F5 power amplifier

[bobodioulasso]

When you double the load value , the power halves.

[sekess]

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]

In a voltage point of view, about 20 volts are available. 20 x 20 /2 x16 = 12.5w
The current capability does not overpass this law.

However, voltage is going to limit that number. I think?

right
Then , with a 16 ohms load the bias current can be halved to suit the voltage capabilities.

[sekess]

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

[a.wayne]

Where to buy Kendeil caps in the US ?

[sekess]

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

[bobodioulasso]

1-The mosfets when turned on still have a residual voltage across them and some voltage is lost across the source resistors.(what Jacco was formulating)

2- Power is calculated from Vmax/1.414

[a.wayne]

Originally Posted by bobodioulasso

1-The mosfets when turned on still have a residual voltage across them and some voltage is lost across the source resistors.(what Jacco was formulating)

2- Power is calculated from Vmax/1.414

supply out or amplifier out ?

[sekess]

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

[bobodioulasso]

Efficient voltage is Vmax/<2 (what you read on a voltmeter) = 20v in our case
<2 = root mean square of 2

P=(Vmax/<2)(Vmax/<2) /R= (Vmax x Vmax)/2R