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Trying to Understand Full-Duplex Intercom Circuit

Stevens R. Miller

May 19, 2016
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Hi All,

I'm new here and hoping I can work with some of you to understand this circuit. I studied electronics a long time ago (the '70s), and have built a few things over the years. I have a ham radio license (WA4LDA), and know some things about electronics. Alas, I don't know as much as I thought I did, at least not where this circuit is concerned.

The schematic is for a full-duplex intercom ("full-duplex" meaning that everyone can talk, and be heard, at the same time, like a telephone party line). Some things about it make perfect sense to me, but others are baffling. I've included a .png of the schematic in this post, but a detailed .pdf is available online.

My first question is about how the signal from IC1 is mixed with the signal arriving from the intercom audio bus through C6. It looks to me like IC1 is part of an inverting amplifier that amplifies the signal from the microphone. Now, I think that the voltage at Pin 1 of IC1 is always positive, the sum of the potential from the "half supply rail" and the amplified audio signal. Assuming the supply rail's potential over ground is 12 volts, that would put the half rail at 6 volts, and the output from IC1 would center on 6 volts, with a peak-to-peak variation of, let's say, 3 to 9 volts (no idea if that's what it actually will be, but it has to vary from some amount 6-dV to 6+dV, right?). Now, I am thinking that C5 passes the audio from that signal, but filters out the DC. Thus, a -3 to +3 signal is applied to R9 (and to the much larger impedance of R8). This AC signal is simply added to the AC signal passing from the intercom bus through C6.

And that's where I go crazy, because I can't make sense of how the intercom signal both gets mixed with the output from IC1, so that it can be fed to IC2 (and heard in this unit's headphones), and gets mixed with the intercom bus, so that it can be fed to to everyone else (and heard in every other unit's headphones). Is it becaues the voltages all just add together, and IC2 is a high-impedance-input voltage amplifier? If so, why is C6 needed?

Also (and here is where I really get confused), what happens if the sum of the AC voltage from C5 and the signal from C6 add together to exceed the voltage from Pin 1 at IC1? That is, suppose Pin 1 is at 3 volts. The AC from C5 would be at -3 volts, so the drop across C5 would be a full 6 volts. What guarantee is there that the sum of the voltages from all the other units wouldn't deliver more than 6 volts through C6? If that were to happen, wouldn't that mean C5 would be "reverse biased?" (Does that phrase even apply to capacitors?) Or, are the AC signal voltages on the intercom bus so small that, absent an improbably large number of them, there is no risk of them adding up to more than the "half supply rail" voltage?

Thanks for any help. Looking forward to learning more on this forum.

Intercom.png
 

dorke

Jun 20, 2015
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Welcome to EP.

About the Capacitors:

C6(DC blocking cap) :

If the entire intercom system was contained in a single closed box,
you could theoretically eliminate cap C6 since you wouldn't have any DC voltage on the "audio bus line",the presence of C5 (DC blocking cap )should be enough.

But in practice, think about putting an intercom system in an apartment block.
Each apartment would have that "small box"with wires coming into it "intercom audio bus line" ground and VDC supply rail.
From some sort or reason a DC Voltage may be connected to the "audio bus line" accidentally.
0r be there if C5 is damaged(shorted).
The presence of C6 will prevent it from damaging the circuit.

C5(DC blocking cap):
The quiescent output voltage of IC1 (btw, pin 6 not 1)is set at the "half supply rail".
That is because the op. amp is powered from a single supply rail and biased properly with"half supply rail".
C5 blocks this output DC from reaching the "audio bus line"and the input of IC2.

C5 is a polarized cap.
if you apply constant DCV in the reverse polarity it will be destroyed.
This will not happen with the audio(AC) voltage coming from the "audio bus line" .
in a "constant" way ,although,it may be created momentarily.
e.g.
at a moment in time the output of IC1 swings low to 0V while the incoming "audio bus line" signal coming from another unit, at that same moment swings to "supply rail".
so C5 momentarily has on it a reverse voltage of "supply rail".

The AC signals on the top point of VR1 are indeed added together.
That will happen even if the input of IC2 would be lower resistance.
The simplest way to realize that is "superposition" ,each "audio source" contributes it's portion of voltage on VR1.
 

Stevens R. Miller

May 19, 2016
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If the entire intercom system was contained in a single closed box, you could theoretically eliminate cap C6... From some sort or reason a DC Voltage may be connected to the "audio bus line" accidentally...
Okay, so C6 is there to make sure only AC gets into the audio amplifier of each unit. Makes sense.

