

Q: Does the 5 k$\Omega$ resistor affect which diode turns on?

No, the diodes are in parallel with each other, so the one with the lower knee voltage will always dictate the turn-on behavior regardless of the series resistance.

Question

At $0.3	ext{ V}$ and $0.7	ext{ V}$, the diodes Ge and Si become conductor respectively. In given figure, if ends of diode Ge overturned, the change in potential $V_0$ will be

Options:

(A) $0.2	ext{ V}$

(B) $0.6	ext{ V}$

(C) $0.4	ext{ V}$

(D) $0.8	ext{ V}$

Sigma Percentile · Accepted Answer

### The Circuit Setup
Imagine you are looking at a water pipe system where the battery is a powerful pump pushing water forward. In our circuit, we have a $12	ext{ V}$ battery connected to two parallel branches. One branch contains a Germanium (Ge) diode, and the other contains a Silicon (Si) diode. These parallel branches then merge and flow through a $5	ext{ k}\Omega$ resistor. Our objective is to determine how the output voltage $V_0$ across the resistor changes when we physically flip the Germanium diode around.

### The Race to Conduct
When we turn on the battery, the positive terminal applies a high potential to the p-sides of both diodes. This means both diodes are initially forward-biased and eager to conduct. However, diodes are not perfect conductors; they require a minimum "toll" or knee voltage to open their gates.

For Germanium, this toll is $0.3	ext{ V}$, while for Silicon, it is $0.7	ext{ V}$. Because Germanium requires less voltage to turn on, it wins the race! As soon as the voltage across the parallel section reaches $0.3	ext{ V}$, the Germanium diode begins to conduct heavily. By doing so, it clamps the voltage across the entire parallel combination at exactly $0.3	ext{ V}$. Since $0.3	ext{ V}$ is less than the $0.7	ext{ V}$ required by the Silicon diode, the Silicon diode remains firmly shut. All the current flows exclusively through the Germanium branch.

### Calculating the Initial State
Now let's put some numbers to it. According to Kirchhoff's Voltage Law, the total voltage supplied by the battery must equal the sum of the voltage drops across the components in the loop.

The battery supplies $12	ext{ V}$. The Germanium diode consumes $0.3	ext{ V}$. The remaining voltage must appear across the $5	ext{ k}\Omega$ resistor. Therefore, the initial output voltage $V_{0i}$ is:

$$V_{0i} = 12 - 0.3 = 11.7	ext{ V}$$

### The Plot Twist: Reversing the Diode
The problem throws a curveball: we are asked to "overturn" the ends of the Germanium diode. By flipping it, the n-side is now connected to the positive terminal of the battery. This places the Germanium diode in a state of reverse bias.

In reverse bias, an ideal diode acts like an open switch. It completely blocks the flow of current through its branch. With the Germanium path now blocked, the current has no choice but to seek an alternative route.

### The Final Calculation
The voltage across the parallel section will now rise until it hits $0.7	ext{ V}$, which is the exact amount needed to turn on the Silicon diode. Once the Silicon diode turns on, it clamps the voltage at $0.7	ext{ V}$, and all the current now flows through the Silicon branch.

Let's recalculate the output voltage for this new state. The battery still supplies $12	ext{ V}$, but now the Silicon diode consumes $0.7	ext{ V}$. The final output voltage $V_{0f}$ across the resistor becomes:

$$V_{0f} = 12 - 0.7 = 11.3	ext{ V}$$

Finally, to find the change in the potential $\Delta V_0$, we simply subtract the final voltage from the initial voltage:

$$\Delta V_0 = 11.7 - 11.3 = 0.4	ext{ V}$$

And there we have it! The change in potential is exactly **$0.4	ext{ V}$**.

Animated Solution for Physics - Semiconductors: At $0.3 V$ and $0.7 V$ , the diodes Ge and Si become conductor respectively. In given figure, if ends of diode Ge overturned, the change in potential $V_{0}$ will be

Select Answer:

Visualized Solution

Circuit Analysis

Forward Bias Characteristics

Current Path (Initial)

Initial Output Voltage

Reversing the Ge Diode

Current Path (Final)

Final Output Voltage

Change in Potential

The Way Forward

The Sigma Insight: P-N Junction Diode

The Circuit Setup Imagine you are looking at a water pipe system where the battery is a powerful pump pushing water forward

The Race to Conduct When we turn on the battery, the positive terminal applies a high potential to the p-sides of both diodes

Calculating the Initial State Now let's put some numbers to it

The Plot Twist

The Final Calculation The voltage across the parallel section will now rise until it hits $0.7 V$ , which is the exact amount needed to turn on the Silicon diode

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