Audio Circuit for *High* Output Impedance

One approach is a passive DI box "backwards" - you send signal in the XLR, and hook the 1/4" ( which is now the output ) to the amp.

These folks:

specialize in this, called "reamping". Since the amp I use has a solid-state input, I usually don't bother with anything that fancy.

-- Les Cargill

"Michael B Allen"

** Electric guitars have a complex output impedance that varies with frequency, the type of pick-ups used and control settings.

Best way by FAR to is to simply place the guitar's output signal in SERIES with your sweep generator by connecting the earth side of the jack to the sweep output.

The signal in the jack tip will than have the same source impedance at each frequency that the guitar does.

...... Phil

S

ch

Wow. This is a great solution.

However, why is it that when I connect headphones to the jack tip, I can still hear the signal? Using my sound card oscilloscope I calibrated the generator level to be roughly the same amplitude as the guitar signal (actually less). It is significantly attenuated but I can hear it fairly clearly.

If the output voltage and output impedance of the above guitar-in- series configuration are the same as that of the guitar by itself, then shouldn't that be sufficient to model the output characteristics of the guitar by itself and thus be too weak to drive the headphones (which is the case when the headphones are plugged directly into the guitar)?

Mike

Err, so what voltage does the guitar put out, into the SAME load impedance as you are using to test here? 'Cuz if it's a low load, no shit you'll have a small signal..

Tim

-- Deep Friar: a very philosophical monk. Website:

"Michael B Allen" "Phil Allison"

Wow. This is a great solution.

However, why is it that when I connect headphones to the jack tip, I can still hear the signal? Using my sound card oscilloscope I calibrated the generator level to be roughly the same amplitude as the guitar signal (actually less). It is significantly attenuated but I can hear it fairly clearly.

** Not that surprising to me.

If the output voltage and output impedance of the above guitar-in- series configuration are the same as that of the guitar by itself, then shouldn't that be sufficient to model the output characteristics of the guitar by itself and thus be too weak to drive the headphones (which is the case when the headphones are plugged directly into the guitar)?

** Got no idea what sort of guitar or pick-ups you have.

Maybe try various resistor values in place of the guitar to see what sort of impedance you are dealing with.

Might be much less than you think.

...... Phil

Someplace on the net I saw a gizmo for sale which is nothing more than a coil of wire placed over the pickup which induces a signal in the pickup by transformer action. You might want to search for a pickup sweeper or anything you can think of along those lines for a DIY solution. bg

Try this page

by

Yes, I saw that [1]. It is interesting and an obviously a more complete and fairly elegant tool chain. But even though I am not exactly poor, I cannot justify buying such a thing as I really am only doing this for fun. I would sooner try to build one of those gizmos and understand it rather than just getting something to work.

Unfortunately impedance is still somewhat of a mystery to me. It seems matching the voltage, impedance and frequency is, by itself, not sufficient to model the unamplified signal emitted by a passive electric guitar. If I plug the headphones directly into the guitar, I cannot hear anything. But with Phil's configuration, I can hear the PC driven signal in the headphones. Why? According to the oscilloscope the voltage is the same (although because it's sound-card based I cannot determine what the actual voltage is). Is there a difference in current?

I think the PC signal in series with the guitar is the basis for a good solution because it changes the impedance with the change in frequency without changing the amplitude of the signal. But, by itself, it just doesn't quite model the unamplified output of an electric guitar. That is what I need to figure out how to do.

Mike

[1]

a

p by

You could use Audacity to record the output of the guitar while being played. Then you'd have a record of a broadband signal, within which any desired frequency could be found by Fourier analysis. You could use Scilab to perform the FFT, and you'd get a graphical output which would amount to the same thing as a sweep. You'd measure both input and output, and later you'd correlate these signals to obtain the characteristics of the unit under test. "Audacity" and "Scilab" are freeware which you can download and use with minimal fuss. Mark

"Michael B Allen"

Unfortunately impedance is still somewhat of a mystery to me. It seems matching the voltage, impedance and frequency is, by itself, not sufficient to model the unamplified signal emitted by a passive electric guitar. If I plug the headphones directly into the guitar, I cannot hear anything. But with Phil's configuration, I can hear the PC driven signal in the headphones. Why?

** Hey pal - you are REFUSING to answer questions.

What is your damn guitar ?

What pick-ups are fitted ?

What are you headphones ?

Have YOU altered anything on the guitar ?

I think the PC signal in series with the guitar is the basis for a good solution because it changes the impedance with the change in frequency without changing the amplitude of the signal. But, by itself, it just doesn't quite model the unamplified output of an electric guitar.

** A guitar has SIX strings that produce six notes with lots of harmonics.

So of course it is NOT the same as a sine wave tone.

