Input impedance

[Laudanum]

I really apologize in advance for this because I have some really basic questions and this may not be the right place to ask ...

I have built quite a few audio kits both tube and non-tube, incuding a tube phonostage, a couple of small PP amps, restored a Scott 299, got a non-working Ah Tjoeb CDP to work, rebuilt and old Kalamazoo bass 30 and every speaker that I have, I built, both single driver and conventional woof-mid-tweet.  I could usually "follow" the circuit (although sometimes I just winged it) but I rarely actually understand the circuit.
Most of the explanation on the web gets too technical and once that happens and I get lost ... I usually get lost even when it's simple.  But despite the failing eyes, Im pretty good with a soldering iron.

Anyway ... here goes a series of dumb questions.  Im going to build Crack when it arrives so I guess I'll ask with that in mind.

- In laymens terms as much as possible, what actually determines input impedance of an amplifier, like Crack.

- If that amplifier has a volume control, how does it relate to the amps input impedance (100K pot I believe if we are talking about Crack).  If that volume pot is designed into the cicuit but is omitted, or value of the pot is changed without any other circuit changes, what affect does it have in terms of the amps input impedance or otherwise?  (Im not planning on doing this, just thought it may help draw out an explanation that would help me understand better).
-A 100K volume pot for example presents 100K ohms when all the way counterclockwise and next to 0 ohms when all the way counterclockwise, correct? Doesnt the source see a vaying input impedance when the volume pot is turned (and it's resistance is lowered or increased)?  

If I understand the answers to these two, I know I'll have a couple more questions.


If there is better place to ask, here or on another forum, please let me know.  I realize that these are very basic questions.

Thanks

[Grainger49]

1) most amps and preamps have a resistor or pot across the inputs.  That is, from the input hot to the ground or signal common.  That resistor value or pot value determines the input impedance.  In Crack that is 100k Ohms.  An odd one is the FP III which has a 33k resistor in series to the input attenuator.  The impedance there varies from the 33k series resistor at zero volume setting to 48k at maximum volume setting.

2) If you omit the volume pot a resistor of similar value will be appropriate.  Just put it on the input RCA jack from the center conductor to the outer conductor.  Ground the outer conductor and run a wire from the center conductor to the grid/grid stopper resistor that the pot would have fed.

[Laudanum]

1) most amps and preamps have a resistor or pot across the inputs.  That is, from the input hot to the ground or signal common.  That resistor value or pot value determines the input impedance.  In Crack that is 100k Ohms.  An odd one is the FP III which has a 33k resistor in series to the input attenuator.  The impedance there varies from the 33k series resistor at zero volume setting to 48k at maximum volume setting.

2) If you omit the volume pot a resistor of similar value will be appropriate.  Just put it on the input RCA jack from the center conductor to the outer conductor.  Run the wire from the center RCA jack conductor to the grid that the pot would have fed.

Ok, so the pot value determines the rated input impedance, I can understand that.  But for an amp witha 100K pot as a volume control, the source sees the same input impedance regardless of whether the pot is rotated to max volume or midway or lowest volume, correct?  
So, If the resistance of the pot varies from 100K at lowest "volume" setting to near 0 ohms at highest "volume" setting, and signal from the source passes through the pot, then how does the source NOT see a varying input impedance?   Is the impedance of a 100K pot always 100K when measured from the input pin to ground regardless of the rotation of the shaft ... and that's what the source actually "sees" rgardless of the "volume" setting?

Thanks

[Grainger49]

Ok, so the pot value determines the rated input impedance, I can understand that.  But for an amp witha 100K pot as a volume control, the source sees the same input impedance regardless of whether the pot is rotated to max volume or midway or lowest volume, correct?  

So, If the resistance of the pot varies from 100K at lowest "volume" setting to near 0 ohms at highest "volume" setting, and signal from the source passes through the pot, then how does the source NOT see a varying input impedance?   Is the impedance of a 100K pot always 100K when measured from the input pin to ground regardless of the rotation of the shaft ... and that's what the source actually "sees" rgardless of the "volume" setting?

