JL slash series amps

[wcbjr]

Did JL do this to allow people to run a variety of setups? Or did they do it to allow people to run higher impedances (increasing damping) and still maintain full power output?

 

What I am trying to ask is, does the amp reduce power when seeing a lower speaker impedance or increase power when seeing a higher impedance (when compared to unregulated power supplies)?

 

Can anyone understand what I am asking?

[PGT]

did you read the part of their site that explains the technology? Tightly regulated power supplies plus some proprietary stuff:

 

http://mobile.jlaudio.com/products_amps_pages.php?page_id=29

[wcbjr]

It didn't answer my question, at least in a layman's way that I can understand.

[PGT]

They output constant power regardless of impedance. Similar concept to McIntosh's PowerGuard circuitry, but probably different in execution.

[wcbjr]

I know that already. How do they do it? Why?

 

Unregulated would double the power with half the impedance. Does this amp simply take away the doubling? I just want to know how...

[PGT]

it has circuitry that measures actual load (dynamic impedance) and adjusts rail voltage for constant power output.

[schwinn]

They basically raise the voltage output so that, at higher impedances, the lower current draw is "compensated" for, resulting in the same power output. This compensation is handled by part of what JL calls "RIPS" technology.

 

These amps are great for mids and highs, but as a subwoofer amplifier, there are a few issues to keep in mind. The bottom line is that a higher-impedance subwoofer on any amp (RIPS or otherwise) will not work as well as a lower impedance sub on an amp.

 

In other words, there is more to this RIPS technology than meets the eye, when you are looking to power subwoofers... read on if you are interested...

 

Warning: Technical discussion ahead

A few basic equations:

P=IV or Power = Current * Resistance

V=IR or Voltage = Current * Resistance

I will assume the reader has a basic knowledge of algebra to mix these two up, resulting in the following:

P=(I^2)*R and I = sqrt(P/R)

P=(V^2)/R and V = sqrt(P*R)

 

Now, lets say your 500W amp (JL or other brand) is running into a 4-ohm speaker. Using the equations above, the amp is driving 44.7V (from V=sqrt(P*R)) at 11.2A (from I=sqrt(P/R)). Most amplifiers can only drive a single rail voltage, so we must assume that a "standard amp" will only be able to deliver 44.7V to the speakers (at max power, of course).

 

Now, if you double the impedance of the speakers to 8-ohms, then see what happens: On the same 44.7V amplifier, we see that the amplifier is now driving I=44.7/8 or 5.6 Amps (not the original 11.2Amps). In power-speak, this means you are driving P=IV or 44.7 * 5.6 = about 250W... or about half the power. Anyone remotely experienced with speakers and amps is familiar with this relationship.

 

The JL amps make up for this by raising the voltage so that you still get 500W, as it says on the amplifier. In other words, in this example, your 8-ohm speaker would actually end up seeing:

P=IV, and using I=V/R (from V=IR) you get P=(V^2)/R, therefore V= sqrt(P*R) = sqrt(500*8) = about 63.2 Volts, instead of the original 44.7V. Taking this further, with that voltage, you get I=V/R (from V=IR) so I = 63.2 / 8 = 7.9 Amps.

 

In list form:

4-ohm speaker will see 44.7V and 11.2A, or 500W

8-ohm speaker on a regular amp would see the same 44.7V, but only 5.6A, or about 250W

8-ohm speaker on a JL RIPS amp would see 63.2V at 7.9A, or about 500W

 

Sounds great, doesn't it? Finally, an amp that delivers full power at higher impedances! But, there's a catch when dealing with subwoofers. First off, just to review, how does a subwoofer work? Subwoofers push air via a magnetic field between the magnet and the coil - period. Without going into a longer technical explanation, current makes magnetic fields, NOT voltage. For this reason, in subwoofer-land, current is king. I am not going to go any further on this, as I will assume that we can agree on this point (though I have had many people argue about this very issue - you can't argue with physics.)

 

So, looking at the list above, if using the RIPS amp on a subwoofer, getting 500W on an 8-ohm speaker only gets you 7.9A, while a 4-ohm speaker is still delivering 11.2A. Since current is what causes the subwoofer to physically move, the "500W" designation is kinda meaningless here. Sure, you are still getting 500W to both the 8-ohm and 4-ohm speaker on this amp, but you won't get the same performance from both, since the current drive is still lower on the 8-ohm speaker.

 

For mids and highs, I am sure that the JL is a great choice, because these speakers are not purely driven by current like subs are... but for a subwoofer application, a quality, low-impedance capable amp and low impedance speaker will still perform better.

