I would've thought you couldn't feed a ground-biased AC signal into a single supply op amp because you don't want the negative component of the AC hitting your op amp input.

Is this treated as ok here simply because the microphone's signal will be too small to do any harm? In other words, you wouldn't set an op amp up as a half wave rectifier the way you described if the AC signal was several volts, right? It's only ok because we expect the signal to be small fractions of a volt. Or am I missing some other reason why this is acceptable?

Many opamps have transient and/or reverse polarity protection diodes at the inputs. Some have a linear input common mode range that extends down below the part's negative rail (GND, -V, whatever). In general, I think it is safe to say that a few millivolts of reverse boas won't do any damage no matter what the input stage looks like. But for a low impedance, higher voltage source, you are correct - the negative currents might cause damage unless dealt with.

ak

 

apcircuits

I used to use apc quite a bit, though I never much liked their quality control, and I really began to dislike their product for small surface mount parts because of the (inconsistent) great mounds of solder on the pads that would make parts placement more of a pain than usual. They didn't level the solder.
They only did one-day turn for standard laminate. Their finished copper weight was often grossly insufficient for my purposes, and turned out to be consistently well below what they claimed it was. I once did a board where I attempted to use a PC track as as current sense resistor in a prototype of an industrial battery charger. I discovered then that their copper weight claim was BS. They told me I was the only person who had ever raised an issue with their copper. Their board cutting was sloppy, so for a lot of things I had to make the board oversize and then trim it, which was a big time waster. They did improve that. Their boards were OK for run of the mill low power stuff. I used Network Circuits in Quebec when I needed heavy copper and thick laminate and simply waited a few extra days.

 

Not quite. In your drawing, the signal "after capacitor" will be sitting on the 2.5 V pedestal. This will be true if the DC portion of the microphone signal is either above or below 2.5 V. The capacitor is charged up (or down) by current from the opamp output through R2 and R1. Since these resistors usually are larger than the 1K pullup resistor at the microphone, they set the time constant and determine how long it takes for the coupling capacitor to match the voltage difference imposed across it in the steady state condition.

The opamp output does whatever is necessary to make and keep the voltages at its two inputs identical. Since it cannot change the + input, it changes the - input through R2. This creates a "virtual ground" at the R1-R2 node, a point that the input signal cannot change directly. The input pumps current into (or out of) the R1-R2 node through R1. Since R1 and R2 form a voltage divider between the microphone and the opamp output, this attempts to change the voltage at the - input. The output voltage changes in the opposite direction to increase or decrease the current through R2 to bring the - input back to 2.5 V. If you put a scope on the - input you will not see the audio signal, yet the output is an amplified version of the input.

ak

Hi , thanks for the reply! What do you mean by" will be sitting on the 2.5 V pedestal." I thought the function of the capacitor is to block the DC component. And how does the opamp change the - input . Ex if i connect the r1 directly to a 5v and if the value of r1 doesnt give a voltage drop of 2.5v wouldnt the - input not be 2.5v. Sorry for my dumbness .

 

Op Amps for Everyone has lots of useful theoretical and practical info
http://www.cypress.com/file/65366/download

Also see HANDBOOK OF OPERATIONAL AMPLIFIER APPLICATIONS
http://www.ti.com/lit/pdf/SBOA092B
(this one has schematics in stupid colors; I hate schematics in stupid colors - the colors are for aiding drawing on the screen, not for publication except as aids to highlighting bits of circuitry)

 

in reply to ebeowulf17 at 39:

The important thing to avoid damaging an amplifier is to limit the current that can flow into an input in the case where you are overdriving it (of course when you aren't overdriving it, the current into or out of an input is just a very small bias current). In general, it is safe to drive up to about 5 mA into inputs. Some parts will specify a maximum of 2 mA. But that is just the damage issue. Remember that any amplifier likely will have current flowing into or out of inputs if the source of its input is powered when the amp isn't, if the source voltage is more than a volt or so above or below zero.

Some op amps, when overdriven by more than a small amount, will exhibit phase reversal - the signal at the output is positive-going when it should be negative-going or vice versa. This can be a really serious problem in closed loop systems because it can amount to positive feedback, which is almost always a bad thing unless it is intended. An error amplifier that says "move that ram with a hundred horsepower behind it out" instead of "pull it back" tends to break things.
The other thing that happens when op amps are overdriven so that the output tries to swing "past the supply rail" is that things in the internal circuit, such as bipolar transistors or frequency compensation capacitors, get saturated. Recovery from saturation can take a long time. The consequences of this depend very much on what the circuit is doing. A recovery delay of a few microseconds may be of no consequence whatever or may be very serious.

Good data sheets will specify things like absolute maximum input current, though many don't. Many data sheets won't say anything about phase reversal unless the amp is guaranteed not to do it. Recovery time from overdrive is often not spec'd.

 

Off Topic

@ebp - Welcome to AAC. Sounds like you have quite a number of years of experience under your belt.

There is a special thread "Who are you?" for introducing yourself to the rest of AAC membership. Would love to hear about your background, eh? Weather in TO is a balmy 4°C and sunny right now.

 

Hi , thanks for the reply! What do you mean by" will be sitting on the 2.5 V pedestal." I thought the function of the capacitor is to block the DC component. And how does the opamp change the - input . Ex if i connect the r1 directly to a 5v and if the value of r1 doesnt give a voltage drop of 2.5v wouldnt the - input not be 2.5v. Sorry for my dumbness .

