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This thread is a follow-up to this one and that one where I asked for solutions to feed a signal from a guitar pedal device rated 10 dBV (9V p2p) and 10 kOhm output impedance, into an Electrosmith Daisy seed. In the first thread I ask for general considerations regarding impedance, particularly with guitar; in the second thread I asked for clarification regarding a ready-made circuit I found for the Daisy Seed.

Combining what I've learnt from both threads, I decided I wanted to let my device rated 10dBV feed into the circuit without having to turn down the volume knob really low and instead leave the volume knob at about noon. I'd still like to avoid frying my Daisy Seed, so I should ensure the output will not damage the 5V tolerant ADC.

So the solution I've opted for is the following:

Scaling 9V p2p down for Daisy Seed

Here is what it tries to achieve:

  • Inverting Op-amp circuit with a gain of 0.38, which would scale the 9 V p2p down to 3.42 V p2p, as the Daisy Seed ADC expect 3.6 V p2p
  • Input impedance of the circuit is 1 MOhm, which is consistent with good practice in the world of guitar pedals: 1 MOhm in, 10 kOhm out
  • Power is achieved with a 5 V buck converter fed with a power supply rated 12 V DC, 2.5 A which is also used to power the Daisy Seed
  • Op-amp is powered with 5 V because according to bobflux who very kindly and helpfully replied in my second thread, the inverting amplifier setup allows it, although the input signal is 9 V p2p, and the added benefit is that the output cannot exceed 5 V and damage the Daisy Seed
  • Bias is also 5 V, the 9 V p2p input should oscillate between 0.5 to 9.5 V
  • U12 and U13 deal with high frequency interference (as learned in my second thread)
  • C1 and C4 are DC blocking, the values I took from this site about guitar pedals
  • C2 and C5 gradually rolls off the high frequencies above the audible range according to same site, I wonder if they are now redundant with U12 and U13
  • Although a guitar is mono I'm adding the option for stereo since there are two inputs, hence the same inverting op-amp setup repeated twice

My question is now: will this work ? I think I've applied everything I've learnt, but I am not sure that it is correct. I am especially unsure about the combination of +5 V /0 V power to the OPA1678 and 5 V bias to the +input of the Op-amp: I thought the 9.5 V peak of the incoming signal will clip the Op-amp and distort the signal. I trust bobflux's knowledge, but any deeper insight on why this works would be very appreciated.

Thank you very much for all the help!


For completion, let me summarize all the activity in this thread.

  • Justme very correctly pointed out I should bias with Vcc/2 and that I should not use a buck converter to power an op-amp intended for an audio application, bobflux also warned me in an earlier thread. A better approach is an LM7805 linear converter.
  • GodJihyo provided a non-inverting setup which should just work, so I validated the answer.

My inverted op-amp setup might just work though, as bobflux was explaining in the comments to his reply on my second thread. I have been mislead and confused a bit by simulations which emulate behaviors of certain op-amps which do not allow operation on the full Vee to Vcc range: some op-amps apparently will start clipping at Vee + 2 and Vcc - 2 and the simulations led me to believe momentarily that bobflux's suggestion wouldn't work since I had 1V left to work with when applying +5V/0V to the op-amp. The OPA1678 recommended by bobflux can output between Vee + 0.8 and Vcc - 0.8, leaving me 3.4V of headroom which is just about what I need, before any clipping happens. I think therefore it would be possible to apply 5V at Vcc and put Vee to ground, while using a voltage divider to divide Vcc by 2 and apply that to the +input of the OPA1678. The result would be indeed an inverted op-amp which scales 9V p2p down to 3.4V p2p and that would saturate at 3.4V p2p.

Here is the diagram:

Inverted op-amp gain 0.4 which clips at 5V

I would like to thank everyone very warmly for giving their inputs.

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  • \$\begingroup\$ You can't bias the positive input to same voltage as the positive supply, because the output cannot go above the positive supply. \$\endgroup\$
    – Justme
    Commented Aug 6 at 19:02
  • \$\begingroup\$ @arno This schematic is hard to read, can you draw it with the actual opamp symbol? \$\endgroup\$
    – Voltage Spike
    Commented Aug 6 at 19:16
  • \$\begingroup\$ @Voltage Spike Hello, I am using EasyEDA to make my schematics with also parts and PCB footprints automatically taken care of, it makes it very convenient for me to make a PCB when I have worked out everything ! Unfortunately I don't control the symbols of the parts added to the database. However you can see the little flag labels, they show you where everything goes and it does fit the real life shape of the component. With a bit of adjustment, you should be able to read it ! I hope that's ok. \$\endgroup\$
    – Arno
    Commented Aug 6 at 19:43

2 Answers 2

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Not sure why you decided to go with an inverting amplifier, but you could try a voltage divider followed by a non-inverting unity gain buffer.

Something like this will give you around 860k input impedance, scale and offset your 9 Vpp 0 V centered signal to just under 5 Vpp 2.5 V centered, and limit the output to 0 to 5 V.

guitar buffer amplifier

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  • \$\begingroup\$ Thank you very much for your reply ! Actually I went with the inverting setup because I did not know how to compute the impedance of the non-inverting setup. I saw the inverting setup's impedance was the first resistor the input sees, so I went with it. In your diagram, I understand 860k = 470 k + 390 k, and so this is the input impedance ? What about the route through C2, Vbias, R5 then ground: doesn't that factor in the input impedance ? I guess the answer must be no, I'm just not saavy enough to know why. Thank you again ! \$\endgroup\$
    – Arno
    Commented Aug 6 at 21:41
  • \$\begingroup\$ @Arno It does, but at high frequencies. It just rolls off the high end a bit. If you use LTspice you can plot the impedance, for the non-inverting input of an opamp it will be quite high, so the divider impedance dominates it. I just picked a couple of standard value resistors to give roughly a 5/9 division and add up to something near 1 meg. \$\endgroup\$
    – GodJihyo
    Commented Aug 6 at 23:20
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No, it won't work.

If you bias the +ve input to VCC of the op-amp, then output bias is also VCC, and it would mean that when input has a negative transition below VCC, the output would have a positive transition above VCC, which is impossible.

Maybe you intended to bias the +ve input to half-VCC instead of VCC.

And for sensitive analog circuits, you might not want to power it with a switch mode converter, at least not without making sure the buck converter ripple is not a problem or that it can be filtered out.

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  • \$\begingroup\$ Thank you for your answer, I posted a corrected schematic in another answer. How does it work, should I mark your answer as correct and let you earn the point ? If it's ok I'd still like to leave my amended circuit below, so this thread also features the solution to my problem. \$\endgroup\$
    – Arno
    Commented Aug 6 at 19:50
  • \$\begingroup\$ My updated answer is now reflected in my initial question. \$\endgroup\$
    – Arno
    Commented Aug 6 at 23:47

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