Class-A calculations

[K-amps]

Based on a few responses on the forum, I rigged an excel sheet to calculate the class-A output using variables like number of devices, emitter resistance value, voltage drop across the emitter resistors, DC rails and came up with the following:

Class-A output Calculation: (Nakamichi PA-7)

Inputs
Number of output devices: 7 Pairs
Voltage rails (per rail): 70 volts
Emitter resistance (per device): 0.22 ohms
Bias voltage per Emmiter resistor: 32 mv
Speaker ohms 8 ohms

Results
Idle bias per device: 0.15 Amps
Total Amplifier bias (both rails) 2.04 Amps
Total Dissipation (per channel) 142.55 Watts
Class-A output: 50 watts

Does this look correct? because for the Adcom 555II at factory biasing, it comes out to be 11 watts class-A... Is that probable?

Adcom GFA-555II.

Number of output devices: 4 Pairs
Voltage rails (per rail): 82 volts
Emitter resistance (per device): 0.22 ohms
Bias voltage per Emmiter resistor: 10 mv
Speaker ohms 8 ohms

Results
Idle bias per device: 0.05 Amps
Total Amplifier bias (both rails) 0.36 Amps
Total Dissipation (per channel) 29.82 Watts
Class-A output: 11 watts

Have I done this right? Any thoughts are welcome.

-K.

[millwood]

Originally posted by K-amps
Total Amplifier bias (both rails) 2.04 Amps
Class-A output: 50 watts
-K.

the "class a" current swing would be from +2amp to -2amp. the peak power output on an 8-ohm speaker would be 2^2*8=32w. RMS is 50% of that, or 16w. so that's not too far from the factory claim of 11w.

now, if their calculation is off 6-ohm speakers, then it would be 24w peak and 12w rms. even closer.

[K-amps]

Millwood,

The 11w figure (for the Adcom) is not a factory claim. It 's the calculation I arrived by using the factory specified 10mv bias voltage across the 0.22ohm emmiter resistors with 82 volt rails and 4 complementary pairs in outputs.

The 2.04amp number was for the Nakamichi PA-7, in which case:

I calculated by taking total amplifier bias current (in this case one channel) multiplied by Dc rails 2.04*70= 142.8 peak watts dissipated. i.e. 71 watts per rail (1.02amps) and the RMS value for this 71 watts is around 50 watts.

My question is did I calculate this correctly?

thanks!

-K

[millwood]

well, then, we have a problem,

7 pairs x 0.15amp = 1amp (give or take a few hundred ma).

Into 8ohm load, that's peak power of 1amp^2*8ohm = 8w. or rms power of 4w.

total dissipation per rail is 1amp*70v=70w. peak dissipation on the transistors happens when there is no signal. lowest (class-a) dissipation on the transistors happens when output 1amp. at that point, it is 1amp * (70v-8ohm*1amp)=62w.

you can do that math with the adcom as well. but my numbers are: peak class a output power 0.36amp^2*8ohm=1w (close enough). total transistor dissipation per rail is 82v*0.36amp=30w.

[K-amps]

Thanks Millwood I think we are getting somewhere. Some more questions: My descriptions may be misleading so let me try again.

1) 7pairs * 0.15amp is 2.1amps (because the 0.15 amp is per device not per pair). Would your calcs still hold true. What I would get based on your calcs is:

7 pairs x 0.15amp = 2.1 A

Into 8ohm load, that's peak power of 2.1^2*8ohm = 16.8 watts or rms power of 8w. ?

2) By the same token the dissipation would be 2.1A* 70v= 147 watts.

3) Conventional wisdom says that for each 1 watts of class-A, about 3 watts are dissipated as heat (in case of a push-pull design) therefore would'nt that be more in line with my calucation of 50 watts of class-A for 147 watts of dissipation?

Please bear with me here!!

[millwood]

Originally posted by K-amps
1) 7pairs * 0.15amp is 2.1amps (because the 0.15 amp is per device not per pair). Would your calcs still hold true. What I would get based on your calcs is:

7 pairs x 0.15amp = 2.1 A

Into 8ohm load, that's peak power of 2.1^2*8ohm = 16.8 watts or rms power of 8w. ?

