Zener Diode as a Shunt Regulator

Zener Diode as a Shunt Regulator

• The simplest shunt voltage regulator circuit uses a zener diode, to regulate the load voltage. Fig. 3.2.1 shows the arrangement of zener diode in a regulator circuit.

• The zener diode has a characteristics that as long as the current through it is between I_Zmin and I_Zmax the voltage across it is constant equal to zener voltage V_Z.

• As zener diode is connected in shunt with the load resistance, the output voltage is equal to the zener voltage.

The value of R can be obtained mathematically as,

R =  [ V_in – V_Z ] / I

I = [ V_in – V_Z ] / R

• While the relation between the various currents is given by,

 I = I_Z + I_L

Regulation with varying input voltage:

• Fig. 3.2.2 shows a zener regulator under varying input voltage condition.

• It can be seen that the output is

 V₀ = V_Z, is constant.

I_L = V₀ / R_L = V_Z / R_L = Constant

• And I = I_Z + I_L

• Now if V_in increases, then the total current I increases. But I_L is constant as V_Z is constant. Hence the current I_Z increases to keep I_L constant.

• But as long as I_Z is between I_Zmin and I_Zmax the V_Z i.e. output voltage V₀ is constant.

• Thus the changes in input voltage get compensated and output is maintained constant.

• Similarly if V_in decreases, then current I decreases. But to keep I_L constant, I_Z decreass. As long as I_Z is between I_Zmax and I_Zmin' the output voltage remains constant.

Regulation with varying load:

• Fig. 3.2.3 shows a zener regulator under varying load condition and constant input voltage.

Fig. 3.2.3 Varying load condition

• The input voltage is constant while the load resistance R_L is variable.

• As V_in is constant and V₀ = V_Z is constant, then for constant R the current I is constant.

I = [ V_in ‒ V_Z  ] / R

constant = I_L + I_Z

• Now if R_L decreases so I_L increases, to keep I constant I_Z decreases. But as long as it is between I_Zmin and I_Zmax output voltage V₀ will be constant.

• Similarly if R_L increases so I_L decreases, to keep I constant I_Z increases. But as long as it is between I_Zmin and I_Zmax' output voltage V₀ will be constant.

• Thus irrespective of changes in the line voltage or changes in the load, the output voltage remains constant.

 

Design of Zener Regulator

• Let us see the design of zener regulator with varying load as well as varying input conditions.

• Let

V_Inmin = Minimum value of input voltage,

V_inmax = Maximum value of input voltage

I_Lmax = Maximum value of load current,

I_Lmin = Minimum value of load current

I_Zmax = Maximum value of zener current,

I_Zmin = Minimum value of zener current

  V₀ = V_Z = Output voltage

• The limiting values series resistance R for a given zener can be obtained as,

• For any value of R between R_max and R_min' the circuit works successfully as a regulator.

• If load current or input voltage is constant and not varying then its both maximum and minimum values must be treated as a constant value in the above equations, to obtain R_max and R_min' The I_Lmin value is generally treated to be zero.

• To satisfy the equations (3.2.1) and (3.2.2), proper zener diode having I_Zmin and I_Zmax values, which can satisfy the required varying load conditions, must be used in the circuit.

• If I_Zmax is not given, but the power dissipation rating of zener diode is given then the maximum zener current can be obtained as,

I_Zmax = P_D / V_Z

 Or

I_Zmax = P_Z / V_Z

Ex. 3.2.1: 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. 3.2.2: 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. 3.2.3: 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. 3.2.4: 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. 3.2.5: 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

 

Review Questions

1. Explain the principle of working of zener diode as a regulator.

2. Discuss the performance of zener diode regulator interms of the source and load effects.