Hello I look to buy GainClone with LM4780 (para) 1x 120w but many people said LM chip not great bass or no bass, blabla.......
what's the modifications to make to have BEST AMPLIFIER ?
What's the best +/- volt it's 30v or 35 v (for 120w 4 ohm load)
This amp have very little power supply Capacitor !
(no power reserve Capacitor for punch bass)
Do you add TANTALUM CAP for more great sound ?
WHAT CHANGES YOU HAVE MADE ??? (with good result only
!
Thanks !!!
Nick
what's the modifications to make to have BEST AMPLIFIER ?
What's the best +/- volt it's 30v or 35 v (for 120w 4 ohm load)
This amp have very little power supply Capacitor !
(no power reserve Capacitor for punch bass)
Do you add TANTALUM CAP for more great sound ?
WHAT CHANGES YOU HAVE MADE ??? (with good result only
!Thanks !!!
Nick
The only way to get good bass is to have big PSU caps. It's mathematical.
I'd suggest you to use 1000 or 2200 µF on the chips + 2*10 000µF on the PSU board.
You can calculate the needed capacity with this formula :
P is the power you need
R is the impedance of the load
U is the RMS voltage at the terminal of the load
P= U²/R
Once you've got U, you calculate the equivalent voltage Ur of the supply rails :
Ur = U*root(2).
I'd suggest you to use the highest possible rail voltage.
Then you calculate C :
Uc is the voltage at the terminals of the cap
t is the period of the discharge cycle (time during which the capacitor is discharging)
Uc = Ur-Ur*exp(-t/(R*C))
With 50Hz mains, you have t = 1/100
Then you can calculate for a given C and a given Ur the real voltage Uc at then end of the cap's discharge cycle.
So you deduce the real uncompressed power you can get from your PSU :
Preal = Uc²/R
From these formulas, you can see that your maximum output power is limited by the rail voltage and the caps of the PSU.
Don't forget that the amp also takes juice and has a voltage dropout, so increase the capacitance value by a 1,5 or 2 factor.
I have LM4780 amps in bridged mode and they work very well.
I'd suggest you to use 1000 or 2200 µF on the chips + 2*10 000µF on the PSU board.
You can calculate the needed capacity with this formula :
P is the power you need
R is the impedance of the load
U is the RMS voltage at the terminal of the load
P= U²/R
Once you've got U, you calculate the equivalent voltage Ur of the supply rails :
Ur = U*root(2).
I'd suggest you to use the highest possible rail voltage.
Then you calculate C :
Uc is the voltage at the terminals of the cap
t is the period of the discharge cycle (time during which the capacitor is discharging)
Uc = Ur-Ur*exp(-t/(R*C))
With 50Hz mains, you have t = 1/100
Then you can calculate for a given C and a given Ur the real voltage Uc at then end of the cap's discharge cycle.
So you deduce the real uncompressed power you can get from your PSU :
Preal = Uc²/R
From these formulas, you can see that your maximum output power is limited by the rail voltage and the caps of the PSU.
Don't forget that the amp also takes juice and has a voltage dropout, so increase the capacitance value by a 1,5 or 2 factor.
I have LM4780 amps in bridged mode and they work very well.
Hi all,
I queried the recommendation of using <=+-2.2mF on the chipamp as the sole smoothing/decoupling cap for a wideband amplifier in another thread.
All to a man/woman insisted that these wideband amps sounded best when the caps are kept small even suggesting that +-1mF was optimum.
It appears our thread starters has been following this advice.
Now I see the completely opposite recomendation if one requires bass rather than the more onerous wideband response.
What goes?
Are there two camps in the Forum, [big smoothing+decoupling] and [big decoupling]?
I queried the recommendation of using <=+-2.2mF on the chipamp as the sole smoothing/decoupling cap for a wideband amplifier in another thread.
All to a man/woman insisted that these wideband amps sounded best when the caps are kept small even suggesting that +-1mF was optimum.
It appears our thread starters has been following this advice.
Now I see the completely opposite recomendation if one requires bass rather than the more onerous wideband response.
What goes?
Are there two camps in the Forum, [big smoothing+decoupling] and [big decoupling]?
I'd avoid a large 'lytic... and rack up 20 or so 1000uF, with a little extra bypass.
I'd also include some small chokes (high amp, low DCR, inductance doesnt need to be colossal - you're using it to hammer RF grunge) in all 3 rails (+, - and Ground) - make the star earth point after the chokes.
I would also adopt a 'Pass' trick - run some of the capacitance across between the + and - rail.
Local bypassing of the chip is useful if you have marginal stability, or RF grunge present on the rails, to stop anything that might excite oscillations.
Have fun!
Owen
I'd also include some small chokes (high amp, low DCR, inductance doesnt need to be colossal - you're using it to hammer RF grunge) in all 3 rails (+, - and Ground) - make the star earth point after the chokes.
I would also adopt a 'Pass' trick - run some of the capacitance across between the + and - rail.
Local bypassing of the chip is useful if you have marginal stability, or RF grunge present on the rails, to stop anything that might excite oscillations.
Have fun!
Owen
AndrewT :
NOTE : i have made an error in the previous formulas. They applied to the charge cycle of the cap, which doesn't interests us.
I will consider these states in my explanation :
t0 is the moment at which the voltage at the capacitor's terminals is maximal (fully charged)
t is the time elapsed from t0
t1 is the time elapsed during one discharge cycle of the cap, that we will consider being equal to the time elapsed between two voltage peaks (pessimistic, and simpler)
Ur is the maximal voltage from one rail to the ground. It is the rail voltage as we would like it to be in a perfect world with infinite caps.
