| sawreyrw |
| The schematic says it is to reduce the inductance. This doesn't make any sense at audio frequencies.!! However, the power dissipation per resistor is reduced. This make some sense. |
|
|
| jarthel |
| quote: | Originally posted by sawreyrw
The schematic says it is to reduce the inductance. This doesn't make any sense at audio frequencies.!! However, the power dissipation per resistor is reduced. This make some sense. |
that is true
but look at the way he suggested on how to parallel them. I think that's the heart of the my question.
he suggested that the bands (not sure of the term) must be in opposite directions. |
|
|
| rpapps |
Hi Jarthel
There are two components to this theory.
Firstly, resistors are trimmed to their final value by cutting a spiral groove into the surface. This gives rise to some inductance.
By placing two inductances in parallel, the total inductance is halved. Like wise four in parallel would quarter it.
Secondly, by reversing one resistor, I think he is trying to get the fields to cancel each other. However this would only work if the resistors were in close proximity (mutual coupling) and they were all manufactured so that the painted bands were always put on with the same orientation to the spiral cut. I don't know if that could be guaranteed.
But ,as the previous poster mentioned, I doubt it makes any appreciable difference at audio frequencies.
Cheers
Rob |
|
|
| AndrewT |
Hi,
paralleling will reduce inductance. It has absolutely nothing to do with the usable frequency.
The effect of reduce inductance will be influenced by the usable frequencies.
Wire wounds have inductance. There is a very simple way to wind a WW to reduce the self inductance to near zero.
Unfortunately putting two spiralled resistors side by side cannot use this inductance reducing property if both resistors have the spiral turns in the same direction. Batches of resistors will always have the spiral turns in the same direction.
However, some manufacturers may construct their resistors consistently from one handed spirals and other manufacturers may consistently spiral in the opposite hand. Here we have a solution that will reduce the inductance more than just the paralleling effect.
Select two equal value spiral formed resistors ( it matters not whether they are metal film, metal oxde or wire) made from opposite handed spirals. Placing them very close together and side by side will cause some mutual coupling of the fields produced by the spirals. The mutual coupling will reduce the overall inductance.
In a non inductive wire wound that is wound in spirals the first half of the resistor is wound in one hand and then halfway along the spiral the hand of the spiral is reversed until the end is reached. If the two halves are electrically and physically identical then the inductance from the two halves is almost exactly cancelled by the mutual inductance from the two end to end inductors. The remaining inductance in this type comes mostly from the lead out wires.
Who is going to reseach the spiral directions of the different manufacturers for us and put it into a Wiki? |
|
|
| pinkmouse |
From what I was told about resistor manufactur by a friend who worked in component distribution, there is no correlation between banding and spiral direction.
The resistor slugs are made, dumped into a big bin, then taken to the cutting machine, where the spirals are cut, then the slugs are dumped into another bin. Then they have the leads attached, and are then tested, coated, and marked but as you can see, they go will through that machine in a completely random manner. Of course, this doesn't meant to say you couldn't measure each resistor individually, but the markings are not indicative. |
|
|
| jarthel |
| But is it a good recommendation that "paralleled resistors with the spirals opposite to each other" have good effects? (assuming a person is able to tell where the spiral is) |
|
|
| macboy |
| Rather than trying to find some mystical reverse-wound resistor which is otherwise identical, wouldn't it be easier to just connect two identical resistors so that the current flows in opposite directions and therefore creates cancelling magnetic fields? |
|
|
| AndrewT |
Hi Mac,
NO!!!
Read my post again.
turning a sprial cut/wound resistor end for end does not change the hand of the winding.
Go and have a think about it.
