How do you calculate the output signal resistance?

[ninjaaron]

Trying to figure out how to best attenuate the frequencies I want gone, but it's hard... when you don't know how. At the moment I'm working with JFETS, but I could reasonably expect that I will want to try it with BJTs, opamps, or whatever.

I'm guessing this will have something to do with E=I*R... like probably E/I=R

But I don't know which values to plug in.

[sethj]

output of what?  an amplifier (gain stage) in an effect?

[Sir H C]

BJTs or FETs (j or not)

Whatever the load resistor can be viewed as the output resistance.

Opamps, look at what they say for the device.

[petemoore]

Trying to figure out how to best attenuate the frequencies I want gone, but it's hard... when you don't know how.
  It'd be easier to explain if I knew what the relevant frequencies are here.
  Understanding what a capacitor does [attenuate increasingly as frequency gets lower], and how that can be used to create a HF or LP filter [high pass and low pass filter] is a good start, in series with signal path...the increasing resistance of LF's...cut LF's [reduce bass], between the signal path and Gnd. the increasingly reduce HF's by blocking the LF's from being shunted to ground.
  At the moment I'm working with JFETS, but I could reasonably expect that I will want to try it with BJTs, opamps, or whatever.
  Most of these devices provide amplification over a very wide or extremely wide frequency band [for our uses], in most applications, BJT, Ge's in particular have the best chance of being lousy enough in the frequency response range to do what you might be expecting.
  First try some capacitor type filters.
  I'm guessing this will have something to do with E=I*R... like probably E/I=R
  You've got most of it, but for 'frequency' you need an 'f' [IIRC that's the abbr.] in the equations.
But I don't know which values to plug in.
  Here's a couple hints:
  Use an input capacitor socket, or splice a capacitor in the input wire [2 caps in series = a smaller uf value].
  Try a LP Filter circuit: Alligator clip >wire>cap>Pot>wire>Clip. Values?See below.. [start with .01uf and 10k, move to .1uf if that don't do it..
  See Jack Ormans AMZ calculator and notebook.
  Don't be afraid to reference materials to get better understanding from your studies at GEO. All Links/google.

[ninjaaron]

output of what?  an amplifier (gain stage) in an effect?

Yes. In my current project I'm working with cascading JFETs, and I want to know what values of resistors are needed in specific cases for effective filtering of bass frequencies using a cap in parallel with a resistor. Because the resistor is in parallel, it's not the resistance I have there that is giving me my R value to plug in.

BJTs or FETs (j or not)

Whatever the load resistor can be viewed as the output resistance.

Opamps, look at what they say for the device.

Which resistor is the 'load' resistor?

Trying to figure out how to best attenuate the frequencies I want gone, but it's hard... when you don't know how.
  It'd be easier to explain if I knew what the relevant frequencies are here.
  Understanding what a capacitor does [attenuate increasingly as frequency gets lower], and how that can be used to create a HF or LP filter [high pass and low pass filter] is a good start, in series with signal path...the increasing resistance of LF's...cut LF's [reduce bass], between the signal path and Gnd. the increasingly reduce HF's by blocking the LF's from being shunted to ground.
  At the moment I'm working with JFETS, but I could reasonably expect that I will want to try it with BJTs, opamps, or whatever.
  Most of these devices provide amplification over a very wide or extremely wide frequency band [for our uses], in most applications, BJT, Ge's in particular have the best chance of being lousy enough in the frequency response range to do what you might be expecting.
  First try some capacitor type filters.
  I'm guessing this will have something to do with E=I*R... like probably E/I=R
  You've got most of it, but for 'frequency' you need an 'f' [IIRC that's the abbr.] in the equations.
But I don't know which values to plug in.
  Here's a couple hints:
  Use an input capacitor socket, or splice a capacitor in the input wire [2 caps in series = a smaller uf value].
  Try a LP Filter circuit: Alligator clip >wire>cap>Pot>wire>Clip. Values?See below.. [start with .01uf and 10k, move to .1uf if that don't do it..
  See Jack Ormans AMZ calculator and notebook.
  Don't be afraid to reference materials to get better understanding from your studies at GEO. All Links/google.

Yeah, I know about the whole f = 1/(6.28*RC)

What I'm not so sure about is figuring out the output inpedance of my amplifier device (currently working on JFETs, but I may put some BJTs in for buffering, if it seems advantageous. The next design I have on the table is opamp based, so I will also need to know that).

Anyway, I know how to calculate the frequences, but I can't do it correctly without knowing the out Z of my device, which I don't know.

The main reason for my difficulty is that I'm trying to do a pre-clipping bass reduction without actually dropping the freqs altogether, so I'm using a cap in parallel with a resistor. I was putting this after the gain control, but I do believe it's causing a lot of interaction between the bass content and the gain settings, so I want to put a buffer before. When I put in the buffer, the filtering doesn't seem to work right, and I am guessing this is because of a significant change in impedance...

but I don't know what the impedance is!!   

