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?
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?
Both...
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.
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.
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).
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.
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.
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.
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.
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.
All right i started to design my output stage. I calculated with 96 volt rails Peak to Peak. my wattage out is 288 watts into 4 ohms.
Im using BJT 2sc3284 and 2sa1303 as a pushpull amplifier
all calculations are for a four ohm load
96^2/8*4 = 288 watts V^2pp/8*Rl
Ic(sat) 48/ 2*4 = 6 amps Vcc / 2*Rl
Vce(off) 48/2 = 24 volts Vcc / 2
Now the transistor 2sc3284 are rated as Vce(sat)
= Ic 5 amps Ib .5 amps 2.0 max volts
Now my transistor wont be able to handle the 6 amps Ic correct?
so i will have to place 2 2sc3284 and 2sa1303 into my circuit correct?
Im using BJT 2sc3284 and 2sa1303 as a pushpull amplifier
all calculations are for a four ohm load
96^2/8*4 = 288 watts V^2pp/8*Rl
Ic(sat) 48/ 2*4 = 6 amps Vcc / 2*Rl
Vce(off) 48/2 = 24 volts Vcc / 2
Now the transistor 2sc3284 are rated as Vce(sat)
= Ic 5 amps Ib .5 amps 2.0 max volts
Now my transistor wont be able to handle the 6 amps Ic correct?
so i will have to place 2 2sc3284 and 2sa1303 into my circuit correct?
ElectronicsTech said:
One pair of these outputs using these voltage rails could probably drive 8Ohms, "linearly", Using two pairs, paralelled should drive 4Ohms. There are a couple of issues that must be addressed here first.
1) In order to ensure equal current sharing between paralelled transistors, they need to be matched as close as posible, beta particularly.
2) Series emitter resistors need to be included on each transistor as well as series base resistors. Adding these resistors makes the percent difference of the internal impeadence of the two paralelled transistors much less. Unfortunately, it also cuts down on eficiency by lowering the overall beta, not to mention that you have two in paralell, and this will require a more beefy driver stage. ---------[0.3or 0.4Ohms for Re, 6.2-15 Ohms for Rb.]
You might overdesign the driver stage just a little to make sure there is enough current to drive the two paralelled transistors to the peak current.
BTW 20RMS volts accross 4Ohms is only 100W.
ElectronicsTech said:
If this were true, the world would be simple.
Unfortunately, transistors may have the same number or be of the same type, but they are far from exact to each other. In fact, you could even match up devices from different manufacturers but this I would do only if necessarry. When you order complementary pairs, you have the option to order MP parts, or matched pairs. They are usually matched within 10% of each other. This would be considered fairly close I suppose. The trick is to design your circuit so that it is not Beta dependent. In otherwords, the steady state operation is not affected by the Beta of one device being slightly different from anoter. This way if you replace one transistor with the same type or replacement component, the circuit operation will not be affected. However, when parralelling devices, they need to be fairly close. To match them, just make a simple DC circuit that has one series base resistor to turn it on, use a large value so that it doesn't saturate. A collector resistor will have a certain current flow, or voltage across it...Ib*Beta=Ic. You will still have to add the base and emitter resistors to fine tune the difference, so to speak. There is nothing wrong with having a larger driver stage, I mean output transistors don't have large beta values anyway....unless they are darlingtons and I don't recomend darlingtons for audio outputs although they will work. It's a temperature coefficient thing relating to stability.
ElectronicsTech said:
Exactly! This is because the circuits are designed NOT to be beta dependent, to an extent. The steady state of the circuit will be achieved even if the beta value(and other charactoristics) is just close. In some cases, even a different kind of transistor can be substituted if the charactoristics are close enough to the original. This of course depends on function of the device and circuit design. When designing your own circuit, the process to make the amp circuit beta dependent or were bias is determained by gain of a device, might work well with one transistor but would not reproduce desired results if another device of the same type is put in the circuit.
Beta changes very much with temperature so even perfectly matched transitors can have different betas when operating, and the beta of one transistor changes with different operating conditions(like how much power the amp is putting out. So designs need to be as beta independent as possible, espesially in power applications, fortunatley most Outputs are Emitter followers which are exactly this.
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