Thanks

To the first question, almost all the time.  As I mentioned the FP III wires the attenuator and series resistor differently.  Most amps have the signal coming to the "top" of the pot.  The wiper taking the signal to the next stage and the "bottom" of the pot wired to common.  In these cases the input impedance is the value of the pot regardless of the wiper position.

A "normal" volume control, as described above, will have the lowest volume when the wiper is at, or near, the common lug.  The highest volume is attained when the wiper is at the top, nearest the input wire.

To the second question, the resistance of the pot is a constant value from the top to bottom lugs (input to common for a normal pot).  The resistance changes from the wiper to the top and from the wiper to the bottom as you move the wiper.  The reason that the wiper moving doesn't change the overall impedance of the input is that the next thing attached to the wiper is a very, very high impedance.  It doesn't make a measurable difference to what the source sees as an input impedance.

For instance the Crack's 100k pot paralleled with the 100M ohm input of the tube (probably higher, I'm swagging here) calculates to 99,900 ohms.  In other words 99.9% of the starting value.  So it doesn't matter to the source.

[Laudanum]

Ok, that helps so much, thank you.

So when standard pots are wired as shunt pots thats most commonly when the source ends up seeing a variable impedance, like with your Foreplay III example, correct?

Also, in a case where the volume is too loud near the bottom of the pots range, it could be necessary to add a series resistor between the signal input and pot input ... in this case, the source would also see a varying impedance, correct?   

[Grainger49]

Yes, look at VoltSecond's site:

http://www.siteswithstyle.com/VoltSecond/12_posistion_shunt/12_Position_Pure_Shunt.html

Both of the shunt mode wirings he shows give a variable impedance to the source.  With the lowest impedance being the shunt resistor value.  The FP III wiring is not shown on the link.

And, Yes again.  A series resistor will "waste" some of the volume you have allowing you to get the wiper away from the bottom of the pot and have greater resolution with your volume control.  As Doc and Paul have allowed in teh FP III you can tune each input to the output of the device attached.  I have one value resistor for my Ack! dAck!, another value for my SACD player and no resistors for my Seduction and Tuner in my FP 2.  These resistors are between the selector switch and the volume controls.

[Laudanum]

Yes, look at VoltSecond's site:

http://www.siteswithstyle.com/VoltSecond/12_posistion_shunt/12_Position_Pure_Shunt.html

Both of the shunt mode wirings he shows give a variable impedance to the source.  The FP III is another example not shown on the link.

And, Yes again.  A series resistor will "waste" some of the volume you have allowing you to get off the bottom of the pot and have greater resolution with your volume control.  As Doc and Paul have allowed you can tune each input to the output of the device attached.  I have one resistor for my Ack! dAck!, another for my SACD player and no resistor for my Seduction and Tuner.  These resistors are between the selector switch and the volume controls.

Did you add a selector switch to the Crack itself ?   I imagine a simple selector could be added to the Crack itself with the inputs feeding the selector with appropriate resistors for each source and then on to the the volume pot.    I'll build it as stock with maybe just a few inexpensive passive upgrades and go from there I think.

Thanks again, you've been a great help!

[Grainger49]

No, I haven't built the Crack.  But this holds up regardless. 

And, Yes, you could put a selector switch and additional RCA jacks on a Crack to allow you more versatility.

[Laudanum]

Thank you again.  Much appreciated.

[sbelyo]

this thread is perfect...

I am going to digital volume control on my dac and I wondered the same thing. 

So, if I understand correctly when taking a volume control out of the circuit I should replace it with a resistor of equal value or a value that matches something that my ears are comfortable with?

[Paul Joppa]

A volume control is a potentiometer. A potentiometer is TWO resistors.

If it's turned all the way up, then one of the resistors is zero, and you can replace the pot with a resistor of equal value.

If you want a reduced signal level, then you have to replace the pot with two resistors, whose sum is the equal value. One goes from the input to the first stage input (grid), the other goes from the first stage inpur (grid) to signal ground. The ratio between the resistors determines the attenuation, their sum is the input impedance.

