jaya000 Post subject: suggestions for circuit design Posted: Tue
Feb 20, 2007 8:10 am Captain Joined: Tue Feb 20, 2007
5:43 am Posts: 14 To start learning RF i planned for one small
signal 2 stage antenna preamplifier. Frequency range 88108 Mhz.Mhz.
Accordingly i hv done the following according to the bias condition
required. . I hv selected input transistor with following S parameters.
f S11 S21 S12 S22 GHz MAG ANG MAG ANG MAG ANG MAG ANG
0.10 0.71 49.5 18.61 139 0.023 65.8 0.834 22 Feeding this data
into a software vierpol (two port in English) but in old version. Then
i got following Zin, Zout. . For input/output impedance BJT @100
Mhz. Calculated on Vierpol in Elekta Polar Rect Input impedance
Zi = 102.100 65.3 42.622 92.668 Output Impedence Zo = 233.22 64
102.237 209.617 The output transistor with following S parameters.
FRE S11 S21 S12 S22 K MAG1 (MHz) MAG ANG MAG ANG MAG ANG
MAG ANG (dB) 100 0. .45 78 26.73 125 0.01 69 0.68 31 0.77 34.3
For Input output impedance BJT @100 Mhz. Calculated on Vierpol in
Elekta Polar Rect Input Impedance Zin = 58.490 47.8 39.289
43.33 Output Impedance Zout = 148.82 52.5 90.596 118.067
I got it in polar but also coverted into rectangular. All this
is with 50 ohm input output. I am newbie in RF. Is this
ok till now? If this is ok then i will like to know from RF experts/Gurus
here that which type of circuit i must use for input/output & interstage
matching? Circuit input output impedance will be 75 ohms.
Top fred47 Post subject: Suggestions for circuit designPosted:
Tue Feb 20, 2007 1:05 pm General Joined: Wed Feb
22, 2006 3:51 pm Posts: 104 Hi! Two words: Smith Chart.
The Smith Chart is a graphical representation of the sparameters,
with a grid representing impedances relative to the system reference
impedance (usually 50 Ohms, but in your case, I'd use 75 Ohms). You
can still buy paper versions, but you can download a program called
QuickSmith and let your computer do all the boring work. It's free,
and a nice piece of work. Good Luck! Fred Top
jaya000 Post subject: Posted: Tue Feb 20, 2007 10:54 pm
Captain Joined: Tue Feb 20, 2007 5:43 am Posts: 14
Thanks fred47, As suggested by u, I hv downloaded the Quicksmith.
Now just guide me how to get started. Should i devide the zin zout
by 50? But then What to do for impedance of quicksmith chart? Should
it be also normalised to 1? I am newbie so needs help to start with.
I hv not seen any document about smith chart matching with 75 ohms.
Top mhtplsh Post subject: Posted: Wed Feb 21, 2007
1:20 pm Captain Joined: Wed Feb 21, 2007 12:50 pm
Posts: 18 Fred47, I hv downloaded & tried Quicksmith. It
is nice software. I am stucked here. The noise parameter window
asks for the following parameter. G0 ( Mag) G0 ( Deg)
Rn (Ohms) Tell me how to get or calculate this. If i set
the smith chart to 75 ohms do i need to change the s parameters &
zin Zout to 75 ohm as asked in some other post here. Top
fred47 Post subject: QuickSmithPosted: Wed Feb 21, 2007
1:59 pm General Joined: Wed Feb 22, 2006 3:51 pm
Posts: 104 jaya000: Good observation  if you were using paper,
you'd need to normalize by dividing by the system impedance (50 or 75
Ohms). For QuickSmith, you set the system impedance  the Center or
"Prime" point  and use ordinary impedances. Just enter the values
 QuickSmith assumes 50+j0 Ohms for the Center "Prime" point unless
you tell it differently with the "toolbox" (which you get to by clicking
the wrench icon at the top of the chart page). I've posted conversion
details (50 to 75 Ohms) in another thread in this subject. But the impact
of that is minimized by a program like QuickSmith. mhtplsh: I'd
be a genius if I could adequately summarize lownoise amplifier design
in a short post. I'd suggest Guillermo Gonzales' book on microwave amplifier
design. I'll see if I can come up with a quick procedure, but I'm not
sure if I've got the time. Good Luck! Fred Top
jaya000 Post subject: Posted: Thu Feb 22, 2007 2:16 am
Captain Joined: Tue Feb 20, 2007 5:43 am Posts: 14
Thanks Fred47, Cannot find ur post of 50 to 75 ohm conversion.
