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PA/driver with Diff. Pair - RF Cafe Forums

Because of the high maintenance needed to monitor and filter spammers from the RF Cafe Forums, I decided that it would be best to just archive the pages to make all the good information posted in the past available for review. It is unfortunate that the scumbags of the world ruin an otherwise useful venue for people wanting to exchanged useful ideas and views. It seems that the more formal social media like Facebook pretty much dominate this kind of venue anymore anyway, so if you would like to post something on RF Cafe's Facebook page, please do.

Below are all of the forum threads, including all the responses to the original posts.


kris
Post subject: PA/driver with Diff. Pair Posted: Wed Dec 28, 2005 1:46 pm
hi guys

i need to clarify a design issue for a PA using differential pair...for a simple single ended design the output power is Pout = i^2 * RL where RL is the optimum load impedance. but how do u characterize output power Pout when using a differential pair...let's say I need 5 dBm= X Watts of output power do I just divide this X watts by 2, take the current flowing in each port (line) and find the optimal RL for rough calculations and then just transform 50 Ohm to this RL for each port (line). Meaning:


Collector/output of Transistor 1 of diff pair------i1--->RL <---MN--50 Ohms

Collector/output of Transistor 2 of diff pair------i2--->RL <---MN--50 Ohms

MN stands for matching network.

Please provide your feedback and clarify this issue

-kris


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Guest
Post subject: Posted: Wed Dec 28, 2005 2:13 pm
Differential output would be X watts, were the single ended would be X watts/2


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kris
Post subject: Posted: Wed Dec 28, 2005 3:57 pm
hi guest...ok but again what exactly defines differential output power...if differential power is X watts how do i relate it to optimal RL...

I'm thinking Diff. output power in Watt = (vo1 - vo2) * RL where vo1 and vo2 are the collector/output voltages of the transistors in the pair and RL is the optimum differential impedance but my confusion then comes when designing the matching network...do i transform 50 Ohms to RL/2 for each port(line) of the differential signal? I need to match to 50 ohms but differentially it sees 100 ohms, right? Meaning, each port(line) needs to be matched to 50 Ohms but again transform 50 Ohms to what? RL/2 or RL or something else? I would think RL/2.

This is will also help me to see how to setup the loadpull analysis

Thanks

-kris


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kris
Post subject: Posted: Wed Dec 28, 2005 5:00 pm
please try to answer the post above this but i guess another way to ask the question would be what defines differential output power?? does each node of the diff. signal provides X/2 watt if differential output power is X?


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Guest
Post subject: Posted: Wed Dec 28, 2005 6:08 pm
Hi kris,

I guess that your amplifier is Push-Pull. In this case differential output means that the output current is flowing in each transistor separately at each half cycle of the input signal and not simultaneously in both of the transistors. So in the end-effect each transistor 'sees' an output impedance of 50 ohm (Out of the 100 ohm load). So you will need to match this output impdeadnce to the load of the next stage...

Hope this helps!


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kris
Post subject: Posted: Wed Dec 28, 2005 10:02 pm
in fact, my PA is a differential pair but i see your point but again in PA you are looking for optimal output load to present to transistor output/collector. i agree that each transistor sees 50 Ohms but the thing is that for PA design I need to tranform this 50 Ohms back down to some small value using a matching network....

and as i said earlier if i assume that the DIFFERENTIAL omptimal output impedance is X then for each transistor it will be X/2? Having X as my output impedance differentially will give me a certain amount of output power...if i say that each transistor has X/2 as the "optimal" output impedance, would it be right to say that?


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Guest
Post subject: Posted: Thu Dec 29, 2005 2:02 am
Hello again kris,

Yes it would be right to say that X/2 si the optimal impedance for each stage. But I don´t understand why you don´t use a combiner? You will need to combine these differential outputs into a total output power and by that to achieve twice the output power that each stage provide. This will definitely solve your problem, because then you will have a single output that you will need to match.

Or am I wrong in my idea?


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kris
Post subject: Posted: Thu Dec 29, 2005 4:38 pm
Thanks for your help. Yes, I can use a combiner but I need to keep the output differential--i guess it will feed a differential input of a component. But here's where my confusion is this: (in fact in another post some one asked a similar question but I didn't get it)

Let's say I know my differential optimal impedance but how do I match it DIFFERENTIALLY to 50 ohms? Do you see my point? How do you do differentia match in case where you have to match it to 50 ohms. Do I take my one part (node) of differential signal with optimal impedance X/2 and match it to 50 Ohms and do the same thing with the other part (node) of differential signal. Is this how differential match done in the case here?


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IR
Post subject: Posted: Thu Dec 29, 2005 6:53 pm

Site Admin


Joined: Mon Jun 27, 2005 2:02 pm
Posts: 373
Location: Germany
Hello kris,

I was one of the guys who posted to you in the previous replies.

Yes, you will need to match each transistor to 50 ohm separately. I would try to use a transformer (Step-up or step-down) as part of the matching network, depending on the type of your impedance transformation and frequency range. This will make the matching network more wideband and less complicated to implement.

Good luck!

_________________
Best regards,

- IR


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kris
Post subject: Posted: Thu Dec 29, 2005 7:01 pm
really appreciate your reply IR...one last thing: considering that I match each transistor with 50 Ohm to its optimal load, after i use a balun to convert diff. to single ended, the single ended termination should be 100 or 50 AND why for the purposes of s-param simulations?? meaning, 50 ohm termination is the right setup or 100 ohm termination is the right way to see if your differential pair output is matched to 50 ohm properly or not? what's exactly meant by that for diff. signal if you match to 50 Ohm you see 100 Ohm differentially.


