Consider the circuit below:

enter image description here

So we know that the op-amp will try to keep \$V_{+}\$ equal to \$V_{-}\$ (Inverting and non-inverting voltages of the op-amp) . Knowing this we can calculate the output voltage:

$$ V_{+}=15\times \frac {1k}{9.1k+1k}=1.48\; V $$ $$ V_{-}=V_{out}\times \frac {R_{2}}{R_{1}+R_{2}}=V_{+} $$ $$ \Longrightarrow V_{out}= V_{+}\times (1+\frac {R_{1}}{R_{2}})= 1.48\;V\times(1+\frac {R_{1}}{R_{2}}) $$

So as we can see, the output voltage is not dependent on the load resistor, but in practice we know that our circuit is not perfect and if we make a significant change in the amount of load resistor, there will be a small change in the output voltage, why does that happen?


1 Answer 1


There are two types of regulation specified in data sheets for regulators ...

Line regulation - change of output when the input changes
Load regulation - change of output when the load changes

You are talking about load regulation.

Your equations for the output voltage assume infinite gain in the opamp. Once you've written down the full equations including a finite gain, you'll find that the voltage at the junction of R1 and R2 will have to vary to provide the varying current output, and there's your change of output voltage.

  • \$\begingroup\$ I see, by the way i ran a simulation using hspice assuming R1=R2=1 which gives \$ V_{out}\approx 3V \$ and i found out the output voltage is only equal to 3v for input voltages greater than 3v , does it mean this regulator only works as step down? And what changes should be made to this regulator for it to be able to have a constant output voltage equal to 3v for input voltages lower than 3v? \$\endgroup\$
    – Ali Nategh
    Jun 1, 2020 at 7:34
  • 1
    \$\begingroup\$ This is a linear regulator. They work by dropping some of the input voltage across an active element, in this case Q1. If you want a higher output voltage, you typically need a SMPS (switched mode power supply), aka boost converter. So your regulator 'changes' would consist of throwing this one away, and replacing it with a boost converter. \$\endgroup\$
    – Neil_UK
    Jun 1, 2020 at 8:56

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