Then on the web i got the 50 ohm to 75 ohm conversion .xls sheet.
I can fill my 50 ohm S parameter, input output impedance & can get
75 ohm conversions. Elekta software can give direct 75 ohm input output
impedance from 50 ohm S parameters. I will just confirm it today.
I can set quicksmith to 75 ohms for input matching circuit design.
Further there is Zin & Rx(L) on smithchart. I think Zin to be
set to 75 ohms & Rx(L) will be input impedance of the device in
75 ohms. will this be ok to proceed. Top mhtplsh
Post subject: Posted: Thu Feb 22, 2007 2:39 am Captain
Joined: Wed Feb 21, 2007 12:50 pm Posts: 18 Fred47 sir,
I got the info about the above mentioned parameters from a book.
The manufacturer NEC hv given noise figure for 500 Mhz minimum. I want
to design premp for VHF. Hence cannot do anything about NF input in
Quicksmith. I am thinking of practical thoughtful solution. So to
give some input NF in quicksmith with careful study of the datasheet,
I hv come to conclusion that i can take noise figure of 1.25db for 500
mhz at 10V 7 ma & use it for 100 Mhz at 6.5V 13.5 ma. There
might some minor difference. Further I hv checked the another
transistor package with same chip for 200 Mhz & noted the noise
figure. It is around 1.10db @ 200 mhz to 1.20db @ 500 Mhz at low 4.5V,
3ma. But for practical purpose will this be ok? Top
fred47 Post subject: ckt designPosted: Thu Feb 22, 2007
1:32 pm General Joined: Wed Feb 22, 2006 3:51 pm
Posts: 104 Hi mhtplsh! Your NF assumptions look safe to me
 even a bit conservative. As a side note, at one time I needed
a very low noise figure in the lowMHz range  the best I could do was
with a [b]microwave[\b] transistor, suitably rolled off so it didn't
oscillate. (And yes, it was an NEC transistor!) Good Luck!
Fred Top fred47 Post subject: 50to75 OhmPosted:
Thu Feb 22, 2007 1:34 pm General Joined: Wed Feb
22, 2006 3:51 pm Posts: 104 Hi jaya000! Here's a copy
of the post you couldn't find: Hi! To get 75 Ohm sparameters
from 50 Ohm parameters, convert to impedance (for example, by plotting
on a Smith Chart normalized to 50 Ohms, or by formula), then convert
to 75 Ohm sparameters (for example, by plotting on a Smith Chart normalized
to 75 Ohms, or again by formula). A reminder about the Smith
Chart: it's normalized to a value of unity for the center point ("prime
center", etc.). That means that every impedance you plot is divided
by the actual impedance. Philip Smith did that so that you'd always
be able to use the same chart. Straight impedances, like those you get
from a data sheet, aren't normalized, so if they give a value like "10+j5
Ohms" you can normalize and plot like always. That means that
all the techniques that you've learned for 50 Ohms transfer directly
to 75 Ohms. Formula: the impedance matrix can be calculated from
the scattering (s) matrix by this equation: Z = (1+S)(1S)^1
and the scattering (s) matrix from the impedance matrix by
S=(Z1)*(Z+1)^1 A reminder that if S has eigenvalues of
+1 or 1, Z doesn't exist. (This isn't usually a problem, but just in
case...) OK, an example (strictly made up, doesn't correspond
to any likely part!): S11 = 0.5+1.0j S12 = 0.1 + 0j S21 = 3.0
 0.5j S22 = 1.0 + 0.1j The normalized Z matrix is 1.0 +
0.5j 0.5 + 0j 15 + 2.5j 3.5 + 5.0j and the 50Ohmdenormalized
Z matrix is 50 + 25j 25.+0j 750 + 125j 175 + 250j Now,
normalize to 75 Ohms by dividing by 75. The 75Ohmnormalized Z matrix
is 0.66667 + 0.33333j 0.33333 + 0.00000j 10.00000 + 1.66667j
2.33333 + 3.33333j Finally, convert back to the scattering (s)
matrix: 0.16578 + 1.15508j 0.12834 + 0.03209j 4.01070 + 0.32086j
1.09626 + 0.16043j I got the answers with the help of the GPL
program Octave  makes life a lot easier than using a calculator!