[sbelyo]

Thanks Paul.

I'm wondering if I should abandon the digital volume control, or maybe not I'm not sure.  I was trying to have a central volume control.

The DAC has balanced and single ended outputs and I am planning on using both and I was trying not to have several high end volume pots.

Right now I have all single ended gear with no volume pots and a 2 in 2 out passive preamp with a 50K alps blue pot in between the source and amps

[Laudanum]

A volume control is a potentiometer. A potentiometer is TWO resistors.

If it's turned all the way up, then one of the resistors is zero, and you can replace the pot with a resistor of equal value.

If you want a reduced signal level, then you have to replace the pot with two resistors, whose sum is the equal value. One goes from the input to the first stage input (grid), the other goes from the first stage inpur (grid) to signal ground. The ratio between the resistors determines the attenuation, their sum is the input impedance.

Paul ... I found this at the Goldpoint site ...   http://www.goldpt.com/mods.html   ... Per your explanation of the two resistors,  is this essentially what Goldpoint is doing here to reduce gain with what they are calling "pre attenuation" using the two resistors, except of course, while also retaining the pot/attenuator?
Looking at the table, they are using different value resistors for R1 and R2 but the total value essentially equals the pot value with the exception of the smallest attenuation levels (-6db) resistor values which total about 50% over the pot value.  The -10db attenuation resistor values are slightly over the pot value.  

This would still be a type of shunt attenuator, correct?  
Im not sure I understand why the total resistor value is 50% higher for the -6db configuration (and a little higher for the -10db) though.

[Paul Joppa]

Yes, the Goldpoint circuit is what I was talking about.

To calculate the attenuation, notice that Rp2 is in parallel with the control. For example, if the control is 100K and you use Rp2=100K, then the resistance of those in parallel is 50K. Using Rp1=50K means the voltage at the attenuator "IN" is half the Input voltage - which is 6dB attenuation.

Rp1 and Rp2 are what is called an "L-pad." (If you draw it with Rp1 horizontal, then the two resistors make a L shape.) Long story, but if you chose the resistors correctly the impedance at the input terminal is the same as the impedance of the control alone. This is what is done on the Goldpoint site.

The level control is still a potentiometer, not a shunt-mode control. That's yet another long story ...

[Laudanum]

Yes, the Goldpoint circuit is what I was talking about.

To calculate the attenuation, notice that Rp2 is in parallel with the control. For example, if the control is 100K and you use Rp2=100K, then the resistance of those in parallel is 50K. Using Rp1=50K means the voltage at the attenuator "IN" is half the Input voltage - which is 6dB attenuation.

Rp1 and Rp2 are what is called an "L-pad." (If you draw it with Rp1 horizontal, then the two resistors make a L shape.) Long story, but if you chose the resistors correctly the impedance at the input terminal is the same as the impedance of the control alone. This is what is done on the Goldpoint site.

The level control is still a potentiometer, not a shunt-mode control. That's yet another long story ...

Of course, like an L pad in a passive crossover.  When one is not particularly tech savvy, such as myself,  one tends to not be able to put 2 and 2 together ...
I was playing around with the suggested resistor values and a pot I have on hand.  I didnt have exact values, I think I used 47K (in place of the 49.9) and 100K.  I assumed that would be close enough.
Measured just under the 100K value of the pot just as you and the Goldpoint site describe.  Learn something new every day if you want to.

Im getting ahead of myself here anyway.  I dont know if I need attenuation until I know if I need attenuation.

I do have another couple input impedance related questions though.  I would like to use a digital music player as a source.  It has a dedicated line-out that is spec'd at 2V out.  There is no output impedance spec given for the line-out however.  It works very well with a small opamp based headphone amp that I built which has an input impedance of 10K ohms.  Is there any reason why it wouldnt work well with the Crack and its 100K ohm input impedance?  

Also, what determines the choice of the 100K pot/input impedance?  Does it lend itself better to the overall amplifier design or could, for example,  a 10K pot/input impedance (or other value) have been chosen just as well?

Thanks again