If that is too much, then get a bit of single core wire and wind a spiral for yourself and check what it looks like when you turn it end for end. |
|
|
| macboy |
Andrew,
I understand that turning a spiral around has no effect. But look at what I have done. I didn't turn the resistor around, I turned the current around. The current flows in the opposite direction through two coils/spirals which are the same. That means the the magnetic fields are opposite to each other. |
|
|
| AndrewT |
Hi Mac,
imagine two bar magnets side by side but with opposite poles adjacent to each other. They are still magnets (try pulling them apart if your not sure).
The magnetic lines of force leave one pole curve around to enter the opposite pole. A magnetic field circuit has been set up.
Now take your spiral resistors, they too create a magnetic field, albeit a weak one, and they too create magnetic poles. Placing two spirals of matching hand adjacent to each other but with each carrying current in opposite directions will create opposing magnetic fields but they will link at each end to form that magnetic circuit. They actually reinforce each other, rather than eliminate the field.
No, as far as I know, reversing the current has not reduced the inductance. |
|
|
| jneutron |
The band orientation will have no affect.
The act of splitting the current into two paths will, though. However, there are better ways.
Cheers, John |
|
|
| macboy |
| quote: | imagine two bar magnets side by side but with opposite poles adjacent to each other. They are still magnets (try pulling them apart if your not sure).
The magnetic lines of force leave one pole curve around to enter the opposite pole. A magnetic field circuit has been set up.
|
Yes, a magnetic field circuit has been set up. Exactly. By coupling the two opposite magnetic fields you reduce the inductance. It's true! If you put an equal but opposite current through to magnetically coupled inductors you will get zero inductance. Now, the two resistors are hardly perfectly magnetically coupled - far from it. But their magnetic fields will interact to some extent. So we will only get partial concellation of inductance. This is the same reason that you should twist your DC power supply wires (supply and ground/return) together; individually they each introduce inductance to your power supply lines (bad), but when coupled together, the inductances cancel nearly 100% because the currents in the two conductors are equal and opposite.
Try sticking your double-magnet thing to a plate of steel. It won't stick (not well anyway) because you've coupled the magnetic fields together and there is nothing left to have external influence. That's the point.
I realise now that I should not have said "cancelling magnetic fields" because that is not what happens. The magnetic fields couple and interact. |
|
|
| poobah |
Seems to be an assumption here, there is not much coupling between the resistors in the first place.
In fact, the inductance is probably not worrying about in the first place.
Reversing the current in one of the resistors may very well increase the inductance because a magnetic loop is being created... much here would depend on the geometry of the whole deal.
Relax Jarthel... assuming the designer actually knows that a reduction in inductance is necessary at all, the arrangment won't be so critical.
Truth be told, if a parasitic inductance was actually a factor here, reducing it by a factor of only 2 seems pretty silly. Keep this in context with the author's assertion that flipping the resistor reverses the helix (it doesn't) and most of this is just absurd.
;) |
|
|
| macboy |
| quote: | Truth be told, if a parasitic inductance was actually a factor here, reducing it by a factor of only 2 seems pretty silly. Keep this in context with the author's assertion that flipping the resistor reverses the helix (it doesn't) and most of this is just absurd.
|
I completely agree. It is absurd.
And I think that the parasitic inductance of the extra traces necessary to reverse-wire the second resistor will add more inductance than the weak cross-coupling will cancel out. Still, I wanted to make the point that two mutually-coupled coils with opposite currents effective cancel their inductances.
But maybe the author of the original schematic, actually put the resistors in opposing pairs like that because resistor sound is directional. Using two in opposite orientation restores symmetry! Oh wait, that is absurd too (but still believed amoung some of the hardest of the hard core). |
|
|
| poobah |
oh yeah... you betcha.
Whenever I restore an old tube amp, I always mark the resistors I remove so I can put them back in the same way... don't want to reverse the current after they've broken in for so long!
:D |
|
|
| sawreyrw |
poobah,
Ya, I agree. Sort of like putting 2 50 nH in parallel. The impedance of 1 uH is 6.28 ohm at 1 MHz, .628 ohms at 100KHZ. So what?
