# Class B Amplifier

Joined Jun 22, 2013
96
The circuit of Figure corresponds to a class B amplifier, where Q1 and Q2 have VBE = 0.7V, VCEsat = 1.2V and RL = 8Ω.

1)Explain why this circuit introduces crossover distortion (crossover) in the output signal.

I think this circuit introduces crossover distortion in the output signal as the two bases of the transistors are directly connected to each other and thus the transistors only starts conducting when Vi> 0.7 V or Vi <-0.7 V, rigth?

2)Determine the average power dissipated in RL when the voltage Vo has the maximum possible extent.(Considering Vi a sine wave with amplitude 9V)

Can i calculate this in this way:

$$Il=(9sqrt(2))^2/8=20,25 W$$

Thanks

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#### Jony130

Joined Feb 17, 2009
5,155
I think this circuit introduces crossover distortion in the output signal as the two bases of the transistors are directly connected to each other and thus the transistors only starts conducting when Vi> 0.7 V or Vi <-0.7 V, right?
Yes, you're right
2)Determine the average power dissipated in RL when the voltage Vo has the maximum possible extent.(Considering Vi a sine wave with amplitude 9V)

Can i calculate this in this way:

$$Il=(9sqrt(2))^2/8=20,25 W$$

Thanks
Hmm, isn't this 9V already a peak value of a sine wave?

#### Ramussons

Joined May 3, 2013
813
The circuit of Figure corresponds to a class B amplifier, where Q1 and Q2 have VBE = 0.7V, VCEsat = 1.2V and RL = 8Ω.
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You will have to fine tune the values.

Ramesh

Joined Jun 22, 2013
96
Yes, you're right

Hmm, isn't this 9V already a peak value of a sine wave?
So can i calculate this by doing:

Average dissipated power$$=(9/sqrt(2))^2)/(8)= 5 W$$

Joined Jun 22, 2013
96
So can i calculate this by doing:

Average dissipated power$$=(9/sqrt(2))^2)/(8)= 5 W$$
Is the above calculation correct?

Determine the potency dissipated by each transistor on rest and on case in that Vo it is max.

For the rest PT=VCE*IC=VCE*0=0 w-->This must bew wrong because class b amplifiers work closely to cut refgion but they reach the cut region...

When Vomax=VCEsat*ICsat=0.5*(12/8)=0.75 W-->This must be wrong because class b amplifiers do not work on the saturation region...How can i calculate IC and VCE for this situation?

#### t_n_k

Joined Mar 6, 2009
5,455
Is the above calculation correct?
If you ignore the base-emitter diode drop in the conducting transistor then your answer for load power is an indicative approximation with the relatively large signal input of 9V peak.

Joined Jun 22, 2013
96
If you ignore the base-emitter diode drop in the conducting transistor then your answer for load power is an indicative approximation with the relatively large signal input of 9V peak.
I have another question.

How can i determine the powerdissipated by each one of the transistors when they are at rest and in the case that Vo is maximum. (for Vi 12 V)

For the rest $$PT=VCE*IC=VCE*0=0 W$$-->This must bew wrong because class b amplifiers work closely to cut refgion but they reach the cut region...

When $$Vomax=VCEsat*ICsat=0.5*(12/8)=0.75 W$$-->This must be wrong because class b amplifiers do not work on the saturation region...How can i calculate IC and VCE for this situation?