Good Luck! Fred Top jaya000 Post subject:
Posted: Thu Feb 22, 2007 10:11 pm Captain Joined: Tue
Feb 20, 2007 5:43 am Posts: 14 To learn i decided to first design
with 50 ohm input output. I tried to design in quicksmith with all
50 ohm parameters. Opened Quicksmith & selected amplifier design.
I inserted following S parameters with polar input selected
f S11 S21 S12 S22 ! GHz MAG ANG MAG ANG MAG ANG MAG ANG 0.10 0.71
49.5 18.61 139 0.023 65.8 0.834 22 Got K = 0.783 Delta = 0.0426
Potentially unstable transistor Then entered noise parameters
NF GO ANGLE Rn 1.15 .18 126 7.5 In the literature it is given
as Rn/50 = .15 So I multiplied it with 50. Hence entered 7.5 is this
ok? Set chart parameter to 50 ohms. Then on the circle menu
selected stability circles. Got the message : For both Source plane &
Load plane : The stability region is outside the circle Then
selected available gain circle. It shows that MAG is 29.078 but i
selected 20db gain as transistor was unstable K =0.783 Then selected
the noise circle as 1.5 db. as the 1.2 db circle was too small.
Then selected Source impedance & Load impedeance from transfer
I got the following readings : GammaS 1.056 <49.289 Zs
= 4.086+j70.525 Ga = 20db NF 1.5db GammaL = 1.014<49.281
ZL = 2.015+j108.975 Then i clicked back button. Guide me
now how to go further. My goal is to design 88108 mHz input filter
with mimimum .1db or less ripple in the required bandwidth.
Top IR Post subject: Posted: Fri Feb 23, 2007 3:29
am Site Admin Joined: Mon Jun 27, 2005 2:02 pm
Posts: 373 Location: Germany As a general procedure you should
design your matching to satisfy stability, Ga and NF requirements.
When you display on the Smith Chart the Gain, Noise circles
then at some areas these circles are intersecting together. These intersection
points will satisfy all of the required parameters. Thes point on the
Smith Chart are the required impedances. Of course that you have to
assure that these impedances are also at the region stable to avoid
oscillations! Your matching networks should convert 50 ohm to
these impedances... Top fred47 Post subject:
Amplifier designPosted: Fri Feb 23, 2007 12:58 pm General
Joined: Wed Feb 22, 2006 3:51 pm Posts: 104 Thanks IR
 that's a good terse statement! Hi jaya000, 0.1 dB gain
flatness seems rather tight for the broadcast band, where antenna matching
is hardly that good. There's a lot unmentioned that you need
to think about: 1. What does the response need to be outside the
88108 MHz band? Do you need a tight bandpass response to eliminate
interference, or does it not matter at all? 2. What kind of return
loss do you need at the input and the output? (That is, how accurate
a 50 or 75 Ohm impedance do you need?) In general, you can't
have both good filtering and good impedance matching over a broad frequency
range. Do you have a good university library available to
you? (In other words, would it do any good to give you the names of
some books on the subject of amplifier design & impedance matching,
or would that be unhelpful?) Good Luck, Fred Top
jaya000 Post subject: Posted: Fri Feb 23, 2007 3:45 pm
Captain Joined: Tue Feb 20, 2007 5:43 am Posts: 14
Thanks fred47 & IR, I can get the book i want due to my good
friend circle. Even one friend ( not a RF man ) told me that he
can arrange his company computer with serenade 8 on weekends if i want
to learn. So just guide me, i will do it with all my energy &
resources. I will try to look into the suggestion of IR. I got
stuckup at the point as mentioned above. I am not able to move ahead.
The gain circle & noise circle r not intersecting anywhere.