"7 pairs" means to me 14 devices: 7 of them on the upper side and 7 of them on the lower side. so total idle current from the upper side is 7x0.15ma=1amp. the same is true for the idle current on the lower side.

Originally posted by K-amps
3) Conventional wisdom says that for each 1 watts of class-A, about 3 watts are dissipated as heat (in case of a push-pull design) therefore would'nt that be more in line with my calucation of 50 watts of class-A for 147 watts of dissipation?

the 25% efficiency is for SE class A driving a current source. not that applicable here.

in your case, each rail is supplying 70v (for Nak) and 1amp. so it is 70w each rail and 140w total (for both rails).

[Eva]

A pair means one device for sourcing current and other sinking it, isn't it?

With 0.15A per device, total current being cross-conducted between positive and negative supplies would be 7*.15 = 1.05 A

With a resistive load and 0V output we are sourcing 1.05A and sinking 1.05A so there is no net current available to the load

The principle of operation of complimentary class A is that as current sourcing from the upper devices increases, current sinking from the lower devices decreases by the same magnitude [and vice-versa]

If we want the output to source 0.5A, then the upper devices will have to increase its conduction by .25A and the lower devices will have to decrease it by .25A [and vice-versa] to provide a current difference of 0.5A

If we want 2.1A to be sourced, then the devices on the upper side would have to increase its current by 1.05A [giving a total of 2.1A] and the devices on the lower side would have to decrease its current by 1.05A [giving a total of 0A and geting out of class A zone]

So with 1.05A bias actually we have +-2.1A output swing [this is the cool thing about complimentary class A, it has double the efficiency than single ended class A]

With +-2.1A into 6.4 ohms [typical Re of most 8 ohm drivers] we get 13.44V peak that on 8 ohms give 11.3Wrms

[ChocoHolic]

Hi Guys!
I think you are both missing one point.
The definition of class A is that always NPN and PNP
remain conductiv.

Let's calculate it for 2.1A idle current and a 8Ohms speaker.
The output current is difference between the current in the
upper and lower transitor (sum of all currents into one node
is always zero).
At zero A output the upper transistor will push 2.1A into the
node, the lower will push -2.1A into the node (means it pulls 2.1A from the node). 0+2.1A-2.1A=0. Fine.

Let's go on. 2A into the load. This means the upper transistor will have 3.1A (2.1A idle + half load). The lower transistor will have 1.1 A
(2.1 A idle - half load).
The sum in the output node:
3.1A is delivered from the upper, 1.1A is drawn from the lower,
2A are given to the load.
Still both (NPN and PNP) transistors are conductive.

The limit of class A will be when the lower one get no current
any more. This is reached at 4.2A in the upper, 0 A in the lower
and 4.2A into the load.

So the traditional limit for class A is reached, when the
peak value of the output current is twice of the idle current.


I am not sure if I understood your particular design right,
I would think the following.
4.2A peak into 8 Ohms, makes 141W peak or about 70W rms
The total heat dissipation would be 2x70Vx2.1A=294W!!!
If I understood right that you are working with rails of
+/-70V, right?
If this true, then this Amplifier will have a good power margin above
its class A range. Class A will end at about 33.6V peak, but
the amplifier can probably deliver up to more than 50Vpeak (depending on the saging of the rails...).
Looks like you are having a real Monster with you!!!!

The Adcom will only have a very low power range in class as
its idle currents is only 200mA....

...hope I did not add more confusion to this forum..
Markus

[ChocoHolic]

Hi Eva!
Ok, you win. Ladies are simply faster

[ChocoHolic]

...still interested if your Amp really has 7 pairs for each chanel...
meaning 2.1A idle in each chanel...?
meaning around 300W heat per chanel...?
Is it a Monoblock?

If the 7 pairs are for stereo... how would this work.
left chanel 3 pairs??!!! right chanel 4 pairs?.!!
Would not sound reasonable....