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I am trying to solve the 2013 paper set by ISRO for electrical engineers. I wanted to verify if my answer to Question no 18 is correct.

q 18

So \$I_L\$ is being drawn out of the circuit shown. Since the zener diode operates in reverse bias, let \$I_Z\$ be the current that flows into the ground through the zener diode. Since the power rating of the diode is \$0.4 \ W = 5\times I_{max}\$, we know that \$I_Z\$ can at max be \$\frac{0.4}{5} = 80 \ mA\$.

By the current law,

$$\frac{10-5}{50} = 100 \ mA = I_Z + I_L $$

So \$I_L\$ can't go below \$ 20 mA\$ or the power rating of the diode will be broken. It can't go above \$ 100 mA\$ or the diode will no longer be in reverse bias.

So the answer is B? Anything wrong here?

From what I know isn't there some minimum current that has to pass through the zener diode for reasonable regulation? Izk? Should we just assume that the value for that is near zero? Because then the upper limit might not hold.

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  • \$\begingroup\$ Looks good to me. Eventually the load will draw enough current and create enough voltage drop that the diode will turn off. So I don’t think there’s a minimum current required by the diode. \$\endgroup\$ – Joe Mac Mar 6 at 5:36
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    \$\begingroup\$ b is correct.... \$\endgroup\$ – Sunnyskyguy EE75 Mar 6 at 5:36
  • \$\begingroup\$ Your calculations are correct. And yes, a minimum current should flow through zener so that it can stabilize its voltage. The series resistor should allow this minimum current under no-load state. You may not need that minimum current here. \$\endgroup\$ – Rohat Kılıç Mar 6 at 5:37
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You're right with the upper limit.
The function of the zener is limit the voltage to 5V. When the voltage across the zener is lower than 5V, the zener doesnt need to fullfill that job anymore. So, ....

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  • b) is correct.... (0.5-0.4W=0.1W min into load ) and Izt = 5mA is the test current for rated Vz

  • when load= 100mA R= 50 Ohms

  • so even with Iz=0 Vout is still 1/2 of 10V so Izt is irrelevant here. but would be relevant if V+ was say 12V.

other info

  • 60 Ohms is the Zzt at the test current
  • Izk is the knee current at 10% of the test current or 0.5mA and Zzk= 1000 ohms or almost 20 x the Zzt
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  • \$\begingroup\$ Where is Izt = 5 mA specified in OP's question? Although Izt = 5mA is commonly used, I've seen zeners that used Izt = 20mA for their rated Vz. \$\endgroup\$ – Huisman Mar 6 at 7:53
  • \$\begingroup\$ @Huisman Sorry I meant to say that Izt doesn't matter before I quoted the std Izt for 500mW Zeners at 5V, but I agree with you, 20mA is used for higher power Zeners. It is like saying (but they don't) the Vzt for white LEDs is 2.8V at 5% Imax and the Zzt = 1/2Prated +25%/-10%. They use some algorithm is all I'm saying. \$\endgroup\$ – Sunnyskyguy EE75 Mar 6 at 8:27

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