Class A Bias

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Hi everybody,

If with a bridge topology at +4A per rail per side at +/- 42V DC Rail is given, the total power dissipation is about 672W.
Let's say it is roughly a little over 400W peak Class A power.
Given a 220mF capacitor bank on each rail, how can I correlate the idle power dissipation at the primary with the power dissipation of the secondary, keeping in consideration that the charging current for the reservoir bank brings brings peak current of 46A at the secondary after rectifier thus peak current of over +/- 22A at the primary (this will significantly increase the RMS current thus power consumption) ?
That would suggest that an amp with less capacitance will idle less power for a same given amount of Class A bias(?!?)
Is this statement true?
 
Hi, I'm pretty sure its not true in any significant sense, rgds, sreten.

I can't follow the argument to arrive at the suggested conclusion.

The argument is that if for a given amount of bias you have more capacitance, your peak current to charge this capacitance will be much higher than if you had less capacitance, which will reflect in an higher RMS current at the primary which ultimately translates into a huge amount if idle power.

The XA200.5 consumes roughly 700W idle but yet has 100mF per rail and 3.6A bias, which seems to be a pretty low figure compare to what I measure on my amp.
Maybe going from 100mF to 200mF causes the consumption to go from 700W to 1.2KW.
But then looking at the specs for the XS300 series which boosts twice as much capacitance as the XA, power consumption is still roughly 3 times the power rated power. :magnify:
 
well , Physic can became idiotic only in idiotic company ......... :devily:

so , when two of us are combined , even poor Physic is Dodo

however , I know how dumb I am ......... so , now I'll leave you to squeeze someone else ;

unfortunately , I can't take Physic out of thread , with me
 
I just don't understand how the XA200.5 can idle only 700W

It means you do not understand how the XA200.5 operates.

A regular 200W Class A push-pull amplifier would have a quiescent current level of ~3.5A (<= 400W peak Class A in 8 ohms)
With 10% rail losses added, a minimum dissipation of ~440W.

A balanced/bridged design has twice the rail losses, let's say 20%.
Makes the dissipation ~480W.

Which means that 700W power consumption of the XA200.5 is pretty damn high, instead of ''only''.
(aka the difference of 220W is not all lost in the power supply)
 
What I am wondering though is, accordingly to the 20% estimated loss of a bridge topology, referring to the stock 200.5 with 3.6A bias, takes to about declared 700W power consumption which should be about 6A at the primary if you are running in America...and roughly 3A if you are in Europe.
However cranking some calculations also supported by measurements, to charge up a load of capacitors brings currents to very high peaks, therefore the IRMS figure at the primary is much larger than the expected 6A....and here is where I get confused because all this extra power must be burnt on the rectifiers.
If somebody could explain this I would really be thankful.
 
It means you do not understand how the XA200.5 operates.

Also If you can keep the tone down to a nicer tone would be really appreciated as well!...might also help this post to flow in a nicer way!!
It seems most of people on the pass labs forum are bitter for some reason! I start thinking it is pointless to post interesting question on this forum.
Beyond the simplistic first condiscendent answers I get, I notice that most of the time the main point gets missed and lost on stupid useless debates!
This question brings up an interesting point on how big power amp construction it carried on.
If the nature of the question is not understood...instead of giving judgments and/or first hand answers...why don't just ask me for clearer explanation rather than simply assuming that the question is stupid?!

Anyway if anybody has an answer to my original question I would really appreciate it!
😎
 
Sorry, I don't know if I follow but you let me know if this is not what you ask.

You will have a large load peak when you switch the amp on, corresponding to the large capacitor bank getting charged.
The idle current has nothing to do with this peak. The idle current is what the amp draws from the power supply in a constant way and basically depends on the bias current at which the amp runs.

To soften the capacitor bank peak charging current install any suitable soft start circuit at the transformer/s primary. If this is your concern then this is just the way to solve it, and not reducing the total capacitance value of the bank.
 
Sorry, I don't know if I follow but you let me know if this is not what you ask.

You will have a large load peak when you switch the amp on, corresponding to the large capacitor bank getting charged.
The idle current has nothing to do with this peak. The idle current is what the amp draws from the power supply in a constant way and basically depends on the bias current at which the amp runs.

To soften the capacitor bank peak charging current install any suitable soft start circuit at the transformer/s primary. If this is your concern then this is just the way to solve it, and not reducing the total capacitance value of the bank.

thanks Pablo for your reply. No I wasn't saying that. I have a soft start circuit.
Thanks for your intervention anyway.
I probably need to put some picture to support my question.
 
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