Zener Diode as Shunt Regulator

Zener Diode Symbol and Equilibrium Circuit

VZ voltage is set across the diode in reverse condition if diode works in breakdown region i.e. voltage crosses the level VZ across the diode.

Zener Diode Rating (Zener Diode Specification)

Zener Voltage (VZ): It the voltage across Zener diode from cathode to anode when it is biased to operate in Zener breakdown region. This is the value of reverse voltage across diode at which breakdown occurs. This voltage depends on the material used and doping level in the diode. Specified values of VZ are 3.1V, 4.7V, 5.1V, 6.8V, 9.1V, 12.1V, 15.1V, 18.1V, etc.

Power dissipation (PZ): The Zener is normally operated in the reverse biased manner. So it needed to calculate power dissipation which is given by     PZM = VZ × IZ    

where VZ is Zener voltage and IZ is Zener current in present state.

Maximum Power dissipation (PZM): The Zener is normally operated in the reverse biased manner. So it needed to calculate power dissipation which is given by         PZM = VZ × IZM   

where VZ is Zener voltage and IZM is maximum Zener current which can be flow in the diode without damaging it because of thermal changes in the diode.

Maximum Zener Current (IZM or IZ,max): IZM is maximum Zener current which can be flow in the diode without damaging it because of thermal changes in the diode. This value is different for different diodes and depends upon the material used and doping level in the diode.

Minimum Zener Current (IZ,min): IZ,min is minimum Zener current which is required to flow in the diode to keep it in breakdown region. It is also defined as the minimum current  required by the diode to bring it in breakdown region. It is also known as Zener knee current Izk or break-over current Ibo. Normally this value is 2mA.

Dynamic Zener Resistance (rZ): It is defined as the ratio of change in voltage across diode to change in Zener current.  Ideally voltage across diode should be constant or change in voltage across diode should be 0V. So ideally rZ should be zero and practically very low (few Ω).

Mathematically        rZ = dVZ/dIZ

Zener Diode as Voltage Shunt Regulator    

We know that a voltage equal to the breakdown voltage is set across the junction when breakdown occurs in the diode. So if larger than VZ voltage is applied across the diode, we get constant voltage across the diode. So if we use Zener diode in shunt to the load, then due to parallel combination, voltage across load becomes regulated (constant) which is equal to the set voltage across the diode i.e. VZ. This the basic concept of designing a shunt regulator using Zener diode. In figure, RL is load resistance and Zener is connected in shunt  to load.

Vi is the input voltage applied to the regulator circuit and VL is the voltage applied across the diode in reverse condition.

For limiting the current in the circuit, we require a series resistance RS. The value of RS depends on the requirement of output current to the load.

Calculation of Parameters: Vi and VZ of diode is normally known. Any one of IZM or PZM will also be given. IL is the load current where IL = VL/RL , where RL is load resistance. We can see that RL and IL are reciprocal to each. It means IL increases on decreasing RL and vice versa. VL depends on the situation of calculation, discussed later. Voltage drop or voltage across the series resistance RS , VR = Vi – VL ,

VR can also be calculated as  VR = IRRS or IR = VR/RS , where IR is current in RS.

Zener current is calculated as follows:

Applying KCL at cathode node of diode

IR = IZ + IL        or        IZ = IR – IL

If any one of IZM or PZM is given, then other can be calculated as

PZM = VZIZM      or        IZM = PZM/VZ

Case 1: RL and Vi both are fixed:

If load as well as input supply is fixed, then our aim will be to check whether diode is ON or not means diode is working in breakdown region or not.

We know that VL is the voltage across the diode. So if we calculate VL in the absence of diode (by removing the diode), we can find its value and compare it with the required voltage to make diode ON that is VZ.

When diode is removed, VL can be calculated by potential division rule

So    VL = ViRL/(RL + RS)

If VL < VZ , diode is OFF hence open

If VL ≥ VZ , diode is ON hence replace with a voltage level VZ

If VL < VZ , diode is OFF hence open (the diode is then removed permanently)

So, diode current      IZ = 0 mA

From IR = IZ + IL ,        IR = IL      and can be calculated as      

                                     IR = IL = VL/RL

Power dissipation in diode

PZ = VZ IZ = VZ × 0 = 0 mWatt

Voltage drop across series resistance

VR = IRRS         or         VR = Vi – VL

If any one of IZM or PZM is given,

then other can be calculated as

PZM = VZIZM      or        IZM = PZM/VZ

If VL ≥ VZ , diode is ON hence connected in the circuit and replaced with a voltage level VZ which is in shunt with the load resistance.

In parallel circuit,    VL = VZ ,   (Old value of VL is removed from calculation)

The voltage drop across RS

VR = Vi – VL = Vi – VZ

So   IR = VR/RS       

              and             IL = VL/RL = VZ/RL

So         IZ = IR – IL 

So         PZ = VZIZ

If any one of IZM or PZM is given,

then other can be calculated as

PZM = VZIZM      or        IZM = PZM/VZ

Numerical:  (a) For the Zener diode circuit, determine VL, VR, IZ, Pz and IZM.

(b) Repeat part (a) with RL = 3 kΩ.

Solution: Part (a)

RL = 1.2 kΩ,

R= 1kΩ,

Vi = 16V

First remove the diode

Applying potential division rule

VL = ViRL/(RL + R)

VL = 16×1.2/(1.2+1) = 16×1.2/2.2 Volt

Or VL = 8.727 Volt

But  VZ = 10 Volt (Given in figure)

So we can see that VL < VZ ,

so diode remains in OFF condition.         

