# How independent is the output voltage from the load resistor in this linear regulator?

Consider the circuit below:

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?

• 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? Jun 1, 2020 at 7:34