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For the Zener diode shown in the network below, determine, V L , V RIz and V z . Determine the range of I L and R L that will result in V RL being maintained at 10V. Determine also the voltage rating of the diode.

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2 parts
b(i)

8.73 V8.73 \, \text{V}

7.27 V7.27 \, \text{V}

0 A0 \, \text{A}

b(ii)

0 mA≤IL≤24 mA0 \, \text{mA} \le I_L \le 24 \, \text{mA}

500 Ω≤RL≤∞500 \, \Omega \le R_L \le \infty

12 V12 \, \text{V}

Step-by-step solution

7 steps

Here are the solutions for Q7:

Part a(i)

A clamper circuit

**State the primary function of the following

A clamper circuit adds a DC offset to an AC signal, shifting the entire waveform either above or below a specific DC level without altering the shape or peak-to-peak amplitude of the input signal.

Part a(ii)

A clipping circuit

A clipping circuit limits or "clips" a portion of an input signal that exceeds a certain voltage level, effectively removing peaks or troughs of the waveform to prevent the output voltage from exceeding a predetermined value.

Part b(i)

8.73 V8.73 \, \text{V}

For the Zener diode shown in the network below, determine VLV_L, VRIV_{R_I}, IZI_Z and VZV_Z.

Given circuit parameters:
Input voltage Vin=16 VV_{in} = 16 \, \text{V}
Series resistor RS=1 kΩ=1000 ΩR_S = 1 \, \text{k}\Omega = 1000 \, \Omega
Load resistor RL=1.2 kΩ=1200 ΩR_L = 1.2 \, \text{k}\Omega = 1200 \, \Omega
Zener voltage VZ=10 VV_Z = 10 \, \text{V}

  1. Determine if the Zener diode is in breakdown (ON) by calculating the voltage across the load if the Zener were not present (open circuit).

    Vopen=Vin×RLRS+RLV_{open} = V_{in} \times \frac{R_L}{R_S + R_L} Vopen=16 V×1200 Ω1000 Ω+1200 ΩV_{open} = 16 \, \text{V} \times \frac{1200 \, \Omega}{1000 \, \Omega + 1200 \, \Omega} Vopen=16 V×12002200V_{open} = 16 \, \text{V} \times \frac{1200}{2200} Vopen=16 V×1222=16 V×611V_{open} = 16 \, \text{V} \times \frac{12}{22} = 16 \, \text{V} \times \frac{6}{11} Vopen≈8.727 VV_{open} \approx 8.727 \, \text{V}

  2. Compare VopenV_{open} with the Zener voltage VZV_Z.

    Since Vopen(8.727 V)<VZ(10 V)V_{open} (8.727 \, \text{V}) < V_Z (10 \, \text{V}), the Zener diode is not in breakdown; it acts as an open circuit (OFF state).

  3. Calculate the required values assuming the Zener is OFF.

    If the Zener is OFF, no current flows through it, so IZ=0I_Z = 0. The circuit becomes a simple series circuit with RSR_S and RLR_L.
    The load voltage VLV_L is the voltage across RLR_L.

    VL=Vopen=8.73 VV_L = V_{open} = \boxed{8.73 \, \text{V}}

    The voltage across the Zener diode VZV_Z is the same as the load voltage VLV_L since they are in parallel.

    VZ=VL=8.73 VV_Z = V_L = \boxed{8.73 \, \text{V}}

    The total current flowing through the series circuit is ISI_S.

    IS=VinRS+RLI_S = \frac{V_{in}}{R_S + R_L} IS=16 V1000 Ω+1200 Ω=16 V2200 ΩI_S = \frac{16 \, \text{V}}{1000 \, \Omega + 1200 \, \Omega} = \frac{16 \, \text{V}}{2200 \, \Omega} IS≈0.007273 A=7.27 mAI_S \approx 0.007273 \, \text{A} = 7.27 \, \text{mA}

    The voltage across the series resistor RIR_I (which is RSR_S) is VRIV_{R_I}.

    VRI=IS×RSV_{R_I} = I_S \times R_S VRI=0.007273 A×1000 ΩV_{R_I} = 0.007273 \, \text{A} \times 1000 \, \Omega VRI≈7.27 VV_{R_I} \approx \boxed{7.27 \, \text{V}}

    The Zener current IZI_Z is zero because the diode is OFF.

    IZ=0 AI_Z = \boxed{0 \, \text{A}}

Part b(ii)

0 mA≤IL≤24 mA0 \, \text{mA} \le I_L \le 24 \, \text{mA}

Determine the range of ILI_L and RLR_L that will result in VRLV_{RL} being maintained at 10V. Determine also the voltage rating of the diode.