The quiescent output voltage of IC1 (btw, pin 6 not 1)...
It actually does say, "6," I think. That "1" is a pin on the electrolytic. There's a "6" just to the left of the "1," right on top of the junction just to the right of the output from IC1.

C5 is a polarized cap.
if you apply constant DCV in the reverse polarity it will be destroyed.
This will not happen with the audio(AC) voltage coming from the "audio bus line" .
in a "constant" way ,although,it may be created momentarily.
So I pretty much got that part right? It can happen that the electrolytic is subjected to the wrong potential difference, but never long enough to blow it.

Thanks! This is coming together for me. I really appreciate the help.

Going on: What does R5 do for me? Seems like R2 and R3 are a simple voltage divider, which should hold Pin 3 of IC1 at "half supply." Makes sense, but, assuming the input impedance of IC1 is high, why add another 10k ohms there?

Similarly, what is R1 for? The notes for this circuit say, "R1 and C1 decouple the microphone amplifier from the main supply. This amp has quite high gain, and can easily go into oscillation (with itself, or with the earphone amplifier, together). The decoupling helps to prevent that." That's a bit opaque, to me. I can guess that C1 further helps purify the DC supplied to the microphone amplifier and to the divider but, if that's a good idea, why isn't IC2 also taking its power from the decoupled rail? (What does "decoupled" mean, here?)

Regarding IC1, if I understand op amps correctly (and I probably don't), the gain it provides is set by the ratio of the drop across R7 (tiny little C4 is there just to filter out high frequencies, according to the notes) to the drop across the impedances between Pin 2 and ground (well, if that ratio is high, otherwise it's 1 + Rfeedback/Rground). In this case, R7 is 2M Ohms, but what's the impedance of the path from Pin 2 to ground? Ignoring C3 (on the assumption that it is big enough to pass AC without meaningful drop), I'm thinking it is the sum of R6 and the microphone's impedance. Am I right about that? That is, if, say, the mike's impedance is 2k Ohms, then the amplifier gain will be 2,000,000 / 4,000 = 500. Is that correct? That seems about right, if the microphone is a dynamic type and putting out, say, 15 mV, peak-to-peak. That would mean IC1's output would run 7.5V, peak-to-peak, nicely within a nine-volt supply range.

How am I doing so far? (And thanks again!)
 

dorke

Jun 20, 2015
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You are doing great:)
Your questions are not basic at all they have to do with fine details of the circuit.
A note:
Don't assume that the designer of this circuit(and any other for that mater) did everything in the best and optimal way.
Sometimes things are done "just in case" or because there is a habit or a rule of thumb to follow etc.

To the point:
IC1 ,the input mike amplifier is configured as a band pass filter in the audio range.
The High "knee" is defined by R7 and C4 .
The Low "knee" is defined by R6 and C3 .
In the middle of the band pass the gain is like you described correctly,
assuming the mic resistance is 2K(<<56K).

R1 and C1 form an RC L.P.F so that the supply rail to IC1 will be "cleaner" than that of the IC2 Power amp supply rail.
In that sense the "power supply lines " are de-coupled.
Noise generated/found on one will not reach the other.

That is further "cleaned" by another RC filter consisting of C2,
to create the bias for the (+) input of IC1which is the most sensitive point to noise coming from the supply lines,(remember a gain of 500).

IC2 could have been decoupled, the designer chose not to.
I would have inserted a decoupling capacitor from supply to ground on IC2 ,
but ,no resistor because of the relatively "high current" IC2 draws.

About R5,
At first site your assumption is correct "functionally" it is not needed.

What follows is "more advanced practical stuff ";)

Consider what happens when the power supply is switched-off
We have C1 charged to VCC and C2 charged to 0.5VCC.
As the Power supply voltage drops towards zero the voltage off C1 and C2 discharges.
this should happen through the supply line to ground.

C1 has a low resistance path through R1=100,that is fine,not a problem.

But C2 has a very high resistance path through R2=10K.
If R5 was not present.
Since R2*C2 >>R1*C1 we will get to the state where the supply voltage of IC1-pin7 is lower voltage than on the (+) input of IC1.

