Speaker Attenuation...

jakelm

Can someone give me a few, if any, reasons why a resistor in series would not work? Cant be the ohms, because I'm raising the ohms.

jakelm

8ohm in series with a 3.5ohm resistor = 11.5ohms

11.5ohms paralell with a 6ohm speaker = 3.9ohms. (4ohms)


My HK should not have a major problem with that. Because without the resistor, its 3.5ohm, and it runs fine.

ShinAce

Check the resistor with an ohm meter.

Make note that a lot of headphone amplifiers use resistors in the range of 20-120 ohms on the output, sometimes bypassable.

jakelm

ShinAce wrote: »

Check the resistor with an ohm meter.

I did 3.2ohms

Make note that a lot of headphone amplifiers use resistors in the range of 20-120 ohms on the output, sometimes bypassable.

Im not quite sure how a headphone amp fits in to my problem.

ShinAce

You asked: "Can someone give me a few, if any, reasons why a resistor in series would not work?"

My answer is no, this should work just fine. It is a fairly common practice.

jakelm

ShinAce wrote: »

You asked: "Can someone give me a few, if any, reasons why a resistor in series would not work?"

My answer is no, this should work just fine. It is a fairly common practice.

Thanks Ace. I wish there some literature on the subject, so I wouldnt have to keep bothering you guys. But I do appreciate all the help.

I have read and know about adding resistance and ohms law, with adding speakers. I cant see a difference between adding a speaker and adding a resistor.

jakelm

Ok..I couldnt stand it any more. So I tryed it again. Seems to be working ok, although I havent cranked it yet. From the readings, the 3.5ohm resistor seems to have a -4db @ ~60dB , effect on the 8ohm speaker. I dont hear anything sonicly changed, but of corse I havent listened to it at higher volumes. But so far so good and the HK hasnt acted up yet. I really dont know what happened last time and why the HK would cut out.

ShinAce

If you're really brave, you can try some 60 Hz square waves with and without the resistor in an attempt to hear the difference. There is a difference, but very minor and increasing with attenuation.

To know the attenuation, you can measure the voltage across the speaker and across the resistor on a sine wave. If they share the voltage 50/50, it's a 3 dB drop. 60% resistor and 40% speaker is 4 dB.

jakelm

ShinAce wrote: »

If you're really brave, you can try some 60 Hz square waves with and without the resistor in an attempt to hear the difference. There is a difference, but very minor and increasing with attenuation.

I measured with Pink noise. If you dont mind me asking, why 60hz? Would another sine wave work? Like 100hz. Since I have my xover setting @80, I dont think measuring 60 would work so well.

To know the attenuation, you can measure the voltage across the speaker and across the resistor on a sine wave. If they share the voltage 50/50, it's a 3 dB drop. 60% resistor and 40% speaker is 4 dB.

Ok, if I play a 100hz sine wave. I place the multimeter on what setting? And place the +tive from multimeter to +tive on speaker post. Then -tive to -tive? This measurement is takin with the resistor OUT of the loop. Then take another measurement with JUST the resistor? Or with the resistor and speaker together?

Sorry such dumb questions, just want to make sure I do this right. I am truely a newbie when measuring voltage from speakers.

jakelm

Just took and official measurement. The 4's are 3.8ohms and the 7's are 7.7ohms. Combined total in paralell, 2.4ohms. Wow, and the HK never missed a beat. That realy suprized me. I drove the system hard with both pairs stacked (2.4ohms), never clipped , never shut down.

I did an A/B comparison with just the 4's, with 3.5ohm resistor and without. Major sound quality difference. I romoved the resistor from the 4's and they opened up alot. I now have removed the resistor and the 4's from the loop. 4's are on 2 chanel computer duty now.

Systems

jakelm wrote: »

I now have removed the resistor and the 4's from the loop. 4's are on 2 chanel computer duty now.

Wise descisoin,IMO what you were attempting with two totally different speakers would comprimse your 2 channel stereo performance.

ShinAce

jakelm wrote: »

I measured with Pink noise. If you dont mind me asking, why 60hz? Would another sine wave work? Like 100hz. Since I have my xover setting @80, I dont think measuring 60 would work so well.