Rick |
|
|
| hermanv |
jnuetron is right (damn I hate it when that happens :)), paralleling two inductors cuts the inductance in half, ignore the spiral issue.
Whether there is any benefit to cut inductance this small (nano or pico henries?) in half is an altogether different question.
I personally don't think that deposited metal film resistors have the best sound. |
|
|
| sivan_and |
well guys therz a clean method to reduce inductance further in wire wound resistors.
Had to tap the center of the WW resistor as shown... |
|
|
| AndrewT |
Hi Sivan,
you're obviously a man that understands. |
|
|
| sivan_and |
| Men r we disscussing RF circuits ? :D |
|
|
| jneutron |
| quote: | Originally posted by sivan_and
well guys therz a clean method to reduce inductance further in wire wound resistors.
Had to tap the center of the WW resistor as shown... |
Physically, it's too big. Your trying to use a minivan at the indy 500..sheesh.
Now THIS is what ya gotta do. This baby is 50 watts, air cooled, has an inductance of about 60 picohenries (although I must admit, I have only been able to confirm it is less than 250 pH due to test limitations).
The current flows in the same direction for all the resistors, and the current returns through the wire between each resistor, this pic is just prior to soldering all the leads to their common.
This baby has no external field.. and no internal field (well, ok, the external falls off as 1/R 19 and internally as 1/R 18. But , close enough.
btw, this is rev B...rev C will be designed for lower inductance.
Cheers, John |
|
|
| cpemma |
| Why not really confuse little Henry and tombstone alternate resistors? :clown: |
|
|
| audioservice |
| At the risk of creating hard feelings there is no physical law that could apply to improve an audio amplifier's perfromance by installing parallel resistors backwards to each other. First, the inductive reactance of a spiral cut film resistor is so low that its own leads have more inductance than the resistive element itself. This has been amply studied using a VHF impedance bridge by someone highly qualified to do such a study who shall remain nameless here (not me, either). Therefore unless an audio amplifier is also intended to be an RF driver at TV frequencies resistor self indcutance is not a player. The circuit traces and wiring make up 99.99% of the parasitic inductance in an audio amplifier circuit. Second, turning a resistor end for end does not reverse the rotation of the spiral cut. Check out a screw if you want to understand the geometry better. |
|
|
| EC8010 |
You fellows are looking at this from the wrong angle. The inductance of a resistor can be an issue at audio frequencies. Try measuring the power output of a loudspeaker amplifier using a big 8 Ohm wirewound resistor at 1kHz using V2/R and you'll get more power than you expect. Why? Because you measured the resistance of the dummy load at DC, but measured the AC voltage across it at 1kHz. And it can cause a significant error, not merely 1% - it can be 10%! (Been there, done that.)
The two important factors are the value of the resistor and the resistive material.
Carbon has much higher resistitivity than manganin (resistance wire) so a shorter path gives the same resistance. Since resistors are commonly wound as a helix, that means fewer turns and reduced inductance. That's why we use carbon resistors for valve and FET grid/gate stoppers.
The other issue is the actual value of the resistor. Who cares if a 1M resistor has a few uH of inductance? You would never notice it compared to 1M. But you would compared to 1 Ohm. It turns out that even wirewound resistors (which are more inductive because of the lower resistivity of their material) only begin to show significant inductance once their value falls below 1000 Ohms. For metal oxide resistors (with their higher resistivity), I'd expect to only see inductive effects below 100 Ohms.
A practical 100k resistor has no significant inductance. (Although it has measurable shunt capacitance.)
