# another question about PA design - RF Cafe Forums

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nickc

Post subject: another question about PA design Posted: Wed Jul 26, 2006 2:14 am

Captain

Joined: Tue Jul 18, 2006 9:52 am

Posts: 6

hey

i have another question about PA class biasing. usually people talk about collector current and collector-emitter voltage when discussing different classes. i understand that but i am having trouble understanding what happens at the input: take class AB for example, if it is biased at 25% of Imax then there is no collector current for some of a cycle but doesn't that no current situation also correspond to vbe swing going below the threshold or VBE(on)? i think that is right---then why nobody discusses or says anything about biasing the base or base-emitter properly so that the swing does go below threshold?

is biasing Ic and Vce sets the base or base-emitter biasing? if so, can someone tell me what is the relationship, how they are related? thanks

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fred47

Post subject: PA BiasingPosted: Wed Jul 26, 2006 10:13 pm

General

Joined: Wed Feb 22, 2006 3:51 pm

Posts: 104

Hi!

Yes, the base-emitter voltage goes below the turn-on voltage for part of the cycle - that's what is required for the collector current to go to zero for part of the cycle, to give other than Class A behavior.

Setting Ic(ave) and Vce(ave) for a given drive level can only be done by adjusting the base-emitter (bias) voltage.

The relationship between Vbe and Ic is (perhaps over-simply)

Ic = Beta * Ib = Beta * Io e^(q*Vbe/(n*k*T) )

so the collector current depends on junction temperature T, the physical structure of the device (which determines n and Io), and a bunch of other high-level phenomena as well. Beta is also more of a variable than a constant...

Hope this helps,

Good Luck!

Fred

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venkataravi

Post subject: Output - Back offPosted: Thu Jul 27, 2006 8:19 am

Captain

Joined: Thu Jul 27, 2006 7:36 am

Posts: 11

Location: Bangalore --India

Hi,

Can any one tell what is the meaning of Output - Back off with reference to PA design -- Yvravi

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fred47

Post subject: PA BackoffPosted: Thu Jul 27, 2006 11:44 am

General

Joined: Wed Feb 22, 2006 3:51 pm

Posts: 104

Hi!

Up to a certain point, the output power of a PA increases with the input drive power. With increased drive and output, there is increased current draw, heat generation, and distortion. So sometimes, to get a better thermal situation and linearity, we reduce the output power, usually by reducing the input drive. This is output back-off - moving away from the maximum-output-power rating of the PA.

Hope this helps!

Good Luck,

Fred

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nickc

Post subject: Posted: Thu Jul 27, 2006 12:21 pm

Captain

Joined: Tue Jul 18, 2006 9:52 am

Posts: 6

hey,

thanks fred....that helps and that is what i now understand. but that raises another follow up question: i was doing simple biasing for a class A PA but i wasn't paying attention to ac swing at the input and as is taught normally i was just thinking about Ic and Vce swing and load lines. so...i did the biasing and ended up with VBE of .9 volts (emitter grounded)....this design is not PCB but IC but it's all the same concept.

but now my question is wouldn't i be going below turn-on voltage at the input if my ac input voltage swing is greater than Vpp=.6 peak-to-peak considering threshold (VBE,on) is .6V (driving the PA harder)? meaning .9-(Vpp.2) = .6; if so, then i don't have class A anymore!

i know driving PA's harder cause compression but i thought the compression is only on voltage swing (due to supply limit) but current swing will remain relatively the same (for class A), but if my input swing is going below turn-on voltage then the current goes to zero too for part of a cycle....any suggestion/feedback?

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IR

Post subject: Posted: Thu Jul 27, 2006 2:55 pm

Joined: Mon Jun 27, 2005 2:02 pm

Posts: 373

Location: Germany

Hello,

If Vbe will go to zero then the base current Ib will also go to zero and therfore there will be no output current, Ic.: You are right there is no Class A anymore when this happens. The impact it will cause is distortion and hence degraded linearity - like in Class B operation. The exact class of operation is determined by the part (In degrees) that the output signal goes below zero, i.e. no input signal.

_________________

Best regards,

- IR

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venkataravi

Post subject: PA designPosted: Fri Jul 28, 2006 8:44 am

Captain

Joined: Thu Jul 27, 2006 7:36 am

Posts: 11

Location: Bangalore --India

Hi Fred,

Thankyou very much, for providing the reply and this is another question to you

What is the meaning of

Shoulder distance or Shoulder Height; Digital Preccorrection

Also please provide ur email ID

Regards

Ravi

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venkataravi

Post subject: Spec relatedPosted: Sat Jul 29, 2006 7:29 am

Captain

Joined: Thu Jul 27, 2006 7:36 am

Posts: 11

Location: Bangalore --India

Hi Fred,

I am basically a CDMA engineer and presently working on Amplifier design

I want to know regarding "spurii carrier dependent " & Spurii carrier independent

With Regards

Y.V.Ravi

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fred47

Post subject: PA QuestionsPosted: Mon Jul 31, 2006 11:20 am

General

Joined: Wed Feb 22, 2006 3:51 pm

Posts: 104

Hi Venkata Ravi,

I'm not a real expert in CDMA amplifiers - you'd do better to post your question on the main forum, not in the reply section for another question. CDMA amplifiers have specifications unique to that application, that other uses do not have.

Also, that would let other people find both the question and whatever replies you get much more easily. I'm sure IR and Kirt would both appreciate that!

Good Luck!

Fred

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nickc

Post subject: sorry, couple of more questionsPosted: Fri Aug 04, 2006 9:42 pm

Captain

Joined: Tue Jul 18, 2006 9:52 am

Posts: 6

hey,

i still need to clarify two things about the BJT operation.

* in PA, VCE is biased at VCE=VCC usually and you can swing from 2VCC to zero (ideally). when Vce=0, Ic is max and also Vbe should be max. in such a situation, wouldn't the transistor be in saturation as the the B-C junction will become forward bias as well, assuming collector is ~.7 volts below the base voltage and the PA is being driven hard?

* is Vce,saturation (Vce,sat) the same value as VBC(on) ~=.7?

* In reality, Vce doesn't swing all the way to zero but swing down to Vce,sat and when the PA is driven hard, the Vce swing clips at 2VCC and at Vce,sat. I understand why it clips at the high point, because of power supply limit, but what determines that it won't go below Vce,sat as the PA is driven hard? why doesn't it go to zero and then clip? in other words, what makes the Vce swing not go below Vce,sat?

hopefully, someone ca answer these questions. it will help me out much.

Posted  11/12/2012