andre tchmil

Re: Studer 169 EQ in API 500 format
« Reply #180 on: July 17, 2011, 07:06:15 PM »
In case someone needs Neutrik NTL1's for this project, I'm selling some in the black market .


dmp

Re: Studer 169 EQ in API 500 format
« Reply #181 on: July 17, 2011, 09:11:24 PM »
Quote
I did notice that when it is switched in i lose about 6db in volume. i remember reading one of AndyP's post mentioning using a 2:1 transformer to make up the 6db loss of the DOA, but it was sort vague and no one else has mentioned it so far. Any ideas?
My pair do not have a 6dB loss... maybe double check that the are connected w / balanced in / out correctly.

andre tchmil

Re: Studer 169 EQ in API 500 format
« Reply #182 on: July 19, 2011, 05:59:18 PM »
Few Neutrik trannies left.

JamesW

Re: Studer 169 EQ in API 500 format
« Reply #183 on: August 13, 2011, 10:55:45 AM »
Thanks for this project sounds great. Here is a pic of the extra caps on daughter boards that worked out quite well. I used knobs and "L" bracket from Classic API.
Subject to change with out notice

Indecline

Re: Studer 169 EQ in API 500 format
« Reply #184 on: August 22, 2011, 08:50:14 PM »
Thanks for this project sounds great. Here is a pic of the extra caps on daughter boards that worked out quite well. I used knobs and "L" bracket from Classic API.
that's pretty nifty! How are you liking the extra frequencies?

JamesW

Re: Studer 169 EQ in API 500 format
« Reply #185 on: August 22, 2011, 09:39:59 PM »
They get used for tracking and the extra frequencies get used.
Subject to change with out notice

reboot

Re: Studer 169 EQ in API 500 format
« Reply #186 on: September 05, 2011, 08:02:24 AM »
hello

sorry for that question ...
if i want to use doa , do i have to remove the amp op under them ?
i put everything and my doa's burnt . :-\
thank you.

Indecline

Re: Studer 169 EQ in API 500 format
« Reply #187 on: September 05, 2011, 07:43:22 PM »
hello

sorry for that question ...
if i want to use doa , do i have to remove the amp op under them ?
i put everything and my doa's burnt . :-\
thank you.
ouch! Yea the 5534 needs to be pulled out if you're using doa's

reboot

Re: Studer 169 EQ in API 500 format
« Reply #188 on: September 06, 2011, 03:51:47 PM »
ok , i'll keep you informed ..

thank you .

rmaier

Re: Studer 169 EQ in API 500 format
« Reply #189 on: September 25, 2011, 09:25:29 PM »
Newbie question here, so I apologize in advance...

I'm planning on racking a pair of these in a 1u case to use for stereo applications, so I'd like to use stepped switches throughout. Can I use single-gang throughout, including the mid sweep (P4)? If so, do I need to recalculate the resistance?

Thanks.
"Even if you don't know where the destination is of the sounds that you originate, it's worth getting it right at the start." - Rupert Neve


MikeClev

Re: Studer 169 EQ in API 500 format
« Reply #190 on: September 25, 2011, 11:04:41 PM »
Mid frequency sweep (P4a and P4b) needs to be dual gang. The others can be single gang. If you look at the schematic, 'HI' (Dual 10k Linear) and 'MID' (Dual 10K Linear) are wired so each gang/track is in parallel, so Hi (P2A and B) is working like a 5k single track linear pot, Mid (P3A and B) is also like a 5k single track linear pot. Only one half of P1 is used so it is acting like a single track 22k linear pot. P4 is the only one that really takes advantage of both tracks fully.
Soldering burns: No pain, no gain.

If my opamps can't be discrete, I at least try to keep them discreet.

rmaier

Re: Studer 169 EQ in API 500 format
« Reply #191 on: September 26, 2011, 12:22:39 AM »
Thanks, Mike.
"Even if you don't know where the destination is of the sounds that you originate, it's worth getting it right at the start." - Rupert Neve

rmaier

Re: Studer 169 EQ in API 500 format
« Reply #192 on: January 03, 2012, 06:42:46 PM »
Another question... Is it possible to limit the amount of cut/boost by simply changing the value of a few resistors, as in the sontec mod? I'm not likely to need more than 5 or 6db.
"Even if you don't know where the destination is of the sounds that you originate, it's worth getting it right at the start." - Rupert Neve

rmaier

Re: Studer 169 EQ in API 500 format
« Reply #193 on: January 06, 2012, 07:02:28 PM »
No?
"Even if you don't know where the destination is of the sounds that you originate, it's worth getting it right at the start." - Rupert Neve

