L-Pad internal impedances

[TheWinterSnow]

I am designing a constant load, variable power device using transistors in an L-Pad configuration and cannot find any information on the internal impedance of the series and shunt resistors.  I need the maximum values the two resistors would be in any configuration, acknowledging that they can depending on the setting be zero ohms, but what has confused me is choosing the maximum value for the shunt resistor given the speaker's impedance. 

With fixed attenuates is makes sense that the shunt resistor should be infinite (open) but given the fact that restive variable L-Pads have some maximum value for shunt resistance I have no idea about how to find out that value to choose.  This is mainly because I did some calculations and it seems like the shunt resistance inside a variable L-pad is logarithmic so a proper shunt value relative to some controlling voltage would have to be properly chosen to keep minimal mismatching across the range of the output power.

Or is it even possible to control a shunt resistor in a non-linear/exponential way with just a potentiometer and fairly simplistic voltage conditioning?

[uncle_bob]

Hi

Ok so let's back up a bit: (all based on fixed load and source impedances)

1) Simple one resistor pad: You get attenuation and that's it. You have one variable (attenuation) and one unknown to solve for (the resistor. You can solve for either a series or a shunt resistor. 

2) Two resistor pad: You now have two unknowns, you get to pick two variables. The normal ones are port match (source or load) and attenuation. You pick the side you want to match and the attenuation. Out come the resistors. The resistor configuration is pretty much fixed (three locations, they must go in these two of them). In this case (and the next one) match really means set.

3) Three resistor pad: With three, you can match both ports *and* get an attenuation. You have two configurations (pi and Y). Once you have picked the config, the resistors are directly dependent on the variables.

In a system with a zero source impedance on one side, the three resistor setup does not make a lot of sense. Matching zero ohms is silly (and it is a corner case where the numbers don't work). An infinite load is a similar sort of thing (numbers blow up).

So, you now can run the math and get your results.

(hint: matching the speaker load is not the best way ...).

Bob

[Audioguru]

An L-pad was used with a vacuum tubes amplifier 60 years ago to match the high output impedance (8 ohms) of the amplifier's output transformer.
Modern amplifiers do not have an output transformer and have an output impedance of 0.04 ohms or less for good damping of the resonances of speakers.
An L-pad used with a modern speaker and amplifier will cause a boomy sound.

[uncle_bob]

Hi

How much of an issue you get from damping factor depends on the range of the L pad. If you set it up to provide a constant 4 ohm load to the amp, the speaker will be driven from between a 1 and 2 ohm source at the worst setting. Exactly what the number is depends on the impedance of your speaker. No that's not quite a damping factor of 1,000. As you increase the attenuation past 6 db, the damping factor will continuously improve. To get your 1,000 number, the shunt element in the attenuator would need to go to 0.008 ohms (on an 8 ohm speaker). Do you need 1,000 or is 10 good enough? That depends on what you are after.

Bob

[TheWinterSnow]

(hint: matching the speaker load is not the best way ...).

For tube amplifiers it is the only way without getting large and expensive.  My design is for tube amplifiers, guitar amplifiers to be exact.  I need a constant load, variable power attenuation to the speaker.  My issue comes from the design differences between fixed and variable L-pads given variable L-pads have some maximum shunt resistance and given how the shunt resistance should approach infinity as the series resistance approaches 0, again not quite how variable L-pads work. 

The extremely exponential impedance of an ideal shunt resistance versus what is implemented in variable L-pads and subsequent trade-off/errors that come along with each is what is confusing me.  I mean I could easily just choose a shunt resistance two orders of magnitude higher than the series resistance and call it a day but I want to be more precise than that.

[uncle_bob]

(hint: matching the speaker load is not the best way ...).

For tube amplifiers it is the only way without getting large and expensive.  My design is for tube amplifiers, guitar amplifiers to be exact.  I need a constant load, variable power attenuation to the speaker.  My issue comes from the design differences between fixed and variable L-pads given variable L-pads have some maximum shunt resistance and given how the shunt resistance should approach infinity as the series resistance approaches 0, again not quite how variable L-pads work. 

The extremely exponential impedance of an ideal shunt resistance versus what is implemented in variable L-pads and subsequent trade-off/errors that come along with each is what is confusing me.  I mean I could easily just choose a shunt resistance two orders of magnitude higher than the series resistance and call it a day but I want to be more precise than that.

Hi

Ok, here's the formula:

The input to the L pad (load on the amp) will always be exactly 8 ohms regardless of the attenuation. Your speaker is assumed to be exactly 8 ohms. For zero attenuation you have a zero ohm and an infinite ohm resistor in the pad.

You want to increase the attenuation. The shunt resistor goes to 8 ohms. The series resistor goes to 4 ohms. The amp now sees an 8 ohm load. The speaker has a 6 db pad.