The noise cicle is small & nearly in the center. The gain circle
is almost at outside of smith chart. Filter ripple is not important
for me. But able to design with ur guidance is important. The amplifier
is to boost fm signals for my car. waiting for ur new suggestions
on how to proceed further with Quicksmith chart. Top
jaya000 Post subject: Posted: Sat Feb 24, 2007 4:55 am
Captain Joined: Tue Feb 20, 2007 5:43 am Posts: 14
Then entered noise parameters NF GO ANGLE Rn 1.15 .18 126 7.5
In the literature it is given as Rn/50 = .15 So I multiplied it with
50. Hence entered 7.5 is this ok? I admitted Rn= .15, then the
noise circle became big & intersected the gain circle. Now
let me know what u would hv proceeded further to design 88~108 mhz.
This will help me to go further from where i got stucked.
waiting for reply, Top fred47 Post subject:
amplifier designPosted: Tue Feb 27, 2007 12:50 am General
Joined: Wed Feb 22, 2006 3:51 pm Posts: 104 Hi jaya000,
Two very popular books on amplifier design: Gonzalez, Guillermo,Microwave
Transistor Amplifiers: Analysis & Design, 2nd edition. Abrie,
Peter,The Design of Microwave Amplifiers and Oscillators. BTW,
I don't get the same numbers you got  K = 0.325 was what I got, and
Delta = 0.692; NFmin came out with 2.9 dB NF, not 1.15. I don't know
what the difference is. You might want to find a transistor that's
not quite so "hot"  that's a lot of gain available! Good Luck,
Fred Top jaya000 Post subject: Posted: Tue Feb
27, 2007 2:33 pm Captain Joined: Tue Feb 20, 2007 5:43
am Posts: 14 Respected Fred47 : Thank u very much for ur
cooperation & help till now. U said : "BTW, I don't
get the same numbers you got  K = 0.325 was what I got, and Delta =
0.692; NFmin came out with 2.9 dB NF, not 1.15. I don't know what the
difference is. " Yes u r right. I hv one typing decible mistake.
I hv corrected & got the same K = 0.325 and Delta = 0.692. But
i hv NF data from the manufacturer as below. Noise parameter
Nf =1.15 GO .18 ang. 126 Rn/50 = .15 Hence i did the following
to go further. The Maximum Gain available is 29.080 but the
stability facor K is very poor .325 Hence decided for 16 db gain
so that i will finally get 13~14 db. The noise circle with 1.2db
is inside the gain circle. Then increased the noise upto 1.33db so that
noise circle intersects gain circle. Here i got the following :
GammaS = 1.026 < 132.591 Zs = 0.758+j21.949 GammaL
= 1.555 <10.463 Zl = 267.655+j335.299 Now i want to design
one matching Bandpass filter for 88~108 Mhz. The ripple can be .25db
to .5db at the input. For output i will prefer to use 3 pole low pass
filter or just impedance matching LC circuit. Guide me how to proceed
further? I bought the Gonzalez, Guillermo,Microwave Transistor Amplifiers:
Analysis & Design, 2nd edition. book. It is quite expensive. it
costed me US$ 95.00. If this is ok, then guide me for final
stage of the BP filter design help. Top fred47
Post subject: Amplifier designPosted: Wed Feb 28, 2007 3:41 am
General Joined: Wed Feb 22, 2006 3:51 pm Posts:
104 Hi jaya000, OOPS!!! STOP RIGHT THERE!!! IT'S NOT WORTH
GOING ON! A negative real part of the input or output impedance
means you almost certainly have an oscillator, not an amplifier.
You wrote: GammaS = 1.026 < 132.591 Zs = 0.758+j21.949
/\ GammaL = 1.555 <10.463
Zl = 267.655+j335.299 /\
The arrows point to the negative real impedances you report.
I'd seriously suggest a different transistor! Good Luck!
Fred Top jaya000 Post subject: Posted: Wed Feb
28, 2007 11:32 pm Captain Joined: Tue Feb 20, 2007 5:43
am Posts: 14 Thanks fred47. I got it now. It is due to u i
learned a lot. For me this will be a new start. I will now restart
with some other transistor as suggested by u. with regards Jaya000
Top jaya000 Post subject: Posted: Thu Mar 01, 2007
2:31 am Captain Joined: Tue Feb 20, 2007 5:43 am Posts:
14 I hv seen this transistor used in Japanese vhf boosters. Hence
i once more played with quicksmith. Due to help from fred47 i came to
know the problem. Hence i increased the NF circle & got the following
readings. Gamma S = 0.987 < 172.904 Zs = 0.335+j3.100
Ga = 16.00db Fi = 1.45db Gamma L = 0.953 < 6.701
ZL = 299.09+j732.347 Now this r positive. Suggest me now how
to go ahead. Top fred47 Post subject: circuit
design suggestionsPosted: Fri Mar 02, 2007 12:18 am General
Joined: Wed Feb 22, 2006 3:51 pm Posts: 104 Hi jaya000!
