Amp In A Box input & output impedances

[POTL]

I built it with a switchable output buffer, in principle the effect on the sound is almost not noticeable, I will leave the final version without buffer, I'll make the buffer separately =)
Thank you for helping everyone who responded!

[POTL]

OK one more question
Input impedans on this schematic is 500K(R3//R4) or 330K (R2//R3//R4).
Or R2 is not taken into account?
Or R2 it Input impedance?

[POTL]

I found the answer
On electrosmash website
The value will be 330K
Correct if wrong =)

[Phoenix]

Input impedance is ((R4+R5+R19)||R3)||R2 = 333k898 or ~330k.

[antonis]

Το extend Greg's analysis a little further:

We can also parallel C3 is series with C4//R5 in series with parallel combination of R8 & R7 is series with R10+R20 and so on...
(but obviously, it should be an overkill..)

[POTL]

Phoenix and Antonis Thanks

Yeach and one more question& About output buffer.


I think that setting the buffer would be a good idea if you set the buffer after the potentiometer of the volume, to avoid its effect on the output impedance
I like the scheme of Jack Orman
It has an output resistance of several hundred ohm
But whether it is necessary to put a shunting resistor (since the volume knob will be up to the buffer,
Or a resistor at the output of the circuit is optional?
Or I can put a resistor (say 1K) in series with the capacitor without connecting it to ground (like in BOSS buffers)

[antonis]

Yeach and one more question& About output buffer.


I think that setting the buffer would be a good idea if you set the buffer after the potentiometer of the volume, to avoid its effect on the output impedance
I like the scheme of Jack Orman
It has an output resistance of several hundred ohm
But whether it is necessary to put a shunting resistor (since the volume knob will be up to the buffer,
Or a resistor at the output of the circuit is optional?
Or I can put a resistor (say 1K) in series with the capacitor without connecting it to ground (like in BOSS buffers)

Yes to all... 

Jack's buffer - as it is - has an output impedance of about Source resistor value..

Placing a shunt resistor usually lowers total output impedance..
(as seen from whatever is driven by the buffer - but it also "loads" Source in AC in predictable way..)

It could be optionall but in case you place one there you have to take in mind the HPF created by out cap and shunt resistor..

You can also place a resistor in series with the capacitor if you worry about the current drown by the next stage..

[POTL]

Yeach and one more question& About output buffer.


I think that setting the buffer would be a good idea if you set the buffer after the potentiometer of the volume, to avoid its effect on the output impedance
I like the scheme of Jack Orman
It has an output resistance of several hundred ohm
But whether it is necessary to put a shunting resistor (since the volume knob will be up to the buffer,
Or a resistor at the output of the circuit is optional?
Or I can put a resistor (say 1K) in series with the capacitor without connecting it to ground (like in BOSS buffers)

Yes to all... 

Jack's buffer - as it is - has an output impedance of about Source resistor value..

Placing a shunt resistor usually lowers total output impedance..
(as seen from whatever is driven by the buffer - but it also "loads" Source in AC in predictable way..)

It could be optionall but in case you place one there you have to take in mind the HPF created by out cap and shunt resistor..

You can also place a resistor in series with the capacitor if you worry about the current drown by the next stage..

Thank you
1) As I understand, the default output impedance is 2K7
2) I can add a shunt resistor and the output impedance will be equal to the parallel connection of the resistors, I understood correctly? Example RS 2K7 // RC 1K8 = 1K Impedance
At the same time, a frequency filter is formed in which I can calculate or pick up a capacitor by ear (this is not important).
3) I did not quite understand what would happen if I added a resistor to a series with a capacitor? Rather, how will the impedance be calculated? Is the impedance equal to this added resistor or the sum of the resistor at the source or will it also be the parallel value of the resistor at the source and the added resistor?

[Phoenix]

Actually, the output impedance of that buffer, if using the same BS170 used elsewhere, is ~5Ω. Output impedance of a source follower is ≈1/gm BS170 has a typical transconductance of 200mS/mmho, so that's 1/0.2=5. That's in parallel with the source resistor though, so if you need to calculate with greater precision, it's Rs/(gm.Rs+1) = 2700/(0.2×2700+1) = 4.99.....Ω small enough a difference to ignore.

[POTL]

I here invented a hybrid Boss and Jack Orman
It has a filter similar to BOSS / Ibanez / MXR
Input Impedance 1M and Output Impedance 833ohm
If there is a VR in the circuit, then you can delete R2 and R1 connect to VR


If you delete C2 and R4, then the output impedance is 900 ohms
But will this be the correct scheme?