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IR
Post subject: Posted: Fri Dec 30, 2005 2:09 am

Site Admin


Joined: Mon Jun 27, 2005 2:02 pm
Posts: 373
Location: Germany
Hello kris,

You will have to match the output of the Balun to the impedance that is being used in the rest of the sytem from that point and on. In other words, if the output of the Balun is connected to a device that has impedance of 50 ohm, you will have to match it to the 50 ohm.

To answer your second question, here is a link with a description of matching concepts for several types of amplifiers. I hope it will be useful for you.

http://www.rane.com/note124.html

_________________
Best regards,

- IR


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rficdude
Post subject: Posted: Sat Dec 31, 2005 9:47 am

Lieutenant

Joined: Sat Dec 31, 2005 9:19 am
Posts: 4
For a differential amplifier the differential output impedance is 2x the impedance of either output single ended. For example, if the real part of the impedance is 50 ohms for one output (at a specific frequency) then the real part of the differential impedance is 100 ohms.

Aproaches to matching to the next stage depends on what the load is. If the load is differential then you could simply match each single ended output to the single ended input of the next stage (or you could solve it differential to differential if you like).
If the load is single ended then you must consider the balun in the matching network too. For instance, if the balun is an ideal 1:1 transformer then the single ended load impedance will be transformered to a differential impedance of the same value. The single ended impedance at each terminal of the balun (on the differential side) will be half the load impedance. So, you would consider matching each single ended output to half the load impedance to achieve the desired match to the load on the other side of the balun. Another example is if you have a sqrt(2):1 turns ratio transformer then the single ended load impedance will transformed up to a differential impedance that is two times the load impedance. Now each single ended output on the differential side will be equal to the load impedance.

You have to be careful when looking at baluns to since sometimes they are specified as impedance ratios instead of turns ratios.

Hope this was helpful.

_________________
RFICDUDE


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kris
Post subject: Posted: Sat Dec 31, 2005 3:24 pm
Hi RFICDUDE

thanks a lot for you explanationa and IR's too. you really nailed my question/concern. It really helped but still my simulation is not coming out right. I want to clarify something you said. First, let me tell you in my design I want to keep it differential output and as you said my idea is to match each transistor/single-end itself. I am matching each side to 50 Ohms so differentially I have 100 Ohms. But ONLY for simulation purposes (so that I have clear s11 and s22, etc. results) I am using an ideal 50-ohm reference Balun that will just convert my differential to single ended. I won't be using the Balun for my 'actual' design.

Now, please clarify this to me: If I am using an ideal balun for s-param simulation what port impedance do I terminate my single ended side of balun? 50 or 100? Remembering that I matched each side/transistor of my differential output signal to 50 Ohms (or 100 Ohms differential match). So, what I am saying is that I have my diff. outputs, I match each side/transistor with 50 Ohms using matching networks and then I use balun to convert the diff. signal to single-ended and I want to know what port impedance to use to terminate the single-ended side of the ideal balun. I think and by reading your post I need to terminate with 100 Ohm but someone told me, no terminate in 50 Ohms. Answering this question will clarify things. Also, imaginary part of a differential impedance is 2X the impedance of one side? Thanks again.


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IR
Post subject: Posted: Sat Dec 31, 2005 9:45 pm

Site Admin


Joined: Mon Jun 27, 2005 2:02 pm
Posts: 373
Location: Germany
Hello kris,

You should match the single ended-port of the Balun to the impedance that runs in the rest of your circuit. I.e. if the single-ended port of your Balun connects to a device with characteristic impedance of 50 ohm, then you should match it to 50 ohm.

_________________
Best regards,

- IR


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kris
Post subject: Posted: Sat Dec 31, 2005 11:26 pm
hi IR....i think you didn't get my question. I might not be clear here. I don't need the balun in my design at all...i'm only using it in simulation just for the sake of looking at matching single endedly on both side....my actual design ends at the differential signal matching network for each transistor....now, after the matching network I want to just put a simple plan ideal balun and would like to know when i'm running s-param simulations how should i terminate the single end side of balun, with 50 or 100 ohms, considering each transistor output in the differential signal is matched to 50 ohms. i think i terminate with 100 ohms....thanks


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kris
Post subject: Posted: Sun Jan 01, 2006 5:28 am
hi guys...a related but different sort of question. if i am not combining my output from a differential pair (having a diff. output) how does one define the spec for output power? let's say i need only 15 dBm of output power but how is it defined in case of differential pair...does it mean 15 dBm=.031 W from each transistor or half (.031/2) from each transistor. my understanding is that since i'm not combining my output (and only one transistor is on at a time), requirement of .031 W specification means .031 W of output power from each transistor. is that right??


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rficdude
Post subject: Posted: Wed Jan 04, 2006 11:04 pm

Lieutenant

Joined: Sat Dec 31, 2005 9:19 am
Posts: 4
If both outputs are matched to 50 ohms single ended then an ideal balun will transform the 100 ohms differential to 100 ohms single ended on the other side.

Now you have a choice of just using a 100 ohm port impedance or you could choose to match the 100 ohm impedance to 50 ohms and use a port. Ultimately you want to analyze the amplifier when it is matched to the real load you plan on presenting to the amplifier output.

_________________
RFICDUDE




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