Copyright: 1996 - 2024
BSEE - KB3UON
RF Cafe began life in 1996 as "RF Tools" in an AOL screen name web space totaling
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design engineer. The World Wide Web (Internet) was largely an unknown entity at
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PA/driver with Diff. Pair - RF Cafe Forums
RF Cafe Forums closed its virtual doors in 2012 mainly due to other social media
platforms dominating public commenting venues. RF Cafe Forums began sometime around
August of 2003 and was quite well-attended for many years. By 2010, Facebook and
Twitter were overwhelmingly dominating online personal interaction, and RF Cafe
Forums activity dropped off precipitously. If the folks at
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sign-in from the major social media platforms, I would resurrect the RF Cafe Forums,
but until then it is probably not worth the effort. Regardless, there are still
lots of great posts in the archive that ware worth looking at.
Below are the old forum threads, including responses to the original posts.
|-- Amateur Radio
Gripes & Humor
-- CAE, CAD, &
Test & Measurement
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). Meaning:
of Transistor 1 of diff pair------i1--->RL <---MN--50 Ohms
Collector/output of Transistor 2 of diff pair------i2--->RL <---MN--50
MN stands for matching network.
Please provide your
feedback and clarify this issue
Post subject: Posted: Wed Dec 28, 2005 2:13 pm
output would be X watts, were the single ended would be X watts/2
Post subject: Posted: Wed Dec 28, 2005
hi guest...ok but again what exactly defines differential
output power...if differential power is X watts how do i relate it to
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
This is will also help me to see how to setup the loadpull
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?
Post subject: Posted: Wed Dec 28, 2005 6:08
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!
Post subject: Posted: Wed Dec 28, 2005 10:02 pm
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?
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?
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?
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.
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.
Post subject: Posted: Fri Dec 30, 2005 2:09
Joined: 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.
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
Posted: Sat Dec 31, 2005 9:47 am
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
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.
subject: Posted: Sat Dec 31, 2005 3:24 pm
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.
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.
Posted: Sat Dec 31, 2005 9:45 pm
Mon Jun 27, 2005 2:02 pm
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.
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
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??
Post subject: Posted: Wed Jan 04, 2006
Joined: Sat Dec 31, 2005 9:19 am
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.