This is not good, the "remaining" voltage on C2 may find a discharge path through the (+) input of IC1 to pin(7) ,that can go through the input protection/parasitic diode of IC1 and destroy it.
Having R5=10K will prevent this discharge path through IC1.

Intercom.png
 
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AnalogKid

Jun 10, 2015
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Some things to consider...

1. C4 and R7 form a 5.3 kHz lowpass filter. However, the circuit is set up for 60 dB of closed loop gain, and a 741 cannot achieve that. Typical gain-bandwidth product is 1.5 MHz for the A part, and around 1 MHz for the standard part. Divided by a gain of 1000, that is a bandwidth of only 1 kHz. So at 5 kHz the gain already is reduced by over 12 dB. Also, at that frequency there is no negative feedback so the output impedance and crossover distortion both increase with frequency.

2. A typical small dynamic or electret mic element does not make 15 mVpp. A better assumption is half that.

3. When operating into a 2K load (and your intercom audio bus impedance will be less than that), a 741 is spec'd at a minimum of +/-10 V when operating on +/-15 V. That is 10 V p-p of lost output voltage swing, and it does not scale down with the power rails. That is, if the delta between the maximum positive peak output voltage and the positive power supply pin is 5 V at Vcc = 15 V, it will still be approximately 5 V when Vcc = 6 V, which is the effective positive supply voltage in a single-rail 12 V circuit. So the output will make only about +/-1 Vpp before clipping. Most chips are better than the datasheet worst case, but how much is unknown. More that the input stage noise and distortion, this is the main reason the 741 is not a good audio device. The 5532/34 is vastly superior.

4. Loading. Because each intercom station looks like a 1 K resistor to the intercom bus, the bus audio volume level with decrease with each station that is added. The solution to this is to convert the mic preamp to a current source output rather than a voltage source output. The first time I saw this was in an IVC9000 studio camera. It was a B-priced camera that thought it made A-quality pictures. It didn't, but it had the best intercom ever.

5. The circuit has no sidetone cancellation. When you speak, your voice will come out of the local speaker at full volume. Both the mic preamp and speaker amp are inverting, so the net phase shift is zero degrees, meaning that the circuit almost certainly will squeal (British term: howl-around) if the mic and speaker are near each other. This is less of a problem with headsets, but even they can squeal given decent coupling from the earphone to the skull. The solution is sidetone cancellation, where a part of the local mic signal is inverted and summed into the speaker driver where it cancels the mic signal on the intercom bus. If we call the local mic signal A, the inverted A signal -A, and the intercom bus B, then at the top of VR1 you have A+B. If you also feed A into pin 3 of the LM386, it is effectively subtracted from the signal on pin 2. So inside the 386, the output of the input differential amplifier is (A+B) - A = B, and what comes out each stations speaker is the sum of all stations audio minus each station's local mic signal. When using a headset, a small amount of sidetone is desirable because it helps people regulate their own speaking volume. I know that is not a clear explanation. Search for hands-free intercom circuits and telephone hybrid circuits to see many examples.

http://e-project4u.blogspot.com/p/blog-page_13.html about 1/4 down the page
http://www.redcircuits.com/Page78.htm
http://www.eeweb.com/blog/extreme_circuits/full-duplex-intercom

These are three variations of the same idea. A transistor with equal resistances in the collector and emitter is a phase splitter. The audio amplitude is approximately equal on both legs, but 180 degrees out of phase. Only one phase goes onto the intercom bus, but both phases go to the speaker amplifier, where they cancel out each other to prevent local squealing.

6. The schematic note about adding a 22 pF cap to the 5534 is incorrect. That is needed only for circuit gains of less than 3.

ak
 

Stevens R. Miller

May 19, 2016
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You are doing great:)
Thanks! You guys are an immense help. I suppose I could just go ahead and build two of these things on faith, but I hate to plug in my soldering iron until I feel I know how a circuit works, so I can deal with any problems. Also, I'm just kind of compulsive about that.

IC1 ,the input mike amplifier is configured as a band pass filter in the audio range.
The notes (I linked to them in my first post) also say C3 is to stop DC from reaching a dynamic mike (which I guess it would as the mike is in the feedback path from the output of IC1, right?)

I would have inserted a decoupling capacitor from supply to ground on IC2 ,
but ,no resistor because of the relatively "high current" IC2 draws.
And IC2 is not an ordinary op-amp, is it? The spec sheet I found indicates it has a "default" gain of 20, which can be changed with a resistor network. Is it an op-amp with its own network built in? Very convenient little gizmo to know about.

