Inverting & non-inverting

YOU WILL BE ABLE TO

Derive the inverting amplifier's gain from the golden rules.
Derive the non-inverting amplifier's gain.
Compare input impedance and available gain ranges.
Recognise the follower as a limiting case.

Two configurations account for most op-amp circuits, and both derive from Lesson 1's two lines in about three steps each. Start with the inverting amplifier.

The inverting amplifier: the − input is a virtual ground, so the input current has nowhere to go but through R_f.

The non-inverting input is grounded, so rule 1 puts the inverting input at $0$ V too — a VIRTUAL GROUND: a node held at zero volts by the feedback, without being connected to ground. Then the current through $R_{in}$ is $v_{in}/R_{in}$ , and rule 2 says none of it enters the op-amp, so all of it continues through $R_f$ to the output. Ohm's law on $R_f$ finishes the job:

A_v = \frac{v_{out}}{v_{in}} = -\frac{R_f}{R_{in}}

inverting amplifier

Now move the signal to the non-inverting input and ground the bottom of the input resistor instead. The circuit is the same skeleton with the signal injected in a different place — worth noticing, because the two circuits look far more different than they are.

The non-inverting amplifier: same skeleton, signal moved to the + input and R_1 grounded.

Rule 1 puts the inverting input at $v_{in}$ . That node sits at the top of $R_1$ , whose bottom is grounded, so the current through $R_1$ is $v_{in}/R_1$ — and rule 2 sends all of it through $R_f$ . The output is therefore $v_{in}$ plus the drop across $R_f$ :

A_v = 1 + \frac{R_f}{R_1}

non-inverting amplifier

WORKED EXAMPLE

Both configurations, same parts

Take $R_f = 100$ kΩ and the other resistor $= 10$ kΩ.
Inverting: $A_v = -100/10 = -10$ . Input impedance $10$ kΩ.
Non-inverting: $A_v = 1 + 100/10 = 11$ . Input impedance essentially infinite.
Same two resistors, gains of $-10$ and $+11$ — and note the non-inverting gain can never be brought below $1$ , no matter what the ratio is.

-10

and

+11

WORKED EXAMPLE

Designing to a specification

Required: amplify a $0$ – $100$ mV sensor signal to $0$ – $3.3$ V for a Pico ADC, without inverting (the ADC cannot read negative voltages).
Needed gain: $3.3/0.1 = 33$ . Non-inverting, since the sign must stay positive and the sensor is weak.
$1 + R_f/R_1 = 33 \Rightarrow R_f/R_1 = 32$ . Choose $R_1 = 1$ kΩ, $R_f = 32$ kΩ; nearest stock $33$ kΩ gives $A_v = 34$ .
Check the output range: $34 \times 100\ \text{mV} = 3.4$ V — slightly beyond the $3.3$ V rail, so trim to $R_f = 30$ kΩ for $A_v = 31$ , giving $3.1$ V full scale with headroom.
That final check is not optional: an amplifier whose output cannot reach its computed value clips, and Lesson 4 explains what that looks like.

Non-inverting,

R_1 = 1

kΩ,

R_f = 30

kΩ,

A_v = 31

PRACTICE

An inverting amplifier uses $R_{in} = 4.7$ kΩ and $R_f = 47$ kΩ. Find its gain and its input impedance.

Design a non-inverting amplifier with a gain of exactly $5$ using stock values.

Why can't a non-inverting amplifier have a gain of $0.5$ , and what would you use instead?

Explain what "virtual ground" means and why the inverting amplifier's input impedance equals $R_{in}$ rather than something enormous.