changing circuit of class A output stage to a class B amplifier

Thread Starter

fantabulous68

Joined Nov 3, 2007
51
could someone please help me:

1. Give the circuit diagram for a class 'A' output stage
2. Describe how one could change the circuit to make it a class 'B' amplifier
 

hobbyist

Joined Aug 10, 2008
892
First what do you ubderstand about classes of amplifiers.

1. A class "A" amplifier is biased at cutoff, (TRUE or FALSE) ??
2. A class "B" amplifier conducts only half the time of a input excursion. (TRUE or FALSE) ??

In answering these questions, will determine to what extent the people on this forum can help you, with..
 

Thread Starter

fantabulous68

Joined Nov 3, 2007
51
I know that in a class A amplifier:
current must run in both bjts at all times

Each device conducts 180 degrees --> 360 degrees of cycle

Class A amplifiers are biased so that variations in input signal polarities occur within the limits of CUTOFF and SATURATION. In a PNP transistor, for example, if the base becomes positive with respect to the emitter, holes will be repelled at the PN junction and no current can flow in the collector circuit. This condition is known as cutoff. Saturation occurs when the base becomes so negative with respect to the emitter that changes in the signal are not reflected in collector-current flow.

Biasing an amplifier in this manner places the dc operating point between cutoff and saturation and allows collector current to flow during the complete cycle (360 degrees) of the input signal, thus providing an output which is a replica of the input.
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I know that in a class B amplifier:
current must ONLY ever run in 1 bjt at any instant

Each device conducts only 180 degrees

Amplifiers biased so that collector current is cut off during one-half of the input signal are classified class B. The dc operating point for this class of amplifier is set up so that base current is zero with no input signal. When a signal is applied, one half cycle will forward bias the base-emitter junction and IC will flow. The other half cycle will reverse bias the base-emitter junction and IC will be cut off. Thus, for class B operation, collector current will flow for approximately 180 degrees (half) of the input signal
could someone please help me:

1. Give the circuit diagram for a class 'A' output stage
2. Describe how one could change the circuit to make it a class 'B' amplifier
BJTs CANT BE CLASS B AMPS.

to be class B---->NO quiescent current(Iq)
NO Iq---->VBE=0,

if VBE=0, it shorts the bases of the 2 BJTs that are connected together-------->hence no current in both bjts

and as Audioguru said..........class B is much more distorted than class A


Am i RIGHT?
 

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hobbyist

Joined Aug 10, 2008
892
I could not open your attachment,

I was basing my questions on a single transistor stage.

I don't know what you drew up but your explanation about a Class "A" your using the word they, so I'm assuming this drawing has 2 transistors in totem pole like a class "B" config.

A class "A" uses only one transistor, and is biased 'on' at the center of it's load line, so it conducts the full input excursion, 360 degress of input signal, and has an inverted output.

A class "B" could use a single transistor, where it is 'not' biased into conduction, and relies solely on a input signal, to bring it into conduction. So it conducts half the input wave form, 180 deg.

So to change a class "A" amplifier to a class "B" is to remove the base bias resistors. For a single transistor stage, that is.
 

Thread Starter

fantabulous68

Joined Nov 3, 2007
51
Wow thanks Audioguru, a picture is worth a 1000 words.
Would love to know the source from which you got the above schematic to broaden my knowledge....???
Thanks everyone for your assistance....
 

Jony130

Joined Feb 17, 2009
5,487
Audioguru drew these diagrams by himself.
And of course you can get class A amplifier with push pull output stage.
 

Audioguru

Joined Dec 20, 2007
11,248
Wow thanks Audioguru, a picture is worth a 1000 words.
Would love to know the source from which you got the above schematic to broaden my knowledge....???
I simply used my knowledge of transistor amplifier classes and manipulated your sketch in Microsoft Paint program to show them.
 

Thread Starter

fantabulous68

Joined Nov 3, 2007
51
Q1} would the attached picture be a class A output stage? and if i wanted to change the picture to a class b amplifier, i should replace the collector resistor and bjt by a pair of bjts?


Q2}On positive swings the voltage across the collector resistor reduces because the collector voltage gets closer to the supply rail?
VRC=VCC-VCollector-VPeak
 

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Audioguru

Joined Dec 20, 2007
11,248
Q1} would the attached picture be a class A output stage?
No.
It might melt. The collector resistor would need to be 1.7 ohms for a 700mA peak output current then at idle it would dissipate 9.4W and the transistor would also dissipate 9.4W. usually a constant current source is used instead of a collector resistor so it doesn't dissipate so much heat.

if i wanted to change the picture to a class b amplifier, i should replace the collector resistor and bjt by a pair of bjts?
Maybe.
Usually a class-AB circuit is used in an audio amplifier.

Q2}On positive swings the voltage across the collector resistor reduces because the collector voltage gets closer to the supply rail?
VRC=VCC-VCollector-VPeak
Yes.
Then the current in the resistor must be 700mA. The at idle the current will be much more.
 

Thread Starter

fantabulous68

Joined Nov 3, 2007
51
Q1) Define the conditions that would cause the amplifier to be on the limit of Class A.

Using a voltage follower for the output stage:
Its when 1 bjt is saturated and has 0V across it whilst the other bjt is just about to run out of current????

conditions:
1) IQ=delta
2) peak load current= 2 IQ

ARE there any MORE conditions to be on the LIMIT?


Q2) Given that the stage is running on the limit of Class A, calculate the maximum input power and hence determine the efficiency at:
1. Full load
2. 5% output voltage

1. At full load Pin=(VCC^2)/RL #### Pout=(VCC^2)/2RL #### efficiency=Pout/Pin=50%

2. Help with this pls....
 

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