Ok, if I play a 100hz sine wave. I place the multimeter on what setting? And place the +tive from multimeter to +tive on speaker post. Then -tive to -tive? This measurement is takin with the resistor OUT of the loop. Then take another measurement with JUST the resistor? Or with the resistor and speaker together?

Sorry such dumb questions, just want to make sure I do this right. I am truely a newbie when measuring voltage from speakers.

It's always better to ask that to risk a short somewhere.

Set the multimeter to voltage. Try to do this at very low volume(barely audible if possible) so you'll want to be measuring millivolts. Leave the resistor in circuit at all times.

You would measure the AC voltage across the resistor and speaker, and then just the resistor. The easiest way to do this is to measure the voltage at the amplifier on the speaker output. Polarity does not matter for sine waves(AC). Now measure the voltage from both leads of the resistor.

The voltage across the speaker is the total voltage(as measured on the amp output) minus the resistor voltage.

Use a 60 Hz sine wave as all meters are calibrated for 50/60 Hz use.

mulveling

ShinAce wrote: »

If you're really brave, you can try some 60 Hz square waves with and without the resistor in an attempt to hear the difference. There is a difference, but very minor and increasing with attenuation.

To know the attenuation, you can measure the voltage across the speaker and across the resistor on a sine wave. If they share the voltage 50/50, it's a 3 dB drop. 60% resistor and 40% speaker is 4 dB.

If the resistor and speaker share the voltage 50/50, then the resistor should give a 6db drop in theory. For 60%/40%, the drop should be nearly 8db. Here's the theory:

By the definition of the decibel, we have:
db2-db1 = 10*log(P2/P1)
Where db1 and P1 are the unattenuated audible decibels and audio power provided by the system at a given volume setting; db2 and P2 are the decibels and audio power with the attenuator in place assuming all other settings have remained the same. The logarithmic function is log base 10.

Now, a standard law of physics tells us that
P = I*V where P is power in watts, I is current in amps, V is voltage
We also have:
V = I*R where R is resistance in ohms

These two laws give us:
P = V^2/R

So now we can write the decibel/power equation in terms of voltage:
db2-db1 = 10*log(V2^2/V1^2) where V1 is the voltage across the speaker without attenuator in place, and V2 is the voltage with attenuator.

By a property of logarithms we can reduce it to:
db2-db1 = 20*log(V2/V1)
So you see, doubling the voltage has twice the effect on output decibels compared to doubling the audio power (watts), because power is proportional to the SQUARE of voltage (given that resistance in the circuit is a constant).

Now, computing the effect a given attenuator has on the voltage across the speaker (assuming output from the amp remains the same) is a simple linear interpolation equation:
V2 = V1*Rs/(Rs+Ra) where Rs is the resistance of the speaker and Ra is the resistance of the attenuator, in ohms.

Substituting, we reach our final equation:
db2-db1 = 20*log(Rs/(Rs+Ra))

SO, for example, if we have a resistor of 3.5 ohms and speaker of 8 ohms the result is:
20*log(3.5/11.5) = -3.15218 decibels
Now unfortunately a speaker's actual impedance varies as a function of frequency, which means the speaker probably isn't 8ohms at the measured frequency. That's why the 3.15218 db calculated above doesn't match up with jakelm's measured 4db drop - the speaker's actual impedance at that frequency must be lower than 8ohms. Actually we can solve for this; since we know from the measurements that:
4.0=20*log(Rs/(Rs+3.5)) => 10^0.2*(Rs+3.5)=Rs => (1-10^0.2)*Rs=3.5 => Rs = -5.98339
So, jakelm's speakers had an actual impedance of nearly 6 ohms at whatever signal frequency (or composite of frequencies) he used when he measured that. Knowing that he can use the equations above to determine almost exactly what value resistor to use to get a desired decibel drop for that input signal. Keep in mind 6db-10db is about the minimum to make a non-trivial, perceptible difference (perceived as about half as loud by the human ears).

