Last year I was experimenting with a design for an audio distortion to be used on line level musical instruments (+4dBu with Zout 100-600Ω by my findings) without the knowledge of impedance, its importance, and how to "use" it. After learning of impedance bridging, I returned to the design and saw that, although it worked as I intended, my impedance was "backward," having a lower Zin (approx 1.7kΩ), and higher Zout (8.2kΩ).

So, concerned about damaging my equipment, I did quite a bit of research, and found lots of information on how impedance affects the signal, and that a low Z input causes loading on the source, but nothing really clear on how the source itself is affected.

So my questions are:

  • If one were to hook up two devices in this fashion - feeding an audio signal into a load with equal or lower impedance than the source - are there any long term adverse effects on the equipment itself?

  • Understanding that 10x the source is the rule of thumb, and that something like 4Ω is nearly a short, what ratio to the source would begin to be damaging for a longer duration than something like a plug insertion?

I've learned to adjust my future designs, but am still curious.

Thank You.

Note: This question has been completely rewritten, so for some of the answers/comments to make sense, see the edit history, where you may also find more detailed information on my specific situation, which upon further thought, only brought about the curiosity that is more simply state above, and is mostly irrelevant.

Thanks again.

  • \$\begingroup\$ @Olin Nope, just the first. The rest was to attract some of the famous EE.SE abuse. I actaully spend time trying to organize my questions, provide adequate information, and move (possibly helpful) "babble" after the data I believe actually matters - just in case it would help a volunteer help me. Thank you for your input, and taking the time to let me know that will be of no help. Very friendly and professional. \$\endgroup\$
    – Jay
    Jul 13, 2017 at 18:36
  • \$\begingroup\$ Do you really think that the section of missing components might be so important you want to keep it secret. For this reason alone I stopped reading and writing...... \$\endgroup\$
    – Andy aka
    Jul 13, 2017 at 18:48
  • \$\begingroup\$ Hi @Andyaka. It's more a matter of unimportance, discretion, and a few lesser factors. I've seen a few "secret schematic" Qs that went unchallenged. It annoys me too, but as an intellectual property holder in other areas, I do understand. I did post an analog of my I/O for reference/discussion. Figured that was ample, since the question isn't about the stuff in between. This is an application with hints of (limited) commercial viability. If so, I'd like to benefit before the plastic clone makers start making plastic clones. If not, heck, I'll share it with the world. I'll edit later. Thanks. \$\endgroup\$
    – Jay
    Jul 13, 2017 at 21:26
  • 1
    \$\begingroup\$ The impedance at the emitter affects the impedance looking into the base. There, that's my last word on this. \$\endgroup\$
    – Andy aka
    Jul 13, 2017 at 21:29
  • \$\begingroup\$ @Andyaka I was actually aware of that, but thank you non-the-less for your help. It really is appreciated. Edit: I just thought it was negligible enough for rough figures. \$\endgroup\$
    – Jay
    Jul 13, 2017 at 21:36

3 Answers 3


If I play into this thing I created for longer periods than the comparatively brief audio tests I've performed, will it burn up the instrument's output?

It can't be said in general. It depends on the design of the output.

But if by “instrument” you mean something like an electric guitar of what I understand is the basic design (no internal power source, just pickup coils and passive components), I would expect that there is just not enough power available to damage any of the components even if short-circuited (i.e. connected to an extremely low input impedance), because the power generated by the pickup coils is very small and non-semiconductor components (switches, capacitors, coils) are going to be robust enough to handle the maximum voltage/current simply because you'd have to try hard to make one that isn't.

Also, in practice:

  • It is also very common to have effect boxes that deliberately load the input for the sake of the effects on the sound. Thus I would expect even an instrument with internal amplification to be designed with a robust output circuit.

  • Every TS/TRS plug (a.k.a 1/4" or 3.5 mm) and jack short contacts while being plugged or unplugged (which is why you hear multiple pops when doing so). Therefore, devices which use such connectors should be designed to tolerate their outputs being shorted, whether they are instrument, line-level, or headphone outputs — and you've got something quite far from a short.

If the output is an active device then it is quite possible for its output to be designed such that it would destroy itself with excessive current, but that is not a good design for a line output since incidental misconnections happen. Of course, there's no reason a device can't be badly designed while also being expensive to fix.

Here's a suggestion to be safer: modify your circuit so that it has a 100 Ω series resistor on the input. I just picked that value out of mostly thin air, but it is higher than the input impedances of a lot of headphones. So you can be assured that if the thing you're connecting it to wouldn't be damaged by headphones, they won't likely be damaged by your circuit either.

(Disclaimer: I am not a musician or an analog audio designer. This is just tidbits I've picked up. You should maybe wait for a second opinion.)

Will the high Zout cause my device to destroy itself?

If it would, then your device would also destroy itself if it was powered on while the input is unplugged (infinite impedance). You've probably done that already.

There are things that can be damaged that way: for example, a 'floating' input picking up random electromagnetic noise can damage digital logic circuits expecting a robust 1 or 0 (see "shoot-through"), or in general cause whatever the input is controlling to over-exert itself by trying to respond to noise.

(And an output can possibly require a matched load, but that is more often a matter for RF circuits than audio-frequency ones.)

  • \$\begingroup\$ Thanks for reading my "babble" Kevin. That pretty much answers my question. The intention is for line-level instruments - primarily synths - and there is no logic circuitry present. All analog. In bullet one, did you mean deliberately load the output? That's my concern - (over)loading the outputs of either device to the point of damage - which, from your explanation, it sounds like it will not. Thanks again. \$\endgroup\$
    – Jay
    Jul 13, 2017 at 17:58
  • 1
    \$\begingroup\$ @Jay I've added a few more words on the subject of active outputs, and made a concrete suggestion: add some resistance to the input. \$\endgroup\$
    – Kevin Reid
    Jul 13, 2017 at 18:06

Impedance matching costs you 50% of the input voltage. Don't waste your signal.


Most opamps will have output current limiting built-in, so they can withstand shorts on the output. However, they will heat up.

If our opamp has thermal limiting built-in (or datasheet specifies it will withstand shorts all day) then we're good.

Some opamps like OPA1642 specify "output short circuit: continuous" in the max ratings... but only for one opamp in the package!


  • A SOIC opamp has a thermal resistance Rthja=160°C/W.
  • Assume 50°C air in the box
  • Assume max 130°C junction temp
  • Max dissipated power is (130-50)/160 = 0.5W

Now we have 2 opamps, so that's 0.25W per opamp. With a 15V supply, this corresponds to 16mA into a short. Most opamps can output more current, so if they don't have thermal limiting then could fry if shorted for too long.

Let's add a resistor R on the output. It is always a good idea to do so anyway, to isolate the opamp from the cable's capacitive load, which will reduce its phase margin and possibly make it unstable. 50-100 ohms is a good starting point.

Let's use a RC4580 opamp and R=100 ohms on the output. A quick look at the datasheet page 5 "Output Voltage Swing vs Output Current" reveals that in a 100 ohms load, we'll get about 60-70mA max output current.

  • There will be 7V on the resistor, it will dissipate 0.5W... better not use a 0805 SMD resistor then...
  • The opamp will dissipate 0.5W (per opamp). Still too much!

Now, if we only expect to output a maximum signal level of, say, 3V then 100 ohms would be okay. Same if it is powered from +5/-5V rails.

If there is a DC coupling cap on the output, then we're also safe, since it will limit the duration of the high-dissipation event.

But if we use the full output swing of the opamp, have high supplies, have no coupling cap, and expect shorts, a higher resistor would protect it better. Say, 330 ohms would be a nice value.


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