So IZ = 0 mA 

And PZ = VZ×IZ = 10×0 = 0 mW

                VL = 8.727 Volt.        

So            VR = Vi – VL = 16 – 8.727    or   VR = 7.273 Volt

IZM = PZM/VZ = 30mW/10Volt        or       IZM = 3mA  

Part (b)

RL = 3 kΩ, R= 1kΩ, Vi = 16V

First remove the diode

Applying potential division rule

VL = ViRL/(RL + R)

VL = 16×3/(3+1) = 16×3/4 Volt

Or VL = 12 Volt      

But  VZ = 10 Volt

So we can see that VL > VZ , Diode is ON

So voltage across diode is set as 10Volt       

hence       VL = VZ = 10 Volt

Now   VR = Vi –VL = 16 –10 = 6 Volt,

Or      VR = 6 Volt    

So      IR = VR/R = 6V/1kΩ = 6mA

So      IL = VL/RL = 10V/3kΩ = 3.333 mA

So      IZ = IR – IL = 6 – 3.333 = 2.667 mA

So      IZ = 2.667 mA

So      PZ = VZ×IZ = 10× 2.667 = 26.67 mW         or          PZ = 26.67 mW

Now   IZM = PZM/VZ = 30mW/10Volt                     or           IZM = 3mA

Case 2: RL is fixed and Vi is variable:  

In this case, our aim is to calculate the range of Vi for which output voltage is maintained or regulated or constant.

Means we have to calculate Vi,min and Vi,max (Range of Vi) for which VL = VZ.               

First we have to calculate the minimum required input voltage with the help of potential division rule keeping output voltage VL just equal to VZ as follows

So      VL = VZ = Vi,minRL/(RL + RS)       

Or      Vi,min = VZ(RL + RS)/RL 

Now calculate maximum input voltage:

Maximum Zener current IZM is known.

So       IR,max = IZM + IL

So       VR,max = IR,maxRS           (1)

But      VR,max = Vi,max – VL        (2)

Equating equation (1) and (2), we get

             Vi,max – VL = IR,maxRS

or         Vi,max = IR,maxRS + VL

So range of V_i        V_{i, min} = V_Z (R_L + R_S)/R_L            and         V_{I, max} = I_{R, max} R_S + V_L

Numerical:  Determine the range of values of Vi that will maintain the diode in ON state for the given network in figure.

Solution: VL = VZ = 20V, IZM = 60 mA

RS = 220Ω = 0.22 kΩ,     RL = 1.2 kΩ

Vi,min = VZ(RL + RS)/RL

        = 20×(1.2+0.22)/1.2

        = 20 ×1.42/1.2  =  23.667 Volt

So    Vi,min = 23.667 Volt

Now     IL = VL / RL = 20V / 1.2 kΩ = 16.667 mA

So        IR,max = IZM + IL = 60 mA + 16.667 mA = 76.667 mA

So        Vi,max = IR,maxRS + VL

                      = 76.667 mA × 0.22 kΩ + 20 V = 36.867 Volt

So    Vi,max = 36.867 Volt                       

                                                                So,  Range of Vi = 23.667 Volt to 36.867 Volt

Case 3: Vi is fixed and RL is variable:  

In this case, our aim is to calculate the range of RL for which output voltage is maintained or regulated or constant.

Means we have to calculate RL,min and RL,max (Range of RL) for which VL = VZ.   

First we have to calculate the minimum required load resistance with the help of potential division rule keeping output voltage VL just equal to VZ as follows

So      VL = VZ = ViRL,min/(RL,min + RS)      

Or       VZRL,min + VZRS = ViRL,min

Or       ViRL,min – VZRL,min = VZRS

Or       (Vi – VZ)RL,min = VZRS

Or       RL,min = VZRS /(Vi – VZ)

And    IL,max = VL/RL,min = VZ/RL,min

For calculating RL,max , we should proceed as follows

Maximum Zener current  IZM is known. Remember VL = VZ

Now calculate VR = Vi – VL = Vi – VZ

So         IR = VR/RS      (Constant)

Now      IR = IZ + IL       or    IL = IR – IZ.            So         IL,min = IR – IZM

So         RL,max = VL/IL,min

Or         RL,max = VZ/IL,min

So range of RL        RL,min = VZRS /(Vi – VZ)                 to        RL,max = VZ/IL,min

And range of IL       IL,min = IR – IZM = VR/RS– IZM       to        IL,max = VZ/RL,min

Numerical:  For the network of figure, determine the range of RL and IL that will result in VRL being maintained at 10V. Also calculate maximum wattage rating of diode.

Solution: VL = VZ = 10V, IZM = 32 mA

RS = 1 kΩ,     and    Vi = 50V

Now,   RL,min = VZRS /(Vi – VZ)

                     = 10×1/(50 – 10) = ¼ =0.25 kΩ

So,      IL,max = VZ/RL,min = 10/0.25 = 40 mA

So      RL,min = 0.25kΩ  and  IL,max = 40 mA

Now    IR = VR/RS = (50 – 10)/1 = 40 mA

So       IL,min = IR – IZM = 40 – 32 = 8 mA

And     RL,max = VZ /IL,min = 10V / 8mA = 1.25 kΩ

So       RL,max = 1.25kΩ  and  IL,max = 8 mA

Answer: Range of RL = 0.25 to 1.25 kΩ   and   Range of IL = 8 mA to 40 mA