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Attenuator Bandwidth

DonH56

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This article is to explore how line-level attenuators might affect the signal bandwidth. An attenuator is to reduce the signal amplitude from a source before it is applied to the next component. This could be between a DAC and preamp, or between preamp and power amp. Etc. The reason is usually to optimize the gain range, including noise and distortion as well as range of volume, through the signal chain (see the long thread on signal gain chains here: https://www.audiosciencereview.com/...opagate-through-my-system.33358/#post-1165118). Ideally the attenuators would do nothing but change the signal level, but in the real world they can affect the frequency response as well. And other things, but added noise and distortion from the attenuator itself is usually negligible.

A passive audio attenuator is usually comprised of two resistors, one in series with the source and the other to ground. This divides the signal going to the next component. For this discussion I’ll only consider fixed single-ended (RCA) attenuators; variable and XLR versions work the same way so the general analysis is still valid.

Consider a 20 dB attenuator as shown below. Audio sources are usually low in impedance, perhaps 100 ohms for a SS preamp and 1 k ohms for a tube (with a cathode follower). Audio inputs are usually high-impedance, ranging from perhaps 10k ohms to 100k ohms or more. This means we can neglect cable resistance since it is usually well below 1 ohm for a short interconnect cable (e.g. ~0.2 ohms for 10’). I chose 100k and 10k to provide about 20 dB attenuation; the ideal value is 20*log10[10k/(100k+10k)] = -20.8 dB. With ideal source (0 ohms) and load (infinite ohms, an open) we get 20.8 dB attenuation with no change in bandwidth as expected.

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Now let’s make it more interesting by adding source and load impedances plus a length of cable. I’ll use 100 ohms for the source (no capacitance) and 100k ohms plus 10 pF for the load, and assume 6’ of cable with 25 pF/foot of capacitance for 150 pF total (lumped as a single capacitor since cable resistance is negligible) and negligible inductance (at audio frequencies).

First, a trial without the cable. The attenuator is in the box, with the preamp driving it from the left and amp loading it on the right. The attenuation is changed just a little, to 21.6 dB from 20.8 dB (0.8 dB, about 9% lower than ideal) and the bandwidth is rolled off just a tiny bit (about 0.012 dB at 100 kHz, and only 0.0005 dB at 20 kHz). So far, no impact.

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There are two choices for adding the cable. Since the cable’s capacitance is in parallel with the attenuator, usually you put it where the impedance is the lowest, which would be after the preamp (source) since it is only 100 ohms. The result shows almost no change from the previous simulation, with low-frequency attenuation the same, and the amplitude down just 0.012 dB at 100 kHz and 0.0005 dB at 20 kHz as before.

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Here is the result if we add the cable after the attenuator, which means after the large series 100k resistor. Now we see the bandwidth is reduced, to -2.31 dB at 100 kHz, though only -0.12 dB at 20 kHz so still almost no impact in the audio band (note that the y-axis scale is different). The roll off at 1 MHz is about 18 dB, which could actually be useful if you are worried about RF noise (interference, RFI). For typical solid-state (SS) sources, it does not appear to be a significant problem no matter where you place the attenuator.

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Now repeat, but this time using 1k to approximate a tube preamp’s output. Note my old tube preamp is closer to 300 ohms or so, but I have seen specs as high as 3k, so it depends upon the design. With the attenuator at the preamp’s output, the low-frequency attenuation is now 21.7 dB, a tiny 0.1 dB (1%) difference. Attenuation at 100 kHz is 0.05 dB, again negligible though a hair larger than for the SS preamp. At 20 kHz, attenuation is 0.002 dB, again slightly larger than the SS preamp’s 0.0005 dB but still negligible.

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Placing the attenuator at the amplifier’s input has a bigger impact as expected, but still small. The series 100k-ohm attenuator’s impedance swamps the 1k-ohm source output so the result is very similar to that of the 100-ohm SS preamp. Attenuation at 100 kHz is -2.31 dB, and at 20 kHz -0.12 dB, essentially the same as for the SS preamp.