The next step is to do the impedance matching. QuickSmith
makes that relatively easy  it will transfer your calculated information
to the Smith Chart. Then you observe where you are with respect to the
prime center, and arrange to get there by moving along the reactance
and susceptance (1/X) circles, depending on whether you are adding a
capacitor or inductor in series or in parallel. You need to do
this for both input and output. Then you check to see if your
design meets your bandwidth requirements. Again, QuickSmith provides
the "Sweep" function so you can look at S21 (the "forward gain"  the
spec of most interest to most people.) If you're lucky, it's
OK. If not, then you're up for either using a precanned program such
as Eagleware Genesis, AWR Microwave Office, or Agilent HFSS, etc. Those
are generally pretty pricy  you might want to see if you can get access
to one via a friend or a professor. The alternative is to learn
how to do broadband  and that's a whole 'nother book, my friend. (Specifically,
Carlin's Wideband Circuit Design <grin> ). I do seriously hope
that you're lucky! Fred Top jaya000 Post
subject: Posted: Sat Mar 03, 2007 11:13 pm Captain Joined:
Tue Feb 20, 2007 5:43 am Posts: 14 thanks fred47. I got different
idea to calculate input/output filter. just let me tell u about
input filter. As the s parameters r 50 ohms, and i want 75 ohm input
I will calculate 75ohm to 50 ohm bandpass filter first by software.
then i will calculate simple LC filter for 5o ohms to Zs = 0.335+j3.100
as impedance matching. Then combine this two circuits. This
way i will hv control over frequency response. Waiting for ur
guidance & opinion, Top fred47 Post subject:
Amplifier inputPosted: Sun Mar 04, 2007 1:12 am General
Joined: Wed Feb 22, 2006 3:51 pm Posts: 104 Hi jaya000!
Your proposal sounds good  but it really doesn't work that way.
Think about how a filter works for a moment. The energy
at the frequencies you wants goes through, the energy at the frequencies
you don't want doesn't go through. What happens to that energy?
The short answer is that it gets reflected. The longer answer is
that a filter is (usually) made up of only reactive parts  coils and
capacitors  and not dissipative parts like resistors. So the energy
can't go through the filter, can't disappear in the filter  it must
be reflected. For energy to be reflected, the impedance must
change  if there were no change in impedance, the energy would just
continue on forward. The specification that tells you this is called
"Return Loss", and for a filter, outside of the passband, that's close
to zero dB  all the energy hitting the filter is reflected.
For this to happen, the filter must be properly terminated  and that
usually means "with the characteristic impedance". In your case, it
almost doesn't matter whether that's 75 or 50 Ohms. Your "simple
LC filter" to convert from 50 ohms to 0.335  j3.1 (did you really mean
50 x 0.33f  j3.1 ?) will not look like a 50 Ohm resistor, regardless
 so it will dramatically affect the performance of your filter.
It's really all ONE problem, which you can't "divide and conquer".
That's why Agilent, Eagleware, Advanced Wave Research, etc. can charge
so much for their computer programs  some problems are just really
HARD. You can sometimes get OK results from "cutandtry" with
a simulator, such as Linear Technology's free SwitcherCAD/LTSpice. Of
course, no simulator gives perfect results at those frequencies, due
to unmodeled aspects of the components. But it helps get you in the
right ballpark, at any rate. Good Luck! Fred Top
jaya000 Post subject: Posted: Sun Mar 04, 2007 7:31 am
Captain Joined: Tue Feb 20, 2007 5:43 am Posts: 14
Respected FRED47, Thanks for ur nice answer. It is painful to
know that u hv got cold. Hence reply when u r ok. U said :
For this to happen, the filter must be properly terminated  and
that usually means "with the characteristic impedance". In your case,
it almost doesn't matter whether that's 75 or 50 Ohms. Your "simple
LC filter" to convert from 50 ohms to 0.335  j3.1 (did you really mean
50 x 0.33f  j3.1 ?) will not look like a 50 Ohm resistor, regardless
 so it will dramatically affect the performance of your filter.
