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I'm trying to make a simple circuit that will allow for the combining and recording of both sides of the conversation by tapping into the connection between a cell phone and a headset.

For convenience sake, I want to also use a spare cell phone as the recording device due to portability,etc. I am using iPhones for the cell phones in this application, although that shouldn't matter.

NOTE: The iPhones provide approximately 1.8V to the microphone to power it, thus my decoupling capacitors on both the input and output. The 4.7K on the output is to trigger the recording iPhone to switch over to external mic mode, otherwise it will only record the internal mic and ignore the input signal.

I've tried designing this to the best of my knowledge and understanding, however I am not getting anything on the outputs. I've checked the input with an oscilloscope and I'm getting a good signal at the opamp input pins, but nothing on the other side. I've tried changing out the IC and still nothing, which I'm guessing there's a flaw in my design since I don't fully understand each component's role/effect in the circuit beyond basic theory. Any help would be awesome and much appreciated.

cell phone recorder circuit

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  • \$\begingroup\$ The sleeve on the TRRS connectors is usually ground. I see by your schematic that you are using the sleeve as an audio input and one of the rings as ground. Are you sure that is correct? It looks wrong to me. \$\endgroup\$
    – JRE
    Commented Oct 7, 2015 at 7:21
  • \$\begingroup\$ That is the mic input on an iPhone. \$\endgroup\$
    – Daniel
    Commented Oct 7, 2015 at 7:28
  • \$\begingroup\$ Why have you "mixed the two sides of the conversation together"If you rout the talk and listen signals seperately to left and right channels of your recorder you will be able to seperate the sides better to work out who said what .I think that the unity gain opamps could go also. \$\endgroup\$
    – Autistic
    Commented Oct 7, 2015 at 10:07
  • \$\begingroup\$ JRE - There's two main standards for TRRS, OMTP and CTIA. Apple follows the CTIA, which makes the sleeve the MIC+ instead of ground as Daniel stated. cablechick.com.au/blog/understanding-trrs-and-audio-jacks @Autistic - Ideally I would've done just that, but the mic input for an iPhone is mono, therefore they must be combined. That is my reason for the volume pots per channel before mixing them together. \$\endgroup\$ Commented Oct 7, 2015 at 13:52

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At the very least, you need to bias the + input of your opamp to the virtual ground you want to use. Use a resistor divider between your 9v rail and ground to get 4.5V. This will bias the input to the midpoint of your rails.

In this configuration, you will need two separate biasing dividers for the two op-amps.

A more rational approach might be to configure them as inverting op-amps, and you can use a single voltage divider to both + inputs of the op-amps.

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  • \$\begingroup\$ Hmmm, ok. I've seen what you're talking about explained and it's much like biasing a transistor if I understand it correctly. What is the difference between a non-inverting vs inverting configuration besides the signal being 180 degrees out of phase? In other words, what's the practical advantage/disadvantage of each in order to decide which one is better depending on the application? \$\endgroup\$ Commented Oct 7, 2015 at 14:13
  • \$\begingroup\$ If you don't care about phase (probably not) and you want to simplify the circuit (probably do) you can use the inverting configuration to reduce the number of resistors. As it stands, you have to provide SOME voltage to the floating + lead through a resistance, and setting it to Vcc/2 will give you maximum room for the signal to swing from ground to Vcc. If it goes outside that range, the op amp will clip and won't work. \$\endgroup\$
    – Daniel
    Commented Oct 7, 2015 at 14:17
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The TL07x op amps are not rated for rail-to-rail operation. In connecting pin 4 to ground, you are forcing the inputs much closer to V- than they can support. The data sheet specifies limits on the common-mode voltage input (which means the input voltage in this case) as 4 volts above V- to 4 volts below V+, which means that it won't work as you've connected it. The simplest change is to get another battery, and connect + to ground, and - to pin 4. This will provide +/- 9 volts to the amp. Also, I recommend a 0.1 uF to ground at both pins 4 and 7. This is called decoupling, and you should always do it. You already have a C3 of 1000 uF, but this is much bigger than you need, and will not respond quickly enough to keep you out of trouble. 0.1 to 10 uF is much better.

Also, it's possible that the iPhone outputs are AC coupled. If so, your inputs will drift around, and may be unusable. I recommend connecting your + inputs to ground with a 10k to 100k resistor.

Finally, there's no need for the 22k resistors. Just connect the output of each op amp to the - input.