Uc is the voltage at the cap's terminals, which is the real voltage of the rails during the discharge cycle
Uc = Ur*exp(-t/(R*C))
exp is the exponential function. exp(x) = 2.72^X
At the end of the discharge cycle, with 50Hz mains and a full wave rectifier,
t1 = 1/(50*2) ) = 1/100
So finally we have :
Uc = Ur*2.72^(-0.01/(R*C))
For example :
I have a 25-0-25V trafo
with infinite caps, I get Ur = 25*root(2) = 35V
with 2*10000µF caps
4 ohms speaker
50Hz mains
fullwave rectifier
I get, using the formula above :
Uc = 35*2.72^(-0.01/(4*0,02)) = 31V
NOTE : i have made an error in the previous formulas. They applied to the charge cycle of the cap, which doesn't interests us.
I will consider these states in my explanation :
t0 is the moment at which the voltage at the capacitor's terminals is maximal (fully charged)
t is the time elapsed from t0
t1 is the time elapsed during one discharge cycle of the cap, that we will consider being equal to the time elapsed between two voltage peaks (pessimistic, and simpler)
Ur is the maximal voltage from one rail to the ground. It is the rail voltage as we would like it to be in a perfect world with infinite caps.
Uc is the voltage at the cap's terminals, which is the real voltage of the rails during the discharge cycle
Uc = Ur*exp(-t/(R*C))
exp is the exponential function. exp(x) = 2.72^X
At the end of the discharge cycle, with 50Hz mains and a full wave rectifier,
t1 = 1/(50*2) ) = 1/100
So finally we have :
Uc = Ur*2.72^(-0.01/(R*C))
For example :
I have a 25-0-25V trafo
with infinite caps, I get Ur = 25*root(2) = 35V
with 2*10000µF caps
4 ohms speaker
50Hz mains
fullwave rectifier
I get, using the formula above :
Uc = 35*2.72^(-0.01/(4*0,02)) = 31V
So, we now know the minimal supply rail voltage.
If we are in bridged mode (which I don't recommend you with 4 ohms), we add the voltage of both rails and get 60V p-p
The equivalent RMS voltage is 60/root(2) = 42V R.M.S
So in 4 ohms, you obtain :
42²/4 = 450W
What you can see is that 4 ohms bridged is totally impossible with one chip. As you want to do, we can parallel the two amps in the LM4780, or use an 8 ohms driver.
Using an 8ohms driver :
P = 42²/8 = 225W
Paralleling the amps (or using only one, that's the same). This is the safest solution, and the less heating :
P = (31/root(2))²/4 = 110W
Keep in mind this is theoretical. The amps have a voltage dropout, and you will always need extra margin for transcients.
Note that it is very interesting to increase the supply rail to their maximum voltage.
E is the energy stored in a cap
C is the capacitance of the cap
Uc is the voltage at its terminals
E = C*Uc²
You can see that the energy is proportional to the square of the voltage !
More juice for no cost (or little if you need to increase the cap's rating)
Hope it helps,
cheers !
If we are in bridged mode (which I don't recommend you with 4 ohms), we add the voltage of both rails and get 60V p-p
The equivalent RMS voltage is 60/root(2) = 42V R.M.S
So in 4 ohms, you obtain :
42²/4 = 450W
What you can see is that 4 ohms bridged is totally impossible with one chip. As you want to do, we can parallel the two amps in the LM4780, or use an 8 ohms driver.
Using an 8ohms driver :
P = 42²/8 = 225W
Paralleling the amps (or using only one, that's the same). This is the safest solution, and the less heating :
P = (31/root(2))²/4 = 110W
Keep in mind this is theoretical. The amps have a voltage dropout, and you will always need extra margin for transcients.
Note that it is very interesting to increase the supply rail to their maximum voltage.
E is the energy stored in a cap
C is the capacitance of the cap
Uc is the voltage at its terminals
E = C*Uc²
You can see that the energy is proportional to the square of the voltage !
More juice for no cost (or little if you need to increase the cap's rating)
Hope it helps,
cheers !
Is this the one to which you refer?
An externally hosted image should be here but it was not working when we last tested it.
Yes, I could run a similar test but on this one I can clearly hear what I'm not getting. I generally use a warble generator and calibrated microphone with my speaker in a controlled setting to perform my response test.
The LM3886 performs well and has good bass response. The 4780 clearly sounds like you took it to the Vet for an operation.
Like I mentioned....your board is very nice and I clearly don't have a problem with it. I am questioning the 4780 chip. Maybe I need a bank of 100,000 mfd caps and dual 650 va transformers to wake it up. Maybe there is a problem with the chips. Heck I don't know.
It would seem the chips are working correctly as I have DC offset of 3.7mv one channel and 5.7mv on the other. I get a clean sine wave on both channels.
At this point in time I surely wouldn't recommend it to anyone. I especially wouldn't try to use it for a subwoofer amplifier.
burnedfingers said:
It is very, very uncommon that a bass problem has ANYTHING to do with supply caps. Even with 1000uF people have reported good bass. The fundamental mistake here is to relate the cap discharge of the supply caps to bass response. The only relation is to max output power at bass frequencies, but that's not the issue. The issue (if indeed you have lack of bass) is that the 'gain' at bass frequencies is lower than that at mid/high frequencies. This is a freq response problem.
Possible culprits: input coupling capacitor, feedback decoupling cap, etc.
Without a schematic diagram with values this is impossible to troubleshoot. Give us the circuit!
Jan Didden
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