tinitus said:
Tini,thanh1973 said:
The bias is set. It remains unaltered, if the temperature compensation is accurate, irrespective of the load impedance.
The bias determines the maximum ClassA current.
In single ended the bias =Imax
In push-pull bias =IClassAmax /2
Than,
My calculation only used the +20dB & +10dB ratios.
The speaker efficiency did not enter anywhere.
and yes if one chose 100dB/W/m speakers to go with an F5 then you will have higher SPLs available then if one, somewhat stupidly, selected 85dB/W/m speakers to go with the F5s
pro said:
I thought you were talking about what I saw from Borbely -- recommended a small cap from gate-drain cap of 33pF for the Servo 50 when using the Hitachi (Renesas) Lateral MOSFETs. I think that one of my Haflers has a similar arrangement.
The driver board of the Servo 50 has a 100nF cap across the gates:
http://www.borbelyaudio.com/adobe/servo50.pdf
I haven't done this for an F5 -- will try 100nF and see if it makes any difference.
thanh1973 said:
So, you are saying that into 4ohm, idle bias will be only half (0.6A)
A stereo amp will consume only 30watt pr channel, instead of 60watt pr channel, at idle bias
This situation would mean that one could use smaller heatsinks, for that particular use
And if staying within the classA limit, smaller trafos as well
I guess you are right
I dont question that, but sounds strange to me
C'mon
That has nothing to do with bias current.
Remember power is voltage times current (=idle bias); for 4 ohm speakers you need only half the voltage to drive the same current through the speaker (Ohm's law), so power into 4R is halfed.
On the other hand, if you have 16R speakers, you get twice the power into 16R (as you need twice the voltage to drive the same current into the speaker).
Don't forget though, that power dissipation in the amp itself is little affected by that in the real world. No chance to get away with smaller transformer etc.
Have fun, Hannes
As I understand it staying in Class A is the ratio of idle current to output current, as Mr. Pass speaks about in several of his articles. Once the output current is greater than 2x the idle current the amplifier starts to leave class A. So following the formula that Papa uses 1.3A bias is 2.6A max output into class A per output device..
The power formula for 4OHM load is 2.6A x 2.6A*4ohms = 25W peak or 12.5W RMS (half of peak)
For 8 Ohm load is 2.6A X 2.6A X 8Ohms = 54W peak or 27W RMS
The power formula for 4OHM load is 2.6A x 2.6A*4ohms = 25W peak or 12.5W RMS (half of peak)
For 8 Ohm load is 2.6A X 2.6A X 8Ohms = 54W peak or 27W RMS
AndrewT said:
stupidly like a fox: my mains are 91db and what I am using now... but when I was testing this amp early on I used 85.5 db/w/m speakers. I was able to drive the amp directly from my DAC. One straight wire with no knobs. The combo filled my large room beautifully, but the wife missed the knob
Andrew, thanks for the thoughts on how much output I may actually need for the transients.
Garcia and Grisman are pick'n through my F5
tinitus said:
We can break this down to make it simpler to understand.
Let's look at only half of the amp. Your rail voltage is set at around 24V so if you have a 4v drop across the fet you have 20v available at the output. The current available is set by the bias current of 1.3A, and that is set by the source resistor. You now have two limiting factors in your circuit, your maximum output voltage is fixed, as is your maximum current. To understand this you must forget about the rest of the amp and just consider this as an electrical circuit and not a part of an audio amp.
The optimum load for max power is easily found, E(20v)/ I(1.3a) = 15.4 ohms, easy enough. Max power is 26Watts
If we use a 20 ohm load then we have 20v/20ohms = 1A which means there is a reserve of current still available but we are at the limit of available output voltage. Max power is 20 Watts
If we use a 10ohm load then we have 20v/10ohms= 2A, but we only have 1.3A available so we are now current limited. Max power is about 17 Watts
This is a basic electrical circuit, please do not try to read anything else into it.
Best, Bill
Well, in that case
I understand that a load into half the impedance means only half the voltage is needed
But is that also equal to whats available
Im really trying to understand your point, but dont
This is really not my strong side, and has nothing to do with "not reading"
But I have always thought that current would rise in low impedance load
Nelson Pass said:
25 watts Class A into 8 ohms implies a 1.25 A bias current.
I understand that a load into half the impedance means only half the voltage is needed
But is that also equal to whats available
Im really trying to understand your point, but dont
This is really not my strong side, and has nothing to do with "not reading"
But I have always thought that current would rise in low impedance load
Its obvious we cannot expect 50watt classA into 4ohm
Perfectly clear
Though to me it will stay 25watt classA, no matter what you say about that
Bias is fixed, supply voltage is fixed
To me its simple logic
And I will build with the suggested trafo rating requested, "more than 10A peak"
Banging my head with better knowing has never worked on me
Obviously I will never understand your calculations
Im even more happy now that all information is easily available, in clear text
Perfectly clear
Though to me it will stay 25watt classA, no matter what you say about that
Bias is fixed, supply voltage is fixed
To me its simple logic
And I will build with the suggested trafo rating requested, "more than 10A peak"
Banging my head with better knowing has never worked on me
Obviously I will never understand your calculations
Im even more happy now that all information is easily available, in clear text
Hi Tinitus
Staying in class A: two times the bias current in the load is the limit.
12.5w into 16 ohms / bias 0.625A
25w into 8 ohms / bias 1.25A
50w into 4 ohms / bias 2.5A
At 1.25 bias and 4 ohms load we leave class A at 12.5w
when load current reaches 2times the bias current
P peak = 2Prms.
The only formula is P=RxI^2
Is that clear for you?
Staying in class A: two times the bias current in the load is the limit.
12.5w into 16 ohms / bias 0.625A
25w into 8 ohms / bias 1.25A
50w into 4 ohms / bias 2.5A
At 1.25 bias and 4 ohms load we leave class A at 12.5w
when load current reaches 2times the bias current
P peak = 2Prms.
The only formula is P=RxI^2
Is that clear for you?