In a given circuit diagram, a 5 V Zener diode along with a series resistance is connected across a 5

Q: Why do we assume the load current is zero?

To find the *minimum* safe resistance, we must consider the *maximum* possible current that could flow through the Zener diode. If a load is connected, it draws some current, reducing the current through the Zener. The most dangerous situation (maximum Zener current) occurs when the load is completely disconnected ($I_L = 0$).

Q: What happens if we use a resistance smaller than $500\ \Omega$?

A smaller resistance would allow a total current greater than $90\text{ mA}$ to flow from the source. If the load is disconnected, all of this current would be forced through the Zener diode, exceeding its $90\text{ mA}$ safety rating and causing it to burn out.

Q: Why is the voltage across the series resistor exactly $45\text{ V}$?

The Zener diode, when operating in its breakdown region, maintains a constant voltage of $5\text{ V}$ across itself and anything in parallel with it. Since the total supply is $50\text{ V}$, Kirchhoff's Voltage Law dictates that the remaining $45\text{ V}$ must be dropped across the series resistor.

Q: Does the value of the load resistance $R_L$ matter for this calculation?

Not for finding the *minimum* series resistance. The minimum series resistance is dictated entirely by the Zener diode's maximum current rating and the worst-case scenario (no load). The load resistance would only matter if we were calculating the *maximum* allowable series resistance to maintain regulation.

Q: Can the Zener diode regulate voltage if the supply drops below $5\text{ V}$?

No. A Zener diode requires the supply voltage to be greater than its Zener breakdown voltage ($V_z$) to operate as a regulator. If the supply is less than $5\text{ V}$, the diode will not enter breakdown, and it will act like an open circuit.

Question

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

Sigma Percentile · Accepted Answer

### The Role of the Bodyguard Resistor

Imagine you are designing a power supply for a sensitive electronic device. You have a raw $50	ext{ V}$ source, but your device needs exactly $5	ext{ V}$. Enter the Zener diode!

The Zener diode is a fantastic voltage regulator, but it has a vulnerability: it can only handle so much current before it burns out. In our circuit, the maximum safe current for the Zener diode is $90	ext{ mA}$.

To protect it, we place a series resistor $R$ between the high-voltage source and the Zener diode. Think of this resistor as a bodyguard. It takes the hit—the excess voltage—so the Zener diode and the load don't fry.

### Analyzing the Worst-Case Scenario

According to Kirchhoff's Current Law, the total current $I$ flowing from the source splits at the junction. Part of it goes through the Zener diode ($I_1$), and the rest goes through the load ($I_2$). So, $I = I_1 + I_2$.

To find the **minimum** required value for the series resistance $R$, we must consider the **maximum** possible current that could flow through it. When is the Zener diode in the most danger?

The worst-case scenario occurs when the load is completely disconnected. In this state, the load current $I_2$ is zero. Consequently, all the current $I$ rushes straight through the Zener diode. Therefore, to keep the Zener safe, the maximum total current $I_{\max}$ must not exceed the Zener's maximum rating of $90	ext{ mA}$.

### The Master Equation

Now, let's apply Ohm's law to our bodyguard resistor. The voltage drop across this series resistor is simply the supply voltage minus the regulated Zener voltage.

$$V_R = V - V_z$$

Substituting our known values, the voltage drop is $50	ext{ V} - 5	ext{ V} = 45	ext{ V}$. This $45	ext{ V}$ must be dissipated by the series resistor.

Using Ohm's law, the minimum resistance $R_{\min}$ is this voltage drop divided by the maximum allowable current:

$$R_{\min} = \frac{45	ext{ V}}{90 	imes 10^{-3}	ext{ A}}$$

### Final Calculation

Let's crunch the numbers. Dividing $45$ by $90$ gives us exactly $0.5$.

$$R_{\min} = 0.5 	imes 10^3\ \Omega$$

Multiplying by one thousand, we arrive at our final answer:

$$R_{\min} = 500\ \Omega$$

This $500\ \Omega$ resistor perfectly balances the system. If we were to use a smaller resistance, say $400\ \Omega$, the total current would exceed $90	ext{ mA}$. If the load were ever disconnected, that massive current would flow entirely through the Zener diode, destroying it instantly. Always design for the worst-case scenario!

Animated Solution for Physics - Semiconductors: 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

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$O h m^{'} s L a w f or S er i es R es i s t or$

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The Sigma Insight: P-N Junction Diode

The Role of the Bodyguard Resistor

Analyzing the Worst-Case Scenario

The Master Equation

Final Calculation

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