is current in speaker load to be anyway closer to psu current ratings? - Page 2 - diyAudio
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Old 27th November 2012, 03:30 PM   #11
AndrewT is offline AndrewT  Scotland
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Quote:
Originally Posted by rhythmsandy View Post
......................the output voltage is about 60v peak to peak and when the imp = 4 ohms then according to ohms law i=v/r so total is 60/4 which is 15 amps
No !

Learn that Vdc and Vac and Vrms and Vpk and Vpp all mean different things when taking measurements.
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Old 27th November 2012, 03:34 PM   #12
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Not necessarily fully true.

Music is quite dynamic and there is never given full rated power continuous.

But if we look at it as for the example here, I would have gone for a 1000VA transformer.
The capacity would be around 17,5Amps continously, wich would match Your amp near perfectly.

But bear in mind: This is MY opinion.
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Old 27th November 2012, 03:36 PM   #13
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Originally Posted by AndrewT View Post
No !

Learn that Vdc and Vac and Vrms and Vpk and Vpp all mean different things when taking measurements.
Good.
Thats something to take in concideration too.

But such answers without anything to explain Your statement wouldn't be at any help, would it?
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Old 27th November 2012, 03:54 PM   #14
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sorry it was vrms which i was talking about what will be the output vrms with the psu +/-78v which would give more clear conclusions..
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Old 8th December 2012, 05:11 AM   #15
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i was looking at the nelsons article where he states about for 200watts he recommends using 700w and for stereo to have a 2kva and sounds pretty sufficient..

but i have a practical example for 300
I=sqrt(P/load)

sqrt(300/8)
= 6.12amps

now here the voltage ratings of the trafo is 56-0-56
so considering this when compared to the psu ratings it has to be 2 x 300 = 600W as common mode of selection which will result in
Arms in trafo = 600/(56*2) = 5.35 amps...

now the question is when there is 6.12 amps in speaker then where does the extra current come from? or 50% has to convert into heat so we need 6.12 amps more to burn into heat...

so the ratings of the trafo has to be atleast min of 12.24amps which results as (56*2)x12.24 = 1370VA and with 5-10% losses its roughly 1500VA as standard size...

if that is right?
when we take a 4 ohm load into 500W the same amplifier which works in 300watt in 8 ohm.. the the current in speaker is 11.18 amps and when we double this its 22.36
so (56*2)x 22.36 = 2504VA

now it looks like its 5 times the ratings are the right ones to deliver the continuous rated output..
are these calculations are right?
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Old 8th December 2012, 01:06 PM   #16
AndrewT is offline AndrewT  Scotland
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Go look at PSU design.
The output to the amplifier comes from the smoothing capacitors.
The transformer secondaries are switched OFF for ~90% of the time. These switched OFF secondaries cannot supply any current to the amplifier.

When the secondaries are switched ON, the current from the transformer must recharge the smoothing capacitors.
The charging current is a pulse that can be ~10times the average current supplied to the amplifier.

If your amplifier draws an average current of 5A, then the secondaries may have to charge the smoothing capacitors with upto 50Apk during the 10% duty cycle.

As I said at the beginning of this POST.
Go and read and LEARN.
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Old 8th December 2012, 05:23 PM   #17
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so if an amp draws 10 amps continuously then the transformer has to deliver 100Apeak or 70amps continuous? this heavy trafo? but the strange thing is that in bridge rectifier mode i accept that all the diodes stop working at one instance but not 90% how is it 90% off?
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Old 8th December 2012, 06:33 PM   #18
AndrewT is offline AndrewT  Scotland
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Originally Posted by rhythmsandy View Post
so if an amp draws 10 amps continuously then the transformer has to deliver 100Apeak or 70amps continuous?...............
No.
I told you to go and read and learn.
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Old 8th December 2012, 06:40 PM   #19
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sir i didnt understand can you explain how does this work? there is no proper material on net on this aspect and it actually requires years of expertise to get to a conclusion... can you please tell me how does it work actually.. whats wrong in my calculation?
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Old 8th December 2012, 06:47 PM   #20
AndrewT is offline AndrewT  Scotland
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Read this Forum
Learn from the experts on this Forum

Ignore me since I am alleged to be unhelpful
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