When you try to buy a capacitor ; look at the datasheet of it and find ripple current.Buy the highest ripple current one.Computer grade caps have higher ripple current and they are much more expensive than the others.There are audio capacitors too(Like nichicon gold and elna) but they have less ripple current than computer grade ones.Some say they sound better.I do not know how they sound better(Maybe their distortion is lower).If someone tell me the differences between them I will be glad.Thank you
I think you would waste your money on Mundorf after Kendeil caps !
Those are better than Mundorf .
If you have a lot of money try BHC slit foil or T Network !
BHC - AudioCap Limited
Also Siemens Sikorel great caps to .But BHC slit foil or T network my favorite .
Greets
Those are better than Mundorf .
If you have a lot of money try BHC slit foil or T Network !
BHC - AudioCap Limited
Also Siemens Sikorel great caps to .But BHC slit foil or T network my favorite .
Greets
Why do you think Mundorfs are bad?I have some mkp caps(MCAP RXF) and i am happy with them.Though i did not try any electrolytic mundorf cap.
I am in dilemma to use low esr computer grade cap or audio caps on class A amps(like elna nichicon gold etc)Nelson sometimes says he did not use nichicon etc on F5.He says it is not important to use an audio cap.But he says elna silmic 2 sounds good.I do not know which of them sounds better(low esr or audio cap).And i can not see anything about sound quality on datasheets.There are not any graphic about sound quality.I can learn which of them more quality only on magazine news.
I am interested in elna silmic large capacity caps.I want to try them.But they are very rare!
I am in dilemma to use low esr computer grade cap or audio caps on class A amps(like elna nichicon gold etc)Nelson sometimes says he did not use nichicon etc on F5.He says it is not important to use an audio cap.But he says elna silmic 2 sounds good.I do not know which of them sounds better(low esr or audio cap).And i can not see anything about sound quality on datasheets.There are not any graphic about sound quality.I can learn which of them more quality only on magazine news.
I am interested in elna silmic large capacity caps.I want to try them.But they are very rare!
I didn't wrote Mundorf is bad , but you use Kendeil now which is much better .Also those caps BHC Slit foil and T network are much better than Mundorf .
Check the website in my previous post ..
BHC Aerovox forget it .
Mundorf foil caps are great , electrolytic not the best (top) !
If you buy Mundorf please remember my advise .
You wrote you want something good !
Greets
I did not have an idea about mundorf electrolytics.I only know that they have very low esr and much ripple current(about 100A)
But i do not think that i will buy them.Too expensive
Are there 3 types of BHC caps ? Aerovox ,slit foil and T network ha ?
In my city there are some BHC caps.I think they may be Slit foil ones.If i can get them in good price I will buy them.Thank you
But i do not think that i will buy them.Too expensive
Are there 3 types of BHC caps ? Aerovox ,slit foil and T network ha ?
In my city there are some BHC caps.I think they may be Slit foil ones.If i can get them in good price I will buy them.Thank you
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power output into 16 ohm load
Just curious -- what would be the max class A power output available into a 16 ohm load?
If we have 2.6 amps into a 16 ohm load the peak power should be 2.6 * 2.6 * 16 = 108 Watts peak (so, 54 watts avg).
However, at 108 peak watts into 16 ohms, we would need a voltage of 42v (V=I*R with I=2.6 and R=16). So, we may be voltage limited. We have +/-24 volts at the rails. So, there is 48v available from the rails. But, what is the max voltage output that the circuit will allow?
Thanks,
Steve
Just curious -- what would be the max class A power output available into a 16 ohm load?
If we have 2.6 amps into a 16 ohm load the peak power should be 2.6 * 2.6 * 16 = 108 Watts peak (so, 54 watts avg).
However, at 108 peak watts into 16 ohms, we would need a voltage of 42v (V=I*R with I=2.6 and R=16). So, we may be voltage limited. We have +/-24 volts at the rails. So, there is 48v available from the rails. But, what is the max voltage output that the circuit will allow?