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Attenuators do have an impact, and the components and cables used influence that impact, but in the audio band their use should not generally be noticeable except for the reduced signal level. Of course, other combinations of source, attenuator, cable, and load impedances will yield different results, and some combinations might have much greater impact, but those would have to be pretty special cases IME/IMO.

HTH – Don
 

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DonH56

DonH56

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Here is a simulation with larger amplifier input capacitance. I previously assumed a small capacitor modelling the input stage, but many real-world amplifiers add a larger capacitor to filter out RF noise (and perhaps ultrasonic noise above the audio band). Two scenarios are shown: a solid-state system with 100-ohm preamp output and 10k-ohm amplifier input resistance; and, a tube system with 1k-ohm preamp output and 100k-ohm amplifier input, both having 100 pF input capacitors. These match a couple of amplifiers I have around the house. In the schematic, SS is on top, tube on the bottom. The 150-pF cable capacitance and 100k/10k (20 dB) attenuator are the same as before.
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The SS amp’s input is now the same as the attenuator’s output leg, leading to more attenuation and thus lower signal to the SS amp. Attenuation to the SS amp is 26.5 dB compared to 21.7 dB to the tube amp, about a 5 dB difference. This is one of the things to watch when using passive attenuators – you do not always get the attenuation you expect. There is still excess bandwidth for either amp, however: the -3 dB point is 339 kHz for the SS circuit and 186 kHz for the tube circuit. The tube circuit has about half the bandwidth of the SS amp due to the tube’s higher driving and load impedances, but it is still well above the audio band.
 

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DonH56

DonH56

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Now let’s look at a passive volume control, which is essentially a variable attenuator. Since it is passive, the resistance seen by the preamp and amp varies with the control’s setting, which in turn affects the frequency response (among other things). I’ll use two controls, a 10k-ohm potentiometer (pot, volume control) for the SS circuit and a 100k-ohm pot for the tube circuit. The attenuation is stepped from 0 to “infinite” in a few steps so we can see how the frequency response changes with volume using a passive control. The attenuation steps are about 0 dB, 2 dB, 8 dB, 21 dB, 40 dB, 60 dB, 80 dB, and 140 dB. The corresponding series or “top” resistance value of the SS control in ohms are 0, 1k, 5k, 9k, 9.9k, 9.99k. 9.999k, and 9.999999k. The tube control has the same steps multiplied by 10k-ohms since it uses a 100k-ohm control. Notice how nonlinear the control resistance becomes; this is why most volume controls use some sort of logarithmic taper so you have fine control over the lower range of volume.

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While the roll-off is still above the audio band for this example, both circuits have varying frequency response as attenuation (volume) is changed, markedly more so for the tube circuit due to the higher impedances used. The lowest corner (roll-off) frequency occurs when the control is at the halfway point where the attenuation is about 8 dB (roughly 6k from the control and another 2 dB due to the other resistances in the circuit). At that point, the SS bandwidth (-3 dB frequency) is about 794 kHz and the tube bandwidth is about 79.4 kHz reflecting the 10:1 difference in resistances. As the control moves past the 50% point, further attenuating the signal, the control increasing the series resistance but at the same time the output resistance (to ground) decreases. That in turn decreases the impact of the amp’s input capacitance since the effective driving resistance is lower.

As you turn the pot, starting from full (maximum) volume the input (series) resistance starts at 0 ohms and the output resistance of the control is 10k (100k). At the halfway point, the control’s input resistance and output resistance are the same (5k or 50k). As you turn the control to further decrease the volume (attenuating the input signal to the amp), the input resistance continues to increase, and the output resistance reduces, so the control acts as a divider to reduce the signal. When the SS control is very low, the input series resistance is perhaps 9.99k and so the output, in parallel with the amplifier’s inputs, is then only 0.01k ohms, shunting the amp’s input resistance and capacitance to the capacitor has less effect on the frequency response.

HTH - Don
 
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DonH56

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One more simulation, this time with 10’ of 25 pF/ft cable (typical coax) on both sides of the passive volume control. This is the case when you might want to have the control handy so there are cables running from you chair to the preamp and back to the amplifier.

