# Class B Amplifier

Discussion in 'Homework Help' started by AD633, Jun 23, 2013.

Jun 22, 2013
96
1
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

• ###### Document2.png
File size:
11 KB
Views:
56
JasonL likes this.
2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,993
1,116
Yes, you're right
Hmm, isn't this 9V already a peak value of a sine wave?

3. ### Ramussons Active Member

May 3, 2013
562
92
You will have to fine tune the values.

Ramesh

Jun 22, 2013
96
1
So can i calculate this by doing:

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

Jun 22, 2013
96
1
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?

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
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.

Jun 22, 2013
96
1
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?