circuit-fantasia > circuit stories > building circuits > Op-amp RC integrator (roll)

Building Circuits on the Whiteboard...

Op-amp RC Integrator

< prev step - 0 - 1 - 2 - 3 - 4 - 5 - next step >

Starting point. As the inverting input of the op-amp is at virtual ground (0 V) the input current is constant and determined only by the input voltage VIN and the resistor R. The output voltage Vout is a copy of the voltage across the capacitor C.

Well, let's build the circuit on the whiteboard to prove this speculation...

< prev step - 0 - 1 - 2 - 3 - 4 - 5 - next step >

What is the most elementary electrical integrator? Of course, this is the capacitor...

If we drive a capacitor by a constant current source, it functions as an ideal current-to-voltage integrator. Click to view full-size picture.

If we drive a capacitor C by a constant current source SI, it functions as an ideal current-to-voltage integrator with a current input IIN and a voltage output VOUT = VC. Note that the output voltage changes linearly through the time.

< prev step - 0 - 1 - 2 - 3 - 4 - 5 - next step >

Only, we need usually an integrator with voltage input and voltage output (voltage-to-voltage integrator).

Remedy 1: Let's build a compound voltage integrator just connecting a voltage-to-current converter before the integrator:

V-to-I converter + I-to-V integrator = V-to-V integrator

Only, the voltage drop VC across the capacitor C affects the input current and the output voltage changes exponentially trough the time. Click to view full-size picture.

Remedy 2: Remember that we have to pass a constant current through the capacitor; i.e. we have to drive the capacitor by a constant current source. But a suitable current source does not exist in nature; so, we may build it just connecting a voltage-to-current converter after the input voltage source:

V-source + V-to-I converter = I-source

In order to build an integrator with voltage input, connect a voltage-to-current converter before the circuit.

Problem. No matter what is the viewpoint at the circuit, a problem arises here - the voltage drop VC across the capacitor C "enervates" the input voltage thus decreasing the input current. As a result, the output voltage changes exponentially through the time.

Contradiction. There is a contradiction here: from one side, the voltage drop VC has to be as small as possible, in order not to disturb the input voltage; from the other side, it has to be as high as possible, in order to give a significant output signal. Then what do we do?

< prev step - 0 - 1 - 2 - 3 - 4 - 5 - next step >

Remedy. What do we do in real life when an obstacle stands in our way? We remove it by an equivalent useful "antidisturbance".

Following this recipe, we may remove the "harmful" voltage VC by an "antivoltage" -VC. That means to connect an additional supplementary battery BS and to adjust its voltage so that VS = -VC. As a result, a "harmful" voltage VC disappears and the point A has become a virtual ground! The compound current source VIN-R is "fooled": it doesn't "understand" that there is a capacitor; it thinks" that its output is shorted.

In order to build an integrator with voltage input, connect a voltage-to-current converter before the circuit.

Actually, the additional source helps the input source injecting exactly as much voltage as it drops across the capacitor. Let's redraw the circuit diagram in a more suitable form, in order to show this phenomenon.

Actually, the additional source helps the input source injecting exactly as much voltage as it drops across the capacitor

Note that the two voltage sources are connected in series, in one and the same direction (+ -, + -) so that their voltages are added.

< prev step - 0 - 1 - 2 - 3 - 4 - 5 - next step >

Taking an output. Only, where to take the output? We have three choices:

In order to build an integrator with voltage input, connect a voltage-to-current converter before the circuit.

1. VOUT1 = VA - using the old output. But we have already destroyed this voltage!

2. VOUT2 = VC - using the "original" voltage as an output. It is possible but bad solution to connect the load across the capacitor for two reasons: the load has to have a differential output; the load will shunt the capacitor thus affecting the current.

3. VOUT3 = -VC - using the "copy" voltage as an output. Great idea! First, the load will be connected to the common ground; second, it will consume energy from the supplementary source BS instead from the input one!

In order to build an integrator with voltage input, connect a voltage-to-current converter before the circuit.

< prev step - 0 - 1 - 2 - 3 - 4 - 5 - next step >

Finally, we have only to replace the "manual" op-amp with a real one. Now the op-amp doses the voltage of the power supply thus producing a compensating voltage -Vc. Thus, the combination of an op-amp and a steady battery acts as a regulated battery BS.

Finally, replace the 'manual' op-amp with a real one and use its compensating voltage as an output of the integrator.

The op-amp "observes" the potential of the point A (the difference between the two voltages) and changes instantly its output voltage so that the point A stays always at zero volts (acts as a virtual ground). Doing that, the op-amp compensates the "harmful" voltage drop across the capacitor by copying and adding it to the voltage of the input source (the op-amp helps the input source).

Finally, remove all the "unnecessary" components of the picture (voltage bars, current loops, power supplies etc.) and you will get the classic circuit diagram of the op-amp integrator. It will be simple, small, easy remembering but nonunderstandable. Now, you may place it at any recipe book on electronics:).

Related links

Building op-amp integrator - an interactive flash movie,

How I revealed the secret of parallel negative feedback circuits

circuit-fantasia > circuit stories > building circuits > Op-amp RC integrator (roll)