Operational Amplifier Voltage Regulator

Operational Amplifier Voltage Regulator:

Fig. 4, shows how an operational amplifier is used in a series voltage regulator. Resistor R₁ and R₂ form avoltage divider that feeds a voltage proportional to V_O back to the inverting input: V^- = V_O R₂/ (R₁ + R₂)). The zener voltage V_Z on the noninverting input is greater than V^-, so the amplifier output is positive and is proportionate to V_Z -V^-. If V_O decreases, V^- decreases, and amplifier ouput increases. The greater voltage applied tothe base of the pass transistor causes it to conduct more heavily, and V_O increases. Similarly, an increase in V_O causes V^- to increase, V_Z - V₁ to decreases, and the amplifier output to decrease. The pass transistor conducts less heavily and V_O decreases.

Fig. 4

The operational amplifier in fig. 4 can be regarded as a noninverting configuration with input V_Z and gain

Therefore, neglecting the base to emitter drop of the pass transistor, the regulated output voltage is

R₃ is chosen to ensure that sufficient reverse current flows through the zener diode to keep it in breakdown.

Example - 4

Design a series voltage regulator using an opertional amplifer and a 6V zener diode to maintain a regulated output of 18V. Assume that the unregulated input varies between 20V and 30V and that the current through the zener diode must be at least 20 mA to keep it in its breakdown region.

Solution:

From the output voltage expression 18 V = ( 1 + R₁ / R₂) 6 V.
Thus, (1 + R₁ / R₂) = 3.
Let R₁ = 20KΩ, Then,

The current through R₃ is

Since the current into the noninverting input is negligibly small, the current in R₃ is the same as the current in the zener dode. This current must be at least 20 mA, so the smallest possible value of V₂(20 V):