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I'm putting together a kind-of sort-of theremin (this one in particular http://www.popsci.com/diy/article/2008-04/build-pocket-theremin-cheap) and I'm looking to make a bit of a change to it.

What I would like to do is add in an output jack so I can connect to a small amplifier. I have a 1W Marshall mini guitar amp to play around with. My question is how big of a resistor should I be looking to put in series with the output jack? From what I've been finding, guitars output around (1 * 10^-7A) and 100mV to 1V. So if I wanted to drop the 9V of the theremin (which will end up being a bit less after all the circuitry and the volume knob) to around maybe .25V I think I'd be looking at...

\$R = \dfrac{V}{I} = \dfrac{.25V}{1\cdot10^{-7}A} = 2.5M\Omega \$

I really just need someone to confirm this for me and let me know if I'm proceeding properly here. It's been a while since I've done any circuit analysis and I don't think nodal analysis would work here so I'm a bit in the dark. I don't know if impedance matching is going to be an issue here either, so if someone could point me in the right direction for that as well, that would be great!

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up vote 1 down vote accepted

In modern audio circuits, no one worries about impedance matching. The usual technique is impedance bridging, which is in short, making the source impedances small compared to the input impedances. This minimizes the current the source must drive. It also makes parasitic capacitances in the cable and in the circuit color the sound less.

Consequently, you don't really need a resistor. There are two things that could be a problem, though:

  1. This theremin could be much louder than a guitar

  2. When the outputs become shorted, you could break whatever is driving them

On audio equipment, the outputs will become shorted. Sometimes this happens when the plug is being inserted into a jack, or someone might drop the cable on something metal. You should look at the datasheet for the driving element (the 555 IC) and see what what it says about problem #2. Some ICs say explicitly in the datasheet that they can deal with a short circuit on the output (for example, the TL072 op-amp). If they don't, they probably say what the maximum output current is. If the output is shorted to ground, a very large current could flow, defined by Ohm's law: \$I = E/R\$, where \$E\$ is the ouput voltage, and \$R\$ is the (small) resistance of the wires connecting the output jack. Find a value for \$R\$ such that the current can't be more than the rated maximum current for the 555, put it in series with the output, and you should be safe.

So what about problem #1? Well, maybe it's not a problem at all, in which case you are set. If not, a series resistor isn't going to work very well to reduce the voltage, and consequently the volume of your theremin. The reason is that your guitar amplifier is a high-impedance input. A high-impedance input means low-current. Consider Ohm's law: \$ E=IR \$. If \$I\$ is small, it doesn't much matter what \$R\$ is, \$E\$ will also be small. Here, \$E\$ represents the voltage, and thus the volume, lost over the resistor. Incidentally, this is precisely why impedance bridging is desirable.

The solution is a voltage divider. This will increase the output impedance of your theremin, but since the guitar amplifier presents an even higher impedance input, this shouldn't matter much. Also, this solves problem #2. If you want to adjust the volume, you can use an adjustable voltage divider, otherwise known as a potentiometer. The \$5k\Omega\$ pot already in the circuit should do nicely.

DC bias

There is one more potential issue: the output of this device will have a non-zero DC bias. That is, the average voltage it outputs will not be 0V. Since the 555 will either output 9V or 0V, you can see there's no way this can average 0V. Technically, this is a problem even if you don't connect it to a guitar amplifier: this DC bias serves to push the speaker's cone away from the center, on average, which maybe reduces the dynamic range and increases the distortion. But this isn't Stradivari violin, so probably no one cares.

The usual solution for audio circuits is capacitive coupling. Essentially, put a capacitor in series with the output. Your guitar amplifier likely has capacitive coupling on the inputs already, so it may not matter. And, if you have limited the maximum output current, it's unlikely you can do much damage even if this is not the case. So, maybe you want to keep things simple anyway. In any case, this would be best practice for a quality product.

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Hmm, ok. So from the datasheet, it looks like the max current of the 555 is 15mA. By Ohm's Law I'll be good with roughly a 600Ohm resistor? As for the #2, There is a 5k potentiometer for volume in the circuit diagram already. Will that suffice? – MGZero Jan 18 '13 at 0:12
@MGZero \$600\Omega\$ sounds correct. As for the other issues, I guess I should have read the schematic more carefully. See edits. – Phil Frost Jan 18 '13 at 0:27
Ok great, you've been very helpful! One last question, that 600Ω resistor. Does that go on the output of 555 #1 or 555 #2? #2 would be the one connected to the 5K potentiometer and speaker. – MGZero Jan 18 '13 at 0:40
@MGZero between pin 3 of U2, between U2 and the potentiometer. – Phil Frost Jan 18 '13 at 0:47
Thank you again! – MGZero Jan 18 '13 at 1:32

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