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beginner here, please be nice :)

I'm interested in the following DI-style box for my guitar, which I'm going to build, and I'm working on analyzing it. http://www.generalguitargadgets.com/pdf/ggg_ssim_m_sc.pdf

So far I've done reasonably well with figuring out the 4-stage op-amp filters. I'm considering filtering out the DC at the end (I think it's at ~4.5v?), and reducing the gain by modifying R2... but there are some basics that I am not sure on.

The top part of the schematic uses a voltage divider to power the op-amps, it seems so that it can amplify the AC signals from 0..9v, centred on ~4.5v.

  • C10/C11 seem to be part of low-pass filters to strip AC from the power lines
    • why would we bother doing this if we assume the power supply is clean, or use a battery?
    • how can C10 filter the +9v rail without a resistor? wouldn't the 9v just snap to the power source and bypass the capacitor?
    • My calculations put the C11/R13 low-pass filter very low, at 1/3 hz. I guess that's basically strict DC?
    • assuming the op-amps take their power from the power rails, wouldn't this modify the voltage divider, making the DC offset change dynamically based on power consumption?

I assume these are basic / beginner questions for AC circuits, so I hope they're easy to answer. I appreciate all the help in advance.

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  • C10 is indeed a filter capacitor, or local bulk capacitor for energy storage. It is quite important capacitor as it does many things. The DC coming from a battery may in itself be noise-free, but real-world batteries and wires are not ideal. The battery has internal resistance due to their chemistry and construction, and 9V batteries have quite high ESR, in the order of 10 ohms. So if you have varying current consumption, it means for each 1mA of step in current, you get 10mV step in the voltage, at battery terminal. The wires from battery have some resistance too, but much smaller compared to the battery. Wires have inductance, which also limits how fast a change of current required by load can be provided from the source. Op-amps like TL07x here can drive voltage steps with a rate of 13V in 1 microsecond, and it can work with signals up to 3 MHz, so they need to have current available quickly when they need it. If the C10 is a standard electrolytic capacitor, it will have internal ESR in the order of 0.1 ohms. So for fast current needs it is 100x better than the battery, as it has much smaller ESR, and since it is closer to the load, there is less wiring inductance from the capacitor to the load. If load requires high frequency currents, the capacitor provides them, and gets charged with lower frequency currens from the battery.

-So the C10 capacitor does acts as a filter even if there is no separate resitors or inductors drawn, they are just built-in unwanted characteristics of components such as batteries and wires.

-C11 is used as a filter too. The resistor divider creates a mid-supply or half-supply reference voltage for biasing the op-amps. Very little current is drawn from it so it is not used as a power supply, so the resistors can have large values (10k) to save battery consumption. But because the resistors are large, the node has high impedace of 5k, it is susceptible to noise, and any small change in current drawn by op-amp input affects the voltage, so the capacitor is put there to lower the impedance. It is also necessary that when power is applied that the voltage smoothly rises over long time period to final value to avoid sudden snap in the audio.

-The lowpass is not simply R13 and C11, but the parallel resistance of both resistors R13 and R14 in the voltage divider, so frequency is double what you calculated (RC time constant half). 0.68 Hz.

-Op-amps take virtually no current from their inputs, the input impedance, or the nanoamps of bias it requires is not enough to significantly load the half-supply node with 5k of DC output impedance. 1 microamp of current will load it by 5 millivolts.

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  • \$\begingroup\$ Thanks for the details. As a beginner, all the sources I have read have treated power sources as perfect, so I hadn't considered (nor heard of...) ESR of a source. The comparison of source vs cap ESR helps a lot to understand. The 9v cap doesn't seem to filter that low, but I see how it can be a ready current source, and thanks for the correction on the 4.5 filter - current would flow through both so it's the combined amount. Thanks! \$\endgroup\$ – jyoung999 May 8 at 13:47
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Guitar gadgets are fun, aren't they?

C10/C11 seem to be part of low-pass filters to strip AC from the power lines why would we bother doing this if we assume the power supply is clean, or use a battery?

You answered your own question here but you don't realize it. Key word "assume". You are right, if the power source is an 'ideal voltage source' you would not need C10 or C11. But it's not ideal. There's no such thing. It can bounce around a little due to a number of reasons (example, suppose you're jamming next to a radio transmitter...). The caps make an effort to filter out noise. Capacitors are cheap insurance.

how can C10 filter the +9v rail without a resistor? wouldn't the 9v just snap to the power source and bypass the capacitor?

The 'resistor' you don't see is in inside the power source because it is not ideal ("output impedance").

My calculations put the C11/R13 low-pass filter very low, at 1/3 hz. I guess that's basically strict DC? assuming the op-amps take their power from the power rails, wouldn't this modify the voltage divider, making the DC offset change dynamically based on power consumption?

Great question and insight ---

-- there's no power being consumed on the 4.5V rail. It only goes to the input of the two opamps, and the inputs, for practical purposes, are "infinite resistance". i.e. there's no load on the 4.5V line regardless whether you're taking a smoke break or rocking to an old Iron Maiden tune.
-- C10 keeps the 9V rail steady under any reasonable dynamic conditions and C11 keeps the 4.5 rail steady, even if C10 fails to fully filter the power to DC.

The truth is, none of the rails will be truly DC ever. The trick is to get them clean enough that the noise is not objectionable. These circuits would not cut it on a NASA probe, but they work perfectly fine for guitar purposes.

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