Perhap's its just me, but I seem to be seeing inconsistency in National Semiconductor's specs. and info regarding the LM3886.
I would really appreciate some clarity from the all-powerful DIY guys of this site! I did a search but no clear answers...and before I resort to calling National Semi....
Problem: data sheet AN-1192 specifies: (for PA-100 parallel amp design)
"...the equation does not change for total Pdmax
where Pdmax = VccTOT^2 / (2*pi^2*Rl)"
Assuming my parameters of:
Application: car, 4 ohm typical, but 2 Ohm possible
load = Rl = 4 Ohm, letting Vcc = +37V, Vee = -37V (VccTOT = 74V) gives:
Pdmax = (74)^2 / (2*pi*4) = 69.35W
...and for PA100 example, National recommends:
Table 1. "Maximum Supply Voltages"
_______ 2 Ohms 4 Ohms 6 Ohms ____ ....
PA100 +/-28V +/-37V +/-37V
==HOWEVER!==
Using National's Overture_Design_Guide13.xls spreadsheet, we enter the same data, and have:
Enter: LM3886, Rl = 4, Vcc/Vee = +- 37, # of ICs = 2, default feedback
Results in "PARALLEL MODE POWER CALCULATION":
Pd/IC = 69.35 W
Parallel Load = 2 Ohm
Parallel Power = 243.36
Warning: "Required heat sink is too large!"
...which seems contradictory, as the data sheet states:
" Using National's LM3886 in a two device parallel solution driving a 4 Ohm load will typically provide 110W of output power." -AND-
"So for a 4 Ohm solution using two amplifiers will result in each amplifier seeing and 8 Ohm load."
...and recommends values for 2 & 4 Ohm in Table 1.
I'm really confused about the following:
1. What is really the effective load impedance, of the the parallel amp config.?
2. What is the real output power configuration? Should Po (output power) instead be 2*Vcctot^2 / (28pi^2*Rl) ?
3. Any idea where 243.36W or "2 Ohms" came from in spreadsheet calculations?
Also the AN-1192 states PA design supports 2 Ohm load, yet spreadsheet refuses to calculate using Rl = 2. I believe the data sheet/app. note is not clear.
I seem to be really confused about the purpose of the PA configuration. Now I'm believing it's only to share dissipation between LM3886 ICs, not increase power to load. Am I wrong?
Thanks so much if you can clear this up, especially mathematically!
I would really appreciate some clarity from the all-powerful DIY guys of this site! I did a search but no clear answers...and before I resort to calling National Semi....
Problem: data sheet AN-1192 specifies: (for PA-100 parallel amp design)
"...the equation does not change for total Pdmax
where Pdmax = VccTOT^2 / (2*pi^2*Rl)"
Assuming my parameters of:
Application: car, 4 ohm typical, but 2 Ohm possible
load = Rl = 4 Ohm, letting Vcc = +37V, Vee = -37V (VccTOT = 74V) gives:
Pdmax = (74)^2 / (2*pi*4) = 69.35W
...and for PA100 example, National recommends:
Table 1. "Maximum Supply Voltages"
_______ 2 Ohms 4 Ohms 6 Ohms ____ ....
PA100 +/-28V +/-37V +/-37V
==HOWEVER!==
Using National's Overture_Design_Guide13.xls spreadsheet, we enter the same data, and have:
Enter: LM3886, Rl = 4, Vcc/Vee = +- 37, # of ICs = 2, default feedback
Results in "PARALLEL MODE POWER CALCULATION":
Pd/IC = 69.35 W
Parallel Load = 2 Ohm
Parallel Power = 243.36
Warning: "Required heat sink is too large!"
...which seems contradictory, as the data sheet states:
" Using National's LM3886 in a two device parallel solution driving a 4 Ohm load will typically provide 110W of output power." -AND-
"So for a 4 Ohm solution using two amplifiers will result in each amplifier seeing and 8 Ohm load."
...and recommends values for 2 & 4 Ohm in Table 1.
I'm really confused about the following:
1. What is really the effective load impedance, of the the parallel amp config.?
2. What is the real output power configuration? Should Po (output power) instead be 2*Vcctot^2 / (28pi^2*Rl) ?
3. Any idea where 243.36W or "2 Ohms" came from in spreadsheet calculations?
Also the AN-1192 states PA design supports 2 Ohm load, yet spreadsheet refuses to calculate using Rl = 2. I believe the data sheet/app. note is not clear.
I seem to be really confused about the purpose of the PA configuration. Now I'm believing it's only to share dissipation between LM3886 ICs, not increase power to load. Am I wrong?
Thanks so much if you can clear this up, especially mathematically!
