LinkedIn Insight Summer 2016 - Q.2 - Grad Plus

Summer 2016 – Q.2

2. (a) Explain with V-I characteristics the working of Zener diode as a voltage regulator. [6M]

i) Zener Diodes are widely used as Shunt Voltage Regulators to regulate voltage across small loads.
ii) Zener Diodes have a sharp reverse breakdown voltage and breakdown voltage will be constant for a wide rang of currents.
iii) Thus we will connect the zener diode parallel to the load such that the applied voltage will reverse bias it. Thus if the reverse bias voltage across the zener diode exceeds the knee voltage, the voltage across the load will be constant.

iv) In the above circuit diagram excess voltage (Vin – Vz ) will drop across Rs thus by limiting the current through Zener. For the proper designing of the regulator we should know,

  • Unregulated Input Voltage Range
  • Required Output Voltage
  • Max Load Current Required

v) The value of resistance Rs should satisfy the following conditions,

  • The value of Rs must be small enough to keep the Zener Diode in reverse breakdown region. The minimum current required for a Zener Diode to keep it in reverse breakdown region
  • The value of Rs must be large enough that the current through the zener diode should not destroy it. That is the maximum power dissipation Pmax should be less than IzVz.

Thus we should find Rs min and Rs max. To find the value of Rs min we should consider the extreme condition that Vin is minimum and load current is maximum.

$latex I_s\;=\;I_{z\;min}\;+\;I_{L\;max}$

$latex I_{z\;min}\;=\;Please\;Refer\;Datasheet$

$latex V_{s}= V_{in min}-V_{z}$

$latex R_{s\;min}\;=\;V_s/I_s$

To find the value of Rs max we should consider the extreme condition that Vin is maximum and load current is minimum (ie, no load connected).

$latex I_s\;=\;I_{Zmin}\;+\;I_{Lmax}\;$

$latex I_{Zmax}\;=\;P_{max}/V_z\;$

$latex V_{s}= V_{in min}-V_{z}$

$latex R_{s\;max}\;=\;V_s/I_s$

(b) Define α and β in case of transistor. Derive the relationship between them.
If α = 0.98, Calculate value of β. [6M]

ALPHA (α): It is a large signal current gain in common base configuration. It is the ratio of collector current (output current) to the emitter current (input current).

$latex \begin{array}{l}\alpha=\frac{Collector\;current}{Emitter\;current}\\\\\alpha=\frac{I_C}{I_E}\end{array}$

It is a current gain in CB amplifier and it indicates that the amount of emitter current reaching to collector. Its value is unity ideally and practically less than unity.

Beta (β): It is a current gain factor in the common emitter configuration. It is the ratio of collector current (output current) to base current (output current).

$latex beta\;\left(\beta\right)=\frac{I_C}{I_B}$
normally Its value is greater than 100.

Relation between α and β in a transistor:

$latex \begin{array}{l}\begin{array}{l}\alpha=\frac{I_C}{I_E}\;\;and\;\;\beta=\frac{I_C}{I_B}\\\\I_E=I_B+I_C\end{array}\\\\\frac{I_C}\alpha=\frac{I_C}\beta+I_C\\\\\therefore\alpha=\frac\beta{\beta+1}\;\;or\;\;\;\beta=\frac\alpha{1-\alpha}\\\end{array}$

Given value of α= 0.98
To calculate: beta(β)
By using relation of α and β
By substituting given value of α= 0.98
beta(β)= 49.

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