Thanks,
Steve
umut1001 -- I thoght with +- 24v volt supply - you have 48v. So, 48/16 would give you 3 amps.
bobodioulasso -- the power halves when you double the load value when you are looking at power from the max voltage obtainable (V*V)/R. But if we are looking at power from the current (I*I)*R, Power would theoretically double with doubling of the load. However, voltage is going to limit that number. I think?
thanks,
Steve
bobodioulasso -- the power halves when you double the load value when you are looking at power from the max voltage obtainable (V*V)/R. But if we are looking at power from the current (I*I)*R, Power would theoretically double with doubling of the load. However, voltage is going to limit that number. I think?
thanks,
Steve
bobodioulasso --
From the manual:
"The power of 2.6 amps into 8 ohms is I*I*R, or 2.6*2.6*8=54 watts. This is the peak value, and the nature of an undistorted sine wave is that the peak wattage is twice the average, so this circuit would operate 27 watts average Class A into 8 ohms. At currents above 2.6 amps one of the transistors will shut off, leaving the other to continue to increase beyond 2.6 amps in what is known as Clas AB."
So, if I substitute 16 ohms for the 8 ohms used in the equation above, 2.6*2.6*16=108watts peak and 54 watts average.
However, you are saying that only 20 volts are available (20*20)/2. If this is the case, then you are answering my original question which was: how much voltage is available?
Why are we dividing the voltage rails by 2?
Thanks,
Steve
From the manual:
"The power of 2.6 amps into 8 ohms is I*I*R, or 2.6*2.6*8=54 watts. This is the peak value, and the nature of an undistorted sine wave is that the peak wattage is twice the average, so this circuit would operate 27 watts average Class A into 8 ohms. At currents above 2.6 amps one of the transistors will shut off, leaving the other to continue to increase beyond 2.6 amps in what is known as Clas AB."
So, if I substitute 16 ohms for the 8 ohms used in the equation above, 2.6*2.6*16=108watts peak and 54 watts average.
However, you are saying that only 20 volts are available (20*20)/2. If this is the case, then you are answering my original question which was: how much voltage is available?
Why are we dividing the voltage rails by 2?
Thanks,
Steve
bobodioulasso --
sorry, didn't mean to say "why are you dividing the rail voltages by 2"
what I mean to ask is:
1. why is there only 20 volts available and not the entire rail voltage differential?
2. why are you dividing the voltage squared by 2 in your formula for wattage?
Thanks,
Steve
sorry, didn't mean to say "why are you dividing the rail voltages by 2"
what I mean to ask is:
1. why is there only 20 volts available and not the entire rail voltage differential?
2. why are you dividing the voltage squared by 2 in your formula for wattage?
Thanks,
Steve
ok bobodioulasso,
So, if I'm following you correctly:
-- Let's say you've got +-24 volt rails. That would give you 48 volts total.
-- Let's take off 8 volts for various losses. I'm not sure if this is a good approximation or not. I just chose it so that we have an even 40 volts available at the rails.
-- So, if we have 40 volts as Vmax, we can now divide by 1.414. This would give you 28.29volts.
-- Now, using (V*V)/R as the calculation for Power, we get (28*28)/16 = 49 watts.
So, I'm getting 49watts. Where is my error? Do we lose more than the 8 volts I chose to subtract from the rails?
Thanks,
Steve
So, if I'm following you correctly:
-- Let's say you've got +-24 volt rails. That would give you 48 volts total.
-- Let's take off 8 volts for various losses. I'm not sure if this is a good approximation or not. I just chose it so that we have an even 40 volts available at the rails.
-- So, if we have 40 volts as Vmax, we can now divide by 1.414. This would give you 28.29volts.
-- Now, using (V*V)/R as the calculation for Power, we get (28*28)/16 = 49 watts.
So, I'm getting 49watts. Where is my error? Do we lose more than the 8 volts I chose to subtract from the rails?
Thanks,
Steve