Home Theater Forum  ›  Forums  ›  Home Theater  ›  Members Theaters and HT Projects  ›  Possible to modify a sub amp this way?
Forum Nav More Forums | All New Posts | New Thread

Possible to modify a sub amp this way?

#1
Rating: 0
11/11/03 at 3:00pm
  • Joined: September 2003
  • Post Count: 177
is it possible (probably since it can be done by request) to modify a rythmik audio plate amp with MOSFETs instead of bipolar transistors and to upgrade the opamps??

the reason i ask about the opamps is because i have a 12V power tap on my amp and it is only 100mA MAX. I wanted to be able to get more power out of it (more mA) and its the opamps that will burn out if i draw too much off of them. Also what other effects would i notice from having upgraded opamps?

Thanks
Export to Wiki
#2
Rating: 0
11/11/03 at 6:28pm
  • Joined: July 2001
  • Post Count: 1,591
If it's a subwoofer amp I wouldn't worry about changing the output transistors, even if you did it would probably require other modifications to the circuit. I wouldn't want to put in that much work.

I don't quite understand why you want to upgrade the op-amps... what do those have to do with your power supply?
Export to Wiki
#3
Rating: 0
11/11/03 at 7:57pm
  • Joined: September 2000
  • Post Count: 3,719
Why didn't you just have Brian do the mod when you bought the amp?

Brian Bunge
RAD Home Theater

Export to Wiki
#4
Rating: 0
11/12/03 at 4:22am
  • Joined: September 2003
  • Post Count: 177
cause hes done the other mods, and i have consumed so much of the guys time that i dont want to bother him anymore.

Ive got like 30+ emails from the guy of various topics and subjects. I feel bad

but as for upgrading the opamps, I was told by brian that if i draw too much current from my 12V supply that i will burn out the opamps, so i wanted to put something a little more powerful in there so i could draw more current.
Right now it is 12V @ 100mA but a couple of LEDs and that is maxed. I only want to add a couple of LEDs but i dont want to max it and risk burning out my opamps.
Export to Wiki
#5
Rating: 0
11/12/03 at 5:51am
  • Joined: July 2001
  • Post Count: 569
Anthony,
Changing opamps will have no affect on the power output of an amp. You may be able to effect a subtle sonic improvement in terms of "air" and "sweetness"... but this is a subwoofer amp, for gosh sakes. I would only dabble with upgraded opamps if you are using the high-pass filters in the plate amp for your mains.

Excessive loading on the power supplies will NOT damage the op amps. They are perfectly happy running with whatever voltage you give them. Now if you try to get a larger signal swing than the supply rails will support, they will just saturate and clip the signal. No harm done... but it will sound bad. If you are concerned about loading the power supplies with LEDs (can't imagine this... LEDs draw a couple of milliamps at most), you can use an external supply (wall wart) for the LEDs and use transistor(s) to trigger on the LEDs. Incidentally what are the extra LEDs for??

Finally, why would you want to change the output transistors. MOSFETs aren't inherently "better" than bipolars... just different. And you would need to change some of the other circuitry. If nothing else, Bipolars have a positive temperature coefficient... that is, when they get hot, they draw more current. Bipolar amplifiers have compensation elements to stabilize bias current over temperature. MOSFET outputs, having negative temperature coefficients, require entirely different compensation.

I'd recommend leaving the amp alone or sticking with the mods recommended by Brian.
Export to Wiki
#6
Rating: 0
11/12/03 at 8:45am
  • Joined: September 2003
  • Post Count: 177
OK MOSFETs....no go... no problem

But as for the opamps, Like ive been trying to say, I dont want to upgrade them in hopes of getting better sound or anything.

I was told by Brain that i should not draw more than 100mA off of my 12V TAP (This is 12V unregulated power supply that was tapped into by brian so i could power LED's) or i could burn out the opamps.
I do not want to burn out the opamps, so i assumed that if i changed them to more powerful ones then i could run more LEDs safely without worrying about burning out my opamps.
Ultra-Brights are about 30mA and 3.8V (Ive found a way to run them at 12V without burning them out, but thats another thread) i simply do not want to draw too much current by using too many LEDs, because i was told it would burn out the opamps.

The LEDs will be for added effects (Ground EFX, i used a car audio sub, so i wanted to emulate the look of a car so to speak) I do not know what the specs are on the LED that is built into my speaker, but i know it is 12V since it is meant to run on car power (more likely 14.4V), and probably more than 30mA, so running that, Along with 2 other LEDs will bring the current draw ABOVE 100mA, Which i was told would burn out the opamps. I do not want to burn out the opamps.

Dont ask me what they have to do with power supply ask brian, he is the one who put the power tap on there. Im guessing he tapped into opamps power somewhere along the line which is why they would burn out if i draw too much current from my 12V tap.

