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I'm building a module for a modular synthesizer, according to the Eurorack standard. So this is a module intended to be connected to other modules via patch cables.

While the above standard page does not specify an output impedance for the module outputs, it seems that in general this is in the region \$100\Omega - 1k\Omega\$. Input impedance is specified as \$100\mathrm{k}\Omega\$. As my final output stage is an opamp-based amplifier, I need to specifically drop the impedance, as the op-amp itself would give a very low output impedance. Further, since any outputs and inputs can be connected by the user, I should expect that the output may be shorted to any voltage in the range from \$-12\mathrm{V}\$ to \$+12\mathrm{V}\$ (system power rails) by the user; for example, it is possible for the user to connect two outputs together, and while that won't do anything meaningful, the modules should not be damaged.

Online, I can find two different ways to do this. The obvious op-amp circuit followed by a resistor:

schematic

simulate this circuit – Schematic created using CircuitLab

or putting the resistor in the feedback loop:

schematic

simulate this circuit

In both cases, \$R3\$ sets the output impedance.

The rationale for the latter is that since the feedback is taken from the actual output node, the output is effectively very low impedance under normal conditions, but the resistor nonetheless protects too much current from being drawn: if the power rails are \$\pm 12 \mathrm{V}\$, then at most \$24\mathrm{V}/R3\ = 24\mathrm{mA}\$ (for \$R3 = 1\mathrm{k}\Omega\$) can be drawn before the op-amp saturates (although in practice, the TL07x will saturate earlier, as it can't output that much current).

So there's two related questions here

  1. Is the latter way indeed recommended, and safe for my module and any other (reasonably designed) module it may be connected to? The reason why I have doubts is that looking at modules in the wild, the first way seems to be much more common, so I'm thinking there's some downside which I don't realize. On the other hand, just a direct output from the op-amp seems to be quite common too...
  2. In the latter case \$R3\$ acts really as a current limiter, so I'm inclined to actually use a much bigger resistance, say around \$10\mathrm{k}\Omega\$, so that the maximum current is set by the resistor, not the op-amp output capability. Is this a reasonable thing to do?

Update:

To answer Olin's missing spec: users would not assume that passive mixing by shorting outputs would work (and, indeed, the output impedance of other modules varies, so it is not reliable). So basically any behavior that doesn't damage the modules is acceptable.

On the other hand, since the output of this module is not really usable as a control voltage anyway (due to the nature of the module), a slight loss due to the outside-the-loop resistor doesn't really matter much; for audio that's just a small drop in volume.

Finally, reading this thread, I notice that one potential problem with the latter option is that the op-amp needs to drive any output capacitance directly. In general, modular patch cables are quite short, but there are also wall-sized modulars which may use longer patch cables.

In the end, I think I'm leaning toward the first option, primarily to avoid any problems with cable capacitance, and since the downside (small signal loss) is not really important. But any thoughts or insights are still welcome!

Update 2:

The application note linked by JRE clears things up further, when it comes to capacitive loads: the second circuit in this question is the same as the last one in the app note, except for the capacitor \$C_f\$ in the loop. The app note tells us that this configuration is good for driving a capacitive load, but only if the load capacitance \$C_L\$ is known.

So the conclusion of the previous update still holds, the first circuit is a better bet when we don't know the load.

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    \$\begingroup\$ To add a point to the answers below: take care not to use circuit 1 on the keyboard 1V/octave control voltage. Any error in this signal caused by loading will give a progressive flattening of the pitch as you move up the keyboard. \$\endgroup\$ – Transistor Dec 19 '17 at 18:41
  • \$\begingroup\$ I'm pretty sure that circuit 2 will driveva capacitive load better than circuit 1. \$\endgroup\$ – JRE Dec 19 '17 at 19:16
  • \$\begingroup\$ @JRE can you elaborate a little? The Muffwiggler thread linked to in the question claims exactly the opposite. \$\endgroup\$ – Timo Dec 19 '17 at 20:32
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    \$\begingroup\$ Because of this. I tend to put more trust in information from Analog than from some random guys on a forum called "Muffwiggler." \$\endgroup\$ – JRE Dec 19 '17 at 20:42
  • \$\begingroup\$ Note the options for driving a capacitive load with an opamp. One of them is a dead ringer for your circuit 2. \$\endgroup\$ – JRE Dec 19 '17 at 20:43
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Which circuit you want depends on specifications you haven't told us. The important question is what exactly is supposed to happen when a user connects the outputs of two of these things together?