The standard attenuator design includes a 2nd resistor that shunts to ground so that the amp sees the same load whether the attenuator is in place or not - a different load can change the way the amp behaves in many ways (worst of which is to shut it down). Again, the calculation of the shunt resistor value is complicated because the impedance of a speaker is not constant over frequency. It's really far better to attenuate between the source and preamp, or between the preamp and power amp, because both preamp and power amp have a constant input impedance. Also, there you're dealing with much lower power levels and you can get away with using 1/8th watt resistors (should be able to get MUCH higher quality for a reasonable price; a single Vishay-Dale should be acoustically transparent) instead of the huge beasts you must be using now. Of course you'll need another power amp for that, but you should be using two amps for 2 sets of speakers anyways.

Edit: Of course, this is all ignoring the fact that acoustically it's a very bad idea to place two full-range speakers in near proximity; combing artifacts being the most obvious.

ben62670

Dude disconnect your 4's tweeters. Done, and done. The 4's are no more efficient than the other monitors, its that $hit tweeter. Do that, and it will be kinda like Monitor 10s. I repeat... Do not attenuate, disconnect tweeter!

jakelm

mulveling wrote: »

If the resistor and speaker share the voltage 50/50, then the resistor should give a 6db drop in theory. For 60%/40%, the drop should be nearly 8db. Here's the theory:

By the definition of the decibel, we have:
db2-db1 = 10*log(P2/P1)
Where db1 and P1 are the unattenuated audible decibels and audio power provided by the system at a given volume setting; db2 and P2 are the decibels and audio power with the attenuator in place assuming all other settings have remained the same. The logarithmic function is log base 10.

Now, a standard law of physics tells us that
P = I*V where P is power in watts, I is current in amps, V is voltage
We also have:
V = I*R where R is resistance in ohms

These two laws give us:
P = V^2/R

So now we can write the decibel/power equation in terms of voltage:
db2-db1 = 10*log(V2^2/V1^2) where V1 is the voltage across the speaker without attenuator in place, and V2 is the voltage with attenuator.

By a property of logarithms we can reduce it to:
db2-db1 = 20*log(V2/V1)
So you see, doubling the voltage has twice the effect on output decibels compared to doubling the audio power (watts), because power is proportional to the SQUARE of voltage (given that resistance in the circuit is a constant).

Now, computing the effect a given attenuator has on the voltage across the speaker (assuming output from the amp remains the same) is a simple linear interpolation equation:
V2 = V1*Rs/(Rs+Ra) where Rs is the resistance of the speaker and Ra is the resistance of the attenuator, in ohms.

Substituting, we reach our final equation:
db2-db1 = 20*log(Rs/(Rs+Ra))

SO, for example, if we have a resistor of 3.5 ohms and speaker of 8 ohms the result is:
20*log(3.5/11.5) = -3.15218 decibels

Ecellent info, thanks

Now unfortunately a speaker's actual impedance varies as a function of frequency, which means the speaker probably isn't 8ohms at the measured frequency. That's why the 3.15218 db calculated above doesn't match up with jakelm's measured 4db drop - the speaker's actual impedance at that frequency must be lower than 8ohms. Actually we can solve for this; since we know from the measurements that:
4.0=20*log(Rs/(Rs+3.5)) => 10^0.2*(Rs+3.5)=Rs => (1-10^0.2)*Rs=3.5 => Rs = -5.98339
So, jakelm's speakers had an actual impedance of nearly 6 ohms at whatever signal frequency (or composite of frequencies) he used when he measured that. Knowing that he can use the equations above to determine almost exactly what value resistor to use to get a desired decibel drop for that input signal. Keep in mind 6db-10db is about the minimum to make a non-trivial, perceptible difference (perceived as about half as loud by the human ears).

They actually measured @3.8ohms (4ohm speaker) . Which is why the difference in attenuation.

The standard attenuator design includes a 2nd resistor that shunts to ground so that the amp sees the same load whether the attenuator is in place or not - a different load can change the way the amp behaves in many ways (worst of which is to shut it down). Again, the calculation of the shunt resistor value is complicated because the impedance of a speaker is not constant over frequency. It's really far better to attenuate between the source and preamp, or between the preamp and power amp, because both preamp and power amp have a constant input impedance. Also, there you're dealing with much lower power levels and you can get away with using 1/8th watt resistors (should be able to get MUCH higher quality for a reasonable price; a single Vishay-Dale should be acoustically transparent) instead of the huge beasts you must be using now. Of course you'll need another power amp for that, but you should be using two amps for 2 sets of speakers anyways.

