Gain Pot Woes w/Easy Drive - help!!

Started by fuzzy645, September 20, 2011, 11:42:54 PM

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fuzzy645

First, let me say the Joe Davisson's Easy Drive sounds amazing!  I've been tweaking cap values and experimenting with Diode clipping.  The thing sounds fantastic. Very warm w/lots of overtones.

My only 1 issue is with the gain pot not having enough swing.  Yes, there is some swing, but not much.  It is not really sounding like a "distortion level" control as I was expecting, and there is not a huge difference full up or full down.  It just sounds like a volume control that has been lowered from 10 down to 9 1/2.

The first image below shows you the voltages with different values for the resistor by the 9V battery positive. I used resistor values ranging form 22K  to 67K, with 47K sounding the best.  The second image below shows you the circuit.

Any ideas? 





It seems to me like the diodes are in the circuit 100% of the time so they should always clip even if gain is lowered. I suppose the theory is that if you reduce gain enough the diodes won't clip anymore, which is what I was hoping out of the gain pot.   



Earthscum

#1
Try a 1M input volume? Maybe a 100k with a 100k resistor to ground, give you half signal. The lower you go on the pot value, the more tone sucking you'll get. Feed the guitar signal into the top, other end to ground, and input cap off the center lug/wiper. (actually, 500k would work great if you have one.)

Also, since this is a circuit that relies on your guitar impedance, you may try, first, just using a 100k pot in series with the input cap. To test either of these out, just roll back the volume a bit on your guitar... your drive pot will start to show different ranges.
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fuzzy645

Quote from: Earthscum on September 21, 2011, 12:35:44 AM
Try a 1M input volume? Maybe a 100k with a 100k resistor to ground, give you half signal. The lower you go on the pot value, the more tone sucking you'll get. Feed the guitar signal into the top, other end to ground, and input cap off the center lug/wiper. (actually, 500k would work great if you have one.)

Also, since this is a circuit that relies on your guitar impedance, you may try, first, just using a 100k pot in series with the input cap. To test either of these out, just roll back the volume a bit on your guitar... your drive pot will start to show different ranges.

Are you saying to add a 4th pot, as the diagram already has 3 (vol, tone & gain), or rather moving the gain pot to a different location (Input of circuit)?

Thank you.

jk-fm

The Easy Drive is a common-emitter amplifier. When the emitter is not bypassed by the capacitor, the AC gain is limited to the ratio of the collector resistor divided by the emitter resistor.

Use a 50k gain pot. When the pot is on full, you will have 47/50 ~ about 1 gain. When the pot is at zero, you will be at the transistor's maximum gain.

Ideally you want a reverse-audio taper, that is 50k turned to the left, 5k ~ 9x gain in the middle, and 0k max gain turned to the left.

fuzzy645

#4
Quote from: jk-fm on September 21, 2011, 11:06:29 AM
The Easy Drive is a common-emitter amplifier. When the emitter is not bypassed by the capacitor, the AC gain is limited to the ratio of the collector resistor divided by the emitter resistor.

Use a 50k gain pot. When the pot is on full, you will have 47/50 ~ about 1 gain. When the pot is at zero, you will be at the transistor's maximum gain.

Ideally you want a reverse-audio taper, that is 50k turned to the left, 5k ~ 9x gain in the middle, and 0k max gain turned to the left.

Woo hoo!!   Not only did that solve the problem, but I actually learned something to boot!!!!

There is HUGE swing now (bordelrine too much).  Now that I actually comprehend this (what a concept) I can play with the values a bit.

Another quick question about that bypass cap. Does it need to change in response to the increased pot resistance. I know there is a formula C = 1 / ( 2 * pi * Fmin R)  where the goal is to choose C so as to minimize the capacitive reactance as compared with R for all frequencies above the minimum.  Since humans hear stuff above 20 Hz I selected 20 as my value and came up with

C =  / (2 * pi * 20 * 50000 ohms)

...therefore C = .15 uF (as opposed to the 10 uf i have in there now)

Does that sound right?  It sounds kinda small to me?

Thank you!!!!

PRR

> HUGE swing now (bordelrine too much).

I wuz gonna say 5K. The 50K extreme is so low-gain that why bother? and you don't get a lot of action at 1/2 or 3/4 unless you score that reverse-audio pot.

> that bypass cap. Does it need to change in response to the increased pot resistance.

You can do that, but it is super-awkward. OR you can just make it SO big that it is "big enough for all settings". As you say, 20Hz (or 80Hz) is good enuff for people (or guitar). But so is 2Hz or 0.2Hz, as long as cap cost is affordable.

> 50000 ohms... ...therefore C = .15 uF

No. The 50K varies to zero. AND is in-series with transistor emitter, which acts like roughly 300 ohms.

I'd use 80Hz 300 ohms = 6.9uFd == 10 or 22uFd for practical purpose.

> C = .15 uF  It sounds kinda small to me?

Are you listening to the equation?? Put a point-one or point-two in there and listen to the pedal. We don't necessarily want all the bass to get in and thrash against the melodic notes. 0.15u may be too small, but the ear is the only judge, and the experiment is only pennies.
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fuzzy645

Quote from: PRR on September 21, 2011, 09:30:35 PM
> HUGE swing now (bordelrine too much).
I wuz gonna say 5K. The 50K extreme is so low-gain that why bother? and you don't get a lot of action at 1/2 or 3/4 unless you score that reverse-audio pot.
Thanks for responding PPR. Yes, I agree 100%.  Since I don't have a 5K pot, I tried it with 10K and that was better than 50K.  I will try with a 5K too.


Quote from: PRR on September 21, 2011, 09:30:35 PM

> 50000 ohms... ...therefore C = .15 uF

No. The 50K varies to zero. AND is in-series with transistor emitter, which acts like roughly 300 ohms.
I'd use 80Hz 300 ohms = 6.9uFd == 10 or 22uFd for practical purpose.
Thanks for help with that.  One question though:  How can one determine that a given transmitter emitter (such as the emitter from the 2N5089) behaves like approximately 300 ohms of resistance? Personally, I do find it very useful to plug in numbers in formulas, but to me the only difficult part is figuring out the actual values to plug in (for things like R) .  The super easy part is doing the math.

Quote from: PRR on September 21, 2011, 09:30:35 PM
> C = .15 uF  It sounds kinda small to me?
Are you listening to the equation?? Put a point-one or point-two in there and listen to the pedal. We don't necessarily want all the bass to get in and thrash against the melodic notes. 0.15u may be too small, but the ear is the only judge, and the experiment is only pennies.

Yes, I tried hooking it up last night with 3 cap variations:  22 uf,  10 uf and the .1 uf.  The .1 uf was way too bright.  The 10 uf was nice, but the overall best sound was the 22 uf.