About R5, at first site your assumption is correct "functionally" it is not needed.
What follows is "more advanced practical stuff ";)

Consider what happens when the power supply is switched-off...

Ah, of course! I was only thinking about the circuit in its "steady" operational state. But it does get turned on and off, doesn't it? Your explanation reminds me of a tip I learned 40 years ago, about shorting relay coils in the reverse direction with diodes, for the same reason: as the magnetic field in a relay's coil collapses (after the coil is no longer energized) it would tend to send a shot of current back in the reverse direction. A diode becomes a short in that direction and the current just heats the coil (and the diode, I suppose) briefly, instead of back-biasing whatever drove the relay. Similar to what R5 is doing, if I understand you correctly.

Great diagram and explanation!
 

dorke

Jun 20, 2015
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About the analogy to relay driving.
Not the best one,
relay(coil) driving on/off is driving an inductive load .
When driving it off a "back emf voltage" is generated creating a "relatively" high voltage (higher than the supply line) ,which can damage the driving semiconductor.
The diode "clips" that voltage to a bit above the relay supply voltage which is safe.
In our case the voltages remains the same.

About C3,
Yes, a cap blocks DC both ways .
The DC volatge on IC1 inputs (-) and (+) is the same in this circuit,half VCC.

IC2 is an Audio power amp. that can drive a speaker(very low impedance load),
a regular op.amp can't do that.
 

Stevens R. Miller

May 19, 2016
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Some things to consider...

1. C4 and R7 form a 5.3 kHz lowpass filter. However, the circuit is set up for 60 dB of closed loop gain,
I'd like to make sure I understand why it's 60 dB: if I recall correctly, voltage (and current?) gains are conventionally doubled, as they will be multiplied to obtain power gains (or something like that :rolleyes:). Ignoring the mike's impedance, R7/R6 ~ 1,000, or 10^3, which is 30 as dB, and that would be expressed as twice that, or 60 dB, for voltage gain.

Am I close?

and a 741 cannot achieve that.

And that's puzzling, because he has it labeled as an "LM741," but nearby, and in the notes, he calls for an NE5534. Which is it? Why does he have both part numbers on the schematic?

2. A typical small dynamic or electret mic element does not make 15 mVpp. A better assumption is half that.
Thanks. I've been hollering into microphones for a long time, but this little adventure made me realize I know almost nothing about them. I spent about a half-hour yesterday, trying to get reliable info on the voltage output of dynamic microphones. Not much success with that (I was on the verge of getting one and hooking it up to my 'scope, and just finding out by direct experimentation.) So I was off by a binary order of magnitude (what we physics students used to say when anyone else would have said, "one-hundred percent"). Good to know, but at least my other analysis wasn't rubbish, as a result.

3. When operating into a 2K load (and your intercom audio bus impedance will be less than that), a 741 is spec'd at a minimum of +/-10 V when operating on +/-15 V. That is 10 V p-p of lost output voltage swing, and it does not scale down with the power rails. That is, if the delta between the maximum positive peak output voltage and the positive power supply pin is 5 V at Vcc = 15 V, it will still be approximately 5 V when Vcc = 6 V, which is the effective positive supply voltage in a single-rail 12 V circuit. So the output will make only about +/-1 Vpp before clipping.

Awesome analysis! I love it when circuitry can become math this way. Great explanation of why the output amplitude doesn't scale with the main voltage.

4. Loading. Because each intercom station looks like a 1 K resistor to the intercom bus, the bus audio volume level with decrease with each station that is added. The solution to this is to convert the mic preamp to a current source output rather than a voltage source output.

Got a link or something on how to do that? I get the loading issue, but it's not obvious to me how I would go from voltage amplification to current amplification.

5. The circuit has no sidetone cancellation. When you speak, your voice will come out of the local speaker at full volume. Both the mic preamp and speaker amp are inverting, so the net phase shift is zero degrees, meaning that the circuit almost certainly will squeal (British term: howl-around) if the mic and speaker are near each other.

"Howl-around," I love that. The British term for "tube" is "valve," I believe. I've always preferred that one (or I did, back when electronic gear had a lot of valves in it).

