How to incorporate T-pad + resistor H-pad on Sta-Level output?

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Bowie

Bowie

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Jun 22, 2012
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I've spent hours reading and trying to understand this but I'm at the point that I just have to ask.  The Sta-Level has an insanely hot output and the schematic contains a resistor H-pad + 1k pot as shown below.  This was not satisfactory for me so replaced the pad&pot with a Mallory 600ohm T-pad.  That was a huge help and allows me to get enough attenuation.  However, I can't open it more than 15% before the signal is too hot.  I would like to work the resistor pad (or any resistor pad) back into it but I don't understand how this will affect the output impedance.  If I just place the resistor pad (minus pot) before the attenuator, am I ok?  Any education on this would be greatly appreciated.
 

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Why choose for such a 'difficult' solution?
IMHO a 3 resistor voltage divider would be sufficient to create a balanced pad.
The terminating impedance can still be 600 ohms, without further measures the output impedance of the voltage divider will be lower than 600 ohms.
If this is a problem, you could add two more resistors (in series with the output), to increase the output impedance of the voltage divider.
Anyway, that is how I would do it...
 
> I can't open it more than 15%

So cut-down to 15% *then* do your variable pad.

15% is a lot. Assume you can overlook strict impedance.

15% of 600 Ohms is 90 Ohms. Round-up to 100 Ohms: 16.7% is near to 15% with a bit of excess in case your observation is a bit off.

So we want 600 Ohms total, 100 Ohms out of the middle, and lets say we'll do Balanced. (It's not essential but someone will whine if we don't.)

600-100= 500. We need two equal resistors that add to 500. Say 250 each.

So a U-pad with two 250 Ohm and one 100 Ohm.

If I found 220 or 270 Ohm resistors on my bench, I'd use them.

Dunno how hot your StaLevel is (or if that is normal?). Let's say it might be trying to do +30dBm (hot!). That's 1 Watt. Most of the power will appear in the two 250r resistors. They should be half-Watt each. Power in the 100r is less.

At this point you have 15% less presented from a 83.3 Ohm source. The T-pad is not that critical about impedance (because you are not computing, you are knob-turning). I'd feed it direct. However if you must see 600r here, add two 258 (220, 270) Ohm resistors (making an H-pad).
 
RuudNL said:
Why choose for such a 'difficult' solution?
IMHO a 3 resistor voltage divider would be sufficient to create a balanced pad.
The terminating impedance can still be 600 ohms, without further measures the output impedance of the voltage divider will be lower than 600 ohms.
If this is a problem, you could add two more resistors (in series with the output), to increase the output impedance of the voltage divider.
Anyway, that is how I would do it...
Click to expand...
I know less about voltage dividers than I do pads so I was just going what something somewhat familiar.  Thx for the info though.  I'm going to be reading more about that as I'm trying to get away from just following manuals and actually understanding how and why these things work.


PRR said:
> I can't open it more than 15%

So cut-down to 15% *then* do your variable pad.

15% is a lot. Assume you can overlook strict impedance.

15% of 600 Ohms is 90 Ohms. Round-up to 100 Ohms: 16.7% is near to 15% with a bit of excess in case your observation is a bit off.

So we want 600 Ohms total, 100 Ohms out of the middle, and lets say we'll do Balanced. (It's not essential but someone will whine if we don't.)

600-100= 500. We need two equal resistors that add to 500. Say 250 each.

So a U-pad with two 250 Ohm and one 100 Ohm.

If I found 220 or 270 Ohm resistors on my bench, I'd use them.

Dunno how hot your StaLevel is (or if that is normal?). Let's say it might be trying to do +30dBm (hot!). That's 1 Watt. Most of the power will appear in the two 250r resistors. They should be half-Watt each. Power in the 100r is less.

At this point you have 15% less presented from a 83.3 Ohm source. The T-pad is not that critical about impedance (because you are not computing, you are knob-turning). I'd feed it direct. However if you must see 600r here, add two 258 (220, 270) Ohm resistors (making an H-pad).
Click to expand...

Brilliant!  Thanks for the thorough explanation, it taught me the things I wasn't understanding in the technical write-ups.  I found that my optimal output gain setting is closer to 11% open on most sources so I put 65r of resistance between 2ea 270r resistors in a U, after the attenuator and before the XLR.  Now, my average output gain setting is a usable 25-35% open.  I tried getting more aggressive with the pad but the low end response lost some of it's magic somehow so I went back to the 65/270/270 U-Pad.  Not sure if my impedance is "correct" but the unit sounds better than ever and I now have an acceptable degree of control over the output.  Thanks again.
 
You can stack those pads back to back for whatever attenuation you need.  Take a look at the Gates SA-39 schematics.  Many of them had a bank of 5 to 6db H pads for input and output the owner could choose from. 

The Hs and Ts are essentially constant Z in and out so whatever is most convenient, or, as you have observed, which ever combination doesn't monkey the response too much the wrong way.
 
Have a look here

www.electronics-tutorials.ws/attenuators/t-pad-attenuator.html

You'll be able to build just about any type of attenuator you want. It's got a lot of good info and you'll be able to build a variable one: exactly what you need.
 

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