If they really aren't intended to be connected together, then the resistor is just for protection. In that case, your second circuit is better. You set the resistor value to not exceed the opamp's output current capability under the worst case conditions.

If multiple modules are intended to be tied together and you're supposed to get the average result, then you need to use your first circuit. This would be the case, for example, if it's allowed within the specs to connect the left and right channels together to get mono. In that case, the resistor needs to be whatever the specified output impedance of each module is supposed to be. If they are supposed to average by shorting, then each needs to have a defined and controlled impedance. That impedance must be specified by the standard.

For example, if you were to pick 1 kΩ and someone else picked 10 kΩ, then connecting the two modules wouldn't yield the average as intended. The resulting signal would be 10/11 parts from your module and 1/11 parts from the other module. For the averaging scheme to work, all the impedances need to be equal, and therefore agreed upon ahead of time.

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    \$\begingroup\$ I've never seen a modular synth that is stereo, but they might exist. Generally, modular synth signals are only combined via an active mixer, never using a passive resistor network. One of the biggest risks for this output is that it would be accidentally connected to another output. That other output could be AC (up to audio rates) or DC with a possible voltage range between -10 and +10 and a similar output impedance to this output. The output could also be shorted to ground accidentally. Accidental connection to a very low impedance input is unlikely, although maybe possible. \$\endgroup\$ – Todd Wilcox Dec 19 '17 at 17:25
  • \$\begingroup\$ I guess your second paragraph is the answer in this case. Two modular synth outputs are not intended to be connected together. \$\endgroup\$ – Todd Wilcox Dec 19 '17 at 17:30
  • \$\begingroup\$ I added an update addressing the specification. \$\endgroup\$ – Timo Dec 19 '17 at 18:30
  • \$\begingroup\$ @Timo: Thanks for the update. I think your conclusion is good. \$\endgroup\$ – Olin Lathrop Dec 19 '17 at 18:38
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The two circuits respond differently to an external load to GND. Using an extreme case to demonstrate, assume a 1 K resistor to GND as the load. In the first circuit, the opamp output pin voltage does not change, the circuit gain does not change, but the external voltage decreases 50%.

In the second circuit, you now have a 50% attenuator inside the feedback loop. Before, the right end of R2 (68K) was connected to a zero ohm voltage source. Now it is connected to a Thevenin equivalent voltage that is 1/2 the opamp output voltage, through a equivalent 500 ohm resistor.

So the feedback resistor value is different, which changes the circuit gain, and the feedback voltage is very different, which really changes the gain. The opamp output pin voltage will double (approximately) as it tries to close the loop. The external output voltage will not decrease much until the opamp saturates. Or something like that.

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  • \$\begingroup\$ The intended load is more on the order 100k Ohms, as mentioned in the quesiton. \$\endgroup\$ – Todd Wilcox Dec 19 '17 at 17:28
  • \$\begingroup\$ I used "an extreme case" to make the differences easier to see and discuss. Also, while the intended load is 100K, part of the question was about unintended connections. R3 protects the output in both circuits, but the outputs respond very differently. \$\endgroup\$ – AnalogKid Dec 19 '17 at 18:20
  • \$\begingroup\$ Oh I see. I didn't see the "extreme case". I would say a more extreme case worth worrying about is short to ground. \$\endgroup\$ – Todd Wilcox Dec 19 '17 at 18:33
  • \$\begingroup\$ With a short to GND, one opamp won't care. The other will hard-saturate, but not overcurrent. \$\endgroup\$ – AnalogKid Dec 19 '17 at 20:24
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    \$\begingroup\$ Short to ground is actually going to happen basically every time a lead is inserted, since in Eurorack the connections are on TS minijacks. The tip, which carries the signal, shorts to ground before making contact with the signal in the jack when inserting. \$\endgroup\$ – Timo Dec 20 '17 at 9:51
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Most Op Amps have built-in short cct. protection already which when you use negative feedback is the 2nd method. This is why OA’s have such a low current limit .

The down side is capacitive cable load Ic=CdV/dt requires you to compute supply voltage and. Here to limit to guarantee the design avoids a saturation or slew rate limiting. Thus it is unwise to raise the drive above Zo of the cable.

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I would do the first one because it will load the op amp independently when it is summed into another effect/stage when others are connected in parallel. The added side effect of providing a load on output shorts (due to plugging the output in an out of patch points) is welcomed too. I wouldn't worry about the impedance because it will go back into a hi-z unbalanced input at less than 3 meters.

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