The notes address this, a bit, by saying you can achieve cancellation by feeding the microphone audio into the program input (if you have no program, which I don't; this will be for a community theater, not a TV or other studio environment). But he's a bit vague about where I'd make the connection. I don't think I can just connect at Pin 1 of S1, as that would have me feeding the same intercom audio to both the inverting and non-inverting sides of IC2. That can't be right. So, would I connect from Pin 6 of IC1?

Regarding inversion (and I think you've made a good observation about that), can either IC1 or IC2 be configured as a non-inverting amplifier? If not, how about adding a unity inverting buffer after IC1?


Great stuff, particularly that first one. Interestingly, the notes to the intercom circuit on the second page say the two units are identical, but for a resistor that is only present in one unit, not the other. The notes for the intercom on the third page say this as well (in fact, they look identical), but the resistor in question is present in the schematic for both units on that page. This is why I like to understand a circuit before I start welding. (Having said that, I don't know what that resistor is for, but we'll leave that for another day. I think I'll do this circuit first, as I think I'm getting a good grip on it.)

6. The schematic note about adding a 22 pF cap to the 5534 is incorrect. That is needed only for circuit gains of less than 3.
Got it. When you need it, what's it for?

Really appreciate the guidance, folks. And they said the internet would isolate us all from each other. Not quite how it turned out, eh? :cool:
 

Stevens R. Miller

May 19, 2016
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About the analogy to relay driving.
Not the best one,
relay(coil) driving on/off is driving an inductive load .
When driving it off a "back emf voltage" is generated creating a "relatively" high voltage (higher than the supply line) ,which can damage the driving semiconductor.
The diode "clips" that voltage to a bit above the relay supply voltage which is safe.
In our case the voltages remains the same.

Good point. Radio gear is full of coils, but my "Intro to Electronics" book makes the case that almost everything you might want to do with a coil can be done with RC networks instead, and that a good reason to do so is that you limit your voltages that way. A fully energized capacitor (I have been advised to avoid saying, "charged," as there is no more charge in a capacitor storing a lot of energy than there is in one storing no energy; not sure I'm sold on that advice yet, but it helps me remember how a capacitor does what it does) can't ever exceed the voltage across it when it discharges, now can it?

Either way, though, I wasn't thinking about circuit elements that are there in contemplation of the device being turned on or off. Going to add that to my mental check-list from now on, whenever I am trying to comprehend a new schematic.
 

dorke

Jun 20, 2015
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dB in voltage terms(voltage amp) 20Log 1000=60dB.

The thing is the LM741 GBP (Gain Bandwidth Product) is too low to achieve 60dB closed loop gain at about 5.3Khz.
It will probably have that gain at about 1Khz.

Which "Intro book" do you use?

About Op. Amps here is a great one from TI .
 

Stevens R. Miller

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Which "Intro book" do you use?

"Tab Electronics Guide to Understanding Electricity and Electronics, 2nd Edition, " by G. Randy Slone. I like it because Slone was the first one to be able to help me understand how a common-emitter transistor amplifier works. I'd read a number of descriptions, but something in his version allowed me to have that "oh, so that's it" break-through moment.

I also managed to find one of these for cheap on eBay:
1-PB-503_main_72dpi.jpg

I built Slone's circuit directly from the book into this thing, and it worked great. Immensely uplifting experience (and no solder required). I haven't used it in a couple of years, but this discussion about the intercom circuit is going to change that (that, and the fact that my community theater really needs a cheap intercom).
 

Stevens R. Miller

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Oh, speaking of cheap (and I do, a lot), for testing, if not for use in the theater, I would like to use the headsets we have here for computers/games. Finding it very difficult to get specs on them, to the point where I have to admit I don't even know if they are dynamics or electrets (I am assuming they're not condensers). Is there some site or resource that gathers this kind of info? For example, I have a perfectly good little headset that I use with my iPhone, but all it says on it is "Ativa." There's no model or part number on it that I can find, so I've had no luck in Googling a data sheet on it.

Any suggestions on that front?
 

AnalogKid

Jun 10, 2015
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The vast majority of computer and phone headsets have electret mics, and require a DC bias of 5 V or 3.3 V through a 2.2K to 10 K resistor. The three schematics I linked are almost identical, and all show an electret with DC bias. The reason one half does not have a collector resistor is that that is the audio bus point. If each circuit had a collector resistor, then the loading would increase with the number of stations as in your original circuit. With only one common resistor, the audio bus has a consistent impedance. Each collector is a relatively high impedance point, like a soft current source. If you have a single 12 V power supply that drives 3-wire runs to all intercom stations on shielded twisted pair, then you can locate the collector load resistor at the power supply and all stations have identical components. An advantage of this is that that point has all of the intercom audio, and can be recorded or monitored. There is a 2-wire variation of this circuit, but it takes a more complicated power supply.

