why my power amplifier cannot amplify?huhu..

[marinara27]

im working on analysis of different classes of power amplifier..
can any1 explain to me why my output signal not amplified?
really2 need your help....

thanxs..

 

[abbeyromy]

Are you trying to implement a class AB operation?
The input to second transistor is not directly given your circuit shows that voltage across R3 is being fed to the the bottom pnp transistor.

The values of chosen resistor has any math behind or you have just taken the values randomly.have you understood the class A operation calculations I posted earlier?

 

[keith1200rs]

Well, it's a complementary emitter follower so you won't get voltage gain, just current gain.

Your voltage loss is probably caused by R5 being 1k. Also, normally two diodes would be used for biasing not one.

Keith

 

[marinara27]

abbey

act i want to implement the class b quasi complementary power amplifier operation...

i dont understand ,"the input to second transistor is not directly given your circuit shows that voltage across R3 is being fed to the the bottom pnp transistor."

the value of resistance i get from google search..
how can i calculate all this and make sure this is the correct circuit for class B operation?

 

[keith1200rs]

You cannot drive an 8 ohm load while you have a 1k for R5!

Also, as I mentioned, it is normal to use two diodes (and only one coupling capacitor). You then need to adjust the current through the diodes so the volts drop across them is just less than the sum of the two transistor Vbe drops.

Keith.

 

[abbeyromy]

well if you are talking about quasi complementary push pull amplifier then i remember that the transistors that u need to use should be matched npn transistors. But in the circuit I see one npn and one pnp.

For any amplifier you need to be sure than your transistor operates in linear mode. Find IC and IB saturation values and then apply basic KVL equations to deduce wether your IB is less than IB sat or not. Refer any text book for equations. try this( electronic devices and circuit theory-Robert L Boylestad)
Then your gain is directly dependent on emitter AC resistance Re and your Collector Resistance RC, The output load ( this comes in parallel with your RC and thus can reduce the gain)

well leave my statement earlier I thought you were trying to make simple class B for that all that arrangement is not required.

 

[keith1200rs]

Quasi complementary is different to this because it doesn't use a "complementary" pair. It usually avoids the use of PNP and uses two NPNs, as you say.

Keith.

 

[abbeyromy]

@keith

Correct me if wrong...
Though the Quasi complementary transistor uses pnp the output transistors are still npn. You need pnp to take care of your negative cycle.

Requesting you to please see the attached design I have tried to make for quasi complementary push pull amplifier!

@marinara27
In class AB operation, the push-pull stages are biased into slight conduction, even when no input signal is present. This can be done with a voltage-divider and diode arrangement.

refer attached design.

 

[marinara27]

abbey..
i have tried simulate the circuit you attach and the output just same like yours.. no amplification..

 

[FvM]

no amplification..

As a expectable, a complementary source follower has a gain less or equal unity.
If you want a gain above unity, you have to use a different circuit.

 

[keith1200rs]

@keith

Correct me if wrong...
Though the Quasi complementary transistor uses pnp the output transistors are still npn. You need pnp to take care of your negative cycle.

The quasi-complementary uses what is sometimes called a compound transistor - a PNP and NPN combined to give the characteristics of a PNP but with the NPN carrying the current. That then gives a similar overally characteristic to a complementary emitter follower. Compound transistors are used sometimes in IC design where the PNPs can be a bit feeble - I have used them occasionally.

Keith.

 

[abbeyromy]

@keith1200rs
Thank for the info!

@marinara27
As FvM pointed it out this is a emitter follower topology so the output is almost same as input but the circuit shows what is called as class AB operation.
If you want gain greater than unity try modifying the circuit or you can just add a common emitter stage after this.

 

[marinara27]

okay2.. i will try modifying the circuit and testing the output...

 

[marinara27]

now i have another question..
how to plot efficiency of the power amp using pspice?
is there any method that u can share?
really need help..thanxs..

 

[keith1200rs]

I am not sure with Pspice, but you should be able to plot "power in a device" with Spice. I would then plot the power in the load and the power in the power supply (voltage source). The ratio of those will be the efficiency. If you cannot find how to plot "device power" you can always plot voltage*current to get the same result.

Keith.

 

[FvM]

It works perfectly with PSPICE. As an advantage, you can perform any required calculations off-line in the probe tool by creating arbitrary
expressions, combining the existing measurements. You have to know your definition of efficiency of course.

 

[marinara27]

@keith

do u mean plot a graph of input power and output power? so,the graph must be plot separately is it?

@FvM

eff=Pout(ac)/Pdc
Pout(ac)=Pin*Ap ; Pin=(Vin^2)/Rin

Pdc=IccVcc=2Icq*Vceq

correct me if im wrong...

 

[keith1200rs]

I would run a transient analysis and then plot input and output power separately. Then the ratio of the average of those two values will be efficiency. You could directly create an equation/measurement for efficiency.

Keith

 

[marinara27]

@keith

yes u are right..from equation i can directly get..
but i want to plot the graph..i working on plot the graph of Vin*Iin vs time..but, how can i choose the average value?

 

[keith1200rs]

I assume your software can measure the mean value of a trace.

Keith