LM3886 power & dissipation
National goes by the book on there output and dissipation factors, to simplify things look at it this way, lm3886 package dissipation is 120w
to keep in the SOA of the chip =50% of the dissipation or 60w to the load, so two chip into a 4ohm load at 35VDC+/- supply would give you around 120W, so if you want to drive a 2 ohm load you will need 3 parellel chips and average power to the load will be 180w. Peak power will be higher because of by the book rules.
If you plan on useing the amp to drive a 2ohm load at sustained high output levels I would suggest 4 parellel chips, then you are sure to stay within the SOA of the chips, ecspecially when in the real world when a 2 ohm speaker load can drop much lower then 2 ohms at certain frequency's.
PS this of course takes into consideration proper heat sinking.
And yes it is partially to share dissipation and increase power to load, but it all revolves around the dissipation of each chip.
National goes by the book on there output and dissipation factors, to simplify things look at it this way, lm3886 package dissipation is 120w
to keep in the SOA of the chip =50% of the dissipation or 60w to the load, so two chip into a 4ohm load at 35VDC+/- supply would give you around 120W, so if you want to drive a 2 ohm load you will need 3 parellel chips and average power to the load will be 180w. Peak power will be higher because of by the book rules.
If you plan on useing the amp to drive a 2ohm load at sustained high output levels I would suggest 4 parellel chips, then you are sure to stay within the SOA of the chips, ecspecially when in the real world when a 2 ohm speaker load can drop much lower then 2 ohms at certain frequency's.
PS this of course takes into consideration proper heat sinking.
And yes it is partially to share dissipation and increase power to load, but it all revolves around the dissipation of each chip.
Ok, thanks guys, that helps a little bit.
Of course, it is 4 Ohms I plan to use typically. I do know that the supply V must be lower for lower Rl (as stated in Table 1. above).
I think most of it is cleared up-I suspected what you just said...
...just it would be nice to know the formula used by National for the spreadsheet for arrving at "2 Ohm" from Rl/#ICs and that power dissipation they stated (see above).
Thanks and Happy Halloween!
PS: if anyone needs footprint for the LM3886 or LM4780 for Orcad Layout, let me know.
Of course, it is 4 Ohms I plan to use typically. I do know that the supply V must be lower for lower Rl (as stated in Table 1. above).
I think most of it is cleared up-I suspected what you just said...
...just it would be nice to know the formula used by National for the spreadsheet for arrving at "2 Ohm" from Rl/#ICs and that power dissipation they stated (see above).
Thanks and Happy Halloween!
PS: if anyone needs footprint for the LM3886 or LM4780 for Orcad Layout, let me know.
LM design rules
I just looked over the design guide you are 100% correct something is wrong with there spread sheet. The math dosnt add up no matter what you do.
If you are driveing 4ohm load in parellel you are safe at 35VDC+/- with proper heat sinking.
I run 8 LM3875's in Bridge parellel at 40VDC+/- with no issues even into 4 ohms, thermal becomes an issue at 4ohms over long high output durations.
I just looked over the design guide you are 100% correct something is wrong with there spread sheet. The math dosnt add up no matter what you do.
If you are driveing 4ohm load in parellel you are safe at 35VDC+/- with proper heat sinking.
I run 8 LM3875's in Bridge parellel at 40VDC+/- with no issues even into 4 ohms, thermal becomes an issue at 4ohms over long high output durations.
i'm not overly suprised. The LM4780's datasheet has had a "inverting amplifier" example and explaination that is incorrect and has been so since the inception of the product. I have twice posted bugs on this but to no avail.
edit -- oh and they also have another error that i've found. They stated that an amplifier without feedback is unstable. I forget where that was.
In short, I'd confirm everything from National and not take them at their word. (same for anything else important, but this is the 3rd issue i've heard of from National). And National is still a good company, I just wish they'd revise and correct some of their documents.
edit -- oh and they also have another error that i've found. They stated that an amplifier without feedback is unstable. I forget where that was.
In short, I'd confirm everything from National and not take them at their word. (same for anything else important, but this is the 3rd issue i've heard of from National). And National is still a good company, I just wish they'd revise and correct some of their documents.
There is nothing wrong with calculations it is just you are not understanding things correctly. Here are some tips.
1. When running in bridge mode (BTL) like the BR100 amp in AN-1192 each IC will see 1/2 the load impedance.
2. When running in parallel mode like the PA100 in AN-1192 each IC will see #number of ICs in parallel * load. 2 ICs means double the real load impedance is seen.
3. BPA mode with 4 total ICs means each IC sees the actual load impedance, so 8 ohm real load means each IC sees 8 ohms. For higher chip BPA you have to work it out. First each side of the bridge sees 1/2 the REAL load. Then each side of the bridge with the ICs in parallel sees as #2 above.
