Op-Amp Shunt Voltage Regulator Design
Op-Amp Shunt Voltage Regulator Design
In this blog i will be explaining about the design of a Opamp shunt voltage regulator. Pros and Cons of the shunt voltage regulator will be discussed. The complete explanation can be watched in video below.
Working :
Due to variation in input voltage Vin or load current IL there is a change in output voltage of shunt regulator.
Terminology Used :
Vin = Input Voltage
Vo =Output Voltage
Vz = Fixed Reference Voltage
R1 & R2 = Sample or Voltage monitoring Circuit
Opamp- Error Amplifier
Rs = Input Current Limiting resistor
Vs = Voltage across Rs
Transistor = It is used to regulator the output voltage
RL = Load Resistance
IL = Load current
Is = Source Current
Ic = Transistor collector current
V+ = Voltage at Non Inverting Terminal
V- = Voltage at Inverting Terminal
Case 1: If the output voltage increases due to sudden change in input voltage or load current. Since sample circuit monitor the output voltage using voltage divider R1 and R2. The Voltage at R2 i.e. VR2 is equal to the voltage at V+.
Since Opamp is connected in closed loop configuration and in negative feedback. Due to virtual ground concept the voltage at V+ is always equal to the V- but practically it is not possible. The Voltage at V+ is marginally higher than V- i.e.
V+ > V- (1)
Because of that the error amplifier output becomes more positive than it was so Opamp output drive transistor with effectively more base current. Since
Ic = Beta * Ib (2)
If base current of transistor increases in that case the collector current of transistor increases. If the collector current Ic increases that lead to increase the Source Current Is. Neglecting all other currents in the circuit.
Is = Ic+IL (3)
Vs = Is * Rs (4)
By above formula if Is increases the Vs increases.
Now Apply KVL ,
Vo - Vs - Vo =0
Vin - Vs = Vo (5)
So we can see that if Vs increases the Vo decreases. So that is how a shunt regulator regulate the output voltage in case of if output voltage Vo increases.
Case 2: If the output voltage decreases due to sudden change in input voltage or load current. Since sample circuit monitor the output voltage using voltage divider R1 and R2. The Voltage at R2 i.e. VR2 is equal to the voltage at V+.
Since Opamp is connected in closed loop configuration and in negative feedback. Due to virtual ground concept the voltage at V+ is always equal to the V- but practically it is not possible. The Voltage at V+ is marginally higher than V- i.e.
V+ < V- (6)
Because of that the error amplifier output becomes more negative than it was so Opamp output drive transistor with effectively less base current. Since
Ic = Beta * Ib (7)
If base current of transistor decreases in that case the collector current of transistor decreases. If the collector current Ic decreases that lead to decrease the Source Current Is. Neglecting all other currents in the circuit.
Is = Ic+IL
Vs = Is * Rs (8)
By above formula if Is decreases the Vs decreases.
Now Apply KVL ,
Vin - Vs - Vo =0
Vin - Vs = Vo (9)
So we can see that if Vs decreases the Vo Increases. So that is how a shunt regulator regulate the output voltage in case of if output voltage Vo decreases.
So hope till now you have understood the working of Opamp shunt voltage regulator.
Formula for Output Voltage :
The Voltage at V+ terminal is given by
V+ = Vo * {R2/(R1+R2)} (10)
Due to closed loop configuration and negative feedback opamp will always try to maintain the same voltage at both inverting and noninverting terminal.
V+ = V-
V- = Vz
Vz = Vo * {R2/(R1+R2)}
Vo = Vz * {(R1+R2)}/R2 (11)
The output of opamp shunt voltage regulator depends on three parameters i.e. Vz, R1 and R2. So by change the value of in equation (13) we can change the output voltage.
Pros : Internal Short Circuit Protection
Cons : Poor efficiency due to loos resistance Rs.
Here is Video link for complete explanation:
Please leave your comments for any further queries.
Thanks !!!
