# Why isn't IC1 a comparator?

This is with reference to a circuit i saw in The Art of Electronics.The authors claim that IC1 is an integrator and that the circuit ensures that a current of magnitude |Vin|/(200k) constantly flows through the capacitor.They further explained that only the direction of current changes at regular intervals determined by the Schmitt trigger IC2.

I suspected that the opamp IC1 is a comparator(no negative feedback).So I simulated the circuit with LT Spice (with Vin =6 Volts) giving me the following results.

The waveform in green is the voltage at inverting terminal of opamp IC1.The waveform in red is the voltage at the output of IC1 and the blue waveform is the current through capacitor C1.The simulation proved that the authors were indeed right!! The inverting input was held constant at 3 Volts(for Vin=6 Volts).

This meant that the capacitor somehow completes the feedback loop providing negative feedback and thus ensuring that voltage at non inverting terminal = voltage at inverting terminal. I came to terms with the fact that the capacitor 'completes' the feedback loop.I guessed that since the capacitor conducts throughout the time period ,it must always behave like a varying resistor completing the feedback loop.

But then I had trouble explaining why a capacitor doesn't complete the feedback loop in an astable multivibrator circuit.

The above circuit can definitely be explained by considering the opamp as a comparator switching between Saturation voltages at fixed thresholds.Simply said- the capacitor doesn't complete the feedback loop.

So to summarize my question -why does the capacitor complete the feedback loop in the Voltage Controlled Oscillator circuit but not in an Astable Multivibrator???

• IC1 has negative feedback - the feedback capacitor provides this. – Andy aka Jan 3 '15 at 0:46

## 4 Answers

I guess by "complete the feedback loop" you mean "hold the inverting and noninverting inputs at the same voltage". This is basically the op-amp's only goal in life, and given suitable negative feedback, it will accomplish it. If it can't, then it will drive the output into one supply rail or the other attempting to do so.

So, why can IC1 accomplish this, while the astable multivibrator can not? Let's consider the essential components of each:

simulate this circuit – Schematic created using CircuitLab

Now consider the definition of capacitance:

$$I(t) = C\frac{\mathrm dV(t)}{\mathrm dt}$$

It might make a little more sense algebraically re-arranged:

$$\frac{\mathrm dV(t)}{\mathrm dt} = \frac{I(t)}{C}$$

That says, "the rate of change of current with respect to time is equal to current divided by the capacitance". So, if you put 1A through a 1F capacitor, voltage changes at a rate of 1V/s. If you increase the current or decrease the capacitance, voltage will change faster. To get voltage to change instantly, you need infinite current or zero capacitance.

For IC1, it's easy for the op-amp to respond to any change in the input. The voltage across a capacitor wants to remain constant -- it takes time and current to change it. If in some instant the input voltage increases by 1V, the output can increase by 1V, and instantly the inverting input also increases by 1V, and the two inputs have the same voltage. Mission accomplished.

But what about IC3? Say the input increases by 1V instantly. What can the opamp do? It can increase the output voltage, but the voltage across C2 (and thus, at the inverting input) can not change instantly. To change it instantly would require infinite current. But that's impossible, because the current the op-amp can drive through the capacitor is limited by R1.

So instead, the op-amp will do the best it can and saturate the output at the positive supply rail. Eventually, it will manage to charge C2 to match the voltage at the input, and the output voltage will go to 0V.

To make an astable multivibrator, you add positive feedback so as the output starts to settle to 0V the input voltage also changes. Thus, IC3 (with positive feedback added) can never accomplish its goal. It's always trying to catch up, and every time it succeeds, it starts another cycle.

• "if you put 1V through a 1F capacitor, voltage changes at a rate of 1A/s" I think you got the units reversed there (I can't make an edit of only two letters and don't see anything else to change) – Olof Jan 10 '15 at 14:52

As the first circuit description says, IC1 is acting as an integrator. That capacitor in the feedback path is a DC blocking capacitor that causes the voltage on the output side of the IC to build up, until it rises to a certain point and is then discharged by IC2 and Q1.