The LM386 is NOT an opamp, or a mic preamp despite what many ebay modules say. It is a specialized audio power amplifier, a low-cost mini version of the ground breaking LM380. As noted it has both min and max gains, set without a standard feedback network. All audio power amp chips are weird and need special power supply decoupling and Zobel networks to keep from breaking into oscillation.

Speaking of oscillating - opamps want to. A 741 and most standard opamps have a very small internal feedback capacitor, called a Miler capacitor, that forms a single pole lowpass filter that rolls off the high frequency gain and introduces phase shift to prevent unwanted oscillation. However, audio is a tough application because preamps need both high gain and wide bandwidth. An opamp with a small or no Miler capacitor has much better high frequency performance, but in low gain circuits, where the available bandwidth is the widest, they can oscillate. For the 5534, they bring out connections so you can add an external Miller cap when needed at low gain, or leave it out and get the extra bandwidth at high gain.

The problem with an inverting-input mic preamp is that the circuit gain is dependent on the mic impedance, which varies with mic type, and from unit to unit, and with frequency. For a more universal design, go with a non-inverting mic preamp so you have a known fixed gain.

The notes about feeding the mic signal into the program input are correct. While the right side of R9 is the medium impedance icom summing bus, the left side is the output of the preamp, a theoretical zero-ohm output impedance voltage source. That means that there is no icom audio there, only mic audio. If you put some of it into the LM386 non-inverting input, the 386 will subtract it from the icom bus audio on the inverting input. In the three linked schematics, this is what the pot across the transistor collector and emitter does. The two ends of the pot have mic audio 180 degrees out of phase, but icom audio on one end only. Somewhere in the middle is perfect cancellation of the mic audio, and icom audio attenuated by 6dB but very much alive and well. This invert and add for local cancellation is the basis of all telephone and modem hybrids, first done with a 4-winding transformer in the 1920's. The earliest patent I've found for an active phase splitter is a vacuum tube circuit in 1944.

ak
 

Stevens R. Miller

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The vast majority of computer and phone headsets have electret mics, and require a DC bias of 5 V or 3.3 V through a 2.2K to 10 K resistor.
Would it be realistic, then, to use the (4.5 V) half-supply for that? (And you're right about my headset: in the mic port, it works; plugged into the line port, I get nothing.)

The notes about feeding the mic signal into the program input are correct. While the right side of R9 is the medium impedance icom summing bus, the left side is the output of the preamp, a theoretical zero-ohm output impedance voltage source. That means that there is no icom audio there, only mic audio. If you put some of it into the LM386 non-inverting input, the 386 will subtract it from the icom bus audio on the inverting input.
So I should connect from the point between C5 and R9 to the point between C7 and VR2, and elminate C7, is that right? If so, can I also eliminate R8, as there will be a lower impedance path to ground through VR2?

This is kind of exciting. I think I can actually build this thing now, and have some idea what I'm doing. I have most of the parts already. Will need the op-amps and the LM386s (two each, else how can I be sure they work?).
 

AnalogKid

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So I should connect from the point between C5 and R9 to the point between C7 and VR2, and elminate C7, is that right? If so, can I also eliminate R8, as there will be a lower impedance path to ground through VR2?
Better to connect the left side of C7 directly to the opamp pin 1 output. This maintains independence between the two circuits.

ak
 

Stevens R. Miller

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Better to connect the left side of C7 directly to the opamp pin 1 output. This maintains independence between the two circuits.
Yeah, I like that too. I think that actually covers all of my questions. I've downloaded the Fritzing program. Will start tomorrow on a layout for breadboarding two of these things. Will report back here on how it goes.

Thanks, folks!
 
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Stevens R. Miller

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Well, Fritzing turned out to be a rather frustrating program to work with. Regardless, I've gotten it to confirm that a viable breadboard layout matches the schematic I'm using. Here it is, below. The two big changes from the original are that it sends the amplified microphone output to the non-inverting input of the LM386, so the user can null out the side-tone as much as they want, and I've added a tap off of the half supply rail to the microphone so I can use an electret.

