understanding the PULTEC EQ circuit

Hi,

I'm just layouting a board for a passive Pultec EQ with rotary switches instead of pots.

I'm not really shure about the mounting directions (cw and ccw)...
Each value in "blue" stands for the actual "pot-setting"

I just need to know the correct BOOST / CUT / BANDWIDTH values for A / B / C / D / E

I hope you can help me...





thanks,
mat

Lord Jesus , that better not be a Pat schematic!!!!!

:evil: :evil: :evil: :evil: :evil: :evil: :evil: :evil: :twisted: :twisted: :twisted: :twisted: :evil: :evil: :evil: :evil: :evil: :evil: :evil: :twisted: :twisted: :twisted: :twisted: :evil: :evil: :evil: :evil: :evil: :evil: :evil: :evil: :twisted: :twisted: :twisted: :twisted:

hehe... what's so funny about it ??

I'm sure you both guys can help me...

Mat,

Can't you simply look at a picture of an original Pultec to find this?

Jakob E.

no!! :wink:

of course I know the directions on thr frontpnael...


but my problem is:

I want to use 12x switches that should replace all the pots...


for example:

if you look at B

the 1k pot that is used for hicut has 2 resistances ... one left an one right to the wiper...

if the resistance on the left side is 0k and the resistance on the right is 1k I need to know, if this setting applies full cut (-16db) or no cut (0db) to the circuit...
I'm not fully shure about that.....

thanks,
mat

You turn all the knobs to the right (clockwise) to boost, cut, and sharpen bandwidth, if that is any help.

It looks ok on your shematic exept bass boost and cut.
0 ohm is no cut, and 0 ohm is no boost.
j

I'd like to understand how the circuit works...


so I need to know on which side of the wiper of the pot is boost or cut and what is the resistance for boost or cut..

10k

thank you for the fast reply...

can you also help me with B , C , D and E ??

thanks,
mat

Imagine the pots as strings of resistors...should be easy then.
look three post up for my bass cut/boost comment.
j

It looks ok on your shematic exept bass boost and cut.
0 ohm is no cut, and 0 ohm is no boost.

Click to expand...

hmmm... but on what side of the pot...??

the ohm values in the picture of my first post should stand as an example for the setting of the pot...

but I need to know full boost/cut or no boost/cut

I feel your pain. That schematic is a total POS!
And to think that weasel charges 100 bucks for that piece o crap!
It's totally unreadable from a intuitive/logical point of view.
:twisted:

Maybe this will help:

Just imagine the pot forcing the signal thru the filters when in one position, and by-passing the filters when at the other position.

Or, another way to think about it, which way would you go if you were the signal? Thru the pot, or thru the filter? Which way offers the path of least resistance?

hey CJ

I looked for a moment to your schem and from my memory I remember that the capacitors for the bass cut network will connect not after the 1K resistor but at the common point of the 100k pot and the 1k resistor.

chrissugar

Hmmmm. Hold on....

Well, which schematic are you using? I drew mine from Pat's, which might explain things. Either way, I do not think it will make that much difference. I have a M-ley version at home also. I will check it tonite.

http://home.earthlink.net/~cayocosta/diy_files/eqp_1a/eqp-1a_layout_final.jpg

This one has arrows for +/- directions. Should help some.

OK, now we have five diferent schematics
1-Jakob's
2-the one redrawn by Brian Sowter
3-Pat's
4-the one from the EQP1R
5-the Lang EQ that was the original.

The one from Brian and Pat has the resistor included in the RC network.
The one from Jakob has the resistor after the RC network.
The one from the EQP1R manual nas no 1K resistor at all
The one from Lang has no 1K resistor at all.

I suspect that the 1K resistor should not be in the RC network because for no bass cut you have to short the capacitor to leave the whole audio to pass. Probably if you leave the 1k inside the RC network, 100k pot in no cut position (short) there will be a very small bass cut (0.5dB or something like that)
I will do some freq sweeps.

chrissugar

OK,

I managed to make some sweeps with my mastering version of Pultec. I don't have the 1k resistor included in the RC network but I can simulate it by adjusting the 100k log pot to 1K. Luckily all my pots are 24 position ELMAs so I had to look into my papers to see at what position the stepped attenuator has 1K. It is position 3.

I made some sweeps with the Pultec in bypass mode, active mode but with all the switches at no cut/no boost, and with the low cut in position 3.
The first two sweeps are identical but the third has approximately 0.6dB cut at low freq.

So my conclusion is that there should be no 1K resistor in the RC network (that is how is the Lang and the original Pultec schem). I don't know if the 1K after it matters or not, that should be another test (I suppose it has no significant effect).

chrissugar


P.S. CJ, now you really have a serious reason to hate Pat. :twisted:

thanks chris...!!! good work!!!