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20dBPad

PostPosted: Sun Mar 04, 2007 6:39 pm
by Alain Benoit
Here is the construction of an 'H Pad'.
I began by ridding an NL4MM 'Speakon Barrel' of its NL4MP connectors. I replaced them with NC3FD-L-B-1 and NC3MD-L-B-1 respectively.

Image

PostPosted: Sun Mar 04, 2007 6:42 pm
by Alain Benoit
Using high precision 1/8 watt resistors I then soldered them into an 'H-Pad' for a total signal loss of 20dB.

Image

PostPosted: Sun Mar 04, 2007 6:49 pm
by Alain Benoit
Here is the finished product. Handy tool to have in your kit.

Image

PostPosted: Tue Apr 03, 2007 12:21 pm
by josh porter
is it easy to make one of these pads?

PostPosted: Wed Apr 04, 2007 7:36 pm
by Alain Benoit
Indeed it is.

Damn host!!!

PostPosted: Fri May 11, 2007 9:48 pm
by oddioguy
U1176 wrote:Indeed it is.

Damn host!!!

Should up the pics to Photobucket. It automatically resizes, and rarely goes down. :idea:

PostPosted: Mon Mar 10, 2008 12:53 pm
by Mathieu Benoit
How costly are these to make?

PostPosted: Mon Mar 10, 2008 1:29 pm
by oddioguy
I would guestimate in the $15.00 range

PostPosted: Mon Mar 10, 2008 2:08 pm
by Alain Benoit
Sounds about right.

PostPosted: Mon Mar 10, 2008 2:34 pm
by Mathieu Benoit
Cool, I want to make some for stocking stuffers this Chrismas... :-P

Re: 20dBPad

PostPosted: Fri May 20, 2011 2:55 am
by Alain Benoit
Since the original post Neutrik released this.

Re: 20dBPad

PostPosted: Sat May 21, 2011 2:18 pm
by Malcolm Boyce
Will or will not pass phantom with no ill effect?

Re: 20dBPad

PostPosted: Sat May 21, 2011 7:07 pm
by Mathieu Benoit
Malcolm Boyce wrote:Will or will not pass phantom with no ill effect?


Good question. Andrew and I have used them with condenser mics before with no noticeable effect that I was aware of.

Re: 20dBPad

PostPosted: Sat May 21, 2011 8:23 pm
by Malcolm Boyce
10dB version available?

Re: 20dBPad

PostPosted: Sat May 21, 2011 11:29 pm
by Mathieu Benoit
Malcolm Boyce wrote:10dB version available?

I can't see why not. I actually want Alain to build a pair of 10dBs, 15dBs, 20dBs and 25dBs for here. Right now we have a pair of 20dBs and one 25dBs.

Re: 20dBPad

PostPosted: Sun May 22, 2011 6:43 pm
by Alain Benoit
Malcolm Boyce wrote:Will or will not pass phantom with no ill effect?


Yes and this is why.

When I design my phantom power supplies I use high precision (<1%) resistors to current limit at 14mA per mic input. Using Ohm's law, R=E/I, 48/.014=3429. Due to the power being equally split between both legs the E96 table for 1% resistors gives us 6k81. When I design my pads I also use high tolerance metal film resistors to ensure that all phantom is ignored by CMR. Since theoretically both legs are at the exact same potential there should be virtually no current passing through the bridge resistor. Lastly in a 20 pad with an impedance of 600 ohms the primary values are well below 1k and should have a negligible effect on available current.

Clear as mud?

Re: 20dBPad

PostPosted: Sun May 22, 2011 7:47 pm
by macrae11
Two points of clarification on my end.

1)Isn't Ohm's Law I=V/R? I had to double check my own remembering, so I was just wondering what the E stands for in your equation, or if it's just a typo.

2) CMR?

Re: 20dBPad

PostPosted: Sun May 22, 2011 7:55 pm
by Malcolm Boyce
macrae11 wrote:Two points of clarification on my end.

1)Isn't Ohm's Law I=V/R? I had to double check my own remembering, so I was just wondering what the E stands for in your equation, or if it's just a typo.
I know I would say I=V/R...

macrae11 wrote:2) CMR?
Common Mode Rejection

Re: 20dBPad

PostPosted: Sun May 22, 2011 8:06 pm
by Alain Benoit
macrae11 wrote:Two points of clarification on my end.

1)Isn't Ohm's Law I=V/R? I had to double check my own remembering, so I was just wondering what the E stands for in your equation, or if it's just a typo.

2) CMR?


1) I=V/R, where I is current, V is voltage and R is resistance. Notice that it's not A=V/Ω ? Where A is the unit Ampere and I is current, we use E in the equation for Electromotive Force.
So, I=E/R can be re-arranged to solve for R where R=E/I.

2) Common Mode Rejection.

Still clear as mud?

Re: 20dBPad

PostPosted: Sun May 22, 2011 8:34 pm
by macrae11
Clear as.

Re: 20dBPad

PostPosted: Sun May 22, 2011 9:16 pm
by Alain Benoit
Sorry sometimes I have difficulty breaking things down and drawing proper analogies, way easier in front of a blackboard where I can draw what I'm thinking, that's my teaching style.

Re: 20dBPad

PostPosted: Sun May 22, 2011 9:21 pm
by Jef
Can these be used in tandem, I mean like plugging one into another one? and if so, is the gain reduction of each added together? For instance if a 10db gets plugged into a 20db will the total gain reduction be 30db? I was thinking about building a few of these for my own tool kit and I could probably get by with just a few different sizes if this is the case.

The construction looks easy enough, but I don't know the math involved in determining the size of the resisters per/gain reduction. Any help for a do-it-yourselfer?

Re: 20dBPad

PostPosted: Mon May 23, 2011 2:44 am
by Mathieu Benoit
Alain Benoit wrote:
macrae11 wrote:Two points of clarification on my end.

1)Isn't Ohm's Law I=V/R? I had to double check my own remembering, so I was just wondering what the E stands for in your equation, or if it's just a typo.

2) CMR?


1) I=V/R, where I is current, V is voltage and R is resistance. Notice that it's not A=V/Ω ? Where A is the unit Ampere and I is current, we use E in the equation for Electromotive Force.
So, I=E/R can be re-arranged to solve for R where R=E/I.

2) Common Mode Rejection.

Still clear as mud?


Image

Basically, I=V/R is not the proper way that one would write it is what he's getting at. Technically ''E'' is the symbol for voltage (Electromagnetic Force) not ''V''.

Re: 20dBPad

PostPosted: Mon May 23, 2011 12:36 pm
by Alain Benoit
Yup E, I, R and P represent the functions of Electromotive Force, Current, Resistance and Power while V, A, Ω and W their respective units.

Re: 20dBPad

PostPosted: Mon May 23, 2011 1:05 pm
by macrae11
Mathieu Benoit wrote:
Image

Basically, I=V/R is not the proper way that one would write it is what he's getting at. Technically ''E'' is the symbol for voltage (Electromagnetic Force) not ''V''.


Yeah I totally understand it now, just that most places it's taught, it's still shown at I=V/R. As in all the places that I learned it before here.

Which is awesome! Not that it will probably make any practical difference in my life... ever.