For the forward biased diode characteristics shown in the figure, the dynamic resistance at I_D = 3

Q: Why can't I just use $V/I$ at $3 \text{ mA}$ to find the resistance?

That formula gives you the **static (or DC) resistance**, which is just the ratio of total voltage to total current. For non-linear devices like diodes, we often need the **dynamic (or AC) resistance**, which measures how the device responds to *small changes* in voltage and current. This is why we use the slope of the tangent line ($\Delta V / \Delta I$).

Q: How do I know which two points to pick on the tangent line?

You can technically pick *any* two points on the straight tangent line, and the slope will be the same! However, to avoid calculation errors, it is always best to pick points that lie exactly on the grid intersections, like $(0.65, 1)$ and $(0.75, 5)$.

Q: Why did we multiply the denominator by $10^{-3}$?

The y-axis of the graph is given in milliamperes ($\text{mA}$). To get the resistance in standard Ohms ($\Omega$), all values must be in standard SI units. Therefore, we multiply by $10^{-3}$ to convert milliamperes into amperes.

Q: What would happen to the dynamic resistance if we operated at a higher current, say $7 \text{ mA}$?

As you move higher up the forward-bias curve, the curve becomes steeper. A steeper curve means a larger $\Delta I$ for the same $\Delta V$, which results in a **lower** dynamic resistance.

Q: Is the dynamic resistance constant for a diode?

No! Because the $V-I$ curve is exponential (non-linear), the slope changes at every single point. Therefore, the dynamic resistance is a local property and changes depending on the specific operating voltage or current.

Question

For the forward biased diode characteristics shown in the figure, the dynamic resistance at $I_D = 3 	ext{ mA}$ will be ......... $\Omega$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome to this fascinating problem on semiconductor diodes! We are given the $V-I$ characteristics of a forward-biased diode.

Our primary objective is to find its **dynamic resistance** at a specific operating current, which is $I_D = 3 	ext{ mA}$.

Unlike a simple ohmic resistor, a diode is a non-linear device. This means its resistance is not constant; it changes depending on the voltage applied across it.

### The Master Equation

So, what exactly is dynamic resistance? It is defined as the inverse of the slope of the $V-I$ curve at a specific operating point.

Mathematically, it represents the ratio of a small change in voltage to the corresponding small change in current. We can write this as:

$$R_d = \frac{\Delta V_D}{\Delta I_D}$$

To find this slope, we don't look at the curve itself, but rather at the **tangent line** drawn at our operating point of $3 	ext{ mA}$.

### Extracting Data from the Graph

We need to pick two easy-to-read points on this tangent line to calculate $\Delta V$ and $\Delta I$.

If we look closely at the grid in the provided graph, the tangent line crosses exactly at a current of $1 	ext{ mA}$ when the voltage is $0.65 	ext{ V}$.

Similarly, it crosses a current of $5 	ext{ mA}$ when the voltage is $0.75 	ext{ V}$.

These two points, $(0.65 	ext{ V}, 1 	ext{ mA})$ and $(0.75 	ext{ V}, 5 	ext{ mA})$, are perfect for our calculation because they lie exactly on the grid intersections.

### Substituting the Values

Now, let's substitute these extracted values into our dynamic resistance formula.

The change in voltage, $\Delta V$, is simply the difference between the two voltage readings: $0.75 - 0.65 	ext{ V}$.

The change in current, $\Delta I$, is the difference between the two current readings: $5 - 1 	ext{ mA}$.

**Crucial Step:** We must remember to convert milliamperes to standard SI unit amperes by multiplying by $10^{-3}$.

$$R_d = \frac{0.75 - 0.65}{(5 - 1) 	imes 10^{-3}}$$

### Final Calculation

Let's simplify the numerator and the denominator.

The numerator becomes $0.10 	ext{ V}$, and the denominator becomes $4 	imes 10^{-3} 	ext{ A}$.

$$R_d = \frac{0.10}{4 	imes 10^{-3}}$$

To resolve the fraction, the $10^{-3}$ in the denominator moves to the numerator and becomes $1000$.

Multiplying $0.10$ by $1000$ gives us $100$. Finally, dividing $100$ by $4$ yields exactly $25 \Omega$.

$$R_d = \frac{100}{4} = 25 \Omega$$

### Conclusion

The dynamic resistance of the diode at $3 	ext{ mA}$ is **$25 \Omega$**.

Notice how this resistance would be entirely different if we picked a point lower on the curve where it is flatter. The resistance there would be much higher!

Always remember, for non-linear devices like diodes, resistance is a local property that depends entirely on where you are operating on the curve.

Animated Solution for Physics - Semiconductor Electronics: For the forward biased diode characteristics shown in the figure, the dynamic resistance at $I_{D} = 3 mA$ will be ......... $Ω$ .

Enter Numerical Value:

Visualized Solution

Operating Point at $I_{D} = 3 mA$

Formula for $R_{d}$

Extracting Points $(V_{1}, I_{1})$ and $(V_{2}, I_{2})$

Substituting Values

Simplifying the Expression

Final Calculation

Conclusion

The Sigma Insight: P-N Junction Diode

Analyzing the Setup

The Master Equation

Extracting Data from the Graph

Substituting the Values

Final Calculation

Conclusion

Similar Questions

The circuit contains two diodes each with a forward resistance of $50 Ω$ and with infinite reverse resistance. If the battery voltage is $6 V$ , the current through the $120 Ω$ resistance is ......... $mA$ .

The circuit shown below contains two ideal diodes, each with a forward resistance of $50Ω$ . If the battery voltage is 6 V, the current through the $100Ω$ resistance (in ampere) is

The circuit shown in the figure contains two diodes each with a forward resistance of $50 Ω$ and with infinite backward resistance. If the battery voltage is $6 V$ , the current through the $100 Ω$ resistance (in ampere) is

In connection with the circuit drawn below, the value of current flowing through $2 k Ω$ resistor is ....... $\times 1 0^{- 4} A$ .

The Zener diode has a $V_{z} = 30 V$ . The current passing through the diode for the following circuit is ......... mA.

The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit?

In the following, which one of the diodes is reverse biased?

The reading of the ammeter for a silicon diode in the given circuit is

In a given circuit diagram, a $5 V$ Zener diode along with a series resistance is connected across a $50 V$ power supply. The minimum value of the resistance required, if the maximum Zener current is $90 mA$ will be ......... $Ω$ .

Enter Numerical Value:

Visualized Solution

Operating Point at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ID​=3 mA

Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rd​

Extracting Points .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (V1​,I1​) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (V2​,I2​)

Substituting Values

Simplifying the Expression

Final Calculation

Conclusion

The Sigma Insight: P-N Junction Diode

Analyzing the Setup

The Master Equation

Extracting Data from the Graph

Substituting the Values

Final Calculation

Conclusion

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In connection with the circuit drawn below, the value of current flowing through 2kΩ resistor is ....... ×10−4 A.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Zener diode has a Vz​=30 V. The current passing through the diode for the following circuit is ......... mA.

The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit?

In the following, which one of the diodes is reverse biased?

The reading of the ammeter for a silicon diode in the given circuit is

Operating Point at $I_{D} = 3 mA$

Formula for $R_{d}$

Extracting Points $(V_{1}, I_{1})$ and $(V_{2}, I_{2})$

In connection with the circuit drawn below, the value of current flowing through $2 k Ω$ resistor is ....... $\times 1 0^{- 4} A$ .

The Zener diode has a $V_{z} = 30 V$ . The current passing through the diode for the following circuit is ......... mA.