20dBPad

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20dBPad

Postby Alain Benoit » Sun Mar 04, 2007 5:39 pm

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.

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Postby Alain Benoit » Sun Mar 04, 2007 5:42 pm

Using high precision 1/8 watt resistors I then soldered them into an 'H-Pad' for a total signal loss of 20dB.

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Postby Alain Benoit » Sun Mar 04, 2007 5:49 pm

Here is the finished product. Handy tool to have in your kit.

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Postby josh porter » Tue Apr 03, 2007 11:21 am

is it easy to make one of these pads?
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Postby Alain Benoit » Wed Apr 04, 2007 6:36 pm

Indeed it is.

Damn host!!!
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Postby oddioguy » Fri May 11, 2007 8:48 pm

U1176 wrote:Indeed it is.

Damn host!!!

Should up the pics to Photobucket. It automatically resizes, and rarely goes down. :idea:
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Postby Mathieu Benoit » Mon Mar 10, 2008 11:53 am

How costly are these to make?
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Postby oddioguy » Mon Mar 10, 2008 12:29 pm

I would guestimate in the $15.00 range
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Postby Alain Benoit » Mon Mar 10, 2008 1:08 pm

Sounds about right.
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Postby Mathieu Benoit » Mon Mar 10, 2008 1:34 pm

Cool, I want to make some for stocking stuffers this Chrismas... :-P
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Re: 20dBPad

Postby Alain Benoit » Fri May 20, 2011 1:55 am

Since the original post Neutrik released this.
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Re: 20dBPad

Postby Malcolm Boyce » Sat May 21, 2011 1:18 pm

Will or will not pass phantom with no ill effect?
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Re: 20dBPad

Postby Mathieu Benoit » Sat May 21, 2011 6:07 pm

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.
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Re: 20dBPad

Postby Malcolm Boyce » Sat May 21, 2011 7:23 pm

10dB version available?
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Re: 20dBPad

Postby Mathieu Benoit » Sat May 21, 2011 10:29 pm

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.
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Re: 20dBPad

Postby Alain Benoit » Sun May 22, 2011 5:43 pm

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?
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Re: 20dBPad

Postby macrae11 » Sun May 22, 2011 6:47 pm

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?
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Re: 20dBPad

Postby Malcolm Boyce » Sun May 22, 2011 6:55 pm

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
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Re: 20dBPad

Postby Alain Benoit » Sun May 22, 2011 7:06 pm

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?
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Re: 20dBPad

Postby macrae11 » Sun May 22, 2011 7:34 pm

Clear as.
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Re: 20dBPad

Postby Alain Benoit » Sun May 22, 2011 8:16 pm

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.
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Re: 20dBPad

Postby Jef » Sun May 22, 2011 8:21 pm

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?
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Re: 20dBPad

Postby Mathieu Benoit » Mon May 23, 2011 1:44 am

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''.
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Re: 20dBPad

Postby Alain Benoit » Mon May 23, 2011 11:36 am

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.
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Re: 20dBPad

Postby macrae11 » Mon May 23, 2011 12:05 pm

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.
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