Lazu Iguana
look no further than AKG K270 Studio
cozy 75R, 2 drivers per ear
for cones mated with wood, resort to something really having 2W or so
(try your B1T with phones, it can be fun)
look no further than AKG K270 Studio
cozy 75R, 2 drivers per ear
for cones mated with wood, resort to something really having 2W or so
(try your B1T with phones, it can be fun)
Yeah a buddy and I had a conversation about how 8 ohms became the "standard" loud speaker impedance. Our agreement was, when solid state overtook tube technology, speaker manufacturers loved the lower impedences possible because they are easier speakers to make. Larger mag-wire diameters, looser tolerences, higher manufacturing yields, more durable, etc. Alas, 96dB are usually 10 or 12 in. diameter. 4 x 12 means a couple of Marshall guitar cabs on your desk top. Not exactly small and defininately not point source.🙄 I have a pair of Electro-Voice 101dB efficient drivers, old 15L from ths 80's. Darn, they're 15 in. guitar speakers though and only go to about 4kHz.ZM is right... 2W way more practical.
The ACP does sound really nice with 80 ohm head phones. Just day-dreaming about something mounted in wood
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@cubicincher
Sorry I didn’t mean to be so curt. I meant that I, too, found this sheet and it doesn’t address what @6L6 and I were discussing. The pot I have on hand seems to match NP’s specs, but the documentation for the various part numbers from Alps is really lacking. Thanks
Sorry I didn’t mean to be so curt. I meant that I, too, found this sheet and it doesn’t address what @6L6 and I were discussing. The pot I have on hand seems to match NP’s specs, but the documentation for the various part numbers from Alps is really lacking. Thanks
From this site - loosely translated using Google...
POTENCJOMETR ALPS RK271 STEREO LOG 20KAX2 8110428691 - Sklep internetowy AGD, RTV, telefony, laptopy - Allegro.pl
"Two-channel (stereo) potentiometer with logarithmic characteristics, commonly used for volume control in audio devices Currently, potentiometers are sold from the "906G" series"
20k audio taper. Note... Not all logs are base 10 😉 Example: Exponential expression: 2^3 = 8 => log base 2 of 8 = 3
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@itsallinmyhead
Thanks. That means the part is the drop-in equivalent for NP’s speced part. This is what I needed to know.
Also, yes, I know the math. The vast majority of the things we discuss on this forum I don’t know even a little. math isn’t one of them. My point, apparently not articulated well, was:
1) in a linear taper you have one unit of change for every unit of adjustment is the part;
2) in a log taper you have a decreasing amount of change for every increasing unit of adjustment;
3) in a 10^x taper you have an increasing amount of change for every increasing unit of adjustment
Basically, in 2) when you turn the knob half way you have 10% of the total resistance (assuming a 10% taper). A linear part should have a 1:1 (or at least constant proportion) between the amount of turn in the knob and the resistance applied. So half way through the turn or slide of a linear taper should result in 50% of the resistance, not 10% of the log taper. A log taper should match hearing better than linear since volume amplitude is perceived logarithmically. (Someone who know better please correct me, this is Webber’s law?) I read somewhere that 3) is sometimes used for balance controls, but I’m not sure if this is true.
Thanks
Thanks. That means the part is the drop-in equivalent for NP’s speced part. This is what I needed to know.
Also, yes, I know the math. The vast majority of the things we discuss on this forum I don’t know even a little. math isn’t one of them. My point, apparently not articulated well, was:
1) in a linear taper you have one unit of change for every unit of adjustment is the part;
2) in a log taper you have a decreasing amount of change for every increasing unit of adjustment;
3) in a 10^x taper you have an increasing amount of change for every increasing unit of adjustment
Basically, in 2) when you turn the knob half way you have 10% of the total resistance (assuming a 10% taper). A linear part should have a 1:1 (or at least constant proportion) between the amount of turn in the knob and the resistance applied. So half way through the turn or slide of a linear taper should result in 50% of the resistance, not 10% of the log taper. A log taper should match hearing better than linear since volume amplitude is perceived logarithmically. (Someone who know better please correct me, this is Webber’s law?) I read somewhere that 3) is sometimes used for balance controls, but I’m not sure if this is true.