To the left of the capacitor, the microphone creates the audio signal by changing its impedance. The microphone element and the 1K resistor (probably too small, increase to 3.3K) for a voltage divider. With no sound hitting the mic, the voltage at the mic-R1 node is some DC value between 0 V and 5 V. As audio hits the mic, this DC value "wiggles" up and down according to how the mic translates changes in atmospheric pressure into changes in its own resistance. The wiggles are the audio signal, "sitting" on that DC value. As you increase the value of R1, two things change. First, the audio part of the combined signal gets larger. Second, the DC value of the signal decreases closer to 0 V (Ohm's Law). At some point, the DC value is so low that the negative peaks of the audio signal try to extend below 0 V. This is called clipping, and obviously means that the value of R1 is too large.

So, there is a DC value on the left side of coupling capacitor C1 (THANK YOU for using reference designators!). To the right is the opamp, thanks to the two 10K resistors (missing reference designators - tsk tsk), both inputs to the opamp are sitting at the 2.5 V established by that voltage divider, thanks to the negative feedback loop. So, now we have two DC values, one on each end of the capacitor. You are correct, the capacitor blocks DC, but that is not the same as making the DC at either end 0 V. In your circuit, if the mic output is sitting at 2.0 V and the opamp input is sitting at 2.5 V, the capacitor is blocking 0.5 V of DC between those two nodes.

ak

 

To the left of the capacitor, the microphone creates the audio signal by changing its impedance. The microphone element and the 1K resistor (probably too small, increase to 3.3K) for a voltage divider. With no sound hitting the mic, the voltage at the mic-R1 node is some DC value between 0 V and 5 V. As audio hits the mic, this DC value "wiggles" up and down according to how the mic translates changes in atmospheric pressure into changes in its own resistance. The wiggles are the audio signal, "sitting" on that DC value. As you increase the value of R1, two things change. First, the audio part of the combined signal gets larger. Second, the DC value of the signal decreases closer to 0 V (Ohm's Law). At some point, the DC value is so low that the negative peaks of the audio signal try to extend below 0 V.

On the left side of the capacitor (connected to the microphone), the DC value can be close to zero. Even with R1 at its highest value, the impedance of the microphone never goes below zero, relative to R1. Hence the DC value never goes below zero. The microphone is most likely an electret microphone. That is, it is a piezo or condenser microphone whose signal is amplified with a FET. R1 is the load resistor on the drain of the FET. The AC and DC voltage never goes below zero.

 

Even with R1 at its highest value, the impedance of the microphone never goes below zero, relative to R1. Hence the DC value never goes below zero

True, but if the DC value is very close to 0 V, the mic will become very non-linear and distort the negative peaks of the audio as it runs out of range.

ak

 

True, but if the DC value is very close to 0 V, the mic will become very non-linear and distort the negative peaks of the audio as it runs out of range.

ak

That should be ok for this application which is a clapper circuit. The microphone is used to pick up the sound of a hand clap. It shouldn't matter if we clip one half of the cycle.

 

That should be ok for this application which is a clapper circuit. The microphone is used to pick up the sound of a hand clap. It shouldn't matter if we clip one half of the cycle.

Agree for this application, but the question was about the general voltage relationships in the circuit, and I think the consequences of increasing R1 are important to point out.

ak

 

To the left of the capacitor, the microphone creates the audio signal by changing its impedance. The microphone element and the 1K resistor (probably too small, increase to 3.3K) for a voltage divider. With no sound hitting the mic, the voltage at the mic-R1 node is some DC value between 0 V and 5 V. As audio hits the mic, this DC value "wiggles" up and down according to how the mic translates changes in atmospheric pressure into changes in its own resistance. The wiggles are the audio signal, "sitting" on that DC value. As you increase the value of R1, two things change. First, the audio part of the combined signal gets larger. Second, the DC value of the signal decreases closer to 0 V (Ohm's Law). At some point, the DC value is so low that the negative peaks of the audio signal try to extend below 0 V. This is called clipping, and obviously means that the value of R1 is too large.

So, there is a DC value on the left side of coupling capacitor C1 (THANK YOU for using reference designators!). To the right is the opamp, thanks to the two 10K resistors (missing reference designators - tsk tsk), both inputs to the opamp are sitting at the 2.5 V established by that voltage divider, thanks to the negative feedback loop. So, now we have two DC values, one on each end of the capacitor. You are correct, the capacitor blocks DC, but that is not the same as making the DC at either end 0 V. In your circuit, if the mic output is sitting at 2.0 V and the opamp input is sitting at 2.5 V, the capacitor is blocking 0.5 V of DC between those two nodes.

ak

Hi ! It is starting to make sense . But how does the increase of R1 affect the signal and Dc value? Thank you so much

 

Inside an electret microphone is the electret element and a FET. The electret modulates the voltage on the FET gate, and the corrresponding change in the FET's channel resistance modulates whatever source of current it is attached to. So one way to think of an electret mic is a resistor that has a fixed average value that increases and decreases slightly with the instantaneous air pressure on the diaphragm. This is not a perfect analogy, but it is close enough for now.

Once you think of the microphone as a variable resistor, it should be clear that if it is in series with a fixed resistor up to a fixed positive voltage, the voltage at the junction of the two will have a steady-state DC value with the audio signal impressed upon it. Using Ohm's Law, as the fixed resistor value increases, the node voltage decreases, and vice versa.

ak