You want to go to 20 db. Your series resistor is now 7.20 ohms. Your shunt resistor is 0.888 ohms.

You want more db ... series gets ever closer to 8 ohms. Shunt gets ever closer to zero ohms.

So the series resistor will always be between 0 and 8 ohms. The shunt resistor is limited only by the range of attenuation you are after. Is 0.0001 db close enough to zero db? How about 1 db?  The max shunt resistance just went from megohms down to about 80 ohms.

Stil to much range? Cascade a series of pads. Make up 10 20 db pads rather than one 200 db pad. Do the last 20 db with a variable.

Bob

[TheWinterSnow]

My issue is not how L-pads work but the issue of using transistors as a load and doing the Vg control via potentiometer.  My issue, is the logarithmic curve of a pot going to represent a voltage that correctly corresponds to a correct resistance across that particular transistor.

The issue is (for a 16 ohm load which is what I default to):

When the series resistance is taking the entire load, the shunt resistance should be 0 ohms and the series resistance should be 16 ohms.  At half load and consequently at half the controlling pot's resistance (knob pointing straight up), the series resistance should be 8 ohms and the shunt should be 16 ohms.  The shunt resistance would need to exponentially increase as it it turned up.  I fear that the voltage curve by the  pot would not match the needed Rds of the transistor and as such the resistance relationship between the series and shunt resistances would not be correct and would skew the total load seen by the source.

[TheWinterSnow]

Again my issue is with the resistance curve of the controlling pot, this schematic should make that clear.

https://drive.google.com/file/d/0B-mVz9WHvBJDXzJhVk1PQ2VOQU0/view?usp=sharing

[uncle_bob]

My issue is not how L-pads work but the issue of using transistors as a load and doing the Vg control via potentiometer.  My issue, is the logarithmic curve of a pot going to represent a voltage that correctly corresponds to a correct resistance across that particular transistor.

The issue is (for a 16 ohm load which is what I default to):

When the series resistance is taking the entire load, the shunt resistance should be 0 ohms and the series resistance should be 16 ohms.  At half load and consequently at half the controlling pot's resistance (knob pointing straight up), the series resistance should be 8 ohms and the shunt should be 16 ohms.  The shunt resistance would need to exponentially increase as it it turned up.  I fear that the voltage curve by the  pot would not match the needed Rds of the transistor and as such the resistance relationship between the series and shunt resistances would not be correct and would skew the total load seen by the source.

Hi

Ok, there is no way around the circuit math. Once you constrain your load to be constant, this is what you get. There is no magic bullet.

..... so ... since there are multiple problems doing this with a MOSFET (like bi-directional current flow, pinch off, and linearity) don't do it with a transistor. It's not going to work. You need a setup that will allow you to directly control resistors over a specific range (like relays). You can also use a MOSFET to switch a resistor. That works ok.

If you really need to do it with a MOSFET, you use multiple devices. They did indeed to this back in the "good old days". Each one operates across a resistor or in series with a resistor. You have a few dozen of them to get several decades of operation. The control voltage is done by a series of diodes and op-amps. Each segment of the control is broken out and drives it's own transistor. Typically these monsters took a few months to tweak to full performance. I never saw one that was "compact" in any sense of the word.

A more modern approach:

Digitize the pot with the ADC on a $1.00 MCU. Drive a half dozen relays to a dozen relays on each side of the pad. No weird distortion, all the power goes up in cheap resistors. No big fan cooled heatsinks.

Bob

[TheWinterSnow]

The reason I was looking into transistors was that 150W+ resistors are fairly large and expensive and even if you reduce the power based on the amount of attenuation, it can get bulky without a tapable power resistor.

The sizes of resistors needed for a 120W amp that has 5 attenuation settings would be large as hell.

I have actually realized I could use a two MOSFET setup, tap some of the audio signal and mix it in with the gate voltage so that the gate voltage maintains the desired Rds for that particular setting.

[uncle_bob]

The reason I was looking into transistors was that 150W+ resistors are fairly large and expensive and even if you reduce the power based on the amount of attenuation, it can get bulky without a tapable power resistor.

The sizes of resistors needed for a 120W amp that has 5 attenuation settings would be large as hell.

I have actually realized I could use a two MOSFET setup, tap some of the audio signal and mix it in with the gate voltage so that the gate voltage maintains the desired Rds for that particular setting.

Hi

If you have to heatsink the MOSFETs *and* run them linearly .... they will take up even more room than the resistors. Resistors are a dirt cheap way to dump power. Fans make a lot of noise.

I'm still not quite clear on why the need to attenuate the signal at the output of the amp rather than at the input.

Bob

[edavid]

I'm still not quite clear on why the need to attenuate the signal at the output of the amp rather than at the input.

It's for a guitar amp... you set the distortion by setting the amplifier output level.

