Hello,
I am trying to build a push pull amplifier using TIP3055 and TIP2955 like this one :



For the diode, i am using 1N4002 and i want the output Vin = Vout and the current Iout = 100 mA. What i am questioning is :
1. Is it possible to do that with the schematic i am using?
2. If it possible, what are the value of RB1, RB2, +Vcc, and -Vcc?
(from the info i get, RB1=RB2 and +Vcc=-(-Vcc))
3. If not, please can anyone give me some suggestions.
Thanks

 

Yes this is a perfectly good circuit configuration which works in 'push pull'.

Yes RB1=RB2 and +Vcc=-(-Vcc))

RB1 depends upon the supply +Vcc/-Vcc . The supply depends upon the required output power and the load resistance. You can set this within wide limits from about ±10 to about ±45, but will have to work out the safe limits from the load resistance which defines the load current.

Once you have set the supply, you can calculate the base resistors to establish suitable standing base currents.

I do not understand this comment

i want the output Vin = Vout and the current Iout = 100 mA

 

What i mean is :
- are the value of RB1 and RB2 has something to do with the output?
- in order to create push pull amplifier with specification : Iout = 100 mA and Vout = Vin, what component should i use?

 

If I(out) means the current through the load you need to specify whether this is rms, peak or peak to peak current and you need to specify the load.

 

what i am trying to is Iout(rms) = 100 mA and Vin = Vout. How do you calculate the value of RB1 and RB2 in order to achieve this?

 

I would approach this problem using the fact that the current gain of your transistors is about 50. Therefore, if 100 mA are needed at the output, Ie, then about 100/50= 2mA into the base. But I want want that biasing circuit to have 1mA more than the current going into the base (to keep the diodes regulating the voltage) in order to avoid crossover distortion. So I would bias the circuit that supplies the transistors at 3 mA. This would mean a total Rb = 2Vcc/3mA. Then Rb1=Rb2=Rb/2. This is a minumum current, and a maximum Rb. A maximum Rb means maximum input resistance, which is a virtue.

 

I would approach this problem using the fact that the current gain of your transistors is about 50. Therefore, if 100 mA are needed at the output, Ie, then about 100/50= 2mA into the base. But I want want that biasing circuit to have 1mA more than the current going into the base (to keep the diodes regulating the voltage) in order to avoid crossover distortion. So I would bias the circuit that supplies the transistors at 3 mA. This would mean a total Rb = 2Vcc/3mA. Then Rb1=Rb2=Rb/2. This is a minumum current, and a maximum Rb. A maximum Rb means maximum input resistance, which is a virtue.

Very informative, will try to make the amplifier according to it. Thank you, hope there is not much problems arise.

 

This problem is one that never goes away, so it is a good exercise. I've dealt with several ways on several threads...

http://forum.allaboutcircuits.com/showthread.php?t=18042

http://forum.allaboutcircuits.com/showthread.php?t=10631

 

Interesting threads, Bill. Seems this is quite a common question with a variety of solutions depending on the driver and the load.