[sethj]

the output impedance is what the signal can see looking back into your circuit, in parallel with what ever load impedance you have attached.  it can be a pretty big hassle to calculate it sometimes... it is frequency dependent, so you're going to want to use frequency domain analysis.

replace your capacitors with -j/(wC) and yer inductors with jwL, then look at the whole circuit as a bunch of impedances (frequency dependent resistances) and you should be able to ball-park it pretty quickly.

[R.G.]

We seem to be getting a lot of imponderables these days.

Ninja, no pro would ever do it that way.  That is because the output impedance of an amplifier is ill defined unless it has a lot of feedback. With a lot of feedback, it's forced to either very high (current source) or very low (voltage source). There are a very few instances where feedback is used to force an impedance other than nearly infinity or nearly zero, but I've encountered only a handful in over three decades of circuit design. If you do manage to come up with a design which is well enough specified so you can use the output impedance in it, this will almost certainly involve the characteristics of the JFET, bipolar, MOSFET, or opamp in the equations, and those vary all over the map, so the second one you build will be different from the first one, and from the third one.

The way this is done is to force the frequency dependencies into resistors and caps (not inductors, for cost reasons) by making the output impedance of the amplifier feeding the filter be much lower than the impedance of the filter and then inserting whatever resistance you want for source resistance as a resistor, not as some calculated output impedance of the amplifier. This is incredibly less fuss and bother, and infinitely more reliable in production.

[valdiorn]

To measure output impedance: put a sine wave through your circuit and measure the peak-to-peak voltage at the output, write the number down.
Stick a variable resistor between output and ground, something like a 100 or 1000 ohm pot. measure the peak-to-peak voltage at the output and tune the resistor until you have a signal that is half of what it was before. Now measure how much resistance you were using, that resistance equals the output impedance. Basically, half the voltage is dropping over the output resistance of the JFET gain stage (which is just an inherent feature of the circuit, not an actual, physical resistor), and the other half over the external resistor you just added (called the load)

To calculate the output impedance: I *think* it's calculated as the drain resistor in parallel with the load resistor, the load resistor being the resistance of whatever you are connecting to the output (can be another gain stage, tone control stage, another stompbox, whatever...)

How I would do it: Use larger resistances on the output, you don't really want to be driving alot of power from the output of a JFET stage anyway. You can scale a frequency selective circuit (filter) up or down by multiplying the resistance and dividing the capacitance with the same number. Pick a resistor that's large enough that the output resistance of the JFET stage won't matter (something like 100k should be sufficient), then design your frequency response from there. If that's not good enough, then just stick a buffer in there! (op-amp or a source-follower JFET stage with minimal distortion)

[frank_p]

The way this is done is to force the frequency dependencies into resistors and caps (not inductors, for cost reasons) by making the output impedance of the amplifier feeding the filter be much lower than the impedance of the filter and then inserting whatever resistance you want for source resistance as a resistor, not as some calculated output impedance of the amplifier. This is incredibly less fuss and bother, and infinitely more reliable in production.

To measure output impedance: put a sine wave through your circuit and measure the peak-to-peak voltage at the output, write the number down.
Stick a variable resistor between output and ground, something like a 100 or 1000 ohm pot. measure the peak-to-peak voltage at the output and tune the resistor until you have a signal that is half of what it was before. Now measure how much resistance you were using, that resistance equals the output impedance. Basically, half the voltage is dropping over the output resistance of the JFET gain stage (which is just an inherent feature of the circuit, not an actual, physical resistor), and the other half over the external resistor you just added (called the load)

I was trying to formulate a question but I think I was stuck with too much variables with the method of R.G. .
In the first quote R.G. emphatise on the frequency response in function of the "complex" load that is very big.  In the second quote (Valdrion) you measure the resistance for a given frequency of operation.  If you are operating at other frequecies it may give you a different resistance -in fact impedance- (Am I right ?).  So, I am a bit confused.  I think (but I am not sure) R.G. is telling to adjust the stage of the amp in front of an big "complex impedance" (so as if it were in front of a buffer stage -?- ), and you , Valdiorn, you are giving the method for measuring just the output resistance of the amp at a particular frequency (?).

So if your amp stage is in front of something different than a very big "complex" impedance (I mean a smaller "complex" impedance and you are not sure what it is), how are you going to decide something.  You make a decision on a convenient frequency responce curve to have and you vary the resistor and the capacitor until it gives you that "optimal" response curve ?  I mean, you make multiples frequecy sweep and vary the resistor (and fixed value cap), and start again with the cap (and fixed resistor value found) ?  I think there is something that I am not catching. Or I did not read well. How do you deal with all the variables ?