But OP, forget the "transistors" - MOSFETs are not linear enough to be used as audio power resistors - you will be adding unpredictable additional distortion.

[TheWinterSnow]

Guitar amplifiers are highly non-linear and sound completely different at different volumes.  Attenuating the power amp allows you to get to sound, but control the volume.  Also, attenuating the front end of a guitar amplifier generally does not reduce volume while drastically changing the sound.

1 10W, 4 50W, 1 70W, 1 100W and 2 150W resistors is not only really expensive ($100+) but would take up a lot of space and would require an enclosure that would be equally expensive.  I have found 130W+ heat sinks that will easily fit in my preferred enclosure and don't completely break the bank.  This is also keeping in mind that those resistors would have fixed attenuation where I was hopping to get variable attenuation.

[TheWinterSnow]

But OP, forget the "transistors" - MOSFETs are not linear enough to be used as audio power resistors - you will be adding unpredictable additional distortion.

So having the Vg follow the input voltage with a DC offset to maintain a constant Rds would still add unpredictable distortion?

[uncle_bob]

But OP, forget the "transistors" - MOSFETs are not linear enough to be used as audio power resistors - you will be adding unpredictable additional distortion.

So having the Vg follow the input voltage with a DC offset to maintain a constant Rds would still add unpredictable distortion?

Hi

If you are planing to put an AC signal through the MOSFET ... no that will not work. It will sound awful.

Bob

[Audioguru]

But it is a Geetar amplifier. They want it to sound awful.

[uncle_bob]

Guitar amplifiers are highly non-linear and sound completely different at different volumes.  Attenuating the power amp allows you to get to sound, but control the volume.  Also, attenuating the front end of a guitar amplifier generally does not reduce volume while drastically changing the sound.

1 10W, 4 50W, 1 70W, 1 100W and 2 150W resistors is not only really expensive ($100+) but would take up a lot of space and would require an enclosure that would be equally expensive.  I have found 130W+ heat sinks that will easily fit in my preferred enclosure and don't completely break the bank.  This is also keeping in mind that those resistors would have fixed attenuation where I was hopping to get variable attenuation.

Hi

Ok, I see the issue.

The maximum power in the series arm will be equal to the output power of the amp.

The maximum power in the shunt arm will be equal to 1/4 the output power of the amp.

So, the total power of all the resistors in the series arm is 150W *if* your amp puts out 150W on a continuous basis. The total power in all the resistors in the shunt arm is 37.5W.

As you have noted, the shunt arm goes from infinity down to zero. Most of those resistors do not need to carry any significant power. I would just use wire for the low value ones (it's free).

The series arm goes from zero to whatever your speaker impedance is. If it's 16 ohms, you will need some power there. Typically you would use at least two (and more likely more) resistors in series to get to the 16 ohms. One example would be 8 two ohm resistors. They each need to handle a bit under 19W. The max power resistors in the shunt arm is 8 ohms in this case. Two of the 2 ohm parts will do that pretty well.

At this point we have ten 19 watt resistors and some other bits. But are they 19 watt? That's only the case if you run a sine wave tone continually (like 20 minutes) through the system at max drive. There are a *lot* of things in a typical amp that are not rated that way. A far more normal rating would be a bit lower.

So off we go ...

http://www.mouser.com/ProductDetail/Xicon/280-CR25-20-RC/?qs=sGAEpiMZZMtTURnxoZnJAMzhUi3Jq2hdJcch1PkZdrg%3d

is the first thing a simple search turns up. They are 25W and 2 ohms. They are fine for audio use. They come out to 99 cents each for 10 and 82 cent for 100. If you back of a bit on the power rating, 10W parts would be about half those numbers.

So we are spending $8 on the series arm and let's call it $8 (for power resistors) on the shunt arm.

That's not a lot of money on the resistors.

Bob

[uncle_bob]

But it is a Geetar amplifier. They want it to sound awful.

Hi

There are a variety of types of awful. Firing the intrinsic source drain diode awful is probably *not* what is intended.

Bob

[TheWinterSnow]

Hi

If you are planing to put an AC signal through the MOSFET ... no that will not work. It will sound awful.

Bob

An AC signal with two opposing FETs for voltage blocking in both directions and bi-directional current?  And if you can maintain a constant Vgs so that the FET maintains a constant Rds over the AC voltages range, it should not sound awful as the Rds/resistance never changes for that particular attenuation setting.  Since the FETs maintain their conduction and never turn off as well as the fact they are constantly passing a signal, there should be to my knowledge little to no distortion across said FETs.

But it is a Geetar amplifier. They want it to sound awful.

Hi

There are a variety of types of awful. Firing the intrinsic source drain diode awful is probably *not* what is intended.

Bob

But it is a Geetar amplifier. They want it to sound awful.

If by awful you mean intentional and desirable harmonic distortion then yes, it is highly desirable. 

