Stepped Input Attenuator Rin Value

[gg85]

It is actually quicker to work it out than to scour the Web for a right calc, and *check* that it is right.

48 steps, starting from 0dB, goes to -47dB.

-47dB is 0.004467.

0.004467 times 47,000 is 210. This is your bottom resistor.

Each 1dB step is a 1dB or 1.12202 bigger resistor.

5111 --- top resistor
4555.4
4060
.....
404
362
322.6
287.5
235.6
210 ------ bottom resistor

While some of these numbers look frighteningly exact, that's just math. When stacked in an array, 1% precision will give inaudible (nearly unmeasurable) error.

I'm revisiting this years later and wanted to double-check something...I've got a 24 position switch and want to use a range from -50dB to -38.5dB. I've worked out the resistor values with K factor of 1.05925 so my bottom is 148.62705 and the top is 558.5960469. Do I then just sum all resistors and minus that from 47000 and have 39522.53099 extra resistances before the switch? Obviously, those are exact values and not actual resistor values, but is that thinking correct? Is there an easy way to change the range of the switch to be 11.5dB from say -60dB up to -48.5dB etc. by varying that input resistor?

Thanks!

 

[abbey road d enfer]

I'm revisiting this years later and wanted to double-check something...I've got a 24 position switch and want to use a range from -50dB to -38.5dB. I've worked out the resistor values with K factor of 1.05925 so my bottom is 148.62705 and the top is 558.5960469. Do I then just sum all resistors and minus that from 47000 and have 39522.53099 extra resistances before the switch?

That's correct.

Is there an easy way to change the range of the switch to be 11.5dB from say -60dB up to -48.5dB etc. by varying that input resistor?

The range is not changed when changing the base attenuation. So you can shift the range by 10 dB by increasing the input resistor in such a way that the total resistance increases by a factor 3.16, for a total value of 148 520 ohms. The 39522r resistor must be replaced with a 138042 ohms .

 

[gg85]

That's correct.

The range is not changed when changing the base attenuation. So you can shift the range by 10 dB by increasing the input resistor in such a way that the total resistance increases by a factor 3.16, for a total value of 148 520 ohms. The 39522r resistor must be replaced with a 138042 ohms .

Thanks for the reply, would you mind explaining the 3.16 part please? If I do need to adjust the range, I might start with a 1meg trimmer and dial it in that way, although would the increase in total resistance then mean that my input transformer is loaded incorrectly?

 

[abbey road d enfer]

Thanks for the reply, would you mind explaining the 3.16 part please?

10dB is a ratio of 1:3.16 (actually square root of 10)

If I do need to adjust the range, I might start with a 1meg trimmer and dial it in that way, although would the increase in total resistance then mean that my input transformer is loaded incorrectly?

TBH I don't know the circuit this attenuator is to be paired with. I didn't even know there was a xfmr involved. I'm too lazy to read the old posts. If there's a xfmr, you may have to put a resistor across the secondary.

 

[Matt Syson]

If you are calculating the resistor values for switched attenuators then you must know the impedance of the SOURCE and the 'destination' as both of these will have a significant effect unless the whole attenuator is designed as 'constant impedance for input and output. The maths gets a lot more involved when you take these two extra factors into account and will totally bow away any notion of 'accuracy' although if the attenuator is 'stereo' at least both channels will be similarly inaccurate (so the image will stay correctly centred. Putting a 1 meg trimmer in may get you somewhere near an approximate extra resistance BUT it will upset the original step values. Constant impedance (typically 600 Ohm for 'old style' audio work, or 50 Ohm for RF work allows the switching steps to remain accurate.
Matt S

 

[abbey road d enfer]

If you are calculating the resistor values for switched attenuators then you must know the impedance of the SOURCE and the 'destination'

Indeed, the load ("destination") matters, but here it's a tube grid, which can be considerd as negligible. The circuit emulates a potentiometer, which law does not vary whatever the source resistance, as long as the load is negligible.

 

[gg85]

10dB is a ratio of 1:3.16 (actually square root of 10)

TBH I don't know the circuit this attenuator is to be paired with. I didn't even know there was a xfmr involved. I'm too lazy to read the old posts. If there's a xfmr, you may have to put a resistor across the secondary.

It's for an sta level build, with stepped in/outs

Just trying to get my head around this. So if I wanted to increase the resistance by 20dB am I looking for the square root of 20 (4.472), or the k factor of 20dB (10), and then I multiply 47k by whichever is the correct value? Then minus the resistance of my attenuator to give the new base resistance?

Would a 47k resistor across the secondaries take care of the loading or will it be a parallel resistor calculated depending on the total resistance of the attenuator + base resistor? I could probably use an h-pad before the input transformer but not sure what that might do to the s/n by attenuating before the circuit.

Thanks again for your help with this, I'm trying my best to google as much as I can but I'm not finding as much as I'd hoped for!

