How to design Op-Amp Shunt Voltage Regulator

An op-amp shunt voltage regulator is an essential component in many power supply systems, ensuring that the output voltage remains constant despite variations in the input voltage or load conditions. This regulator uses negative feedback and a voltage reference to maintain a stable output. In this article, we will discuss the operation of the op-amp shunt voltage regulator, provide a design example, and show how to calculate key components such as $R 1$ , $R2$ , and $R_S$ .

How Does the Op-Amp Shunt Voltage Regulator Work?

The circuit diagram of basic op-amp shunt voltage regulator is shown below.

Op-Amp Shunt Voltage Regulator circuit diagram

In a basic shunt voltage regulator, the operational amplifier (op-amp) is used to control the base of a shunt transistor. The zener diode stabilizes the voltage at the op-amp's inverting input. A voltage divider consisting of $R 1$ and $R2$ samples the load voltage and provides a feedback signal to the non-inverting input of the op-amp.

Negative feedback ensures that any change in output voltage due to fluctuations in input voltage or load is corrected. If the load voltage increases, the feedback signal increases, causing the op-amp to drive the transistor harder, which increases the collector current. This helps counter the increase in load voltage by creating a larger voltage drop across $R_S$ , which stabilizes the output.

Key Equation for Output Voltage

The output voltage $V_{out}$ in the op-amp shunt voltage regulator is given by:

V_{\text{out}} = \left( \frac{R1 + R2}{R1} \right) \times V_z

Where:

$V_{z}$ is the zener diode voltage,
$R1$ and $R2$ are resistors in the voltage divider.

This equation highlights that the output voltage is directly proportional to the zener voltage $V_{z}$ and the ratio of $R1$ to $R2$ .

Example Design: Calculation of Resistor Values

Let’s consider an example where we need to design an op-amp shunt voltage regulator with the following specifications:

Input Voltage (Vin) = 11.5V
Output Voltage (Vout) = 5V
Zener Voltage (Vz) = 2.4V
Series Resistor (Rs) = 120Ω

We need to calculate the values of $R1$ and $R2$ for the voltage divider that will set the output voltage at 5V.

Step 1: Rearranging the Equation

The output voltage is related to the zener voltage and the resistor values by the following equation:

V_{\text{out}} = \left( \frac{R1 + R2}{R1} \right) \times V_z

Substituting the known values:

5V = \left( \frac{R1 + R2}{R1} \right) \times 2.4V

Step 2: Solving for Resistor Ratio

To solve for $\frac{R1 + R2}{R1}$ , divide both sides of the equation by 2.4V:

\frac{5V}{2.4V} = \frac{R1 + R2}{R1}

2.083 = \frac{R1 + R2}{R1}

This can be written as:

2.083 = 1 + \frac{R2}{R1}

Step 3: Solving for $\frac{R2}{R1}$

Subtract 1 from both sides:

1.083 = \frac{R2}{R1}

Thus, the ratio of $R2$ to $R1$ is:

R2 = 1.083 \times R1

Step 4: Choosing Resistor Values

Now that we have the ratio of $R 2$ to $R 1$ , we can choose appropriate resistor values. Let’s choose a convenient value for $R1$ , say 10kΩ.

Then:

R2 = 1.083 \times 1kΩ = 1.083kΩ

So, the standard resistor values for the voltage divider would be:

R1 = 10kΩ
R2 = 10kΩ

Short-Circuit Protection in Shunt Regulators

One notable advantage of the shunt voltage regulator is its built-in short-circuit protection. If the load is shorted, the current flowing through the regulator increases, but none of the components are likely to be damaged, as the input current remains relatively low.

In a typical shunt regulator, the input current during a short circuit is given by:

I_S = \frac{V_{\text{in}}}{R_S}

With $R_{S} = 120 Ω$ and $V_{\text{in}} = 11.5V$ , the input current is:

I_S = \frac{11.5V}{120Ω} = 0.095A

This current is not large enough to damage the components, which makes the shunt voltage regulator suitable for applications requiring protection against accidental short circuits.

Efficiency of the Shunt Voltage Regulator

Efficiency is a critical factor in comparing various power supply designs. The efficiency of a shunt voltage regulator is calculated using the formula:

\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\%

Where:

$P_{\text{out}} = V_{\text{out}} \times I_L$ (output power),
$P_{in} = V_{in} \times I_{in}$ (input power).

The difference between the input and output power, $P_{\text{reg}}$ , is the power dissipated in the regulator components.

In our example, assuming a load current of $I_{L} = 0.5 A$ , we can calculate:

P_{\text{out}} = 5V \times 0.5A = 2.5W

The input power is:

P_{\text{in}} = 11.5V \times 0.5A = 5.75W

Thus, the efficiency is:

\text{Efficiency} = \frac{2.5W}{5.75W} \times 100\% = 43.5\%

This efficiency is typical of a linear power supply but may not be as high as a switching power supply, which is more efficient for high-power applications.

Conclusion

The op-amp shunt voltage regulator is an excellent choice for applications that require a stable output voltage, such as powering sensitive electronics or providing a stable voltage for microcontroller projects like Arduino. In this example, we calculated the resistor values for a given input and output voltage, demonstrating how to design and implement a shunt voltage regulator in practical scenarios.

For more details on designing shunt regulators and other types of voltage regulators, explore our additional articles on how to design high output shunt voltage regulators, custom power supply for Arduino, and how to design basic shunt regulators.