Zener Diodes: Ripple Effect & Load Line Analysis

Second Approximation of a Zener Diode

The second approximation includes a zener resistance \((R_Z)\) in series with an ideal battery.
Total voltage across the zener diode:
- \(V_Z\): Breakdown voltage.
- \(V_{R_Z}\): Voltage drop across zener resistance.

\[V_{total} = V_Z + V_{R_Z}\]

Effect on Load Voltage:
- Ideal Case: Load voltage \(V_L\) is equal to the breakdown voltage \(V_Z\).
- Second Approximation: Includes the zener resistance.
  - \(I_Z\): Zener current flowing through zener resistance.

\[V_L = V_Z + I_Z R_Z\]

Change in Load Voltage:
- The change in load voltage from the ideal case:
- Usually, \(R_Z\) is small, resulting in a minor voltage change (in tenths of a volt).

\[\Delta V_L = I_Z R_Z\]

Example Calculation:
- If \(I_Z = 10 \, ~\mathrm{mA}\) and \(R_Z = 10 \, \Omega\):

\[\Delta V_L = 0.1 \, V\]

Effect on Ripple in Zener Regulators

Ripple : AC voltage variations superimposed DC output voltage.
Equivalent circuit to analyze ripple
Three resistances affect the ripple:
- \(R_Z\) (Zener resistance), \(R_S\) (Series resistance), \(R_L\) (Load resistance)
Simplification: Typically, \(R_Z << R_L\).

Voltage Divider Equation:
- The output ripple voltage can be calculated as:

\[V_{R(\text{out})} = \frac{R_Z}{R_S + R_Z} V_{R(\text{in})}\]

Approximation for Troubleshooting:
- In most designs, \(R_S\) is significantly greater than \(R_Z\):

\[V_{R(\text{out})} \approx \frac{R_Z}{R_S} V_{R(\text{in})}\]

Problem-7

The zener diode has a breakdown voltage of 10 V and a zener resistance of 8.5 \(\Omega\). Apply the second approximation to determine the load voltage when the zener current is 20 mA.

\[\begin{aligned} \text{change in load voltage}~\Delta V_{L} & =I_{Z}R_{Z}=(20 \mathrm{mA})(8.5 \Omega)=0.17 \mathrm{V} \\ \text{load voltage (second approx.)}~V_{L} &=10 \mathrm{V}+0.17 \mathrm{V}=10.17 \mathrm{V} \end{aligned}\]

\[V_{R(\mathrm{out})}\approx\frac{8.5 \Omega}{270 \Omega} \times 2 \mathrm{V}=63 \mathrm{mV}\]
What is the approximate ripple voltage across the load, if

Zener Drop-Out Point and Worst-Case Conditions

Zener Regulation Condition:
- For proper regulation, the Zener diode must remain in the breakdown region.
- There must be sufficient Zener current for all source voltages and load currents.

Worst-Case Scenarios:
- Decreasing Source Voltage:
  - Example: For \(V_S = 20\ V\) and \(R_S = 200\ \Omega\):
  - As \(V_S\) decreases (e.g., from 20V to 12V), \(I_S\) reduces to 10 mA and \(I_Z = 0\) mA.
  - If \(V_S\) drops further, regulation is lost.
- Decreasing Load Resistance:
  - Example: \(R_L\) decreases from \(1\ \mathrm{k}\Omega\) to \(200\ \Omega\).
  - Load current \(I_L\) increases to 50 mA, causing \(I_Z\) to drop to zero.
  - Regulation fails when load resistance is too low.
- Increasing Series Resistance:
  - If \(R_S\) increases (e.g., from \(200\ \Omega\) to \(1\ \mathrm{k}\Omega\)), \(I_S\) reduces, and regulation is lost.

\[I_S = \frac{V_S - V_Z}{R_S} = 50\ \mathrm{mA}, \quad I_L = 10\ \mathrm{mA}, \quad I_Z = 40\ \mathrm{mA}\]

Formula for Maximum Series Resistance:
- To avoid failure under worst-case conditions:
- Alternatively:

\[R_{S(\max)} = \frac{V_{S(\min)} - V_Z}{I_{L(\max)}}\]

\[R_{S(\max)} = \left( \frac{V_{S(\min)}}{V_Z} - 1 \right) R_{L(\min)}\]

Problem-8

\[\boxed{ R_{S(\max)} = \left( \frac{V_{S(\min)}}{V_Z} - 1 \right) R_{L(\min)}} \qquad \boxed{R_{S(\max)} = \frac{V_{S(\min)} - V_Z}{I_{L(\max)}}}\]

A zener regulator has an input voltage that can range from 22 to 30 V. Given that the regulated output voltage is 12 V and the load resistance varies from 140 \(\Omega\) to 10 k\(\Omega\), what is the maximum permissible series resistance?

\[R_{\mathrm{S(max)}}=\left(\frac{22 \mathrm{V}}{12 \mathrm{V}}-1\right)140 \Omega=117~ \Omega\]
A zener regulator has an input voltage ranging from 15 to 20 V and a load current varying from 5 to 20 mA. If the zener voltage is 6.8 V, what is the maximum allowable series resistance?

\[R_S(\max)=\frac{15 \mathrm{V}-6.8 \mathrm{V}}{20 \mathrm{mA}}=410~ \Omega\]

Zener Diode - Maximum Power and Current

Maximum Power Dissipation:
- Power dissipated by the Zener diode:
- Ensure \(P_Z\) does not exceed the Zener diode’s power rating (0.25 W to 50 W).

\[P_Z = V_Z \cdot I_Z\]

Maximum Current (\(I_{ZM}\)):
- Maximum current without exceeding power rating:

\[I_{ZM} = \dfrac{P_{ZM}}{V_Z}\]

Zener Diode - Load Line Analysis

\[I_Z = \frac{V_S - V_Z}{R_S}\]
Zener current:
\[I_Z = \frac{20 - V_Z}{1000}\]
: and Example: For
Saturation Point: \(V_Z = 0\), \(I_Z = 20\ \text{mA}\)
Cutoff Point: \(I_Z = 0\), \(V_Z = 20\ \text{V}\)

For \(V_S = 20\ \text{V}\), \(R_S = 1\ \text{k}\Omega\), and \(V_Z = 12\ \text{V}\):
- Intersection point at Q1 with approximately 12 V.
- Changing \(V_S\) to 30 V:
- Intersection point shifts to Q2, but \(V_Z\) remains approximately 12 V.

\[I_Z = \frac{30 - V_Z}{1000} = 18\ \text{mA}\]

Conclusion: Zener diodes maintain nearly constant voltage despite changes in source voltage.