 

For that matter, driven to low impedances, the RIPS amp will perform the same as another quality amp of the same rating, since they both will drive the same current... but why spend money on RIPS when it's not necessary here?

 

That's my (long) take on the matter... feel free to flame away (someone always does).

[PGT]

I'm not reading all that because I know the details. The dude wants cliff notes

[msmith]

Sounds great, doesn't it? Finally, an amp that delivers full power at higher impedances! But, there's a catch when dealing with subwoofers. First off, just to review, how does a subwoofer work? Subwoofers push air via a magnetic field between the magnet and the coil - period. Without going into a longer technical explanation, current makes magnetic fields, NOT voltage. For this reason, in subwoofer-land, current is king. I am not going to go any further on this, as I will assume that we can agree on this point (though I have had many people argue about this very issue - you can't argue with physics.)

 

So, looking at the list above, if using the RIPS amp on a subwoofer, getting 500W on an 8-ohm speaker only gets you 7.9A, while a 4-ohm speaker is still delivering 11.2A. Since current is what causes the subwoofer to physically move, the "500W" designation is kinda meaningless here. Sure, you are still getting 500W to both the 8-ohm and 4-ohm speaker on this amp, but you won't get the same performance from both, since the current drive is still lower on the 8-ohm speaker.

 

For mids and highs, I am sure that the JL is a great choice, because these speakers are not purely driven by current like subs are... but for a subwoofer application, a quality, low-impedance capable amp and low impedance speaker will still perform better.

 

For that matter, driven to low impedances, the RIPS amp will perform the same as another quality amp of the same rating, since they both will drive the same current... but why spend money on RIPS when it's not necessary here?

 

That's my (long) take on the matter... feel free to flame away (someone always does).

 

 

I'm not sure where you're coming from in the last part of your post, Schwinn... Current is the flow of charge (voltage) and is meaningless in and of itself. 1 amp at 50,000 volts is not the same as 1 amp at 5 volts.

 

All speakers operate with current AND voltage, not just subwoofers. Subwoofers just have heavier masses, larger pistons and larger voice coils... but they're essentially no different than a mid or a tweeter from a functional perspective.

 

Since current is simply the flow of charge (voltage)... the same current with a higher voltage represents more power (which is what actually makes a speaker move). A higher current with the same voltage likewise represents more power.

 

Ohm's Law teaches us the following:

 

Power (P) = Current (I) x Voltage (E)

 

and we can easily solve for voltage or current …

 

Voltage (E) = Power (P) / Current (I)

Current (I) = Power (P) / Voltage (E)

 

Yes, current exists in the circuit, yes it is important... but it does not determine the power in the circuit by itself... resistance and voltage need to be considered.

 

Here are a couple more handy formulas:

P = E^2 / R

Power = voltage squared divided by resistance

 

P = I^2 x R

Power = current squared times resistance

 

The whole point of RIPS is to detect the resistance (or impedance in this case) and adjust rail voltage to deliver optimized power into a variety of loads... whether the amp is operating in higher current/lower voltage mode at lower impedances or lower current/higher voltage mode at higher impedances, the net result is the same power... which is what really matters. A conventional amp needs to operate at its lowest safe impedance to produce optimum power.... a RIPS enabled amp gives you flexibility to use a variety of loads without sacrificing power and it doesn't matter whether you are using it on a subwoofer or a tweeter.

 

So, while you can't argue with physics... you do need to consider the whole picture.

 

Best regards,

 

Manville Smith

JL Audio, Inc.

[PGT]

Cliff's Notes:

 

it has circuitry that measures actual load (dynamic impedance) and adjusts rail voltage for constant power output.

[wcbjr]

Ok, why just not buy a comparable amp that puts out the same power at the same load? RIPS seems kinda like gimmicky. Don't get me wrong, JL amps are quality pieces.

[PGT]

load is not static...it's dynamic with frequency. Look at an impedance curve for a speaker....it's far from flat.

 

http://www.audio-ideas.com/reviews/loudspeakers/graphics/studio-lab-ref-1-imp.jpg

[wcbjr]

Ok, ding! Now it hits home. Put that out on JL's site.

[msmith]

Cliff's Notes:

 

 

Technically speaking, it does not directly monitor or measure impedance.

 

The circuit monitors output current and makes a determination of the load connected to the amplifier based on this measurement and then selects from 3-4 rail voltages (depends on the model) to deliver full design power into that load.