So in BTL an 8 ohm load looks like a 4 ohm load to each IC. The LM3886 datasheet states that for 4 ohm load the supply is +/-28V. This works out to about 40W of power dissipation. For parallel with 2 ICs then a 4 ohm load looks like 8 ohms. 3 chips makes this 4 ohm load look like a 12 ohm load to each chip. All of the Overture chips do not work well at 2 ohms, too hot and Spike protection will kick on fast. You will almost ALWAYS get less power at 2 ohms than 4+ ohms. So this means the design of any multichip amp needs to be designed so each IC sees 4 ohms or higher. Then based on what each IC sees you must use the correct supply voltage range to keep power dissipation around 40W.
For the PA100 table recommendation, for 2 chip, 2 ohm load then each chip sees 4 ohms and the datasheet says +/-28V for a LM3886 driving a 4 ohm load. At 4 ohms each sees 8 ohms and 6ohms means each sees 12 ohms. +/-37V is fine at this level (the design guide shows ~ 35W pd so it is fine). The reason it is not higher for parallel 6 ohms is because you do not want to run too close to the maximum rails. National is always a tad conservative.
What is confusing you is that the load impedance you enter in the design guide on the left is NOT the load impedance for bridge or parallel calculations. The impedance for these designs are show on the right where the power calculations are. If you want to see 4 ohms in parallel then enter 8 ohms to the left. This means each load 'see' 8 ohms but the REAL load is 4 ohms for parallel. You need to read the design guide instructions (linked where the design guide is) to fully understand. You can also change the numebr of ICs in parallel and see the PA load impedance change accordingly.
The reason the design won't work is shown clearly on the design guide and given in the warnings. First problem is the heat sink needed is too large, the ICs will have Spike and thermal shutdown kicking in under heavy use or sine waves. The second problem is that the IC cannot put out that much current. You will be thermal and current limited. In an ideal world you are never limited but in the world of chip amps you are ALWAYS liimited by one or more of three things all interelated:
- thermal limits, how much power dissipation is possible with a 'reasonable' heat sink. For the design guide reasonable is 1C/W or higher thermal resistance. If that number drops below 1C/W then you will get the warning.
- voltage limits, chip amps can only operate with so much voltage before the chip is destroyed. For the LM3886 +/-42V (84V on the datasheet) is about as high as you want to continuously operate it driving a load.
- current limits, relates directly to die size, more die, more cost, more current output potential. Having voltage supplies higher than what is needed to get near the current limit does ABSOLUTELY no good, just makes the chip run hotter since PD is affected by the supply voltage SQUARED. A small change in voltage is a big change in PD.
So you need to keep the voltage, the load impedance seen by each chip, need to be kept within the current limits, voltage limits and thermal limits of the IC. For the LM3886 this means no more than about 40W (give or take) of power dissipation per IC AND stay within the current limits and voltage limits in the datasheet. The design guide tries to do this for you by giving warnings when you try to excced recommendations but not so strict you can't get info outside of the recommendations. The LM3886 is a 60W amp and so you should figure you can get around 60W from each IC when configured properly. You can get more, I rean one at +/-37V with a HUGE heat sink and a big fan to see how hard it could go. Into 4 ohms I could only run it a bit to measure then let it cool. I was getting around 122W.
for question #3, the 243.36W comes from doubling the 128W you will see in the upper part of the design guide and subtracting out the power loss in the 0.1 ohm output resistors for parallel operation. BTW, the current design guide is version 1.5 but all the numebrs you have calculated are the same and accurate.
AN-1192 shows ~110W but because output power varies it is a conservative number. It may be that number was using the LM3886 datasheet +/-35V recommended supplies for an 8 ohm load.
Please ask any more questions but DO read the design guide instructions. All numbers are accurate and you will have a very hard time running LM3886s with 70W of power dissipation without some protection turning on.
-SL
1. When running in bridge mode (BTL) like the BR100 amp in AN-1192 each IC will see 1/2 the load impedance.
2. When running in parallel mode like the PA100 in AN-1192 each IC will see #number of ICs in parallel * load. 2 ICs means double the real load impedance is seen.
3. BPA mode with 4 total ICs means each IC sees the actual load impedance, so 8 ohm real load means each IC sees 8 ohms. For higher chip BPA you have to work it out. First each side of the bridge sees 1/2 the REAL load. Then each side of the bridge with the ICs in parallel sees as #2 above.
So in BTL an 8 ohm load looks like a 4 ohm load to each IC. The LM3886 datasheet states that for 4 ohm load the supply is +/-28V. This works out to about 40W of power dissipation. For parallel with 2 ICs then a 4 ohm load looks like 8 ohms. 3 chips makes this 4 ohm load look like a 12 ohm load to each chip. All of the Overture chips do not work well at 2 ohms, too hot and Spike protection will kick on fast. You will almost ALWAYS get less power at 2 ohms than 4+ ohms. So this means the design of any multichip amp needs to be designed so each IC sees 4 ohms or higher. Then based on what each IC sees you must use the correct supply voltage range to keep power dissipation around 40W.
