Audio Circuit for *High* Output Impedance

Started by Michael B Allen August 11, 2009
Hi,

I need an audio circuit that can convert the very low-impedance of a
PC sound card output (<10 Ohms?) to a high impedance output (250K but
adjustable between 40K and 1M would be nice).

I'm trying to simulate an electric guitar so that I can run AC sweeps
through circuits designed to interface directly with the guitar.

Just sticking a big resistor in series with the signal doesn't work
because it also attenuates the voltage. The guitar can actually put
out about 1V if you jam on it just so. The PC seems to put out about
1V also so any attenuation would not be good.

If I amplify the signal 25x using an LM741 and then use a 270K output
impedance resistor, the voltage is about right (as observed by
comparing the PC SIN wave with the guitar using a PC oscilloscope) but
the result isn't quite right because I still can hear the PC SIN wave
with headphones whereas if I plug the guitar into the same circuit I
can't hear anything. I guess the opamp is also amplifying the current?

Any ideas?

Note that I'm just a weekend-worrier so hopefully the answer isn't to
buy some really expensive audio transformer. A discrete circuit
composed of a few transistors would be great. Can I run a transistor
in reverse? I'm not an EE if you can't tell!

Mike
Michael B Allen wrote:
> Hi, > > I need an audio circuit that can convert the very low-impedance of a > PC sound card output (<10 Ohms?) to a high impedance output (250K but > adjustable between 40K and 1M would be nice). > > I'm trying to simulate an electric guitar so that I can run AC sweeps > through circuits designed to interface directly with the guitar. > > Just sticking a big resistor in series with the signal doesn't work > because it also attenuates the voltage. The guitar can actually put > out about 1V if you jam on it just so. The PC seems to put out about > 1V also so any attenuation would not be good. > > If I amplify the signal 25x using an LM741 and then use a 270K output > impedance resistor, the voltage is about right (as observed by > comparing the PC SIN wave with the guitar using a PC oscilloscope) but > the result isn't quite right because I still can hear the PC SIN wave > with headphones whereas if I plug the guitar into the same circuit I > can't hear anything. I guess the opamp is also amplifying the current? > > Any ideas? > > Note that I'm just a weekend-worrier so hopefully the answer isn't to > buy some really expensive audio transformer. A discrete circuit > composed of a few transistors would be great. Can I run a transistor > in reverse? I'm not an EE if you can't tell! > > Mike
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: http://www.reamp.com/ 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"
> > I need an audio circuit that can convert the very low-impedance of a > PC sound card output (<10 Ohms?) to a high impedance output (250K but > adjustable between 40K and 1M would be nice). > > I'm trying to simulate an electric guitar so that I can run AC sweeps > through circuits designed to interface directly with the guitar.
** 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
On Aug 10, 9:38=A0pm, "Phil Allison" <phi...@tpg.com.au> wrote:
> "Michael B Allen" > > > > > I need an audio circuit that can convert the very low-impedance of a > > PC sound card output (<10 Ohms?) to a high impedance output (250K but > > adjustable between 40K and 1M would be nice). > > > I'm trying to simulate an electric guitar so that I can run AC sweeps > > through circuits designed to interface directly with the guitar. > > ** 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 SERIE=
S
> 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 ea=
ch
> frequency that the guitar does.
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: http://webpages.charter.net/dawill/tmoranwms

"Michael B Allen" <ioplex@gmail.com> wrote in message 
news:2c05a317-95b5-43c4-b7ab-94b6ed1e51ce@m20g2000vbp.googlegroups.com...
> Hi, > > I need an audio circuit that can convert the very low-impedance of a > PC sound card output (<10 Ohms?) to a high impedance output (250K but > adjustable between 40K and 1M would be nice). > > I'm trying to simulate an electric guitar so that I can run AC sweeps > through circuits designed to interface directly with the guitar. > > Just sticking a big resistor in series with the signal doesn't work > because it also attenuates the voltage. The guitar can actually put > out about 1V if you jam on it just so. The PC seems to put out about > 1V also so any attenuation would not be good. > > If I amplify the signal 25x using an LM741 and then use a 270K output > impedance resistor, the voltage is about right (as observed by > comparing the PC SIN wave with the guitar using a PC oscilloscope) but > the result isn't quite right because I still can hear the PC SIN wave > with headphones whereas if I plug the guitar into the same circuit I > can't hear anything. I guess the opamp is also amplifying the current? > > Any ideas? > > Note that I'm just a weekend-worrier so hopefully the answer isn't to > buy some really expensive audio transformer. A discrete circuit > composed of a few transistors would be great. Can I run a transistor > in reverse? I'm not an EE if you can't tell! > > Mike
"Michael B Allen"
"Phil Allison"

> > ** 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.
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
On Mon, 10 Aug 2009 21:26:44 -0700 (PDT), Michael B Allen
<ioplex@gmail.com> wrote:

>On Aug 10, 9:38=A0pm, "Phil Allison" <phi...@tpg.com.au> wrote: >> "Michael B Allen" >> >> >> >> > I need an audio circuit that can convert the very low-impedance of a >> > PC sound card output (<10 Ohms?) to a high impedance output (250K =
but
>> > adjustable between 40K and 1M would be nice). >> >> > I'm trying to simulate an electric guitar so that I can run AC =
sweeps
>> > through circuits designed to interface directly with the guitar. >> >> ** 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. > >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
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.
On Tue, 11 Aug 2009 21:32:38 -0700 (PDT), Michael B Allen
<ioplex@gmail.com> wrote:

>On Aug 12, 12:03=A0am, "Phil Allison" <phi...@tpg.com.au> wrote: >> "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? >> >> ** =A0Hey pal - =A0you are =A0 REFUSING to answer questions. > >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? > >> What is your damn guitar ? > >2007 Fender Standard Stratocaster Made in Mexico. > >> What pick-ups are fitted =A0? > >Stock fender single-coils. > >> What are you headphones ? > >Sony MDR-7506 but cheap headphones yield the same result. > >> Have YOU altered anything on the guitar ? > >No. I purchased it new. It's 100% stock. > >> 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.
>> >> =A0 So of course it is NOT the same as a sine wave tone. > >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
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.
JosephKK wrote:
> On Tue, 11 Aug 2009 21:32:38 -0700 (PDT), Michael B Allen > <ioplex@gmail.com> wrote: > >> On Aug 12, 12:03 am, "Phil Allison" <phi...@tpg.com.au> wrote: >>> "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. >> 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? >> >>> What is your damn guitar ? >> 2007 Fender Standard Stratocaster Made in Mexico. >> >>> What pick-ups are fitted ? >> Stock fender single-coils. >> >>> What are you headphones ? >> Sony MDR-7506 but cheap headphones yield the same result. >> >>> Have YOU altered anything on the guitar ? >> No. I purchased it new. It's 100% stock. >> >>> 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. >> 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 > > A guitar is a "high z" (about 50K) source,
Much knottier than that - they peak up close to 1M ohm. 200k is closer to ... "nominal". http://online.physics.uiuc.edu/courses/phys498pom/Lab_Handouts/Electric_Guitar_Pickup_Measurements.pdf
> 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.
A microamp isn't much current to work with there :) -- Les Cargill
"Les Cargill"
>JosephKK wrote: > >> A guitar is a "high z" (about 50K) source, > > Much knottier than that - they peak up close to 1M ohm. 200k is > closer to ... "nominal". > >
http://online.physics.uiuc.edu/courses/phys498pom/Lab_Handouts/Electric_Guitar_Pickup_Measurements.pdf ** 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