[PGT]

thanks for the clarification Manville - it's sort of stated both ways on the website:

 

The "Intelligent" portion of the R.I.P.S. System is a circuit that actually monitors output current to optimize the amplifier's output power over a wide range of load impedances (1.5 ohm to 4 ohm per channel).

 

 

This will not happen with a JL Audio “Slash” series amplifier because the R.I.P.S. System detects the actual impedance being driven and adjusts output rail voltages to deliver optimum output.

[msmith]

Oooops... well... ahem... yeah... whatever. You get my drift.

[PGT]

[wcbjr]

Would it be of any help if the amp could calibrate itself to the speaker by conducting a full sweep of the impedance, basically building its own chart like posted above? Then it wouldn't need to spend time measuring the impedance anymore, just perform a table lookup.

[msmith]

It wouldn't make any sense to optimize that way, wcbjr.

 

The amp needs to be concerned with the minimum impedance of the load, not the variable nature of it. You wouldn't want an amp that reacted to a higher impedance at a particular frequency by raising voltage for that frequency. This would totally screw up the frequency response of the speaker system. The speaker system is essentially reacting to an AC voltage with current varying based on the impedance at any given moment in time. The amp needs to be able to deliver reliable power at the lowest impedance presented by the load.

 

It's kind of interesting to note that amplifier power is typically measured into a purely resistive (not reactive) load and that in a real speaker system whose impedance can easily swing from 4 ohms to 50 ohms or more that the actual power delivered at full voltage output can vary dramatically.

[Boulderguy]

In my experience, and I'm no engineer, Headroom is the missing link in the great debate about how much power can be eeked out of any amp. So many users get caught up in the push to drive the tallest load & get me most amperage from their amp without ever listening to the overall effect.

 

Most amps have a sweet spot where they operate best, where the power output is optimized AND it retains optimum headroom. Its usually 4 ohms for a conventional, unregulated amp. Driving 2 ohms with that amp will in theory double the output, but also cut the headroom in half. More volume at the cost of a much less dynamic sound. That's the beauty of the regulated JL's, you can go for all kinds of bizarre sub wiring schemes but still keep the headroom & thereby the SQ.

[PGT]

It's kind of interesting to note that amplifier power is typically measured into a purely resistive (not reactive) load and that in a real speaker system whose impedance can easily swing from 4 ohms to 50 ohms or more that the actual power delivered at full voltage output can vary dramatically.

 

Yup. I have a pair of Canton's for home use that have a 'nominal' 4 ohm load. Not the easiest speakers to drive...most amps choke and the high frequencies sound harsh in response.

[schwinn]

Your signature implies you work for JL? Very cool... I wanted to get a first-hand agreement to this for quite some time, as I always get people disagreeing with me on this, and refusing to look at the physics/math behind it. Are you an engineer there, perchance?

I'm not sure where you're coming from in the last part of your post, Schwinn... Current is the flow of charge (voltage) and is meaningless in and of itself. 1 amp at 50,000 volts is not the same as 1 amp at 5 volts.

Back to the point, yes, you are correct, for the most part. My point is that 500W, by itself, is meaningless in subwooferland (particularly), since current is what drives the voice coil motion. Hence, to give an extreme example, 1V at 500A is much better for a sub than 500V at 1A - the former will move the coil much better than the latter.

 

For this reason, the RIPS, though an excellent idea for other speakers which are less dependent on moving massive amounts of air, doesn't get you a whole lot for subwoofers.

 

Here's a test that I have been wanting to see a RIPS owner perform on his RIPS amp. The 500/1 amp lists the following specs for rail voltages:

4ohm or higher = 44.7V

2ohm = 31.6V

 

Therefore, with P=(V^2)/R you can determine the max power, which results in:

4ohm = 499W

2ohm = 499W

 

Both are as expected. (Note however that the amp cannot deliver 500W at 8ohms, which I thought it could... at 8ohms, you'll end up with 250W, as the rail voltage cannot go higher.)

 

Anyway, in theory, 500W to both a 2-ohm and 4-ohm speaker should give you the same punch in a sub, but from what I have heard from other experienced users, this is not the case - the 2-ohm setup still has more punch, due to the current being driven in that setup. (And this is what I will explain more mathematically below.)

 

Anyone care to try this? Maybe you (msmith) have access to the equipment to do this?

 

All speakers operate with current AND voltage, not just subwoofers. Subwoofers just have heavier masses, larger pistons and larger voice coils... but they're essentially no different than a mid or a tweeter from a functional perspective.

 

Since current is simply the flow of charge (voltage)... the same current with a higher voltage represents more power (which is what actually makes a speaker move).