For the PA100 table recommendation, for 2 chip, 2 ohm load then each chip sees 4 ohms and the datasheet says +/-28V for a LM3886 driving a 4 ohm load. At 4 ohms each sees 8 ohms and 6ohms means each sees 12 ohms. +/-37V is fine at this level (the design guide shows ~ 35W pd so it is fine). The reason it is not higher for parallel 6 ohms is because you do not want to run too close to the maximum rails. National is always a tad conservative.
What is confusing you is that the load impedance you enter in the design guide on the left is NOT the load impedance for bridge or parallel calculations. The impedance for these designs are show on the right where the power calculations are. If you want to see 4 ohms in parallel then enter 8 ohms to the left. This means each load 'see' 8 ohms but the REAL load is 4 ohms for parallel. You need to read the design guide instructions (linked where the design guide is) to fully understand. You can also change the numebr of ICs in parallel and see the PA load impedance change accordingly.
The reason the design won't work is shown clearly on the design guide and given in the warnings. First problem is the heat sink needed is too large, the ICs will have Spike and thermal shutdown kicking in under heavy use or sine waves. The second problem is that the IC cannot put out that much current. You will be thermal and current limited. In an ideal world you are never limited but in the world of chip amps you are ALWAYS liimited by one or more of three things all interelated:
- thermal limits, how much power dissipation is possible with a 'reasonable' heat sink. For the design guide reasonable is 1C/W or higher thermal resistance. If that number drops below 1C/W then you will get the warning.
- voltage limits, chip amps can only operate with so much voltage before the chip is destroyed. For the LM3886 +/-42V (84V on the datasheet) is about as high as you want to continuously operate it driving a load.
- current limits, relates directly to die size, more die, more cost, more current output potential. Having voltage supplies higher than what is needed to get near the current limit does ABSOLUTELY no good, just makes the chip run hotter since PD is affected by the supply voltage SQUARED. A small change in voltage is a big change in PD.
So you need to keep the voltage, the load impedance seen by each chip, need to be kept within the current limits, voltage limits and thermal limits of the IC. For the LM3886 this means no more than about 40W (give or take) of power dissipation per IC AND stay within the current limits and voltage limits in the datasheet. The design guide tries to do this for you by giving warnings when you try to excced recommendations but not so strict you can't get info outside of the recommendations. The LM3886 is a 60W amp and so you should figure you can get around 60W from each IC when configured properly. You can get more, I rean one at +/-37V with a HUGE heat sink and a big fan to see how hard it could go. Into 4 ohms I could only run it a bit to measure then let it cool. I was getting around 122W.
for question #3, the 243.36W comes from doubling the 128W you will see in the upper part of the design guide and subtracting out the power loss in the 0.1 ohm output resistors for parallel operation. BTW, the current design guide is version 1.5 but all the numebrs you have calculated are the same and accurate.
AN-1192 shows ~110W but because output power varies it is a conservative number. It may be that number was using the LM3886 datasheet +/-35V recommended supplies for an 8 ohm load.
Please ask any more questions but DO read the design guide instructions. All numbers are accurate and you will have a very hard time running LM3886s with 70W of power dissipation without some protection turning on.
-SL
the "inverting" configuration in the lm4780 is not correct. Earlier in the datasheet, they show a bridged application which correctly shows how an inverting configuration is connected.
The issue is that they show a resistor, Rin which they claim sets the input impedance of the amplifier and can be used in the formula Fc = 1/(2pi Rin Cin). the gain being set by Av = -R2/R1. But R1 is really in parallel with Rin, in which case the input impedance is not 47,000ohm, but rather 900ohm. This moves the pole location into the midrange band!
The other thing i've found was an odd blurb on an Analog University blurb over operational amplifiers. There was a section about operational amplifiers without feedback being unconditionally unstable. This isn't correct either. while the output of an opamp without feedback would be heavily distorted, its not unstable. It will latch to one of two stable points as well which makes it effectively bistable. I was going to reccomend that students in my labs use that source, but decided not to when I saw that part.
The issue is that they show a resistor, Rin which they claim sets the input impedance of the amplifier and can be used in the formula Fc = 1/(2pi Rin Cin). the gain being set by Av = -R2/R1. But R1 is really in parallel with Rin, in which case the input impedance is not 47,000ohm, but rather 900ohm. This moves the pole location into the midrange band!
The other thing i've found was an odd blurb on an Analog University blurb over operational amplifiers. There was a section about operational amplifiers without feedback being unconditionally unstable. This isn't correct either. while the output of an opamp without feedback would be heavily distorted, its not unstable. It will latch to one of two stable points as well which makes it effectively bistable. I was going to reccomend that students in my labs use that source, but decided not to when I saw that part.
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.