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The reverse breakdown voltage of a Zener...

The reverse breakdown voltage of a Zener diode is 5.6 V in the given circuit then `I_z` in mA is ___________

Text Solution

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The correct Answer is:
10
Assuming no diode potential drop across `800Omega` resistor = 7.2 V
`implies` Zener breakdown has occurred `implies` Current through `800Ommega` resistor `=(5.6)/(800)A`
`implies` Current through `200Omega` resistor `=(9-5.6)/(200)=3.4/200`
`implies` Current through Zener diode `=(3.4)/(200)-5.6/800=8/800implies10mA`
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Knowledge Check

  • The sharp range of breakdown voltage in Zener diode is

    A
    `0.1` to `10 V`
    B
    `1` to `20 V`
    C
    `0.05` to `0.1 V`
    D
    `20` to `200 V`

  • In the breakdown region, Zener diode behaves as a

    A
    costant current source
    B
    constant voltage source
    C
    constant resistance source
    D
    constant power source

  • Correct graph of voltage across zener diode will be

    A
    B
    C
    D

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