Electronic – the output impedance of a current source be much greater than the input impedance of the load

^{simulate this circuit – Schematic created using CircuitLab}

This circuit is a bit similar with the well known Widlar current source. I only give the small signal diagram, and my result (the calculation procedure is not difficult, but it will be painful to write it here). If you think the circuit is right, then you can do some calculations with it, and compare with my result.

$$ R_{o} = r_{o}[1+(R_{e}||(r_{\pi} + R_{b}||(r_{d1}+r_{d2})))(g_{m}+1/r_{o})] $$

If \$(1/r_{o}) \ll g_{m}\$, then

$$ R_{o} \approx r_{o}(1+g_{m}(R_{e}||(r_{\pi} + R_{b}||(r_{d1}+r_{d2})))) $$

From the equation of \$R_{o}\$, you can analysis how the components can affect your output impedance.

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.)