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The results are similar to before but now the bandwidth roll-off is greater due to the extra cable, especially the cable after the control. The SS circuit is down 3 dB at 229 kHz (compared to 794 kHz before) and the tube circuit’s bandwidth is 23 kHz, just above the audio band. Sometimes cables matter! Though really the effect here is capacitance due to the length with a passive volume control; a $10 cable and a $1000 cable will do the same thing as long as they have similar capacitance.
 
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RayDunzl

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I ran a pair of 10k pots in the middle of a pair of 1 meter RCA cables between CD player and Power Amp for years without complaint.

I guess you proved why above.

I had nothing with which to measure.

One pot on each cable, so channel balance was never a problem, stand in the middle and set by ear.
 
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DonH56

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The question of crosstalk and how large a capacitor was needed to inject significant crosstalk was raised on another thread. Here is a simulation with two controls, both in the SS circuit, to represent a stereo volume control. The upper input is grounded ("off") so the only response on that channel is crosstalk injected by the other side. Coupling is usually from the capacitance of the cables inside the box, and often a large part is the coupling between the two halves (sides) of the volume control itself. To model the coupling I used two 50 pF capacitors, one on each side of the attenuator, and swept the volume as before. Minimal coupling occurs when the volume is at max, and it increases from there, peaking around -45 dB at 20 kHz with the volume at mid-scale (halfway point on the control). I could vary the capacitors around the circuit to change the amount, and move the corner frequency around so it appear flatter or even rolls off at the top of the audio band, but the point here is that very small coupling capacitance in that little box can cause significant crosstalk.

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KSTR

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Nice Analysis, thanks
To model the coupling I used two 50 pF capacitors, one on each side of the attenuator, and swept the volume as before
I've just measured the section-to-section capacitance of the "industry standard" 10k ALPS RK27 which is quite a large pot physically, but ALPS took care to space the two decks with enough clearance:
Any pin of one section to any other pin of the second section was below 4pF, tested a 10kHz and 100kHz frequencies. Worst-case capacitive crosstalk (vs rotation) will be correspondingly very low (as for the sim, it might be better to model distributed capacitance -- 10 taps or so -- between sections alng the track and to "world"/GND).

Even with reduced spacing between decks I would safely assume the cross capacitance is below 10pF for just any dual deck pot. For the concentric types (rarely used in audio, hopefully) it might be higher and could be approaching several 10s of pF, though.
 

pma

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I've just measured the section-to-section capacitance of the "industry standard" 10k ALPS RK27 which is quite a large pot physically, but ALPS took care to space the two decks with enough clearance:
I agree, this Alps is fine. The point is the PCB design - parallel tracks yes/no, groundplane yes/no, etc....
 
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DonH56

DonH56

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Nice Analysis, thanks

I've just measured the section-to-section capacitance of the "industry standard" 10k ALPS RK27 which is quite a large pot physically, but ALPS took care to space the two decks with enough clearance:
Any pin of one section to any other pin of the second section was below 4pF, tested a 10kHz and 100kHz frequencies. Worst-case capacitive crosstalk (vs rotation) will be correspondingly very low (as for the sim, it might be better to model distributed capacitance -- 10 taps or so -- between sections alng the track and to "world"/GND).

Even with reduced spacing between decks I would safely assume the cross capacitance is below 10pF for just any dual deck pot. For the concentric types (rarely used in audio, hopefully) it might be higher and could be approaching several 10s of pF, though.
Good to know! Years ago ALPS was the "go-to" pot for precision/high-end applications. Some of the cheap pots have no shielding and hardly any separation between the two sections leading to much greater coupling. I last measured them many years ago and have no memory of the numbers except a vague recollection of 10's of pF. The worst was one of the cheap "conductive plastic" models that had just one disc ("wafer") with resistive coating on both sides of the disc.

Setting up a distributed coupling along the pot itself would be more realistic but also a pain to set up and I am not sure that is the major coupling source. All I really wanted to do was check the magnitude of capacitance needed (see below).

