Negative Feedback: Cap to Ground between Resistors?

[thehallofshields]

Hey guys,

I really don't get why designers choose to put 2 series Resistors in the Feedback Loop, with a Cap shunting to Ground in between them.

Can anyone point me towards online articles that explain the function of this, and where/why it would be applied.

I'll try to find a simplified picture if anyone is unfamiliar.

[MrStab]

i don't know if it's the case here, but i know that in some cases the post-cap resistor is there to prevent discharge from damaging components further down the line upon switching.

pure speculation, but maybe the first resistor and cap are there to achieve the desired frequency roll-off with commonplace values, whilst the other resistor exists to achieve the desired gain.

you should post an example, even if just naming a particular design. i'm curious too, now!

[R.G.]

I really don't get why designers choose to put 2 series Resistors in the Feedback Loop, with a Cap shunting to Ground in between them.
Can anyone point me towards online articles that explain the function of this, and where/why it would be applied.

The resistors set the DC gain. The capacitor attenuates feedback at frequencies where it plus the two resistors allow this. This makes for a rising gain with higher frequencies as the capacitor steadily reduces feedback.

It's a version of filtering. It's used to tailor the frequency response of the feedback stage.

In a more general sense (and the way that feedback theory is taught formally), the feedback network as a whole, as a circuit lump, from the output to the inverting input has some frequency attenuation that changes with frequency. It's this complex impedance, the combination of all the parts, that is the effective "feedback resistor", although it's really a feedback impedance.

[thehallofshields]

I have this breadboarded with a 4069 chip right now. Gain definitely goes up.

Is this the same effect as placing a cap across a Common Emitter Amplifier's Emitter Resistor?

[antonis]

Is this the same effect as placing a cap across a Common Emitter Amplifier's Emitter Resistor?

Allthough the "mechanism" is different, yes..

In a further point of view, negative feedback has to deal with introducing harmonic distortion (which isn't allways unwanted..)

[R.G.]

I have this breadboarded with a 4069 chip right now. Gain definitely goes up.

In the feedback setup for CMOS inverters, the cap is sometimes made so large that it eliminates all audio frequencies from the feedback path, so the audio components are running open loop, highest available gain. The gain drops back to the DC value to keep the bias point stable.

With the rolloff set in the middle of the audio band somewhere, the cap serves to boost frequencies above the rolloff frequency.

Is this the same effect as placing a cap across a Common Emitter Amplifier's Emitter Resistor?

Conceptually, yes, although the CE emitter resistor feedback is series feedback not shunt feedback. Feedback process is the same, but the details of the mechanism to get it is different, as noted above.

[thehallofshields]

Here is more-or-less what I have on my breadboard.



Does the RC action happen in the direction of Output to Input (right to left), or from Input to Output (left to right)?

[Groovenut]

Is the stage inverting or non inverting?

[R.G.]

That's not the right way to ask the question.

The amplifier has no idea which way the signal flows round the feedback loop. It takes whatever signal voltage is at its input, amplifies that by whatever gain it has, and that appears on the output.
[Note: this discussion will presume that the power supplies on the amplifier are symmetrical around signal "ground" to avoid having to type more about why the single supply operation doesn't matter.]
The input signal feeds through the 10K to the amplifier input. The input of the amplifier has current subtracted from it by the path from the (inverted and amplified) output. Since the input cannot sink or source any current, the currents provided by the input resistor and the path from the amplified and inverted output must mostly cancel.

If the pot wiper is at the output, the cap is simply a capacitive load on the output. If the wiper is at the input, the capacitor and input resistor form a low-pass network on the input. In the middle, the portion of the pot leading to the cap forms an RC highpass filter that is then fed to the input through the remaining portion of the pot.

For analyzing the operation of the circuit, you can conceive of the amplifier input as a "virtual ground". That is, it sits nearly at AC ground, only different from real AC ground by the output voltage divided by the open loop gain of the amplifier. Since this is quite high compared to the signal, you can assume the amplifier input is "ground" and come back later an correct your assumptions.

This also means that the current going into the node at the amplifier input pin must be almost exactly balanced by current subtracted from the input pin by current pulled by the output voltage through the two resistors and cap.

In the circuit as shown, the input is driven by the input voltage through 10K, and by the feedback network. The feedback network is driven by the output voltage driving the output-side 50K and the cap, and that voltage driving current into the input pin.

The RC time constant from the output side is 50K and 1uF, and F = 1/(2*pi*R*C) = 1/(6.28*50k*1u) = 3.18Hz.

