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 Alfetta87 22nd April 2008 11:35 AM

Which value of a potentiometer

Hi all,

In most tube preamps the input potentiometer have a value of 100k, in most solid state types they are 50k, or even 10k.

What a general rules to use which value, for any kind of preamp.

thanks,
Dan

 Sheldon 22nd April 2008 03:40 PM

The answer is; It depends.

Historically, tube amps and preamps were driven by tube sources. Tube sources and preamps generally had fairly high output impedance - several hundred to several kOhm. It's much easier to have a very low output impedance with solid state components - often as low as 50 ohm. The usual rule of thumb is that you want your input impedance to be at least an order of magnitude higher than the source impedance. So, make that 20x or a little more for good measure. But not much higher than you need so as to minimize any high frequency attenuation due to the input capacitance of the driven stage.

Sheldon

 Miles Prower 22nd April 2008 05:29 PM

Re: Which value of a potentiometer

Quote:
 Originally posted by Alfetta87 What a general rules to use which value, for any kind of preamp.
In an ideal world, every amp would have a Zo= 0R, and a Zi= INF, and it wouldn't make any difference. We don't live in an ideal world, and there really aren't any general rules. So the selection for an input pot depends on two criteria: the Zo of the driver stage, and the Ci of the driven stage. In the case of Zo, you want to avoid losing too much signal across the voltage divider that Zo and the input pot forms. In that case, R => 10(Zo) is good.

The other problem is that of high frequency response. The input pot + Ci forms a first order LPF. Make the pot too big, and you lose your high frequencies. As for what to expect, high-u triodes with large voltage gains have a larger Ci, due to Miller Effect. This is why you see lower values of pot resistance with solid state: transistors can have very large voltage gains, and therefore huge Cmiller's. Lower gain triodes, cascodes, and small signal pentodes, not so much. This needs to be decided on a case-by-case basis.

What I generally do is take a 100K linear pot, and parallel it with a 16K resistor. This gives a nice log characteristic, and the resulting resistance works well with just about any CD player, tape deck, tuner, etc.

 Jeb-D. 22nd April 2008 10:23 PM

Everybody does things different and there isn't an absolute right.

I prefer to use a 10k or 20k A-taper (that's logarithmic). Compared to 100k, it is less susceptible to line noise (with a low impedance signal source the difference is minimal) and will have better bandwidth at volume levels below maximum. The down side is that if the source has a fairly high output impedance you’re not going to get the full signal at the input of the amplifier. However, the input impedance is purely resistive in the audio frequency range, so the response will still be linear. Meaning it’s not like driving a loudspeaker with a high impedance amplifier where amplitude will be all over the place. Most decent or good, tube pre's can handle driving a 10k input impedance without problem. After all we are only talking about small voltage swing.

Shunting a linear pot to mimic a log pot will cause the input impedance to fluctuate depending upon the knob position. For most solid state sources this is no problem. But if the signal source is capacitive coupled (like a tube pre-amp) it will largely vary the lower cutoff frequency depending on knob position. Not to mention the log characteristic will be altered, since the input voltage will vary with with source impedance as well.

 Abelma 23rd April 2008 02:50 PM

Hi
I am in the same situation, make the choice of the value of an attenuator. I have the Aikido attenuator, and I will use an Aikido preamp also. My source will be cd players mostly.
Do you think that 20 K would be a good choice?
What happen if I configure the preamp with fairly high gain, I'd says, 20 db. It will not have a "fine tuning" volume this attenuator of 20 K?
Thanks

 Jeb-D. 23rd April 2008 05:22 PM

Another point I didn't mention earlier, a 100k pot will require smaller value coupling capacitors on possible preceding stages. So if you are unsure of how changing the value of the pot will affect your system, it is safer to go with 100k.

 Sheldon 23rd April 2008 06:56 PM

Quote:
 Originally posted by Abelma What happen if I configure the preamp with fairly high gain, I'd says, 20 db. It will not have a "fine tuning" volume this attenuator of 20 K?
"Fine tuning" or resolution is not affected by the attenuator value. It's a function of the smallest proportion of the range between two settings. Therefore it scales with the overall value. For instance, a 10% settin of a 1k attenuator will give the same volume as 10% of 100k. If it's a resistor ladder, it's a function of the number of steps. If it's a traditional pot, it's a function of the mechanical design - resistive element and wiper.

Sheldon

 Abelma 23rd April 2008 08:05 PM

Thank's!
Mmm..So, I would go with the choice of 50 K..The attenuator is from tubecad.com, it is a mix of ladder and series attenuator. I had experience with a dact series attenuator of 10K using it like passive pre. With short and good signal cables, not very good results, lack of dinamics mainly.
So, you think that with 50 K I should not go wrong ? The sources will be always modern cd players. The choice is important before the buy of the resistances, and I am thinking to invest with excellent quality...and also more \$\$ !..So, better do the good choice. Thanks again

 Alfetta87 29th April 2008 10:28 AM

There is another thought I had: since the voltage drop on a 10k pot (just an example) is higher then on a 100k, at full output setting (potentiometer fully open, i.e. in the right side position), I'll have less output voltage with a 10k then with a 100k. Am I wrong?

I'm saying this, because I'm finishing my tube preamp, and have the choice of the value (any value is available). But most preamps have such a high output level, that I have to play with the volume knob between 7 o'clock (zero level) and 9 o'clock (very loud, pratically full power on my amp). First, it's not very accurate to set the right level, second the pot is less accurate for the two tapers then as they are at 12 o'clock.

I am having two outputs on my CD player: one regular direct from the OpAmp, with about 53ÂµF coupling cap, unbalanced, and a second balanced output done with a Lundahl LL1527 transformer, which is "attacked" by the same 53ÂµF cap on the primary (it'a combi of a Silmic II 47ÂµF cap and a 6ÂµF Auricap in //).

Which value of a pot would be the most appropriate in my case (and avoid too much output level on my preamp). Maybe at the ouput of my preamp I add a voltade divider to put down the level?

Thanks,
- dan

 Nordic 29th April 2008 12:56 PM

Carlos actualy suggested a clever setup before... useing trimpots in the lines to the source selector. This can be used to level match te output of various sources... my dvd player puts out 2 to 3 times the voltage the satelite dish does for instance... switching sources can lead to loud surprises...

I think when it comes to adjusting preamp output to fit within amp input sensitivity... there are 2 routes ... reduce gain of amp so reduceing sensitivity or reduce gain of preamp, reduceing output voltage...both normaly only involves some resistor changes..

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