Special Diodes (Design of Zener Regulator): Important Example Solved Problems

Electron Devices

Chapter 3: Special Diodes

 

Important Example Solved Problems

 

Design of Zener Regulator

Ex. 1: For a zener regulator shown in the Fig. 3.2.4, calculate the range of input voltage for which output will remain constant.

 V_Z = 6.1 V, I_Zmin = 2.5 mA, I_Zmax = 25 mA, r_Z = 0Ω

Solution:

 R_L = 1 kΩ, V_Z = 6.1 V

I_L = V_Z / R_L  = 6.1 / 1×10³ = 6.1 mA constant

For V_in(max) I_Z = I_Zmin = 2.5 mA

I = I_Zmin + I_L = 2.5 + 6.1 = 8.6 mA

V_in(min) = V_Z + IR

= 6.1 +8.6×10^‒3×2.2×10³

= 25.02 V

 For V_in(max)  = I_z = I_Zmin = 25 mA

I = I_Zmin + I_L = 25 + 6.1 = 31.1 mA

V_in(max) = V_Z + IR

= 6.1 +31.1×10^‒3×2.2×10³

= 74.52 V

Thus the range of input voltage is 25.02 V to 74.52 V, for which output will be constant.

 

Ex. 2: In a zener regulator, the input D.C. is 10 V ±20%. The output requirements are 5 V, 20 mA. Assuming I_zmin and I_{z max} as 5 mA and 80 mA.

Design the zener regulator.

Solution:

V₀ = 5V

and I_L = 20 mA

R_L = V₀/I_L =  5V/20×10^‒3 = 250 Ω

The maximum and minimum values of current limiting resistance can be obtained with,

I_Lmax = I_Lmin = 20 mA

R_max = ( V_inmin ‒ V₀ ) / ( I_Lmax + I_Zmin )

Now

V_inmin = 10 ‒ (0.2 × 10) = 8 V

V_inmax = 10+ (0.2 × 10) = 12 V

I_Zmax = 80 mA

I_Zmin = 5 mA

R_max = (8‒5) / (20×10^‒3+5x10^‒3) = 120 Ω

R_min = ( V_inmax ‒ V₀ ) / ( I_Lmin + I_Zmax )

= 12‒5 / (20×10^‒3 + 80×10^‒3) = 70 Ω

So series resistance R must be selected between 70 Ω to 120 Ω. So designed circuit is as shown in Fig. 3.2.5.

 

Ex. 3: A 24 V, 600 mW zener diode is used for providing a 24 V stabilized supply to a variable load. If the input voltage is 32 V, calculate

i) The value of series resistance required ii) Diode current when the load is 1200 Q.

Solution:

 V₀ = V_Z = 24 V

 P_Z = 600 mW

V_in = 32 V

I_Zmax = P_Z/V₀ = (600× 10^‒3) / 24 = 25 mA

 R_min = (V_inmax ‒ V₀) / (I_Lmin + I_Zmax )

= (32‒24) / (0 + 25×10^‒3)

= 320 Ω

i) As V_in and I_L are not changing, R = R_min = 320 Ω is sufficient, to work the circuit as regulator

ii) Now

R_L = 1200 Ω

 I_L = V₀/R_L = 24/1200 = 20 mA

while

 I_T = (V_in ‒ V_Z) / R =  (32‒24) / 320 = 25 mA

 I_Z = I_T‒I_L = 25 ‒ 20 = 5 mA

As

I_Z = I_Zmin = 5 mA, circuit will work

 

Ex. 4: A zener diode has a breakdown voltage of 10 V. It is supplied from a voltage source varying between 20 ‒ 40 V in series with a resistance of 820 Ω. Using an ideal zener model obtain the minimum and maximum zener currents.

Solution:

The circuit is shown in Fig. 3.2.7.

I = (V_in ‒ V₀) / R

I = (V_in ‒ 10) / 820

I_max = (V_in(max) ‒ 10) / 820

= (40‒10) / 820

= 36.585 mA

I_Zmin = 0A             …. For ideal zerer diode

Now I = I_Z + I_L

I_max = I_Zmin + I_Lmax

I_L(max) = I_max = 36.585 mA

when

I_L = I_L(min) = 0 A

          ……… Open load terminals.

Under this condition if V_in = V_in(max) then,

 I = 36.585 mA

and

 I_Z = I_Zmax = 36.585 mA

 

Ex. 5: Calculate the output ripple voltage in a zener regulator if the limiting resistance R_{S =} 100 Ω, zener dynamic resistance r_Z = 10 Ω, unregulated input voltage is 12 volts with 2 volts ripple, V_Z = 5.6 V and load current is 10 mA.

Solution:

The circuit is shown in Fig. 3.2.8.

R_S = 100 Ω, r_Z = 10 Ω

V_in = 12 V, ΔV_in = 2 V

V_Z = 5.6 V, I_L = 10 mA

Now

R_L = V_Z/ I_L = 5.6 / 10×10^‒3 = 560 Ω

When input ripple voltage is considered, the corresponding output ripple will be because of r_Z. Hence consider equivalent circuit with with ripple AV_in = 2V is applied at input, as shown in Fig. 3.2.8 (a).

As input is changing and change in output is to be calculated which is due to r_z, V_Z is not considered which is constant.

ΔI_T = ΔVin / [ R_S + (r_Z||R_L) ]

= 2 / [100+(10||560)]

= 18.2108 mA

ΔI_Z = ΔI_T × R_L/(r_Z+R_L) = 18.2108 × ( 560 / 10+560 )

= 17.89 mA

           ...Current distribution

ΔV₀ =  Output ripple voltage = Δ I_Z × r_Z

= 17.89×10^‒3 ×10 = 178.913 mV