If I am using a 12watt 3.5ohm resistor in series, what size resistor should be to ground?

Edit: Of course, this is all ignoring the fact that acoustically it's a very bad idea to place two full-range speakers in near proximity; combing artifacts being the most obvious.

I know you guys think I am nuts, but it does sound great. Stacking them one on top of the other with the tweets in near prox, give an excellent sound stage. The tweeter is not as bad as you think, its alot better than the sl2000.

jakelm

ben62670 wrote: »

Dude disconnect your 4's tweeters. Done, and done. The 4's are no more efficient than the other monitors, its that $hit tweeter. Do that, and it will be kinda like Monitor 10s. I repeat... Do not attenuate, disconnect tweeter!

I did that Ben, mids were overpowering the single Peerless.

I know you all think I should remove the 4's completely, but the soundstage has really increased with them stacked.

I have removed the speakers off of the HK and put them on the Hafler 200. That amp handles the 2.4ohm load with ease.

mulveling

jakelm wrote: »

Ecellent info, thanks



They actually measured @3.8ohms (4ohm speaker) . Which is why the difference in attenuation.



If I am using a 12watt 3.5ohm resistor in series, what size resistor should be to ground?

The following webpage will give you the quick & easy way to calculate ohms and watts for both resistors:
http://www.bcae1.com/lpad.htm

Personally, I like to figure out the theory behind a formula and/or tool when it's within my capabilities, hence the derivations I went through in my post above. Even better when I don't have to reference anything to do said derivations; this is one such case. So we can extend my previous derivations to explain the formula for the 2-resistor shunt attenuator:

Let R1 be the value of series resistor, R2 the shunt resistor, and Rs the speaker's nominal impedance. The speaker and shunt are in parallel, so by Ohm's law their combined resistance (call it Rp) is defined as:
1/R2+1/Rs = 1/Rp

We want the sum of Rp and R1 (sum because they are in series) to equal the speaker's impedance, so that the amplifier sees the same load despite the presence of the attenuator. This gives:
R1+1/(1/R2+1/Rs) = Rs

Since Rs is known, this give us the relation between R1 and R2. Now we'll derive the equation to give us the value of R2 to use for the desired decibel difference, and then we can use R2 to compute R1. Remember that from my previous post:
db2-db1 = 20*log(V2/V1)
Because of the shunt resistor, the linear interpolation that computes V2 in terms of V1 has changed. Now, the relation is:
V2 = V1*Rp/Rs = V1*(1/(1/R2+1/Rs))/(Rs) = V1*R2/(Rs+R2)

So, by substituting we get:
db2-db1 = 20*log(R2/(Rs+R2))
After doing enough reshuffling of this equation you get:
R2 = Rs*C/(1-C)
where C = 10^((db2-db1)/20)

So now let's go through a computation. Say we have a 4ohms speaker and want a decibel reduction of 6db.
C = 10^(-6/20) = 0.50118
R2 = 4*C/(1-C) = 4.01904 Ohms - shunt resistor value
R1 = Rs - 1/(1/R2+1/Rs) = 4 - 1/(1/4.01904+1/4) = 1.99525 Ohms - series resistor value

Plugging the values into the webpage calculator verifies the result is correct; though our manual calculations display a higher precision given any decent hand calculator

Again, I would be cautious of using an attenuator between amp & speaker because of the variable impedance of real-world speakers (meaning that even if you calculate for the nominal impedance it could still alter the overall frequency response) and the fact that AFAIK (and admittedly I have little knowledge of electronics parts) the really good resistors aren't readily available in sufficiently high wattages. I have used custom attenuators between source and preamp for compensating channel imbalances in high end headphones, and it works wonders there. In fact, channel imbalances have an audible effect at about 1.5db and greater - not necessarily obvious as a L/R channel imbalance - and I believe they are much more common in high end systems than audiophiles realize. Many with high-end gear would do well to investigate and correct slight imbalances before say, the next cable upgrade Sorry for the digression...