Thanks
Any time. Enjoy the build! ACP+ is crazy good, IMO.
No worries, I just wasn't sure what you meant.
1) Yes.
2) No. In general, lower rate of change at first, more at the end.
3) I don't know. Conceptually, if you mean an inverse log (exponential) then it would have a more rapid rate of change at first and then taper off at the end. I wouldn't use it for a volume control myself. I've never seen a 10^x pot, and wouldn't know a use for one, but I can work with the concept. Inverse log pots I've seen are not 10^x.
Yes, a linear pot would be at 50% of its resistance 1/2 way through the turn.
Potentiometer taper | Resistor types | Resistor Guide
No worries, I just wasn't sure what you meant.
1) Yes.
2) No. In general, lower rate of change at first, more at the end.
3) I don't know. Conceptually, if you mean an inverse log (exponential) then it would have a more rapid rate of change at first and then taper off at the end. I wouldn't use it for a volume control myself. I've never seen a 10^x pot, and wouldn't know a use for one, but I can work with the concept. Inverse log pots I've seen are not 10^x.
Yes, a linear pot would be at 50% of its resistance 1/2 way through the turn.
Potentiometer taper | Resistor types | Resistor Guide
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Freejazz - Thank you for the photo. That’s a 20k audio taper RK27 and will be perfect for the ACP.
to freejazz00 #843
Hello freejazz00,
everything is o.k.. No worries. 😉
Greets
Dirk 😀
Hello freejazz00,
everything is o.k.. No worries. 😉
Greets
Dirk 😀
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isn't 4 x 8 ohms speakers in series, 32 ohms? Build a nice little line-array of FE103's . . . Then maybe increase the quiescent current a bit.
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@ItsAllInMyHead
Yes and no. First the "no."
In a log taper you have a decreasing amount of change for every increasing unit of adjustment. For example, to go between 1 and 10 on the x-axis you increase one unit on the y-axis. But to advance the subsequent unit on the y-axis you'd need to increase from 10 to 100, or 90 steps. Likewise, to advance from 2 to 3 on the y-axis you need to advance from 100 to 1000, or 900 steps. When this pattern is plotted the resulting line is convex down. Therefore 10 steps on the x-axis gets you less and less change on the y-axis as you move to the right along the line. Conversely, in 10^x you have an increasing amount of change for every increasing unit of adjustment, which results in a convex up line when plotted.
Now for the "yes," which I didn't know before this discussion. Thank you for "making" me look it up.
What we call an audio log taper is 10^x not log x, and, moreover, it is a pseudo 10^x curve. You are correct to write "lower rate of change at first, more at the end." But this is a convex up, so not log x.
For example, see the plot here:
Potentiometer taper | Resistor types | Resistor Guide
So the curve is the inverse of the perception of the magnitude of sound a listener is hearing. The pseudo part comes in when manufacturers build the parts incorporating two linear segments of line to represent the exponential curve. I guess it it cheaper to make it this way. Therefore my previous 2) is correct, at least when using a true log taper and not an audio log taper. That's confusing.
So I continue to have Good math and bad electronics. Now I know.
Thanks,
Yes and no. First the "no."
In a log taper you have a decreasing amount of change for every increasing unit of adjustment. For example, to go between 1 and 10 on the x-axis you increase one unit on the y-axis. But to advance the subsequent unit on the y-axis you'd need to increase from 10 to 100, or 90 steps. Likewise, to advance from 2 to 3 on the y-axis you need to advance from 100 to 1000, or 900 steps. When this pattern is plotted the resulting line is convex down. Therefore 10 steps on the x-axis gets you less and less change on the y-axis as you move to the right along the line. Conversely, in 10^x you have an increasing amount of change for every increasing unit of adjustment, which results in a convex up line when plotted.
Now for the "yes," which I didn't know before this discussion. Thank you for "making" me look it up.
What we call an audio log taper is 10^x not log x, and, moreover, it is a pseudo 10^x curve. You are correct to write "lower rate of change at first, more at the end." But this is a convex up, so not log x.