Given circuit parameters:
Input voltage Vin=60 VV_{in} = 60 \, \text{V}
Series resistor RS=2 kΩ=2000 ΩR_S = 2 \, \text{k}\Omega = 2000 \, \Omega
Zener diode: VZ=12 VV_Z = 12 \, \text{V}, IZM=32 mAI_{ZM} = 32 \, \text{mA}

Assumption: The question states "maintained at 10V" for VRLV_{RL}, but the Zener diode in the diagram is explicitly labeled VZ=12 VV_Z = 12 \, \text{V}. For the Zener diode to regulate, it must maintain the voltage at its Zener voltage. Therefore, we will assume the question intends for VRLV_{RL} to be maintained at VZ=12 VV_Z = 12 \, \text{V}.

  1. Calculate the total current flowing through the series resistor RSR_S when the Zener is regulating.

    When the Zener is regulating, VL=VZ=12 VV_L = V_Z = 12 \, \text{V}.

    IS=Vin−VLRSI_S = \frac{V_{in} - V_L}{R_S} IS=60 V−12 V2000 Ω=48 V2000 ΩI_S = \frac{60 \, \text{V} - 12 \, \text{V}}{2000 \, \Omega} = \frac{48 \, \text{V}}{2000 \, \Omega} IS=0.024 A=24 mAI_S = 0.024 \, \text{A} = 24 \, \text{mA}

  2. Determine the range of load current (ILI_L).

    The total current ISI_S splits into Zener current (IZI_Z) and load current (ILI_L): IS=IZ+ILI_S = I_Z + I_L.
    For the Zener to regulate, IZI_Z must be between its minimum operating current (IZKI_{ZK}, typically assumed to be 0 mA0 \, \text{mA} for calculations unless specified) and its maximum current (IZMI_{ZM}).
    So, IZK≤IZ≤IZMI_{ZK} \le I_Z \le I_{ZM}.
    Assuming IZK≈0 mAI_{ZK} \approx 0 \, \text{mA}:
    0 mA≤IZ≤32 mA0 \, \text{mA} \le I_Z \le 32 \, \text{mA}.

    From IL=IS−IZI_L = I_S - I_Z:

    • Maximum ILI_L occurs when IZI_Z is minimum (IZ=0 mAI_Z = 0 \, \text{mA}): ILmax=IS−0 mA=24 mA−0 mA=24 mAI_{Lmax} = I_S - 0 \, \text{mA} = 24 \, \text{mA} - 0 \, \text{mA} = 24 \, \text{mA}
    • Minimum ILI_L occurs when IZI_Z is maximum (IZ=IZM=32 mAI_Z = I_{ZM} = 32 \, \text{mA}): ILmin=IS−IZM=24 mA−32 mA=−8 mAI_{Lmin} = I_S - I_{ZM} = 24 \, \text{mA} - 32 \, \text{mA} = -8 \, \text{mA}

    Since load current cannot be negative, the practical minimum load current is 0 mA0 \, \text{mA} (open circuit). The Zener will regulate as long as ILI_L does not cause IZI_Z to exceed IZMI_{ZM} or fall below IZKI_{ZK}. In this case, ISI_S is less than IZMI_{ZM}, so IZI_Z will never reach IZMI_{ZM} if IL≥0I_L \ge 0. The Zener will regulate as long as IL≤ISI_L \le I_S. Therefore, the range of ILI_L for regulation is:

    0 mA≤IL≤24 mA\boxed{0 \, \text{mA} \le I_L \le 24 \, \text{mA}}

  3. Determine the range of load resistance (RLR_L).

    Since VL=12 VV_L = 12 \, \text{V} is maintained, RL=VLILR_L = \frac{V_L}{I_L}.

    • When IL=0 mAI_L = 0 \, \text{mA}: RLmax=12 V0 A=∞R_{Lmax} = \frac{12 \, \text{V}}{0 \, \text{A}} = \infty
    • When IL=24 mAI_L = 24 \, \text{mA}: RLmin=12 V0.024 A=500 ΩR_{Lmin} = \frac{12 \, \text{V}}{0.024 \, \text{A}} = 500 \, \Omega

    Therefore, the range of RLR_L for regulation is:

    500 Ω≤RL≤∞\boxed{500 \, \Omega \le R_L \le \infty}

  4. Determine the voltage rating of the diode.

    The voltage rating of the diode is its Zener voltage, which is given in the diagram.
    Voltage rating of the diode = 12 V\boxed{12 \, \text{V}}

Part c(i)

Avalanche breakdown

**Briefly explain the following

Avalanche breakdown occurs in a reverse-biased p-n junction when a high electric field accelerates minority carriers, causing them to collide with atoms and generate new electron-hole pairs. These newly generated carriers, in turn, cause further collisions, leading to a cumulative multiplication of carriers and a rapid increase in reverse current. This process is dominant in Zener diodes with breakdown voltages above approximately 6V.

Part c(ii)

Zener breakdown

Zener breakdown occurs in a heavily doped p-n junction under reverse bias. The heavy doping creates a very narrow depletion region, allowing electrons to tunnel directly from the valence band to the conduction band when the reverse electric field is sufficiently strong. This quantum mechanical tunneling effect results in a sharp increase in reverse current at a specific voltage, typically below approximately 6V.

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