This means u can't combine two filters. My combine means
i was supposed use the bandpass filter from the 75 ohm side. Then at
50 ohm side i would hv connected ( putting in series) it to another
filter with one side as 50 ohms & another side to Zs 0.335+j3.100
with smithchart impedance selected for 50 ohms as mentioned below. This
figures r from quicksmith Gamma S = 0.987 < 172.904 Zs
= 0.335+j3.100 But as there is no termination between the two
filters, so it will not work. U said : u can sometimes get OK
results from "cutandtry" with a simulator Then how about design &
assemble some filter with active circuit & try & play with sweep
generator. My friend has got one. Top nubbage
Post subject: Posted: Mon Mar 05, 2007 5:29 am General
Joined: Fri Feb 17, 2006 12:07 pm Posts: 218 Location:
London UK Hi All contributers to this excellent topic. I have
sat on the side reading progress. I had no experience to contribute
to active ie amplifier, circuits. Now the attention has moved to
the filter elements, it is more in my territory. It is tempting to
use a rapid rolloff filter for applications like this, so an ellyptic
function or a Chebychev type might be selected. However, not only
the impedance varies rapidly at the band edge, but the plot of phase
vs frequency oscillates rapidly. The end effect of the latter is crossmodulation
between channels, and if you have an application where there are many
strong carriers being amplified with some near the band edges, the intermodulation
can be severe. For this reason a slower rolloff Butterworth filter
would be better. One supplementary question I would pose for IR or
Fred is : Is there a clean, wellbehaved way of reducing gain and
improving stability margin with a wideband amplifier using negative
feedback, perhaps overall from output back to input? Intuitively, this
might lead to oscillation outside the band where the relative phase
of S21 cannot be certain, and hence goes from negative feedback to positive.
Top fred47 Post subject: circuit stuffPosted: Mon
Mar 05, 2007 2:05 pm General Joined: Wed Feb 22,
2006 3:51 pm Posts: 104 Hi nubbage, I may be showing my
age, but amplifier neutralization (and even unilateralization) has been
around for a long time. But it's mostly a relatively narrowband technique.
I think negative feedback is great  gain is cheap, and negative
feedback buys lower distortion, more stable characteristics, and some
other benefits as well, but there are a few "interesting points":
1. Broadband circuits can't really use reactive components for the
feedback  the variation in impedance makes truly broadband operation
unreasonable. That leaves resistive feedback  and for a lownoise amplifier,
that's not good. The resistor raises the noise figure. (There is an
exception  see below) 2. For narrowband circuits, the traditional
oscillation avoidance technique of limiting the range over which the
amplifier has greater than unity gain works well. Traditionally, this
has been done with paralleltuned circuit "tanks" as the amplifier output
load. 3. The MMIC gain block amplifiers (like the Minicircuits
MAR1) have resistive negative feedback built in. If I had to create
an amplifier quickly, I might well use those instead of discretes.
4. There exists a set of techniques called "noiseless feedback",
which use either transformers or directional couplers. This is a fascinating
but problemfilled area  parasitics are an amazingly difficult problem.
I have designs in production which use this approach  but there was
a lot of work to get there. Nubbage, I don't quite follow your
statement "the plot of phase vs frequency oscillates rapidly. The end
effect of the latter is crossmodulation between channels, and if you
have an application where there are many strong carriers being amplified
with some near the band edges, the intermodulation can be severe. "
I was under the impression that intermodulation resulted only
from nonlinear effects  and phase shift "distortion" is a linear effect.
("Linear" in this case having the operational meaning that multiplying
the input waveform by a constant doesn't change the output waveform
in any way other than its amplitude). You might be thinking on
a system level, where the amplifier following the filter creates the
IM  but I'm not sure how to interpret your statement. You might also
be thinking about FM demodulation, where the "capture effect" comes
into play. FM demodulation, of course, is also a nonlinear operation.
(The first part of your statement, about the rapid change in phase
shift for ellipticfunction (Cauer) and Chebyshev filters is, of course,
undeniably true, and it's absolutely true that ignoring that aspect
can get you into trouble.) Thanks! Fred
Posted 11/12/2012
