50k rev log in place of a 25k

Would it ok to use a 50k rev log pot in parallel w/a 25k resitor for the gain adj. pot in a 312?

thanks

James

wouldn't that halve the range of the pot?

Is your 312 close to ready? In that case please try it and post the results, i'm curious myself...

You could use two gangss of a 50k dual rev log in parallel...

you chould use a 50k rev log in parallel with a 50k resistor (25k would give you less than 25k at maximum setting) or just plain on its own. YOu'd just get a few more dB on the gain range, that's all.

Keith

I don't think that will work. You can make a custom rev log using a linear pot with a taper resistor between the wiper and the cw pin.

Correct me if I am wrong here.

NWSM

[quote author="nwsoundman"]I don't think that will work. You can make a custom rev log using a linear pot with a taper resistor between the wiper and the cw pin.

Correct me if I am wrong here.

NWSM[/quote]

You can make a reverse log attenuator (voltage devider), but you cannot make a variable resistor. I believe James is looking for a solution for a variable resistor.

You can make a reverse log attenuator (voltage devider), but you cannot make a variable resistor. I believe James is looking for a solution for a variable resistor.

Click to expand...

If you leave the ccw pin floating you would have a varible resistor. I in at cw pin, I out on wiper. I am using this (100k with a 47k taper resistor) on my Quad Eight line amps. You are just changing how much pot is parallel with the taper resistor.

NWSM

http://www.home.earthlink.net/~nwsoundman/id1.html

At the bottom of this page I put a quick picture up of what I am talking about. Just reverse the cw and ccw pins to get the response you want from the pot.

NWSM

Nope.

We keep coming back to this, but it cannot be done. :sad:

There is simply no way to make a reverse log -or a log- variable resistor from a resistor in parallel with a pot.

The definition I am using is the same thing that we need to get a gain control to operate smoothly: Two connection points with 100% of the specified resistance value when set at at one extreme, 0% when set at the other exreme, and 10% of the specified resistance at the mid-point.

If you can do it using a single gang linear pot and a resistor, I'll give you a months wages. You have it in writing, and a stream of eager withnesses. :grin:

What you can do is an "INverse" log, not a "REverse" log. You can have 90% at the midpoint, but since this is not a potential divider, nowhere is there 10% at the midpoint.

Sorry, people keep claiming that it can be done, but it can't.

Keith

ssltech, yer right about making a rev log pot from a linear and a resistor. Really the reason I asked my question was because I was to lazy to really think about it. I posted it while on a break at my studio.

On my drive home I had time to ponder my problem. Please correct me if I'm wrong but isn't the gain of the opamp = to the ratio of the value of the fixed resistor in the feedback path and the value of the variable resistor that's connecting the inverting input to ground? And if that is the case and I use a 50k rev log pot can't I just double the value of the fixed resistor?

Either way I'm still going to use a 50k in place of the 25k so I can at least start testing this pre. I have a Big Band session in about 6 weeks and I would like to use it. (by "Big Band" I mean old school Sinatra/Dean Martin/swing stuff)

Btw this pre is a modded A__P__I 312 with onboard regulation and load isolaters between the output of the opamp and transformer. There is also a DC blocking cap on the output because I will be testing a number of different amps in this design that all have different DC output offsets. I've made a small board that has an 8pin dip and BUF634 with the 990 pinout and there is also an 8pin dip socket on the board itself ala the JLM99MB. I/O trafos are Lundahl LL1578XL and LL1585 respectively. The chip amps I have are the OP604, OPA627, OPA637, and the ubiquitous NE5534. The first DOA I will try is the Avedis 1122 which I will pick up on Monday. If any one is interested in seeing my layout I'd be happy to email them a bitmap of it. Unfortunately I'm using the demo version of ProteusLite and I can't output any image files other than ".bmp".

My final plan for this pre is to have 8 of them in a 2U rack. Four pairs with each pair having a different opamp and the last two channels being the "what do I feel like putting in there today" experimental channels.

One last thing. I was at Brent Averills shop a week or so ago talking with Avedis about this pcb and this great forum and he threw his 1122 into an original 312 w/a 1k sine on the input and a 20 ohm load on the output. With the gain maxxed it showed no dist. at all on the scope at the output. We then placed a 2520 in there and it had some pretty good clipping. This amp can seriously put out current.

cheers, and thanks for all the info. This forum and the people who frequent it are the best,

James
Big Ugly Van Audio

The definition I am using is the same thing that we need to get a gain control to operate smoothly: Two connection points with 100% of the specified resistance value when set at at one extreme, 0% when set at the other exreme, and 10% of the specified resistance at the mid-point.

Click to expand...

That there is the problem. With the setup I am using you can't have 0 ohms at one extreme. In fact I am using a 150 ohm in series with my pot. I don't see a need for my amp to go silent when the gain contol is fccw. Doesn't the amp in question just need a variable resistance between two pins on the edge connector starting out as 25k going lower as you turn the pot cw? If so a linear pot w/ taper resistor will work but not allow the amp to go silent when the pot is fccw.

Help me understand why you can't have 10 - 20% at mid point if you wire pins 1 and 3 backwards.

I learn something new everyday.

