How to lower preamp output?

[BobBeaston]

I am building a guitar overdrive effects box to be placed before a guitar amps preamp. I really only need the output of the overdrive to be slightly higher than the guitars line level.

I have am using 3 12ax7 gain stages, no cathode followers or anything. I am using a simple high roll off for a tone control consisting of a 100K variable resisitor in series with a .0001uf cap to ground after the last triode stage.

I am using a 250K pot for the volume control. It is ridiculously loud as soon as the pot is turned just a hair. The only thing I can think of is placing a large resistor in front of an even smaller pot for the volume control.

Would doing so give me even worse (higher) output imedance?

Is there a better solution?

[kevinkr]

Does the guitar amplifier have an effects loop - if so perhaps your overdrive effects box has enough gain to completely ignore the amplifier's normal guitar inputs.

Otherwise you could pad down the input at the guitar amplifier inputs.

What guitar amp are you using and does it have high gain/low gain inputs and if not can you add a low gain input?

Yes, so many questions..

Ultimately worst case you could put a 100K/200K resistor in series with a 10K/20K pot at the output of your effects box and use that to drive the input of your guitar amp.. These combinations would give you a minimum of roughly 20dB of attenuation and a lot more if you need it, with a worst case source impedance to drive the cable of 9K/18K.. (Use a well shielded cable and keep it really short..) Using a larger pot (as a sort of a final stage master volume control) is probably not a good idea unless you can add a CF to buffer it and then use that to drive the guitar amp input.

[BobBeaston]

Thanks those are some good ideas..

I have multiple amps and was hoping to use it on all of them (only one has an effects loop).

I was under the impression you needed a large pot at the output for a volume control for adequate output impedance. I've never seen a value with tubes lower than 100K

[Merlinb]

Originally Posted by BobBeaston

I am using a 250K pot for the volume control. It is ridiculously loud as soon as the pot is turned just a hair. The only thing I can think of is placing a large resistor in front of an even smaller pot for the volume control.

Would doing so give me even worse (higher) output imedance?

No, in fact, your output impedance would probably get better (lower!)

At the moment, your 250k pot provides a worst-case output impedance of about 1/4 * 250k = 63k, when set to -6dB attenuation.

If you use, say, a 220k resistor followed by a 47k pot, then the max output impedance would be roughly 220k||47k = 39k. The valve still thinks it is driving a 220k+47k = 267k load, so the gain of the valve itself won't suffer.

Another way you can do it is to split the anode resistor into two parts, and connect your 220k pot to the junction of those two resistors. In some ways that may be a better solution.

[BobBeaston]

Thanks for the input.. I used a 220K resistor followed by a 25k pot. No noticeable difference in volume because it is still so hard to control. My band members give me dirty looks when I accidentally turn the volume pot a tad to much. When it is set to about 1/10th full volume and it it is about 3 times louder than if my guitar was plugged directly into the amp. I was hoping to have something like a normal stompbox where the the level is about equal to your guitar level when set in the middle and you can add more for solos..etc.

Now is there any way I can really decrease the volume significantly and decrease the output impedance/boost current with just resistive parts. Could I lower the resistor and the pot (say 5K pot and a 100K resistor)?

[multisync]

not knowing your circuit I can only make a guess but here goes. Most volume pots are logarithmic ie at half volume they are at 10% of their value. If a log pot is 100k it will be approximately 10k at the 12 o'clock position and 1 k at around the 9 o'clock position. Since your pot is 250 k and you only have to turn it up a tad it could be at a value of only 1% or 2.5k try the following. use a 250,000 resistor in series with a 25k( log) pot going to ground. take the wiper of the 25k pot and connect it to your volume pot. Set the 25k pot to min. set your volume to 10-12 o'clock ie halfway or a little less. adjust the 25k pot to get the volume you want. and then use the regular volume control as normal(the 250k pot) if some of your other kit is more or less sensitive you can adjust the 25k pot accordingly. hope this helps

[Merlinb]

I made not have been clear, when I said a resistor 'followed by a pot' I mean this:
Pot.jpg picture by merlinblencowe - Photobucket

Be sure to use a logarithmic-taper pot (may be marked with a 'B').
Simply make the 220k resistor bigger if you find the output is too large.

[BobBeaston]

No you were clear that's what I have the resistor before the pot..also I'm def using a log pot. I really need to know an example of a bad resistor pot combo and why and a good combo and why in regards to output impedance. Thanks again

[Merlinb]

The output impedance will never be larger than the value of the pot itself, and at any setting other than maximum it will be less. For you it is a simple matter of increasing the fixed resistor so that, even with the pot fully up, the output level isn't overpowering. The output impedance will not be greatly affected.

[BobBeaston]

Ok how bout a 1M resistor to a 25k pot