For example, see the plot here:
Potentiometer taper | Resistor types | Resistor Guide
So the curve is the inverse of the perception of the magnitude of sound a listener is hearing. The pseudo part comes in when manufacturers build the parts incorporating two linear segments of line to represent the exponential curve. I guess it it cheaper to make it this way. Therefore my previous 2) is correct, at least when using a true log taper and not an audio log taper. That's confusing.
So I continue to have Good math and bad electronics. Now I know.
Thanks,
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LOL, you're up relatively late, while I'm up relatively early if you're in SoCal. 🙂 Where we differ in our description is "in a log taper you have a decreasing amount of change for every increasing unit of adjustment". So, I'll change terms, just to get on the same page.
The amount (I'd use the term rate, but I think I know what you mean) of change does not decrease over the course of the turn of the pot. It starts relatively low and goes higher, if we are discussing the change in resistance relative to the change in degrees of rotation. You can see in the graph. I think the graph can be potentially misleading, since with attenuation we generally wire the pot to reduce the resistance with a clockwise turn vs. increase, but it's the same principle.
As you've correctly said earlier, because of the way we hear and perceive changes in relative volume, and since we like our hand to move relatively in line with what we perceive to be the change in sound, and we'd like some finer control at lower volume levels (highest attenuation/highest resistance), we use log pots with an audio taper. So, we have the lowest amount of change in resistance (the way we wire our pots) at the earlier degrees of rotation, then as we turn the pot, it increases.
You move X to affect Y. X is the turn of the pot. Y is the resistance. See the graph we both linked, and use the blue line. BTW, I don't love the way they've labeled their graph, but let's assume the pot is wired the way we would use it to decrease the resistance with a clockwise turn. Let's also assume it's a 100k pot. You have to turn to 50% to reduce the resistance from 100k to 90k, correct? Then from 50% to 100% rotation you move it the rest of the way from 90k to 0k.
Perhaps I am simply not understanding how you've written it, perhaps we're saying the same thing in different ways, or perhaps I have it completely wrong, and I'll learn a bit. Either way, thanks to you also. I always enjoy learning and more importantly how to express ideas in a way that more people can understand. That's not my strong suit.
Happiest of holidays!
The amount (I'd use the term rate, but I think I know what you mean) of change does not decrease over the course of the turn of the pot. It starts relatively low and goes higher, if we are discussing the change in resistance relative to the change in degrees of rotation. You can see in the graph. I think the graph can be potentially misleading, since with attenuation we generally wire the pot to reduce the resistance with a clockwise turn vs. increase, but it's the same principle.
As you've correctly said earlier, because of the way we hear and perceive changes in relative volume, and since we like our hand to move relatively in line with what we perceive to be the change in sound, and we'd like some finer control at lower volume levels (highest attenuation/highest resistance), we use log pots with an audio taper. So, we have the lowest amount of change in resistance (the way we wire our pots) at the earlier degrees of rotation, then as we turn the pot, it increases.
You move X to affect Y. X is the turn of the pot. Y is the resistance. See the graph we both linked, and use the blue line. BTW, I don't love the way they've labeled their graph, but let's assume the pot is wired the way we would use it to decrease the resistance with a clockwise turn. Let's also assume it's a 100k pot. You have to turn to 50% to reduce the resistance from 100k to 90k, correct? Then from 50% to 100% rotation you move it the rest of the way from 90k to 0k.
Perhaps I am simply not understanding how you've written it, perhaps we're saying the same thing in different ways, or perhaps I have it completely wrong, and I'll learn a bit. Either way, thanks to you also. I always enjoy learning and more importantly how to express ideas in a way that more people can understand. That's not my strong suit.
Happiest of holidays!
I hope everyone's had a great Christmas. Time for a little Boxing Day soldering.
If I've read the ACP+ article and several of the posts in this thread correctly, then it isn't essential to match the J113 too closely. I ask this as I'm new to the FET matching game and I only ordered two when getting the parts together a while ago. I breadboarded the two I have this morning in order to choose the right value for R4 and settled on 47.5Ω. This gives a reading of 0.495V and 0.510V on each of the J113. This equates to 10.7mA and 10.4mA respectively.