..... Phil

Ahh, I know that when I see an arrogant and rude answer like this you either know what you're talking about or your seeking attention from people who do. Either way, it means someone here probably knows something about electronics which is a good sign. The question is, will you get over being called a dick and still answer my questions?

2007 Fender Standard Stratocaster Made in Mexico.

Stock fender single-coils.

Sony MDR-7506 but cheap headphones yield the same result.

No. I purchased it new. It's 100% stock.

.

Ok. So there's more power in a pure sine wave as opposed to a complex wave form. That makes sense. You may be a dick but you're making up for it.

Mike

Michael B Allen wrote in message ... On Aug 11, 2:09 pm, "bg" wrote:

by

Yes, I saw that [1]. It is interesting and an obviously a more complete and fairly elegant tool chain. But even though I am not exactly poor, I cannot justify buying such a thing as I really am only doing this for fun. I would sooner try to build one of those gizmos and understand it rather than just getting something to work.

Unfortunately impedance is still somewhat of a mystery to me. It seems matching the voltage, impedance and frequency is, by itself, not sufficient to model the unamplified signal emitted by a passive electric guitar. If I plug the headphones directly into the guitar, I cannot hear anything. But with Phil's configuration, I can hear the PC driven signal in the headphones. Why? According to the oscilloscope the voltage is the same (although because it's sound-card based I cannot determine what the actual voltage is). Is there a difference in current?

I think the PC signal in series with the guitar is the basis for a good solution because it changes the impedance with the change in frequency without changing the amplitude of the signal. But, by itself, it just doesn't quite model the unamplified output of an electric guitar. That is what I need to figure out how to do.

Mike

[1]

The strings act like a moveable core. The magnetism from the pickup enters the string which causes the flux to change. If the flux change remained constant, there would be a 6db per octave rise in output voltage from the pickup. You can use any simulation software to hang pots and caps from the pickup to see the result. You can also drive constant current through a wire or coil in proximity to the pickup to generate an output and measure it. The disadvantage to any of this, is that it does not give you an output any where near the real world. The string's effects on the flux changes is very complex and nowhere near a repeatable standard. The gauge of the string has an effect. The proximity of the pickup to the string has an effect. How you pluck the string has an effect . The position of the pickup along the length of the string has an effect. And so on ------- All of these varibles add up to a situation that make it nearly impossible to electronically inject something that will resemble someone actually playing the guitar. But all is not lost. As I mentioned, a wire across the pickup fed with constant current will generate an output from the pickup. You can then analyse the effects of tone pots and caps, cable capacitance, etc. The pickup will output a 6db rise per octave. You will see any resonate peaks, or signal loss from loading , but what you won't see is the combined frequency response of the string and electronics. Do a google search for guitar pickup frequency response - there is plenty of info on how to do this. BTW - a simple audio power amp driving an 8 or 4 ohm resistor will provide constant current through a wire. If there isn't enough current in the wire to get a decent output when it is held in proximity to the pickup, add some turns to the wire. You can in fact add alot of turns to the wire before it develops enough inductance to be an issue.

Just remember, generosity tends to be one sided around here.

It's highly unlikely you need to bother.

Most audio circuits are 'voltage matched' these days and what you suggest would probably not be required ( subject to caveats ).

Just try a direct connection and see if it works. Guitar pickups don't have THAT high an output Z anyway ( they're inductive so a simple resistive figure isn't accurate but try 10k if you feel you need it ).

Graham

Graham

-- due to the hugely increased level of spam please make the obvious adjustment to my email address

sweeps

SERIES

the

each

Not at all. The output from a guitar is actually pretty good. The Navy still uses sound powered 'phones. They have been around for a hell of a long time. They still work well at high microvolt levels. =20

Shucks, just perform the experiment: hook the guitar directly to the headphones, and lay into it.

harmonics.

A guitar is a "high z" (about 50K) source, most headphones are "Lo z" (about 250 ohm) to speaker z (2 to 32 ohms). Try inserting a microphone high z to low z transformer in between.

Much knottier than that - they peak up close to 1M ohm. 200k is closer to ... "nominal".

A microamp isn't much current to work with there :)

-- Les Cargill

** Shame that is 100% irrelevant to how a REAL guitar behaves.

Any REAL guitar has several pickups, various control pots and switches with associated caps and a long capacitive lead connecting it to the amp input. The combined effect of these additional LOADING components means the source impedance seen by the amp never rises above 100kohms in practice.

The average value over the usable range is more like 20kohms.

The value over the range up to 1 kHz is more like 5kohms - ie the DC resistance of the coils.

Even adding a mere 500pF in parallel with any of the pickups shown in that idiotic physics link will drastically alter the numbers.

Pisshead.

.... Phil

If they're stereophones, then you'll have to dick around with the phone plug so that both left and right are making contact with the one output of the gitbox.

Cheers! Rich