About that last point: is it really that simple? When I look online, I mostly see people putting a resistor between the mike and the positive supply (typically 2.2k ohms), and a wide range of supply voltages, from less than 3 up to 12. Does this risk putting the audio signal into Pin 3 of the NN5534? Or losing some audio signal through C2 to ground?

Very eager to wire this up and actually try it, but I don't want to make any easily avoidable mistakes.

Intercom3.jpg
 

AnalogKid

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The R1-C1 time constant works out to 16 Hz. Because that point is powering the preamp, I would add a 100 nF ceramic cap in parallel to C1, and increase C1 to at least 470 uF.
The mic output audio is partially shorted to GND by C2. Leave C2 in place, decrease R4 to 4.7K, and install it between the reference divider and the mic output (just like you saw online).
I still think 60 dB of mic preamp gain is too much.
Change the value of R13 and add a cap in series with it as recommended in the 386 datasheet.
With the pot set to 80% attenuation, the system gain is 2000. With 5 mV input, that's 10 V out with a 9 V battery.

ak
 

Stevens R. Miller

May 19, 2016
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Joined
May 19, 2016
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Thanks for the review. Let's see how much of it I can understand.

The R1-C1 time constant works out to 16 Hz.

You mean "cutoff frequency," right? Interestingly, a more complex version leaves that cap at 100uF, but moves the resistor down to 22 ohms (cutoff frequency of 72Hz). Can't imagine why, but here's that one:

CC2.gif


Because that point is powering the preamp, I would add a 100 nF ceramic cap in parallel to C1,
Guessing: the larger cap isn't so good at passing higher frequencies, so the 100nF is to get rid of high-frequency noise?

and increase C1 to at least 470 uF.
For a cutoff frequency of about 3Hz, for better filtration?

The mic output audio is partially shorted to GND by C2. Leave C2 in place, decrease R4 to 4.7K, and install it between the reference divider and the mic output (just like you saw online).
Because R4 originally held the dynamic mic's DC level at ground, but the electret's DC level will be fixed by the dividers that supply the half rail, and the new value/position of R4. Tying that to ground would simply pass a DC current from the half rail to ground for no useful reason. Close?

I still think 60 dB of mic preamp gain is too much.
As near as I can figure, that second circuit (the "ComClone") is configured for 57 dB. Less, but not a lot less. What do you recommend?

Change the value of R13 and add a cap in series with it as recommended in the 386 datasheet.
Yeah, the datasheet is pretty consistent about that in its example schematics. What's that capacitor for? Noise filtration?

With the pot set to 80% attenuation, the system gain is 2000. With 5 mV input, that's 10 V out with a 9 V battery.

Yeah, that would be quite a trick, wouldn't it? Backing off the preamp gain would solve that, albeit by limiting the output at the earphone, rather than by jacking up the supply to meet what the circuit could do.

How much is enough at an earphone? Seems like a lot of these inexpensive headsets have a 32 ohm earphone. If I delivered 5 volts, that would be 5 x 5 / 32 = 780 mW, which should be plenty, but I think I am ignoring the output impedance of the amplifier in that little calculation. How does one determine the output impedance? Is it set here by R15, at 100 ohms? If so, I am guessing that the power output is 5 x 5 / 100 = 250 mW, which is still pretty high for an earphone. (Power transfer to a 32 ohm earphone would have it dissipating 46 mW, which is a lot for an earphone, I think.)

Maybe this is more complicated than I'm ready for. Starting to realize there's a lot more to this than I thought. Sure would like to have that intercom, though...

For design/education, how do you all feel about SPICE-based Web sites? I found a couple of them, partsim.com and EasyEDA.com. Partsim seems to have bugs (I couldn't get it to simulate a virtual-ground version of an inverting op-amp amplifier) and doesn't seem to have had much activity recently. EasyEDA correctly simulated a couple of op-amp circuits I put into it, but it can be easy to think you've made a connection that looks real in the schematic, only to find you've left a "wire" just dangling in space, not really connected to anything. Beats heck out of wiring something together only to find out it doesn't work (or, worse, survive).

Thanks again for all the help. I'm learning a lot.
 

Stevens R. Miller

May 19, 2016
13
Joined
May 19, 2016
Messages
13
Change the value of R13 and add a cap in series with it as recommended in the 386 datasheet.
Man, you're good. Looking back at the schematic in my OP, that RC pair actually is there. I mistakenly left it out in my Fritzing schematic.

Good catch!
 
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