[uncle_bob]

Hi

If you are planing to put an AC signal through the MOSFET ... no that will not work. It will sound awful.

Bob

An AC signal with two opposing FETs for voltage blocking in both directions and bi-directional current?  And if you can maintain a constant Vgs so that the FET maintains a constant Rds over the AC voltages range, it should not sound awful as the Rds/resistance never changes for that particular attenuation setting.  Since the FETs maintain their conduction and never turn off as well as the fact they are constantly passing a signal, there should be to my knowledge little to no distortion across said FETs.

Hi

As long as you DC bias the FET's with a constant current source they will do fine.

Edit:

=====

Also be careful with the power rating on the MOSFETs. The data sheet 120W number is into an infinite heatsink, often at room temperature. A real heatsink in a real enclosure likely cuts them to 1/4 the data sheet numbers. The same "do we need full power" question applies to the FET just as much as the resistors. You may not really *need* 150 or so watts in a real system. For strange historical reasons, the resistors I linked to don't need heatsinks and dissipate their rated power in a normal enclosure.
(yes it's a crazy world)

Bob

[TheWinterSnow]

Considering an amplifier that has an output rating of 120 Watts and will put out technically more than that when in full saturation, the series transistors will need to be able to handle 60 Watts a piece.  I was actually concerned given an average Junction-Case thermal resistance of 6C/W even with a 0.5C/W heatsink capable of 130W would an 18Wmax transistor be able to stay under the maximum 150C.

[uncle_bob]

Considering an amplifier that has an output rating of 120 Watts and will put out technically more than that when in full saturation, the series transistors will need to be able to handle 60 Watts a piece.  I was actually concerned given an average Junction-Case thermal resistance of 6C/W even with a 0.5C/W heatsink capable of 130W would an 18Wmax transistor be able to stay under the maximum 150C.

Hi

Ok, so back to the resistors. Starting from 120W instead of 150, they will drop down a notch.

Bob

[TheWinterSnow]

Guitar amplifiers are highly non-linear and sound completely different at different volumes.  Attenuating the power amp allows you to get to sound, but control the volume.  Also, attenuating the front end of a guitar amplifier generally does not reduce volume while drastically changing the sound.

1 10W, 4 50W, 1 70W, 1 100W and 2 150W resistors is not only really expensive ($100+) but would take up a lot of space and would require an enclosure that would be equally expensive.  I have found 130W+ heat sinks that will easily fit in my preferred enclosure and don't completely break the bank.  This is also keeping in mind that those resistors would have fixed attenuation where I was hopping to get variable attenuation.

Hi

Ok, I see the issue.

The maximum power in the series arm will be equal to the output power of the amp.

The maximum power in the shunt arm will be equal to 1/4 the output power of the amp.

So, the total power of all the resistors in the series arm is 150W *if* your amp puts out 150W on a continuous basis. The total power in all the resistors in the shunt arm is 37.5W.

As you have noted, the shunt arm goes from infinity down to zero. Most of those resistors do not need to carry any significant power. I would just use wire for the low value ones (it's free).

The series arm goes from zero to whatever your speaker impedance is. If it's 16 ohms, you will need some power there. Typically you would use at least two (and more likely more) resistors in series to get to the 16 ohms. One example would be 8 two ohm resistors. They each need to handle a bit under 19W. The max power resistors in the shunt arm is 8 ohms in this case. Two of the 2 ohm parts will do that pretty well.

At this point we have ten 19 watt resistors and some other bits. But are they 19 watt? That's only the case if you run a sine wave tone continually (like 20 minutes) through the system at max drive. There are a *lot* of things in a typical amp that are not rated that way. A far more normal rating would be a bit lower.

So off we go ...

http://www.mouser.com/ProductDetail/Xicon/280-CR25-20-RC/?qs=sGAEpiMZZMtTURnxoZnJAMzhUi3Jq2hdJcch1PkZdrg%3d

is the first thing a simple search turns up. They are 25W and 2 ohms. They are fine for audio use. They come out to 99 cents each for 10 and 82 cent for 100. If you back of a bit on the power rating, 10W parts would be about half those numbers.

So we are spending $8 on the series arm and let's call it $8 (for power resistors) on the shunt arm.

That's not a lot of money on the resistors.

Bob

So what value are the shunt resistors supposed to be?

[TheWinterSnow]

The max power resistors in the shunt arm is 8 ohms in this case. Two of the 2 ohm parts will do that pretty well.

Huh?

[uncle_bob]

The max power resistors in the shunt arm is 8 ohms in this case. Two of the 2 ohm parts will do that pretty well.

Huh?

Hi

Yes, that was very poorly worded ... sorry about that.

Two will handle the power.

Four will be needed to hit the resistance. Power is not an issue if you use the 2 ohm parts you already bought.

Hopefully that clears it up.

Bob