EDIT: Duh, I could just ratio the capacitance value using a swept (stepped) parameter as I did for the resistor. But I don't think it's worth it, am lazy, have other things that need doing, pick your excuse... ;)

As @pma said, the greater contribution is usually wiring, PCB traces, and so forth. I suspect a number of inexpensive volume controls do not use a PCB, just run wires from input jacks to pots to output jacks. Bundling and routing wires together, particularly unshielded twisted pairs (or even just plain parallel wires), provides a clean look but a lot of crosstalk.

I do not have data, however... Mainly I got nervous after repeated arguments that pF-magnitude coupling could not be significant at audio frequencies though my gut and calculations showed it could be at the impedances and levels shown in Amir's review. I decided to do a quick check after I completed the other sims and figured I might as well post it. The main take-away is one I constantly remind myself: where -40 dB ~ -60 dB coupling in an RF system at 10 GHz can be a challenge, -100 dB at audio is equally if not more so due to the orders of magnitude lower target levels and similarly higher impedances involved.
 
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DonH56

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Based on the data from other kind members I did a quick simulation using 400 pF amplifier input capacitance instead of 100 pF with the volume control at 50% (worst case for bandwidth). The SS circuit's -3 dB bandwidth dropped from 229 kHz to 200 kHz, and the tube circuit from 23 kHz to 20 kHz, a significant but not huge drop. Some amplifiers include an ultrasonic filter with larger capacitors and that will have a larger bandwidth reduction, natch.

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Volume control at 50%:

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d888sp4

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Hi DonH56,
very interesting indeed, congratulations!
Could you post the .asc files for the simulations?
It would be very useful...
Thank you very much.
Albert
 
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DonH56

DonH56

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Hi DonH56,
very interesting indeed, congratulations!
Could you post the .asc files for the simulations?
It would be very useful...
Thank you very much.
Albert
I usually just throw together temporary files, tweak a bit for the plots I want, then toss them so may not have everything. That said, I attached the last version I did, including a bunch of cable and crosstalk components. You can always recreate the schematic yourself from what I posted or use your own system as a model.

Edit: *.asc is not an allowed file type so I changed the extension to .zip. It is not actually a zip file, just change the extension back to .asc.
 

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MRC01

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For a typical attenuator used as a volume control between the source (DAC) and power amp, can't we simply compute 1 / (2 * pi * R * C) for bandwidth?
Where R is the attenuator output impedance (worst case, typically 2500 Ohm for a 10k ladder attenuator) and C is the capacitance of the cables?

For example, when R = 2500 and and cables have 50 pF, bandwidth (-3 dB point) is 1,273 kHz.

Or does this somehow over-simplify the situation?
 

pma

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Yes, it is a trivial Tau = R*C exponential response, where R is a resistance resulting from parallel connection of source output resistance and amplifier input resistance (load resistance). No rocket science here.

Of course in case that amplifier has more complex input filter, like two stage RC, then it is not so simple, but this is not a topic of this thread.
 
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DonH56

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For a typical attenuator used as a volume control between the source (DAC) and power amp, can't we simply compute 1 / (2 * pi * R * C) for bandwidth?
Where R is the attenuator output impedance (worst case, typically 2500 Ohm for a 10k ladder attenuator) and C is the capacitance of the cables?

For example, when R = 2500 and and cables have 50 pF, bandwidth (-3 dB point) is 1,273 kHz.

Or does this somehow over-simplify the situation?
The effective resistance varies with the source, load, and attenuation setting but simple calculations can give you the correct resistance. You may need to take into account where the capacitance is connected (before or after the potentiometer) to generate the time constants. For low source impedance, you can neglect the cable capacitance on the input, so then the time constant becomes the parallel resistance of both sides of the control and load (input) with the capacitance of the output cable (from attenuator to amp) in parallel with the amp's input capacitance.

Example: Assume a 10k pot at midpoint so 5k each side, and 100k amp input yields 2.44k (only a little lower than the two 5k in parallel since the amp input is high). Then you'd add the cable capacitance plus whatever the amp's input capacitance is to get total C and calculate the result.

HTH - Don
 
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