I have to diasgree with you there, regarding subwoofers particularly. Again, reverting to the math, what makes the coil move? Force. What is the governing equation for a force imparted by a magnetic field and current in a wire?

See: http://phoenix.phys.clemson.edu/labs/223/magforce/ equation (5):

F = IL x B (that's a cross-product for vector notation, and F, L, and B are vectors - it's too annoying to replicate this properly on the forum)

 

Bottom line (ignoring the vectors which only tell you direction), force (F) is a function of the current (I) through a length of wire (L) that is in a magnetic field (B). Note there is no voltage in the equation.

 

Sure, you can get voltage in there, as V=IR, or I = V/R... but the point is that it's the CURRENT that moves the wire. The coil in a sub, which makes the force that moves the cone, is driven by this equation, and imparts force on the cone as a result of it.

 

Now, sure, you can't have current without voltage. However, due to practical limitations of speaker coils, you cannot get ohm-levels that are low enough to drive the same power through a coil "without voltage" (or with negligible voltage), nor can you get amps or electrical systems(reasonalbly priced) that can deliver as much current to drive the resultant theoretical current values. This is why my example of 1V at 500A is not "possible"... it's not feasible.

 

In other words, yes you need voltage to make current, but that's only because you have a resistance in the coil. If you had a superconducting coil (ie, near-zero resistnace) you could give up plenty of voltage and get more current in return... but practicality makes this a non-issue.

 

The whole point of RIPS is to detect the resistance (or impedance in this case) and adjust rail voltage to deliver optimized power into a variety of loads... whether the amp is operating in higher current/lower voltage mode at lower impedances or lower current/higher voltage mode at higher impedances, the net result is the same power... which is what really matters. A conventional amp needs to operate at its lowest safe impedance to produce optimum power.... a RIPS enabled amp gives you flexibility to use a variety of loads without sacrificing power and it doesn't matter whether you are using it on a subwoofer or a tweeter.

 

So, while you can't argue with physics... you do need to consider the whole picture.

Tweeters and other higher frequency speakers are not usually governed by their dependency on current. In fact, tweeters don't typically use a voice-coil at all, hence this F=ILxB equation doesn't apply, and that's why I am holding my discussion to subwoofer applications. In tweeter-land, voltage is usually what drives the output (piezo tweeters, for example) so a RIPS amp there would not affect performace as much. (Though, I suppose the reverse argument may apply there, where you want higher voltages (within limits) more than current capability... but that's another topic for another day/different thread).

 

Again, in the middle-of-the-frequency-range, power, as a whole, is probably a valid measure for the most part... but in subwoofers, it's all about pushing air, due to a movement of the cone, which is caused by the force imparted by the coil due to current through the coil which is in the magnetic field.

 

Pat: I was going to point out the same thing that msmith said about your dynamic-posts... but as he said, the amp doesn't work that way.

 

Boulderguy makes a valid point, that's a little beyond my discussion here - headroom. My discussion is related to max power, and headroom is more of a design consideration when choosing an amplifier for a given load. That being said, you are correct, when you drive lower ohm levels, you are eaching the limits of the amplifier's current delivery very quickly, and that makes for poorer sound. Again, this is more of a design consideration (ie don't drive "too low" for your particular amp - you gotta do the math or check the specs to find out what that is) than an amplifier issue, per se.

 

Phew... I hope I haven't totally lost everyone on this long-winded explanation. But in my experience, it's very important to understand the importance of current for a sub (and, implicty, proper system design ala Boulderguy's post).

[schwinn]

I want to point out - I am not against JL, nor do I think they make crappy equipment. The RIPS is a nice idea and all, but it seems more of a gimmick to me, particlularly in a subwoofer environment.

I like JL equipment... I want to make that clear.

[PGT]

you have too much time on your hands.....get outside and get some fresh air.

[msmith]

Schwinn...

 

Yes, I work for JL Audio and I know a fair amount about speakers, although I am not a degreed engineer. I will be happy to share your comments with one of our transducer engineers if you would like a more educated analysis.

 

To address your theory:

 

Let's simplify things... take a woofer in free air (no box): at its resonant frequency (Fs), its impedance is highest, yet it requires very little power to move a great deal. If the impedance is high, which limits the current and lowers the actual power delivered for a given voltage... how does the speaker move so much? How can it just be the current driving the motion if the above is occurring?

 

P.S. Piezoelectric tweeters are a very small minority in audio... most tweeters (domes, cones, ribbons, ring radiators, etc.) use voice coils.