PA/Driver with Diff. Pair - RF Cafe Forums
Post subject: PA/driver with Diff. Pair Posted: Wed Dec
28, 2005 1:46 pm
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).
Collector/output of Transistor 1 of diff pair------i1--->RL <---MN--50
Collector/output of Transistor 2 of diff pair------i2--->RL <---MN--50
MN stands for matching network.
your feedback and clarify this issue
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
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...
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
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?
Post subject: Posted: Wed Dec 28,
2005 6:08 pm
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...
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?
Posted: Thu Dec 29, 2005 2:02 am
Hello again kris,
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.
am I wrong in my idea?
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
Post subject: Posted: Thu
Dec 29, 2005 6:53 pm
Joined: Mon Jun
27, 2005 2:02 pm
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.
Post subject: Posted: Thu Dec 29, 2005
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
Posted: Fri Dec 30, 2005 2:09 am
Mon Jun 27, 2005 2:02 pm
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.
Post subject: Posted: Sat Dec 31, 2005 9:47
Joined: Sat Dec 31, 2005 9:19 am
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
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.
Post subject: Posted: Sat
Dec 31, 2005 3:24 pm
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.
Post subject: Posted: Sat Dec 31, 2005 9:45 pm
Joined: Mon Jun 27, 2005 2:02 pm
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.
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
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??
Post subject: Posted: Wed Jan 04,
2006 11:04 pm
Joined: Sat Dec 31, 2005
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.