Test: Breakdown Diodes - Electrical Engineering (EE) MCQ

Due to zener effect in reverse bias under high electric field strength, electron quantum tunneling occurs. It’s a mechanical effect in which a tunneling current occurs through a barrier. They usually cannot move through that barrier.

The operation of a zener diode is made in reverse bias when breakdown occurs. So, it allows currnt in reverse direction. The most important application of a zener diode is voltage or shunt regulator.

When voltage is increased, the tunnelling at reverse bias increases. The voltage rises temperature. The crystal ions with greater thermal energy tend to vibrate with larger amplitudes.

 If i is the current flowing, then V₀=10i+7
i=(V_I-7)/210. By substituting, if V_I=10V then i=1/70 and V₀=(1/7)+7=7.14V
if V_I =16V then i=3/70 and V₀=(3/7)+7=7.43V.

Here, I_KNEE=10mA, V_Z=5V. I=I_L+I_Z. I= (10-5)/100=50mA
Now, 50=10+ILMAX .
I_LMAX=40mA. R_LMIN=5/40mA=125 Ω.

 Here, I_1MAX=I_ZMIN+I_LMAX.
I_ZMIN =0.5mA, I_1MAX =(V_1MAX-V_Z )/R_S . Putting the values we get , I_1MAX =24.2mA.
So, 24.2-0.5=23.7mA.

With V_I= 10V when maximum, D₁ is forward biased, D₂ is reverse biased. Zener is in breakdown region. V_OMAX=sum of breakdown voltage and diode drop=6.8+0.7=7.5V. V_OMIN=negative of voltage drop=-0.7V. There will be no breakdown voltage here.

Here, V_z =6V, I_ZMIN=5mA.I_S=I_ZMIN+I_LMAX.
80=5+I_LMAX . I_LMAX=75Ma.R_LMIN=V_I/I_LMAX=6/75mA
=80 Ω.

The carriers in transition region are accelerated by electric field to energies. That energies are sufficient to create electron current multiplication. A single carrier that is energized will collide with another by gaining energy. Thus an avalanche multiplication takes place.

The value of reverse breakdown voltage at which zener breakdown occurs is controlled by amount of doping. If the amount of doping is high, the value of voltage at which breakdown occurs will decrease. Better doping gives a sooner breakdown voltage.