I hope this has cleared up why i want to upgrade the opamps.
Export to Wiki
#7
Rating: 0
11/12/03 at 9:44am
  • Joined: September 2000
  • Post Count: 3,719
Why not just run a resistor in series with the anodes of the LED's to get the current draw down?

EDIT: I just realized my post doesn't make sense. Having said that, how is it that adding more LED's is going to increase the current requirements of the op amps?

Brian Bunge
RAD Home Theater

Export to Wiki
#8
Rating: 0
11/12/03 at 10:50am
  • Joined: July 2001
  • Post Count: 569
Ah, I see... your latest message implies that an internal op-amp is actually sourcing the 12V. Makes sense (although 100mA suggests a pretty beefy op amp!).

Quote:
Ultra-Brights are about 30mA and 3.8V
In that case, the solution is simple. Put four "ultra brights" in series and connect them to the 12V source. You will have a bit less than 30mA and should be fine.

If the slight decrease in brightness is unacceptable, run three (11.4V drop) and a 20 ohm, 1/4 watt resistor (to drop the remaining 0.6V @30mA). Of course, you can run two of these strings in parallel (six LEDs total with 60mA total) or even three strings (nine LEDs and 90mA).

Quote:
I do not know what the specs are on the LED that is built into my speaker, but i know it is 12V since it is meant to run on car power (more likely 14.4V), and probably more than 30mA, so running that, Along with 2 other LEDs
It sounds like your goal is to run the LEDs in your sub along with two other "ultra brights". The ultra brights are easy... two in series would be 7.4V drop @30mA. A 150 ohm, 1/2 watt resistor in series with the two LEDs will drop the other 4.6V for a total of 12V.

As far as how much current the sub LEDs draw, that might be kind of tricky. Either you need to measure it or contact the manufacturer and see if they can tell you. As long as it's under 70 mA, you will be fine running that in parallel with your pair of "ultra brights".
Export to Wiki
#9
Rating: 0
11/12/03 at 11:22am
  • Joined: September 2003
  • Post Count: 177

As far as how much current the sub LEDs draw, that might be kind of tricky. Either you need to measure it or contact the manufacturer and see if they can tell you. As long as it's under 70 mA, you will be fine running that in parallel with your pair of "ultra brights".


Is there an easy way to measure it? The manufacturer dont know shit apparently... i asked them already.

I need you to explain the "drop" though.
Ive been reading about resistors and stuff a bit, but why is refered to as a "drop" when it appears that it adds up the voltage? (3.8 * 3 = 11.4)

and a resistor "resists" does it not? so how does that "drop" the voltage to get the remaining 4.6V if using 2?

What you said is easy to follow in terms of what needs to be done and how i should do it.
I just wanted some clarification of the terms.


how is it that adding more LED's is going to increase the current requirements of the op amps?


Adding more LEDs in parallel would require more amperage right?
But not if run in series? then it only needs more voltage? (they arent run in parallel, i was sort of lost, but Dave sort of cleared it up a bit)
Export to Wiki
#10
Rating: 0
11/12/03 at 11:54am
  • Joined: July 2001
  • Post Count: 569
Quote:
Is there an easy way to measure it?
Yes, if you have a digital multimeter. Most DMMs have a current setting that will handle a hundred milliamps or so. Otherwise, you can put a 1 ohm resistor in the line and measure the voltage across it: mV = mA in this case.

Quote:
I need you to explain the "drop"
Drop is the voltage drop through the resistor when there is current flowing through it. Think of the resistor as a restriction in a pipe. There's a certain pressure (voltage) at the inlet to the restriction and a lower pressure (voltage) at the outlet. The difference between these is the "drop". In electrical terms, voltage drop (V) = I * R where I is the current through the resistor and R is the resistor value.

Quote:
how does that "drop" the voltage to get the remaining 4.6V if using 2?
30 mA (.030A) * 150 ohms = 4.5V

Quote:
Adding more LEDs in parallel would require more amperage right?
Right. Currents add in parallel and voltages add in series.
Export to Wiki
#11
Rating: 0
11/13/03 at 7:26am
  • Joined: September 2003
  • Post Count: 177
What i meant when i asked about the drop (and i think i got it now) is, you were talking about the lights werent you?
IE if there is one light, it is running at 12V, with 2 lights, each one would be running at 6V, so there would have been a 6V drop in each light.

Right???
Export to Wiki
#12
Rating: 0
11/14/03 at 3:11pm
  • Joined: July 2001
  • Post Count: 569
The concept of woltage drop applies to resistors, lights, relays, etc. I was using a resistor as an example, but the same theory applies to the others.

Unfortunately, LEDs are a bit of a different animal. They typically have a constant forward voltage drop of about 0.7V per LED. If your "super brights" have 3.8V drop, they are probably actually five or six conventional LEDs in series. So, in the case of LEDs, V = I * R doesn't really apply. V is relatively constant and they just get brighter with increasing I.
Export to Wiki
Forum Nav More Forums | All New Posts | New Thread