That means that the circuit as drawn wipes all the AC from the output side off the feedback above about 32Hz, and so the circuit as shown will run the amplifier open-loop for the entire audio band.

You get different answers as you wobble the pot setting back and forth, of course.

[PRR]

> You get different answers as you wobble the pot setting back and forth, of course.

If I read this right:

It gives the "same" (near enough) response for any wiper position from 0% to 98%: NO audio NFB, "infinite" gain.

In the last few hundred Ohms, the heavy cap load on the opamp makes it take-off and scream radio waves.

If you breadboard this, don't do it near my house. The "infinite" gain will set my dog off, and I don't think my cable/internet can stand heavy RF.

[R.G.]

It does with that cap. With smaller caps, you get an adjustable high frequency take-off point, and an adjustable boost to the stuff below the rolloff, both inextricably mixed.

[thehallofshields]

The RC time constant from the output side is 50K and 1uF, and F = 1/(2*pi*R*C) = 1/(6.28*50k*1u) = 3.18Hz.

That means that the circuit as drawn wipes all the AC from the output side off the feedback above about 32Hz, and so the circuit as shown will run the amplifier open-loop for the entire audio band.

You get different answers as you wobble the pot setting back and forth, of course.

Interesting. So if I wanted high gain, I could use a smaller cap and a trimmer to tune the Bass response.

[thehallofshields]

In the last few hundred Ohms, the heavy cap load on the opamp makes it take-off and scream radio waves.

I don't understand. If you overload an opamp it starts pumping out high-f noise?

[R.G.]

I don't understand. If you overload an opamp it starts pumping out high-f noise?

Imagine taking two #2 lead pencils, one in each hand, and placing the eraser ends together in front of you. Now imagine pushing the pencils together as hard as you can.  The pencils don't move, as long as the forces are balanced, but there's a lot of forces going on there to hold them in place and not moving. That's analogous to an opamp holding a DC level.

Now imagine a force of just a few ounces pressing the junction of the two pencil erasers at right angles to the force holding them together. The force is small, but it misaligns the forces holding the pencils together, and you break a knuckle. That's capacitive loading on a high-feedback amplifier.

Many opamps become unstable with high capacitive loading. Some of them will even tell you this in the datasheet. But pure capacitive loading on the output of an opamp (or even emitter follower) is not a good idea, and they often scream in the RF range.

[PRR]

> overload an opamp it starts pumping out high-f noise?

It's complicated. Way above your need-to-know at this point.

R.G. gave one view of it.

Another: you want to hold your hand in a specific place. But the boat is always rocking. For slow rocks, your hand-speed is ample to react. But now we put a heavy weight in your hand.

Your hand may be fast but not infinitely fast.

A heavy weight can move slowly, but if you try to move it very-fast it gets very-massy.

There will be some rate-of rocking where declining speed and increasing massiness combine to make you VERY wobbly. The out-of-control zone when you try to do (even not-do) something at the limit of your speed.

(Technically, just two fall-offs won't go into infinite oscillation. But in real systems there are always other "minor" speed limits and just a little of this will push a slow-decay wobble into an increasing wobble.)

[thehallofshields]

Awesome guys. You both answered my question perfectly.

Interestingly, I've seen the emitter resistor of a CE Amplifier called the 'negative feedback resistor', which made me think that this cap trick was similar.

[thehallofshields]

One more thing.

As I understand it, we usually simply the Inverting Amplifier gain to Rfeedback/Rseries... R2/R1
In this case, our Zfeedback = ∞, so Gain will always = ∞.

Does the R2/R1 ratio matter anymore? Could I just omit R1 for simplicity, or make R1 huge for high impedance?

[PRR]

> Zfeedback = ∞, so Gain will always = ∞

The maximum gain of the basic device.

[anotherjim]

Ok, but this isn't an op-amp but a CMOS inverter.
My 2uF
You have resistors setting closed loop gain at 10, which is just about as much open loop gain that the inverter has - its a tiny fraction of the open gain an op-amp has. The inverter has high enough output resistance (10-100 times more than an op-amp) that you are getting away with having a cap as large as 1uF on the output when the pot connects it there. It doesn't become unstable.

Op-amp would become unstable. That 1k resistor you have in the output is usually there to buffer an op-amp from the capacitance of the coax cable from the output - otherwise many op-amps will oscillate at high frequency. It won't harm to keep it there in this inverter circuit.

If you add a 1k resistor between the pot end and the output, so that the inverter can't have the cap directly on its output, then you can safely use all the range of the pot. That 1uF cap should probably be sized a few 10's of nano farads - try 22nF.

[thehallofshields]

Thanks.