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Old 11th September 2004, 03:25 PM   #1
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Default Question about audio output transistors and power out

I'm kinda confused. I always thought that the output transistors determines the output wattage into a speaker load. But has i read more i have noticed that the output power is determined by the rail voltages. I also noticed that the output transistors only determines the current through the load. M i right or wrong?

Also When designing the output stage how do you determine the amount of transistors you will need for the rail voltages to get a desired output to the load?
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Old 11th September 2004, 03:53 PM   #2
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Each output transistor has a Safe Operating Area that shows graphically what combinations of voltage and current can be used. Google for some device datasheets and you will see SOA diagrams as part of the specs. This then allows you to work out the number of devices needed for a given output.
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Old 11th September 2004, 04:59 PM   #3
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Thanks for the information about SAO i will look into it.
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Old 12th September 2004, 01:36 AM   #4
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The rail voltage is one of the limiting factors on output power. The collector voltage has to remain higher then the base voltage or your transistor goes into saturation and no longer acts as an amplifier. For the transistors to stay withen the SOA they must limit their output current. As your load drops you draw more and more current for the same voltage output. Adding more output transistors in parallel increases maximum current able to be supplied (must include some emitter resistors to insure equal current sharing).
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Old 13th September 2004, 01:01 AM   #5
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How is this formula changed when you add multiple transistors to calculate the desired output?

Power out = V^2pp/8*R-load
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Old 13th September 2004, 09:52 AM   #6
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Quote:
I'm kinda confused. I always thought that the output transistors determines the output wattage into a speaker load. But has i read more i have noticed that the output power is determined by the rail voltages. I also noticed that the output transistors only determines the current through the load. M i right or wrong?
The instantaneous power dissipated by a device equals the voltage across it multiplied by the current flowing through it. P = V x I.
If the device is a resistor then using Ohm's Law, V = I x R, gives P = V^2 / R.

So the maximum power into the speaker is determined by the maximum voltage across the speaker, which is usually a few volts less than the power supply voltage of the amplifier. Amps are normally rated by average power into the speaker, which, assuming a sinusoidal signal, is half the maximum power or Pavg = V^2/(2R). This is sometimes incorrectly called "RMS power".

Note that the forumla for speaker power has nothing to do with the transistor power or number of transistors.

The output transistor power is worked out in the same way - the transistor power P = Vce x Ic. This is a little complicated to calculate but a rule of thumb is that the output transistors will, together in worst case, dissipate about 40% (in class B) of the speaker power and about 100% (class A). Of course, knowing the transistor power ratings will suggest what the speaker power can be but it is not accurate. The only accurate way is to know what the maximum voltage across the speaker can be.

When an amp has multiple output devices you can usually divide the power among them. Suppose the maximum speaker power of a class B will be 100W average. The total transistor power will be at least 40W. If the amp has 4 output devices they must each dissipate at least 10W on average. But beware that you must consider both average and peak power - the relationship between peak power and average is not simple because the transistor's Vce is not a sinusoid. The peak power per device may be 25W. So you must find transistors that can withstand 25W peak and 10W average AND maximum Ic and Vce.

It is also essential to consider the temperature at which the transistors will operate - their power ratings reduce dramatically as the temperature rises. It would not be unreasonable to have to double the required power to account for temperature. In the example above each transistor would need to dissipate 50W peak and 20W average.

As a rough rule of thumb with some margin for error for a class B design:
Take your maximum average speaker power and divide it by the number of output devices. Double this and you get the necessary power rating of each transistor at 25C.
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Old 13th September 2004, 05:01 PM   #7
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Default Design Procedure

The people here are correct. The design process is to determine what power you want, then determine the rail voltage needed and then spec transistors that will handle the voltage and resultant current and heat dissipation.

A pair of 2N3904 and 2N3906 will gladly try to put out 100W if you put them in the output stage. However, they will last about 1 second.
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Old 15th September 2004, 01:03 AM   #8
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Default Output stage design tid-bits

One factor not mentioned is that a BJT (well any transistor for that matter) has a "linear" operating region. Even though no transistor is actually a linear device, they do have a region of operation that is fairly "linear". This operating region is determained by a graph of Vce vs Ic, and the particular transistor itself. Designing the output stage so that the Q-point moves along the AC load line but still stays within the "linear" region is the key. BTW the AC load line is determained by load resistance. This is where you determain what output impeadence you would like to drive.

If your Q-point moves on the AC load line to have too little Vce, say a 16Ohm on an 8Ohm amp, then your transistor will operate in the saturation region which is very non-linear and introduces symetrical distortion. Likewise, when you put on too low of a load impeadence, say a 4Ohm on an 8Ohm amp, and your Q-point moves on the AC load line to where there is too much Ic(beyond maximum LINEAR operating current), then the transistor will generate symetrical distortion. It still will also have Vce accross it and will burn up.
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Old 15th September 2004, 01:06 AM   #9
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Default Re: Design Procedure

Quote:
Originally posted by dmfraser
A pair of 2N3904 and 2N3906 will gladly try to put out 100W if you put them in the output stage. However, they will last about 1 second.
Do you think they would last that long? 1 sec is an eternity in electronics. I give'em 50uS....
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Old 15th September 2004, 11:40 AM   #10
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yeah i understand the the load line and biasing the base but the outputs are not always biased instead the drivers are biased.
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