Low values of resistance can be made non-inductive by adding a Zobel network across them. (Non-inductive resistor of the same value as the original resistor, in series with an experimentally determined value of capacitance.) |
|
|
| jneutron |
| quote: | Originally posted by EC8010
You fellows are looking at this from the wrong angle. |
No I'm not. 31.76 degrees, that is certainly the correct angle..:D
| quote: | Originally posted by EC8010
The inductance of a resistor can be an issue at audio frequencies. Try measuring the power output of a loudspeaker amplifier using a big 8 Ohm wirewound resistor at 1kHz using V2/R and you'll get more power than you expect. Why? Because you measured the resistance of the dummy load at DC, but measured the AC voltage across it at 1kHz. And it can cause a significant error, not merely 1% - it can be 10%! (Been there, done that.) |
I also have been there, done that. In fact, the resistor I pic'd is an output load resistor, designed specifically to pull current resistively from an amp output at sub nanosecond speeds. While I have no interest in actually making or measuring an audio amp at those speeds, that speed comes about as a side benefit of a different, far more notorious effect. That'd be B dot error.
The voltage created across a wire as a result of the collapse of the load's current. For audio, that can be significant, ESPECIALLY for low impedance circuits.. That resistor eliminates the field collapse error term. It is a pure resistive load with no error out to several hundred Mhz.
| quote: | Originally posted by EC8010
Low values of resistance can be made non-inductive by adding a Zobel network across them. (Non-inductive resistor of the same value as the original resistor, in series with an experimentally determined value of capacitance.) |
Yuck. I like my method better..:D |
|
|
| mrshow4u |
| ....sorry if it's already been posted, but. non-inductive wire wounds are available. They're bifilar wound, to cancel their inductance. That's for power applications. We also used to use metal oxide from Victoreen. They weren't laser spiral trimmed. Good for high freqs. |
|
|
| EC8010 |
| quote: | Originally posted by jneutron
Yuck. I like my method better..:D |
Ah, well, your method genuinely reduces inductance rather than trying to fudge it away. |
|
|
| jneutron |
| quote: | Originally posted by EC8010
Ah, well, your method genuinely reduces inductance rather than trying to fudge it away. |
Actually, it was because I used pretty resistors..notice the interplay of the slant fin copper with the baby blue body color, highlighted by the band colors..and don't forget the seagreen tap wire.. and all embraced by the tech looking angular repetitive aluminum, framing the structure, not unlike a pair of palm trees would a beach chair..with a corona in the middle..
Cheers, John
Edit: verbage of course, consistent with the 10 days vacation I will embark upon tonight... |
|
|
| audioservice |
EC8010, you make a couple of unsupportable statements:
".......It turns out that even wirewound resistors (which are more inductive because of the lower resistivity of their material) only begin to show significant inductance once their value falls below 1000 Ohms. ..."
Just because your method of observation is incapable of detecting the inductance does not mean it is not there. You should correctly state the effect of the network on the circuit it is employed in. A 1 meg resistor in series with a 1uH choke certainly won't have much effect at 1kHz. Go to 10MHz and calculate the impedance to the two parts in series and see what you get. The inductance doesn't go away, it's alway there, even at DC. Now if you had said INDUCTIVE REACTANCE, you might have been closer to the truth.
"......A practical 100k resistor has no significant inductance. ..."
That's an opinion, not a physical fact.
"....Low values of resistance can be made non-inductive by adding a Zobel network across them. (Non-inductive resistor of the same value as the original resistor, in series with an experimentally determined value of capacitance.)..."
You must be referring to Zobel the Magician. :boggled: |
|
|
| EC8010 |
Audioservice, please bear in mind that I started my post by making it clear that I was talking about audio frequencies, not RF. I cannot imagine any RF application that would even use a 1M resistor, let alone put it in series with 1uH. Nevetheless, you are correct in that strictly I should really have said "inductive reactance".
No, my comment about 100k resistors was based on measurement. 2.5pF of shunt capacitance was needed to model a 100k wirewound resistor (tested, as it happens, up to 1MHz), but no series inductance.
I suggest you read up on Zobel networks. It's perfectly possible to correct an inductive resistor back to pure resistance using a Zobel network. |
|
|
|
