inline attenuators?
Mar 6, 2006 at 11:46 PM Thread Starter Post #1 of 10

shimage

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I'm sorry if this has been discussed already, but I want to put together some inline RCA attenuators, and I was wondering what, if anything, I ought to be concerned about. The pot on the amp (M³) is 50k, and I was thinking of using a ~10k divider; would that load the source unduly? I suppose the answer will probably come down to "try it and see", but I figured I might as well ask first anyway.

Yes, I could replace the resistors in the opamp circuit (reduce the gain), but 1) I don't want to remove the board from the chasis and 2) the gain would be reduced to below unity, which as I understand it isn't so great for stability.

Thanks in advance for any suggestions.
 
Mar 7, 2006 at 12:02 AM Post #2 of 10

rickcr42

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"10K divider" at what attenuation level ?

for instance to get a -20dB attenuation into a 50K load (the input pot) you need a 500K Ohm inline resistor
eek.gif
(10:1 division)

If instead you mean making a full voltage divider fronting the volume pot with the parallel resistor being 10K you should be fine as long as you keep the connection between the two very short and also realise you just lowered the input Z to 10K ohms and not the former 50K ohms.

BTW-the same -20dB attenuator into a 10K resistor would be 100K for the series resistor part
 
Mar 7, 2006 at 12:20 AM Post #3 of 10

shimage

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Thanks for the reply! Sorry, I didn't mean a single inline resistor; I didn't want to have to use such large resistor values, as you mentioned. I was thinking of a 10k:1k voltage divider (basically 10k total); the path ought to be short, since I mean to fit it all into something slightly larger than an RCA plug.

My question concerned precisely what you pointed out: that I will have lowered the input impedance to 10k. Will that load be excessive for a typical (and non-portable) source? I gather that this is so, since I've seen 10k pots, but I thought it wouldn't hurt to ask first.
 
Mar 7, 2006 at 12:51 AM Post #4 of 10

rickcr42

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Quote:

My question concerned precisely what you pointed out: that I will have lowered the input impedance to 10k. Will that load be excessive for a typical (and non-portable) source? I gather that this is so, since I've seen 10k pots, but I thought it wouldn't hurt to ask first.


the divider to pot side is fine but the input side would require the source to have a serious drive capability,able to drive a 1K load without any upper frequency attenuation.

Are you sure you need a full -20dB attenuation ?

BTW-I think there may be a link in the DIY links section to an online attenuator calculator but if not just google for it.Many out there
 
Mar 7, 2006 at 1:29 AM Post #5 of 10

shimage

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Thanks, but my computer comes with a calculator
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I might not go the whole 20 dB. From Tangent's website, I gather that the DACT CT2 has 2.5 dB steps (there are 23 or so steps), so that would cover 8 steps or 1/3 of the range, which is roughly what I'm looking for, though I'll probably tweak it a bit before I put it together.

As an explanation, there's a gain switch on the amp, which switches between what I believe are factors of 2 and 8. The gain of 2 is too much for me (at least with the source that's hooked up to it), even with DT880s, so I think at least 12 dB is in order.

As for having to drive 1k, I'm pretty sure that it wouldn't have to drive a 1k load, since it's got that 10k resistor in series with everything else.
 
Mar 7, 2006 at 1:37 AM Post #6 of 10

rickcr42

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Quote:

Thanks, but my computer comes with a calculator


yes but doubful it does Logarithmic dB calculations into an impedance which is what an attenuator is

Quote:

As for having to drive 1k, I'm pretty sure that it wouldn't have to drive a 1k load, since it's got that 10k resistor in series with everything else.


the 1K IS the input impedance because it is the first parallel resistance that will be seen by the driving source.
 
Mar 7, 2006 at 2:49 AM Post #9 of 10

shimage

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Wow, that's a lot of info. Ok, I was under the impression that I knew about circuits, certainly ones as basic as this, but perhaps I made a boo-boo.

The second link you gave has a three-terminal ... thingy, and he mentions XLR, so I'm assuming this is a balanced signal. Since none of my equipment is balanced, this is not directly applicable. I tried out the program (thank god it's dos, or it wouldn't have run on my computer), and it looks like its "Microphone Pad" setting (since I don't want to fiddle with the frequency response as much as possible) is using the circuit from his Fig. 1. Which (I'm not sure, but) I think is not directly usable.

As for the first link, the situation I was thinking of was the so-called "L-pad".
Quote:

Originally Posted by first linky
Practically speaking, the absolute loss is determined by the parallel combination of the shunt resistor and the input impedance of the mike preamp. Since the preamp bridges the shunt resistor, the actual loss is slightly greater than calculated (less than 1dB). For loss values less than 20 dB, you may have to raise the value of the shunt resistor to make the series resistors large enough to not excessively load the source. In that case, the contribution of the preamp's input impedance to the attenuation error will increase.


This is at least consitent with what I thought was happening. If I understood you correctly you implied that increasing the series resistor would not decrease load on the source (since it wouldn't affect the "first parallel resistance" in any way). I could be wrong, and if I am I would appreciate an explanation of why. Either way, thanks a bunch for your help!
 

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