 

[abbey road d enfer]

Just trying to get my head around this. So if I wanted to increase the resistance by 20dB am I looking for the square root of 20 (4.472), or the k factor of 20dB (10),

20dB is a ratio of 1:10
10dB is a ratio of 1:3.16

and then I multiply 47k by whichever is the correct value? Then minus the resistance of my attenuator to give the new base resistance?

Correct.

will it be a parallel resistor calculated depending on the total resistance of the attenuator + base resistor?

Correct.

I could probably use an h-pad before the input transformer but not sure what that might do to the s/n by attenuating before the circuit.

It's just another possibility. The main difference is that when you attenuate, you decrease the level into the xfmr, which will probably result in less distortion. However, in some circumstances, teh source impedance may incraese, which may result in not so good distortion figure.

 

[Gold]

I could probably use an h-pad before the input transformer but not sure what that might do to the s/n by attenuating before the circuit.

I have a general question about putting a pad or resistors (like for a zero field input transformer) before the input transformer. Doesn't that seriously degrade the CMRR even if you carefully match resistors?

 

[gg85]

gg85 said:
and then I multiply 47k by whichever is the correct value? Then minus the resistance of my attenuator to give the new base resistance?

Correct.

Ok great! One last question, if the attenuator is a dual gang for balanced (grid high/low), does that change the desired resistance? It would essentially be 2 47k attenuators, right? And then I would just change my parallel resistor across the secondaries to accommodate?

 

[abbey road d enfer]

I have a general question about putting a pad or resistors (like for a zero field input transformer) before the input transformer. Doesn't that seriously degrade the CMRR even if you carefully match resistors?

It depends. If the attenuator is refrenced to ground (which may be needed in the presence of high CM voltage), then the CMRR almost only depends on resistor pairing.
When the attenuator is floating, the input stage CMRR is generally dominant. For active inputs, it must be evaluated on a case-by-case analysis.

 

[abbey road d enfer]

Ok great! One last question, if the attenuator is a dual gang for balanced (grid high/low), does that change the desired resistance? It would essentially be 2 47k attenuators, right? And then I would just change my parallel resistor across the secondaries to accommodate?

Show us a schemo.

 

[gg85]

Show us a schemo.

Here's the sta level schematic. I'm essentially changing the input dual gang pot to a stepped attenuator. There is a 200k resistor across the secondaries of the input transformer, and then a dual gang 100k pot. Does this mean that it's seeing 100k on the secondaries (200k in parallel with dual 100k pot), or 66.67k?

 

[abbey road d enfer]

Here's the sta level schematic. I'm essentially changing the input dual gang pot to a stepped attenuator. There is a 200k resistor across the secondaries of the input transformer, and then a dual gang 100k pot. Does this mean that it's seeing 100k on the secondaries (200k in parallel with dual 100k pot), or 66.67k?

Am I right thinking you want to ditch the input H-pad and put more attenuation at the secondary?

 

[gg85]

Am I right thinking you want to ditch the input H-pad and put more attenuation at the secondary?

yep that's right

 

[abbey road d enfer]

Then you must know that the actual xfmr response will vary with the source impedance, which may necessit tuning the secondary loading. You would need to input a square wave from a generator with an impedance matching that of your intended source, and tun ethe loading for minimum overshoot, as seen on an oscilloscope.

 

[gg85]

That's correct.

The range is not changed when changing the base attenuation. So you can shift the range by 10 dB by increasing the input resistor in such a way that the total resistance increases by a factor 3.16, for a total value of 148 520 ohms. The 39522r resistor must be replaced with a 138042 ohms .

I've built up my attenuator and seem to be having some issues with testing. I've got a signal generator feeding the attenuator, and the output is picked off with my scope. Reading the rms voltages it doesn't seem like I'm jumping in 0.5dB steps for some reason so I wanted to confirm that I'm doing things correctly. I have the 1k sine coming in through the bottom of the switch (position 24), then ground off the top of the switch (position 1). My 39.2k resistor off the pole, and the scope on the other side of that resistor. The waveform looks correct as I move through the steps but it's not in 0.5dB steps. I thought it might be possible I've populated the switch backwards and the log steps are incorrect, but when I reverse the wiring to accommodate for that, it still isn't 0.5dB. Have I missed something here?

Here's a link to the spreadsheet I've used for resistor values in case that helps!

https://docs.google.com/spreadsheet...ouid=115964726883800851208&rtpof=true&sd=true

 

[abbey road d enfer]

My 39.2k resistor off the pole, and the scope on the other side of that resistor.

I don't get it. Pls draw a schemo.

 

[gg85]

for sure, I hope I've drawn it well enough for you!

 

[abbey road d enfer]

You realize the position you have marked 1 is actually OFF, so there is an infinite step there; is it what you want?
Now the 39.2k resistor should be feeding the string of resistors.