NWSM

Hi James,

If you not can get a 25k reverse log for your 312 card, but you have a 47 or 50k reverse log and you will have same gain control structure as the 25k, you can change the R3, 20k feedback resistor to 47k, change also C4, 120 pF to 56 pF and the R2, 200 ohm maximum gain resistor to 470 ohm.

This will take care of your problem.

--Bo

thanks Bo,

I thought that changing the feedback resistor would work but I didn't realize that I had to change a cap also. Sorry, but I'm quite ignorant on how the components interact. What is the reason for having to change C4 & R2?

cheers,

James

[quote author="nwsoundman"]Help me understand why you can't have 10 - 20% at mid point if you wire pins 1 and 3 backwards[/quote]
You'll still have 90% in the middle even if you mirror-image the whole thing...

The best way to get a grip on what's happening is to sit down for ten minutes with a 100k pot, a few resistance values like 10k, 100k, 1M etc... and an ohm meter. Whether the whole thing goes to absolute zero or not (ignoring 'padding' resistors in series) is of rather less significance than the basic premise that when you swap terminals 1 & 3 on a log pot you flip the graph horizontally making a reverse log, but that when you add a resistor in parallel with a linear, you flip the graph vertically and make a curve that approximates an inverse log.

If you are using it as a potential divider, you can 'fake' the effect, but not as a 2-terminal resistance. I tried for what seemed like years when I was making a few rather nice state-variable EQs a while back, and I struggled with the idea that I wasn't getting the curve that I needed... in fact the actual curve "bend" was going in the direction that made the scale linearity worse rather than better!

Finally I sat down with a pot, a fluke meter and a resistor and eventually accepted that it couldn't be done... but for a good while I was banging away at it in denial! :wink:

Keith

SSL,

I see where I goofed on the 10 - 20% deal. I will need to look at the 312 schematic to see what is required. I figured they work like the QE CA137'S.

I did go and measure the action of the pots I made for my QE stuff. They are 100k linears with 47k taper resistors. I get most change in the last half of rotation. The resistance values are just what I need for my box. Are you saying the most change in resistance should be in the first half or so of rotation?

Maybe I should have started a new thread? I am sure others can benefit from this.

NWSM

Nah... It's been covered, I'm sure!

I just have a reaction to the subject because I 'wasted so much of my life' before I realised that I was looking at it the wrong way! I tend to leap on it like a forest ranger on a smouldering match, because when one person says it can be done, several others start thinking the same way, and for the longest time I couldn't see the simple observation that mentally I was 'flipping it vertically' instead of 'horizontally'...

I'll keep an eye out and leap all over the next person who starts slipping down the slippery slope to madness!

:wink:

However, once you grasp just how violently the resistance increases, it beccomes apparent that a 50k is so close to a 25k it's not even funny... a 50k pot has 25k resistance at about '9' on a linear rotational scale of 1 to 10, so this really is like the classic spinal tap "goes to eleven" joke... Basically, a 47k set to '9' is the same as a 25k set to '10'. -When you look at it like that, and re-read my f1rst suggestion, the idea of using a 47k reverse log if that's readily available to you becomes a highly attractive one...

Let me suggest it this way... a 47k reverse log is as close to a 25k reverse log as a 25k linear is to a 22k linear...

Keith

I took a good hour to think about this, and as much as I hate to it admit it. SSL Tech is right.

The thing I never got about what he's saying is that you can make a reverse log pot out of a linear pot and a resistor in parallel, but it's like the half of a reverse log pot that you don't need. Because a pot always has to complimentary sides, one going up while the other goes down, if it is used as a variable resistor, you need to specify which side you want to use.

So the curve you get by putting a resistor in parallel with a linear pot is worse than linear because it is as far away from linear as you want, in the wrong direction! You'r only other choice is to use a LOG taper pot and have it work backwards.

too bad.

Mark

PS. Sorry i doubted you Keith.

Thanks for clearing this up. Off to buy some real rev log's. Well maybe I'll go the stepped attenuator route instead?



Thanks
NWSM

FWIW 50k reverse log pots are not as hard to find as other values in small quantities - it's a stock Fender spare part, made by CTS. Many guitar parts suppliers have them.

http://www.angela.com/catalog/guitar-amp-parts/FENDER_AMP_PARTS.html

http://www.vibroworld.com/parts/tech14.html#pots

http://www.tubesandmore.com/ look under 'Potentiometers'

:thumb: :thumb:

James,

The R2, 200 ohm resistor decide the maximum gain when the potentiometer are in full CW position, and in this case with a 20 k feedback resistor the max gain are aprox. +40 dB in the 2520 op-amp. (the minimum gain is aprox. 6 dB when the 22 (25) kohm potentiometer are in full CCW position, and aprox. 20 dB in middle position)
You have also aprox. +18 dB gain i the input transformer and +6 dB in the output transformer from pin 4 and 6, so you have a total adjustable range between 30 to 64 dB.
You get same result if you take a 47 (50) k potentiometer with 470 ohm and a 47 k feedback resistor. (I have shoose the nearly regulare resistor value)

The C4, 120 pF are a "lo-pass" filter in the feedback patch and the upper high frequency limit decide of the relationship between the R3 and the C4, so if you want same upper high frequency limit when you double the R3 you must halve the C4.

--Bo

Bo,

You rock man. I guess it's about time I started to bone up on the more technical aspects of electronics.

thanks,

James