[Phoenix]

I here invented a hybrid Boss and Jack Orman
It has a filter similar to BOSS / Ibanez / MXR
Input Impedance 1M and Output Impedance 833ohm
If there is a VR in the circuit, then you can delete R2 and R1 connect to VR


If you delete C2 and R4, then the output impedance is 900 ohms
But will this be the correct scheme?

Input impedance as drawn is R1||R2 = 2M2||2M2 = 1M1.

No NEED to change the biasing arrangement if you're adding a pot on the input, voltage divider like shown is fine, but a voltage reference/VR is always better because it is decoupled from the Vcc and any noise present there. It would be worth considering changing to a Vr biasing arrangement regardless of whether you plan to use a pot.

Output impedance is >1k, something like 1k01 (dependant on transistor choice) because R5 is in series with the output.

C2 has little effect on output impedance until you get to low frequencies, just pick your value for the heaviest load that you expect might be connected (and remember that R4+R5 is in parallel with that load), and the lowest frequency of interest. Say you expect the heaviest load to be 10k, then that's (10k+1k)||10k = 5k238. With C2 at 10u, the -3dB point is ~3Hz, well below any frequency of interest. You could change C2 to 1u and use a film cap, and change R4 to 100k or even 1Meg, and then with a 10k load, your -3dB point would still be ~18Hz.

Removing C2 and R4 does not change the output impedance of the buffer, because they form part of the LOAD, not part of the buffer itself. The loading that they put on the stage will alter its gain (A<1), but unless you choose a very heavy load, it will not be significant.

[POTL]

Then I think it makes sense to replace the capacitor 10u by 100n, and R4 to 1M
The filter will drop the same, and the impedance will be approximately 900 ohms (the online calculator says for the parallel connection of 1K 10K 1M).
Just film capacitors rated at 1u are not always available and very large.

[Phoenix]

the impedance will be approximately 900 ohms (the online calculator says for the parallel connection of 1K 10K 1M).

What calculator is that? Because there appear to be an error with it, as that result is not correct at all. R5 is in series with the output, it is not in parallel with the load.

[POTL]

Series?
Then this is my mistake 
I calculated all 3 in parallel 
(R4+R5)+R3 it's right?

[Phoenix]

Series?
Then its my error 
I calculated all 3 in parallel 
(R3//R4)+R5 Right?

Almost - the output impedance of the circuit block as a whole is actually the output impedance of the source follower [ Rs/(gm.Rs+1) where gm is the transconductance of the transistor used, and Rs is the source resistor, R3] plus the series element R5. So, for the passband, the math for that circuit is R5+(R3/(gm.R3+1).
R4 forms part of the load, so does not alter the output impedance in the passband (though it does effect what the passband is).

[POTL]

Sorry for stupid question,but transconductance of J201(JFET)/2n7000(MOSFET) As indicated in the datasheet?

[Phoenix]

J201 500μmhos/500μS/0.0005


2N7000 100μmhos/100μS/0.0001

[POTL]

Thanks
Then
J201
R5+(R3/(gm.R3+1)=1K+(10K/(500+1)=1K+0,05=1050 ohm

2N7000
R5+(R3/(gm.R3+1)=1K+(10K(/100+1)=1K+0,01=1010 ohm

R4+C2 Remain in the circuit but not taken into account

[Phoenix]

Thanks
Then
J201
R5+(R3/(gm.R3+1)=1K+(10K/(500+1)=1K+0,05=1050 ohm

2N7000
R5+(R3/(gm.R3+1)=1K+(10K(/100+1)=1K+0,01=1010 ohm

R4+C2 Remain in the circuit but not taken into account

Again, not quite. You have to remember your units. μ - mu - micro, is 10^-6, so you were a few orders of magnitude out there with the transconductance figures. And it seems you misread some of the formula and missed a term.

J201
R5+(R3/(gm.R3+1)=1k+(10k/(0.0005×10k+1)=1k006

2N7000
R5+(R3/(gm.R3+1)=1k+(10k/(0.0001×10k+1)=1k002

And yes, when talking about impedance in general terms, we're referring to the passband (the frequencies within the -3dB points). Within those bounds, C2 doesn't have a significant effect. Outside the passband, C2 will start to add its own impedance and that's why it forms a high-pass filter.
R4 NEVER forms part of the output impedance of this circuit, as it is merely a load on the circuit, not a part of the output.

[POTL]

http://www.electrosmash.com/tube-screamer-analysis#output-stage
Thanks, based on this article, but more and more confused, now I understand everything, thanks again

Since in most buffers the resistor in the emitter / source is 10K (more than 90% of the classical circuits), roughly speaking the output impedance is obtained equal (with a difference in a fraction of a percent of the value) R5