Re: Studer 169 EQ in API 500 format
« Reply #194 on: March 06, 2012, 01:05:42 PM »
I am going to build two of these fairly soon and even though I haven't heard them yet, I have a feeling that they might be suitable for a home built mixer application. Any thoughts on building 8 channels of these to run into a neumann summing amp? Just wondering if these are 'universal' enough to use on a multiple channels/sources with or without the neutrik tranny. Any opinions/suggestions?
Thanks.

dmp

Re: Studer 169 EQ in API 500 format
« Reply #195 on: March 06, 2012, 01:25:16 PM »
Quote
Another question... Is it possible to limit the amount of cut/boost by simply changing the value of a few resistors, as in the sontec mod? I'm not likely to need more than 5 or 6db.
Yes. You can always reduced the range of a pot by replacing with a smaller value pot, with resistors on both sides.
For example, if you want to reduce the range of a 1000 ohm pot by 1/2, substitute a 500 ohm pot with 250 ohm resistors on both sides in series. Then your circuit still sees the same total resistance, but your adjustment is reduced.
You can also put a resistor in parallel with a pot to reduce it's value.
If the pot is log scaled, the math is different, but you might still be able to achieve a similar effect.

With an Excel spreadsheet and the series/parallel equations for combining resistances, you should be able to come up with something.

 
 

Indecline

Re: Studer 169 EQ in API 500 format
« Reply #196 on: March 09, 2012, 12:19:28 PM »
My pair do not have a 6dB loss... maybe double check that the are connected w / balanced in / out correctly.

I double checked and they are balanced i/o correctly. i have two modules built, one with xfr and one without and both modules exhibit the same behavior (so the error shouldn't be in the input stage). I tried swapping opamps in the i/o, and i've double checked the component values and all seem okay. It has to be a wrong component value somewhere though, as the symptom is present on both modules.

any ideas of where to look? both are using doa's, maybe i should try swapping to ic chips and see if the clears up the symptom?

dmp

Re: Studer 169 EQ in API 500 format
« Reply #197 on: March 13, 2012, 01:59:02 PM »
A few thoughts -
Losing 6dB means that your signal is dropping in half. 6 dB is also the amount the output balancing stage will add -  why? Because the unbalanced signal (input to r16 on the schamatic) is inverted by the U4 opamp stage to create the (-) for the balanced signal. Think of your unbalanced as a simple waveform - create it's opposite phase - and then the  (+) (-) when seen by a balanced input has the voltage doubled.
So where to look? Is U4 working correctly? You should be getting 6dB of gain from the unbalanced to balanced signal there at the output. 

dmp

Re: Studer 169 EQ in API 500 format
« Reply #198 on: March 13, 2012, 02:29:54 PM »
Also, to help understand the circuit, the input amp, U1, is a differential inverting amp stage. It creates an output based on the differences in the (+) & (-) input. Because the input r1, r2 are 20k, and the feedback r3 r4 are 10k, the gain of the stage is 0.5, so there is a 6dB decrease.

Indecline

Re: Studer 169 EQ in API 500 format
« Reply #199 on: March 13, 2012, 08:39:35 PM »
A few thoughts -
Losing 6dB means that your signal is dropping in half. 6 dB is also the amount the output balancing stage will add -  why? Because the unbalanced signal (input to r16 on the schamatic) is inverted by the U4 opamp stage to create the (-) for the balanced signal. Think of your unbalanced as a simple waveform - create it's opposite phase - and then the  (+) (-) when seen by a balanced input has the voltage doubled.
So where to look? Is U4 working correctly? You should be getting 6dB of gain from the unbalanced to balanced signal there at the output.
Also, to help understand the circuit, the input amp, U1, is a differential inverting amp stage. It creates an output based on the differences in the (+) & (-) input. Because the input r1, r2 are 20k, and the feedback r3 r4 are 10k, the gain of the stage is 0.5, so there is a 6dB decrease.

Thanks for putting me on the right track! I dug back into it and worked from U4 through the circuit. Come to find out, there's a huge difference between 680 and 680k!! :o 

I should be able to get a couple 680k resistors tomorrow, i will let you know!


 

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