As for your positive experiences with running both speakers; I did not really mean to discourage that. While I am a bit surprised it's worked out reasonably well, there are so many complicated variables to acoustics that I trust if you say it's good then it must be doing something right

jakelm

Thanks Mul, then the series resistor is 2ohms and the shunt resistor is 4ohms. The shunt resistor goes after the series right, like in a xover network?

I dont think it has so much to do with using both speakers, as to how I have them placed.

My 7's are 4' high and laying sideways in a bookshelf, about 5' apart.. And my tv is in between them sticking out about 7". Now that is killing my stage because the speakers cannot "see" each other, and are so high off the ground. But unfortunatly, this is what i have to work with.

By adding a second pair of Polks, I am , in a way, filling in the missing pieces. tilting the 4's down, focuses the high end and mid. Like adding a second sub.

Or it might be because, I have listened to my system for so long in one way, that the different sound is appealing to me, only becuase its different.

I will probably go back to using just the one pair of 7's.

But this project of learning how to equalize 2 speakers without the help of the AVR, is very interesting.

Learning this, I will be able to further my xover bulding techincs. In attenuation.

ben62670

mulveling I have a question. I have modded a lot of XO, but never played with the shunt resistor. What effect does it have on the XO in layman's terms.
Thanks
Ben

unc2701

It's going to keep the impedence stable, unlike adding a series resistor- but it'll also do this as you get away from the nominal impedence. For example, with a 3 db attenuation circuit, at a frequency where the driver impedence is 8 ohm, the attenuator circuit will be leave the system impedence right at 8 ohm, but at a frequency where the driver impedence rises to, say, 16 ohm, the system will be at about 11 ohm.

Edit: As others have pointed out, you're always better off doing the attenuation at line level: 10k ohm vs 10,003 Ohm really doesn't change the load you're driving.

ShinAce

Oops, in regards to calculatins attenuation. Half the voltage is a 6 dB drop in volume, not 3 dB. It's only a 3 dB drop in voltage, but you don't hear volts, now do you?

Using an L-pad between a driver and the crossover will have an effect that must be considered. Using the L-pad between the crossover and amp can be neglected.

jakelm

So its like having a speaker in series with a speaker in paralell. 4ohm speaker in paralell with a 4 ohm speaker is, 2ohms. Then I add a 4 ohm speaker in series with the other 2, I am able to bring the ohms back up to 8.


Same would go with a reistor.

A speaker with 4ohms (4's) in series with a 3ohm resisotor, Thats 7ohms, then add a 9ohm shunt resistor to the series resistor chain, and I'm back to total load, that amp sees, 3.9ohms. But have reduced the speakers dB's by 3dB's, correct?

jakelm

My brain hurts now.....^^^^^:mad::mad::(:(

unc2701

Getting warmer, but the shunt generally goes after the series, not before (although you can work it out that way, too)... so the values would be more like 1.2 series & 9.7 shunt

So you get 2.8 ohm for the shunt & driver plus 1.2 for the series= 4 ohm

jakelm

unc2701 wrote: »

Getting warmer, but the shunt generally goes after the series, not before (although you can work it out that way, too)... so the values would be more like 1.2 series & 9.7 shunt

So you get 2.8 ohm for the shunt & driver plus 1.2 for the series= 4 ohm

So with a 4ohm speaker, what's the attenuation? With only a 1.2ohm series resistor, the attenuation shouldnt be much if any.

The calculation is, paralell first, then series?

ShinAce

Power is dissipated by both resistors.

The formulas you'll find are for a shunt resistor between the series resistor and driver.

unc2701

3db. It's the whole system, not just the series resistor.

jakelm

Does this in any way, effect the speaker that does not have the resistance but on the same channel?



BTW, great info..thanks guys..

ShinAce

No effect.

If anything, the fact that the attenuated speaker is drawing less power allows the amp to deliver more of the overall power to the direct connect speakers.

jakelm

ShinAce wrote: »

No effect.

If anything, the fact that the attenuated speaker is drawing less power allows the amp to deliver more of the overall power to the direct connect speakers.

Very good. So my 7's will recieve more power, while the 4's recieve less, being the ohms are lower. Like a see-saw.


I wonder if the older quad stereo used this application? If they were using different speakers for front and back.