General Op-Amp question

[PhiloB]

I built a basic op-amp overdrive last night and got very low output but everything sounded and functioned well otherwise.  I decided to use my audio probe to see where the output dropped and discovered there was no audio at PIN 6 (JRC4558) and very low output at PIN 7.  I decided it was probably the resistor in the negative feedback loop and found out I had some 1k resistors in my 22k bag.  Problem solved.
But now I'm confused as to why I don't get audio with my probe at the input (PIN 6) but I do at the output (PIN 7).  Can someone help me understand why I can't "hear" the signal at PIN6?
To keep it Madbean, here is a similar schematic:

[blearyeyes]

I built a basic op-amp overdrive last night and got very low output but everything sounded and functioned well otherwise.  I decided to use my audio probe to see where the output dropped and discovered there was no audio at PIN 6 (JRC4558) and very low output at PIN 7.  I decided it was probably the resistor in the negative feedback loop and found out I had some 1k resistors in my 22k bag.  Problem solved.
But now I'm confused as to why I don't get audio with my probe at the input (PIN 6) but I do at the output (PIN 7).  Can someone help me understand why I can't "hear" the signal at PIN6?
To keep it Madbean, here is a similar schematic:

http://www.ti.com/lit/ds/symlink/rc4558.pdf

Here's the spec sheet. If the link works. If on your schematic they are using the second opamp in a non-inverted config it might be using pin 5?


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[PhiloB]

I built a basic op-amp overdrive last night and got very low output but everything sounded and functioned well otherwise.  I decided to use my audio probe to see where the output dropped and discovered there was no audio at PIN 6 (JRC4558) and very low output at PIN 7.  I decided it was probably the resistor in the negative feedback loop and found out I had some 1k resistors in my 22k bag.  Problem solved.
But now I'm confused as to why I don't get audio with my probe at the input (PIN 6) but I do at the output (PIN 7).  Can someone help me understand why I can't "hear" the signal at PIN6?
To keep it Madbean, here is a similar schematic:

http://www.ti.com/lit/ds/symlink/rc4558.pdf

Here's the spec sheet. If the link works.


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Link worked.  I'm missing something though.  Why can't I probe at pin 6 and hear what is being fed into the negative input pin?


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[blearyeyes]

You should be able to.


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[blearyeyes]

Try a different chip? Re-flow solder on 6? Try pin 5? That's all I got.


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[midwayfair]

You should be able to.


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You will not hear audio at the inverting input pin of an op amp.

[blearyeyes]

Out of phase?  Cancellation? School me brother.

[midwayfair]

Out of phase?  Cancellation? School me brother.

I honestly don't know the reason.

[PhiloB]

Hmm.  Inquiring minds want to know!


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[PhiloB]

Learnt me something new regardless.  Won't waste a bunch of time next trouble shoot trying hear the signal next time!


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[EBK]

You will not hear audio at the inverting input pin of an op amp.

I was waiting for someone else to say this before posting.  I felt that I was too sleep-deprived to trust myself confidently asserting this. Hopefully, I don't mess this part up:
The currents flowing into that node are such that the voltage is made to equal the voltage on the non-inverting input (it's the magic of op amps)...  You're listening to a voltage that isn't changing, by design.


(Ok, I did mess that up a little...  I'll try again....)

[EBK]

Trying again:

In an ideal op amp no current flows into either op amp input, and the voltages at the inverting and non-inverting inputs are made exactly equal through the magic of the feedback loop. The op amp's output voltage is driven to whatever level will make this happen.  In short, you are listening to the same voltage that is present at the non-inverting input, which in this case, has no audio component.

[blearyeyes]

One might expect the audio to go in an input and come out the output amplified but opamps react to the difference between the + and - inputs. So in general the larger the resister in the feedback loop the higher the gain because the difference between the + and - inputs is larger. You also see caps and resisters forming filters which manipulate the feedback's frequency content so that it amplifies certain frequency ranges to create eq curves.  That's all I know and I'm sticking to it. But I could be wrong.

[reddesert]

The ideal op-amp, as others have said, has a couple of properties:
- Very high input impedance
- Very high open loop gain across the differential inputs. This means that, when a feedback loop is provided, the output changes in such a way as to equalize the inputs.

In the inverting amplifier configuration, the + input is at ground (or Vref) and the - input is at "virtual ground," or virtual Vref, because the op-amp is trying to keep it there. See for example https://www.electronics-tutorials.ws/opamp/opamp_2.html

An op-amp in the inverting amplifier configuration works by putting the inverting input in the middle of a voltage divider between the output, and the original signal.  In the attached Egghead schematic, the original signal is at pin 1 of the first opamp. The voltage divider is R5 and R6, 10K and 22K to the output of the second opamp. I'm going to ignore C4, the 2.2n capacitor, whose effect is to roll of the gain at high frequencies.

The + input, pin 5, is fixed to Vref = 4.5V. So the opamp is going to try to do whatever it can to equalize the inputs, making the - input also equal to 4.5 V constant (so no audible signal).

Let's work out what happens if there is an AC signal relative to Vref at pin 1 of the first opamp. Suppose that we are at the bottom of a 1 volt zero-to-peak sine wave, so that pin 1 is at (4.5V - 1V) = 3.5V.  The output will slew to keep the + and - inputs equal, such that pin 7 will be at some voltage X, pin 6 is at 4.5V, and pin 1 is at 3.5V.  R5 and R6 form a voltage divider with 10K between pin 1 and pin 6, and 32K between pin 1 and pin 7, and we can assume that the same current is flowing through R5 and R6 (because no current goes into the op-amp).  The current in R5 is (4.5 V - 3.5V) / 10K = 0.1 mA.  This means 0.1 mA current through R6, so the voltage drop across R6 is 0.1 mA * 22K = 2.2 V.  So pin 7 has to be 2.2 V higher than pin 6: pin 7 is at 6.7 V.

Similarly, if the signal is at the top of the 1V sine wave, then because pin 6 is lower than pin 1, the current in R6 means that pin 7, the output, is lower still: it will be (4.5-2.2) or 2.3V.

What happened here is that we put in a 1V signal and we are getting out a 2.2V signal with the opposite phase. As in the link above, the gain of the inverting op-amp is -2.2, which you can calculate from -(22K/10K), the ratio of the feedback and input resistors.

This is all a little hard to narrate without drawings and formulae (see the link above for more), but if you follow the logic about the current running through the input and feedback resistors, you can see why an op-amp can have a signal output when the inputs are both sitting at a constant Vref, no audible signal.

[blearyeyes]

I know there is math, but for me, I glaze over about then. I need to get an overview of the principles to fit into my poor ADHD addled brain first. I think you have it in your post. I'll peruse it in the morning when time permits.

Thank you for taking the time to write it up!