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  • \$\begingroup\$ Please forgive me for use of incorrect terminology, but here's my attempt to understand it. 1) Based on what you're saying, even if I bias the input as Daniel suggests, I will only have about 1Vpp of possible amplitude before it falls into shutoff mode, correct? (below 4V would be shutoff, right?) 2) I've seen it both ways on the decoupling pins 4 & 7. Can you explain what this does exactly? I get decoupling, but why would you decouple the ground/- pin on the op amp? \$\endgroup\$ Commented Oct 7, 2015 at 14:06
  • \$\begingroup\$ 3) I'm not sure what you mean by the iPhone output possibly being "AC coupled". Can you explain what that means in this case? \$\endgroup\$ Commented Oct 7, 2015 at 14:08
  • \$\begingroup\$ 4) The 22K resistor, as I understand it, is for setting the gain by giving the op amp negative feedback. Doesn't directly connecting it without any resistance make it function as a 0 gain buffer? I would like a slight gain so the volume levels can be adjusted, especially if one signal is a bit faint compared to the other. \$\endgroup\$ Commented Oct 7, 2015 at 14:10
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    \$\begingroup\$ @Amereservant - 1) Yes. although you may do better than 1 V. The 4 volt numbers only set the limits where the op amp is guaranteed to work, not where it's guaranteed to fail. 2) You decouple signals to ground, so you don't "decouple the ground". This prevents changes in the output from "coupling" back into the op amp through the power supply pin(s). 3) Google AC coupling, 4) When used the way you do, the op amps are called "voltage followers" and the gain is 1. This is independent of the resistor value. To produce gain with your circuit, put a resistor from - input to ground. \$\endgroup\$ Commented Oct 7, 2015 at 17:43
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    \$\begingroup\$ @Amereservant - 1) I mean 2 caps. 1 from pin 4 to ground, and 1 from pin 8 to ground. (Sorry about the pin 7 spec. You're right, Pin 8 is V+, not pin 7. Brain fart on my part.) Use 2 batteries, for heaven's sake. 2) It is exactly because C1 and C2 are there that you must put resistors to ground. If you don't, the inputs may well drift outside the 4 to 5 volt band the inputs must be in. If you must use a single battery, make a voltage divider of 2 1k resistors between +9 and ground, and connect each + input to this 4.5 volt level with a 1k resistor. \$\endgroup\$ Commented Oct 7, 2015 at 22:40
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Alright, so I created a stripped down version on a breadboard using only one set of I/O pins and connected it to a function generator (100mVpp - 1.5kHz) and an oscope. My function generator can go as low as 50mVpp (which is closer to what the input signal will be), however this low of a signal conflicts too much with noise and the oscope, so I chose to work with a higher Vpp for this test. I tested it with and without the voltage divider and the results are self-explanatory. The yellow is the output signal, the blue is the input signal.

The circuit:

enter image description here

Without the voltage divider: enter image description here

With the voltage divider: enter image description here

Notice that there's a drop in the signal output amplitude compared to the input signal and I'm not sure why. (There's also a phase shift occuring, however I'm not concerned about that unless it's indicative of some other problem of concern.) The output amplitude increases when I increase the frequency, however 1.5K is well within the vocal range and this will be a problem. I saw no difference between a resistor across pin 1 & 2 vs a jumper across them, as per @WhatRoughBeast suggests. I guess I don't grasp the negative feedback to set the gain of the opamp.

My reason for choosing an opamp was to simplify amplification (among other benefits of using an opamp) and to increase the amplitude of the input signal for better mixing on the output. Am I just not understanding this correctly?

EDIT:

Ok, I took Daniel's suggestion and swapped it to an inverting configuration and presto! Here's the revised schematic:

Inverted TL072 OpAmp Amplifier Circuit

And, as you can see, the results are as expected:

Oscilloscope Display

Now I don't quite understand why this works and the non-inverting layout doesn't, but the fact it works is what I was trying to achieve. If I understand what Daniel said, I can connect pin 3 to pin 5, using the voltage divider for both Pin 3 & 5, correct?

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    \$\begingroup\$ Those divider resistors should be more like 10k or 100k. That will reduce the filter effect of the capacitor. \$\endgroup\$
    – Daniel
    Commented Oct 7, 2015 at 16:53
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    \$\begingroup\$ If you want amplification, you need to follow one of the many guides to making a gain circuit. This circuit will only duplicate what is on the input. \$\endgroup\$
    – Daniel
    Commented Oct 7, 2015 at 16:54
  • \$\begingroup\$ Ok, thank you very much for your help @Daniel For gain, will this configuration work? i.imgur.com/WXVii3w.png I got that based on the discussion at forum.allaboutcircuits.com/threads/… which it seems I am just missing the divider resistors, correct? \$\endgroup\$ Commented Oct 7, 2015 at 20:06
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    \$\begingroup\$ I think ideally the feedback network would tie to your virtual ground, not 0V at the battery. You can use a divider and another unity-gain op-amp to generate a nice low-impedance voltage at Vcc/2 for this purpose. \$\endgroup\$
    – Daniel
    Commented Oct 7, 2015 at 21:31
  • \$\begingroup\$ You should obtain this book: amazon.com/Amp-Applications-Handbook-Analog-Devices/dp/… You can also download it for free: analog.com/library/analogDialogue/archives/39-05/… \$\endgroup\$
    – Daniel
    Commented Oct 7, 2015 at 21:36

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