Summers and integrators

YOU WILL BE ABLE TO

Analyse a summing amplifier with several inputs.
Explain why the virtual ground makes summing possible.
Derive the integrator's output relationship.
Recognise the differentiator and the practical limits of both.

The virtual ground does more than set a gain — it isolates. Because the inverting node is held at zero volts no matter what happens elsewhere, several input currents can arrive at it without interacting at all. Each source sees a zero-volt node through its own resistor and neither knows nor cares about the others. That independence is what makes the summing amplifier work, and it has no passive equivalent.

The summing amplifier: three currents meet at a virtual ground and leave through one feedback resistor.

Apply KCL at the virtual ground — CA-3's law, at a node the op-amp is holding at zero. Each input contributes $v_n/R_n$ , and their sum must leave through $R_f$ :

v_{out} = -R_f\left(\frac{v_1}{R_1} + \frac{v_2}{R_2} + \frac{v_3}{R_3}\right)

summing amplifier

Choose all the input resistors equal and you get a scaled sum — an averager, if you also set $R_f$ to $R/n$ . Choose them differently and you get a weighted sum, with each channel's weight set by its own resistor. That is a mixing desk's fader, a digital-to-analogue converter's binary ladder, and the summing junction of every analogue control loop, all from one node held at zero volts.

Now replace the feedback resistor with a capacitor. The virtual ground still forces the input current to be $v_{in}/R$ , and rule 2 still sends all of it into the capacitor — but a capacitor's voltage is the integral of its current (CA-5 L1), so the output becomes the running integral of the input.

The integrator: constant input current into the feedback capacitor makes a linear ramp out.

v_{out} = -\frac{1}{RC}\int v_{in}\, dt

the integrator

NOTE

Why real integrators need help. At DC the feedback capacitor is an open circuit, so the circuit has no negative feedback at DC at all — and the op-amp's own tiny input offset voltage (Lesson 4) then integrates without limit until the output parks against a rail. The standard fix is a large resistor across the capacitor, which restores DC feedback and caps the gain at low frequencies, at the cost of imperfect integration below a corner frequency. Swap $R$ and $C$ and you get a differentiator, $v_{out} = -RC\, dv_{in}/dt$ , which has the opposite problem: its gain rises with frequency, so it amplifies high-frequency noise enthusiastically and usually needs a series resistor to tame it. Integrators are common in practice; differentiators are rare and are usually a sign that something should have been measured differently.

WORKED EXAMPLE

A weighted summer

Design $v_{out} = -(2v_1 + v_2)$ using $R_f = 20$ kΩ.
Channel 1 needs weight $2$ : $R_f/R_1 = 2 \Rightarrow R_1 = 10$ kΩ.
Channel 2 needs weight $1$ : $R_2 = 20$ kΩ.
Check with $v_1 = 0.5$ V and $v_2 = 1.0$ V: $v_{out} = -(2 \times 0.5 + 1.0) = -2.0$ V.

R_1 = 10

kΩ,

R_2 = 20

kΩ

WORKED EXAMPLE

The integrator as a ramp generator

An integrator with $R = 100$ kΩ and $C = 1$ µF has $RC = 0.1$ s.
Apply a constant $v_{in} = -1$ V. Then $v_{out} = -(1/0.1)\int(-1)dt = +10t$ volts.
The output ramps upward at $10$ V/s — reaching $10$ V in one second, at which point it hits the rail and stops being an integrator.
That linear ramp from a constant input is the mathematical signature of integration, and it is how sawtooth and triangle generators work.

+10

V/s ramp, until the output saturates

PRACTICE

A summing amplifier has $R_f = 30$ kΩ with inputs through $R_1 = 10$ kΩ and $R_2 = 30$ kΩ. Find $v_{out}$ when $v_1 = 0.2$ V and $v_2 = 0.9$ V.

Design a three-input averager: $v_{out} = -(v_1 + v_2 + v_3)/3$ , using $10$ kΩ input resistors.

An integrator has $R = 10$ kΩ and $C = 100$ nF. A constant $+2$ V is applied. How fast does the output move, and in which direction?

Explain why a summing amplifier's channels don't interact, and why you could not build the same circuit from resistors alone.