Hi folks. I would like to install a voltage divider network (hardwired/resistors) for one of my preamp's rca inputs. I need about 10db of attenuation. Can you guide me on how to calculate and construct this?
thanks very much
Roy C.
thanks very much
Roy C.
Here is an actual practical method of making a 10db attenuator for a preamp input-
Homemade RCA Attenuator : 5 Steps (with Pictures) - Instructables
Homemade RCA Attenuator : 5 Steps (with Pictures) - Instructables
This is my favourite, it points to were I am:
https://content.instructables.com/F...ebp&frame=1&width=1024&height=1024&fit=bounds
https://content.instructables.com/F...ebp&frame=1&width=1024&height=1024&fit=bounds
The first schematic says "unshielded cable" , which need not be the case, and in fact, typically isn't.
The second schematic is just plain wrong.
And the discussion of balanced vs unbalanced is also wrong:
"The main difference between the two is that one is represented by an unshielded cable while the other does not."
Incorrect. Balanced topology includes two signal wires carrying differential signals. The shield is optional, often left off. Grounding both ends of a shield of a long balanced cable will usually create a ground loop.
Then the rest:
"However, in reality the common RCA cable we use every day is made up of two conductors, one positive and one negative that also performs the shielding function against interferences (provided that is not too long)."
The designation "positive" and "negative" are incorrect, this is an AC circuit.
The shield provides electromagnetic and electrostatic shielding regardless of cable length. It does not provide any magnetic shielding, and dictates a ground at each end, which often causes ground loops, AC ground current flow through the shield, which then couples AC signals to the inner conductor in several different ways. But the shield always works for its intended purpose: electrostatic and electromagnetic shielding.
Finally, his calculations for a 10dB pad are wrong. He gets 6.9K and 4.7K for 10dB, but that's actually 7.8. For 10dB using 5% standard values, you need 6.9K and 3.3K (9.8dB)
A final thought: this pad would be best placed at the input to the receiving device, not at the output of a preamp. The pad increases the source Z driving the cable and load. Putting the pad at the input of the receiving device also will attenuate any noise picked up along the way.
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Why do you say so?
It´s the proper way to do it keeping the balanced signal and with NO reference to ground whatsoever ... as it should.
That it´s unrelated and superfluous in an RCA based discussion is something else, as well as the very sloppy redaction.
A practical way is to use your preamp's input impedance as the shunt value and just compute for the series value and place the series resistor at your preamp input. This way it increases your input impedance and not be affected by a lower impedance voltage divider network.
Rs=(Zin/(10^(-10/20)))-Zin
But that's about the worst approach from a noise perspective, you want as little series resistance as you can get away with, so choose the lowest resistor values your source can drive.
If the preamp input was 47k and you used just a 100k series resistor you are adding 40nV/√Hz of voltage noise. If you use 27k / 16k divider into that 47k preamp you have only 27k of series resistance (6 db less noise), and still looks like 39k to the source.
With a 6k8/3k3 divider you are down to 10k impedance, but a whole 12dB less noise than the 100k series resistor approach.
Ultimately its a compromize between keeping noise low and not loading the source too much.
If the preamp input was 47k and you used just a 100k series resistor you are adding 40nV/√Hz of voltage noise. If you use 27k / 16k divider into that 47k preamp you have only 27k of series resistance (6 db less noise), and still looks like 39k to the source.
With a 6k8/3k3 divider you are down to 10k impedance, but a whole 12dB less noise than the 100k series resistor approach.
Ultimately its a compromize between keeping noise low and not loading the source too much.
Series divider also has the problem that "Input Impedance" is often a rough nominal. In particular it may fall off significantly before the top of the audio band. Not unusual to find 1000pFd of RF filtering across 10k, so any useful series loss will be some dB down at 20kHz. (There are also series capacitors and deliberate bass-cut, but this is much less a problem.)
It's wrong in several ways. It's an H pad, fundamentally intended to match input and output impedances. We haven't dealt with match impedances on balanced audio circuits in many decades. The input legs are designated "Left" and "Right", which in audio refer to stereo channel designations. That's completely wrong for this pad. This schematic is essentially pointless.
Given all the other errors in that page, I don't know why anyone would pay any attention to anything stated, other than perhaps the mechanical design.
A practical way is to use your preamp's input impedance as the shunt value and just compute for the series value and place the series resistor at your preamp input. This way it increases your input impedance and not be affected by a lower impedance voltage divider network.
Rs=(Zin/(10^(-10/20)))-Zin
Assuming, of course, the input impedance is resistive across the audio band. Not always a valid assumption, and without verifying it, not the best approach.
We can assume all we want but unless you give it a try you'll never know. What's a 40nV noise to a 2.83V signal or even to 1V? Even if you multiply it by 20 for your power amp's gain. Also you need a better designed preamp if your nominal Zin changes too much across the audio band.
Thank you
....for all your input (pun intended). I am sorting through this information this afternoon. I appreciate you time(s) and knowledge.
....for all your input (pun intended). I am sorting through this information this afternoon. I appreciate you time(s) and knowledge.
Input measurement
The input impedance of the preamp is1 megOhm. for that series resistance equation, is the "10" in the "(10/20)" the amount of db attenuation?
The input impedance of the preamp is1 megOhm. for that series resistance equation, is the "10" in the "(10/20)" the amount of db attenuation?
Dumb question: why aren’t you using the correct values I already posted? Especially with that high input impedance.
If I use Blues' series equation Rs=(Zin/(10^(-10/20)))-Zin,
Rs=(1meg/(10^-.5)))-1meg
Rs=(1meg/(1.27))-1meg
Rs=.79-1
Rs=-.21meg
it's been a long time since HS math, check me out please....
Rs=(1meg/(10^-.5)))-1meg
Rs=(1meg/(1.27))-1meg
Rs=.79-1
Rs=-.21meg
it's been a long time since HS math, check me out please....
ah.... I didn't see your entire post.
....thank you for figuring that out....for doing the calculations too. I didn't see your entire post. .....and at that preamp voltage level, what rating of resistors would you use?
I appreciate your help.
....thank you for figuring that out....for doing the calculations too. I didn't see your entire post. .....and at that preamp voltage level, what rating of resistors would you use?
I appreciate your help.
40nV/√Hz is 5.5µVrms for 20kHz bandwidth, which put through a power amp with x30 gain becomes 170µV which is usually noticable from sensitive tweeters and obvious in headphones, and is significantly more noise than a good power amp will contribute.