The astable multivibrator works on a different mechanism where C1 is always causing a lag between the IC output and the voltage on the - pin. This (I think, without having simulated it) keeps the input chasing the output, which keeps the output changing.... that is, until the positive feedback is overcome on the + pin, then it switches over.

That's not a perfect explanation, but I think the answer is that they just work differently, using different mechanisms.

It's not a comparator because current flows through the capacitor in such a way that the feedback loop is increased.

Note that for a DC input, if there were no resetting circuitry, the output would eventually reach the rail, and feedback will be broken. At that point, you're already at the rail, so you won't necessarily notice that the amp is behaving like a comparator.

The first circuit is a classic function (both square- and triangle-wave) generator where an integrator (IC1) and a comparator with hysteresis (an op-amp Schmitt trigger - IC2) are connected in a loop. In this relaxation circuit the integrator output voltage swings between the two thresholds of the comparator since, when reaches the threshold, the comparator reverses the direction of the input current... and thus the direction of the voltage changes as well.

The conventional solution is to connect directly the comparator output to the integrator input. But obviously, the authors decided that the IC2 output voltage is not so stable and preffered to drive the integrator by a constant input voltage (Vin)... and, of course, what is more important here, to control the frequency... so a clever trick is used... Let's demystify it.

Note that the IC1's inverting input is "lifted" up to Vin/2 by the voltage divider R2-R4. So, when Q1 is off, a positive input current Vin/2/100k = Vin/200k is injected into the input... and this entering current charges the capacitor. When Q1 is on, it connects the resistor R4 to ground... and it begins "sucking" a negative current Vin/2/50k = Vin/100k from the input. The resulting current is Vin/200 - Vin/100 = -Vin/200... and this exiting current discharges the capacitor. So, really H&H are right that "a current of magnitude |Vin|/(200k) constantly flows through the capacitor".

Here is another (simpler and clearer) explanation. The most general (not obligatory electrical) idea that can be seen in the input circuit part, is to make (reverse) some positive quantity negative by adding a two times bigger negative quantity to it. This idea is implemented here by a summing integrator that sums the positive current "pushed" by Vin through R2 with the double negative current "sucked" by the ground through R4. As a result, the current through the capacitor reverses its direction (but keeps its magnitude unchanged). Now you understand why the IC1's non-inverting input is "lifted" at Vin/2 - to give a chance to ground to create a negative current when Q1 is on... BTW the same powerful idea is used in negative impedance converters (NIC) where either the current (INIC) or voltage (VNIC)... and, as a final result, the resistance is inverted and thus transformed into a negative resistance. I have explained it in another question.

@Kishore Saldanha, now about your main question, "Why does the capacitor complete the feedback loop in the Voltage Controlled Oscillator circuit?" Really, as Phil Frost explained above, the capacitor closes a 100% negative feedback in regards to the instantaneous voltage changes... and the circuit acts like a voltage follower with continuously "shifting" input and output voltage... And, as you are curious enough, I will reveal the secret of this circuit especially for you... I will answer the main question here, "What does the op-amp actually do in the circuit of the op-amp inverting integrator?"

The answer is extremely simple: The op-amp output "copies" the voltage drop across the capacitor (Vc) and adds this voltage (-Vc) in series to it... So the effective voltage (between the two op-amp inputs) is almost zero (the only difference here is that we have to consider the "lifting" voltage Vin/2). Thus the op-amp output voltage neutralizes the variations of the voltage drop across the capacitor and the current through it is constant - |V|/200k... it acts as a kind of a negative capacitor... or a virtual capacitor... See also this Wikibooks story.

The second circuit is a simpler version of the first. Here the perfect op-amp integrator is replaced by a humble RC circuit... and, of course, it is not a voltage-controlled generator. In this circuit actually there is no negative feedback during the transition that is fully controlled by the instantaneous reinforcing positive feedback.