Will this be close enough or should I be ordering a bag full?
If I've read the ACP+ article and several of the posts in this thread correctly, then it isn't essential to match the J113 too closely. I ask this as I'm new to the FET matching game and I only ordered two when getting the parts together a while ago. I breadboarded the two I have this morning in order to choose the right value for R4 and settled on 47.5Ω. This gives a reading of 0.495V and 0.510V on each of the J113. This equates to 10.7mA and 10.4mA respectively.
Will this be close enough or should I be ordering a bag full?
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Perfect!!! That is exactly what you need to do with J113, select R4 to give something close to 10mA. There is no need to have that matched between channels, it’s just a current source. (The music signal is handled by the J74 differential pair.)
Carry on!!
Carry on!!
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I horsed around with 10-mA-CCS-setting-resistors and a big pile of J113s one afternoon. I found that it took only five different resistor values, to suit every single one of my J113s. And I got 10mA plus or minus 1mA, every single time.
The five resistor values were:
There are sophisticated optimizations which speed up the testing quite a bit, and I'm sure you'd use them if you were trying to find five hundred (J113 + source resistor) pairs that each gave 10mA +/- 1mA. But if you only need TWO constant current sources, the procedure above is dead simple, therefore well suited to folks who haven't yet logged hundreds or thousands of hours of electronics laboratory time.
The five resistor values were:
- RA = 150 ohms
- RB = 120 ohms
- RC = 91 ohms
- RD = 62 ohms
- RE = 36 ohms
- Pull out a J113 from the big pile
- Connect the J113 as a CCS using source resistor RA and measure. Is the current 10mA, +/- 1mA? If so, done!
- If not, connect it as a CCS using source resistor RB. Is the current 10mA, +/- 1mA? If so, done!
- If not, connect it as a CCS using source resistor RC. Is the current 10mA, +/- 1mA? If so, done!
- If not, connect it as a CCS using source resistor RD. Is the current 10mA, +/- 1mA? If so, done!
- If not, connect it as a CCS using source resistor RE. Is the current 10mA, +/- 1mA? If so, done!
- Now go back to step 1 and do it again with another J113
There are sophisticated optimizations which speed up the testing quite a bit, and I'm sure you'd use them if you were trying to find five hundred (J113 + source resistor) pairs that each gave 10mA +/- 1mA. But if you only need TWO constant current sources, the procedure above is dead simple, therefore well suited to folks who haven't yet logged hundreds or thousands of hours of electronics laboratory time.
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My thanks to you both.The J113s are now in place. Just the mosfets and caps to put in.
Mark,
I'm going to be using a pair of your PO89ZB filters in series on this build given the comments on that other thread. Rather than a separate box for them, it makes sense to have them in the case from the start.
Also, I'll be doing another order of 30 of the PO89ZB filter boards for anyone who wants them. As before I'm only asking for the price of the postage.
Mark,
I'm going to be using a pair of your PO89ZB filters in series on this build given the comments on that other thread. Rather than a separate box for them, it makes sense to have them in the case from the start.
Also, I'll be doing another order of 30 of the PO89ZB filter boards for anyone who wants them. As before I'm only asking for the price of the postage.
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Thanks for taking the time to get to the bottom of that Mark. One thing I'd love is that if we ever do a second run of the PCBs for this, that we insert a turn POT based on some minimal value so that people can dial this in based on whatever resistor they have within that range.
--Tom
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Once a resistor value is found through simple measurement that value will not need to be changed. Makes it fun a just a touch more hands-on, which will make the total process have greater value as well as more fun.
I like the idea of a bulletproof $0.05 resistor instead of a $3.00 pot that can get funky. Plus the PCB won’t need a revision.
I like the idea of a bulletproof $0.05 resistor instead of a $3.00 pot that can get funky. Plus the PCB won’t need a revision.
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I think so too. When a builder hand-picks resistors her/himself, it makes the project more personal, more *I* did that, more intimately connected. Nelson Pass and his squad trusted *me* to choose the right resistor, and I did, and the result sounds great. *My* ACP+ that *I* built, sounds great.