One of the most common amplifiers in Analog Design is the non-inverting amplifier.

**Figure 1**

Its transfer function is

_{} |
(1) |

How do you derive this function?

Let’s first note that we can consider this Op Amp as ideal. As such, the current in the inverting input is zero (I = 0A, see Figure 2) and the currents through R1 and R2 are equal.

_{} |
(2) |

**Figure 2**

Next, we can write an equation for the loop made by Vout, R2, V and Vin.

_{} |
(3) |

From equation (3) the I2 expression is

_{} |
(4) |

In a similar way we can determine the expression for I1. Equation (5) is the loop equation for R1, V1 and Vin.

_{} |
(5) |

and

_{} |
(6) |

Being an ideal Op Amp, we can consider that the non-inverting input is at the same potential as the inverting input, so V = 0V. This is due to the high gain of the ideal Op Amp. When the output is at a level of a few volts, the differential input can be at a level of some tens of microvolts. Hence, V is very close to zero.

Replacing I1 and I2 in equation (2) and eliminating V, we can write this equation:

_{} |
(7) |

Therefore, the transfer function of the non-inverting amplifier is

_{} |
(8) |

**Q. E. D.**

what is the solution if R1=0 and Vin is applied with a series resistance ?

I am curious. Why would you make R1 zero? What would be the physical accomplishment?