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Author Topic: How to calculate the antenna input impedance of a receiver?  (Read 6040 times)
KC9KEP
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« on: October 27, 2011, 04:56:04 PM »

Hello all,

Can someone help me to understand or guide me to a reference
that will explain how the input impedance of a receiver
is calculated/specified?

I am building a "new" receiver: the "Mighty Midget" as specified in the
late 60's QST and ARRL handbooks.

I wanted to apply a test-signal to the antenna input connection in order to trace through
the first stage of RF amplification.

I used a Heathkit IG-5280 signal generator.  Its maximum output (unloaded)
is about 150 mV P to P.

When I connect the signal generator to the antenna input, the RF signal
from the IG-5289 practically disappears.  Which isn't unexpected because the generator is
driving into about 14" of 26 Gauge wire, or about 6 turns around a 3/4" coil form.

If I'm not mistaken, these ARRL designs are intended for a 50 Ohm antenna.

So, I tried to make the impedance calculation by measuring the inductance and DC resistance.

Xl = 1.23 uH  and DC resistance =  170 m Ohms.

Then I thought I'd try the Pythagorean theorem calculation to get the hypotenuse
of the resulting impedance triangle.

Of course, the DC resistance completely swamps out the inductive reactance, so
my result is still about 170 m Ohms of impedance.

So, how is this calculation made?  Is it a more obscure calculation similar to the
characteristic impedance of a coaxial cable (the square root of "L" over "C"?)

For that matter, what sort of measurement device would one use to "see" such
a small RF signal that, in my case, eventually gets buried in the B+ ripple
when looking at the plate output of a pentode?

Thanks!

73

--KC9KEP
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N2EY
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« Reply #1 on: October 27, 2011, 05:10:56 PM »

Do you have the Radiotron Designer's Handbook? Chapter 23 in the 4th edition explains it all.

If you don't have it, go to the technical books online site and download it free. Lots of other good stuff there.

Btw - is this the receiver you want FT-241A xtals for?

73 de Jim, N2EY
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KC9KEP
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« Reply #2 on: October 27, 2011, 05:48:53 PM »

N2EY,

Yessir, it is :-)  I did find some of the required crystals.  I've build several of these ARRL
designs in order to learn what they "handle" like and how they perform.

I've completed the radio to the point of operation, but it's yet to be mounted into its cabinet.

The crystal filters sound very "band-passy" if there is such a word.  I thought that perhaps
the narrow bandwidth would simply help eliminate the adjacent channel "interference"
(which it does) but it sounds suprising similar to the I.F. amplifier regeneration method.
(Or, Q-multiplier technique.)

I am currently in the process of trying different crystal combinations to achieve various
band pass width .. and, I'm listening to the effect on SSB, AM, CW, etc.

I am still adjusting the LO coil to get proper span on 40m.  My 80m is quite close.
(It cuts off a bit of the lower end of the band right now.)

The receiver works quite well so far .. fairly stable after a bit of warm up (it
utilizes 3 6U8 pentode/triodes)

And thanks greatly for the link to the data which I am seeking!

73!

--KC9KEP
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WB6THE
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« Reply #3 on: October 27, 2011, 06:36:04 PM »

I'll take a shot at answering your questions. I did a goo-goo search and came up with the schematic.
The antenna connection is the primary side of an RF transformer. You've measured the DC resistance
of the primary winding as 170 milliohms and the inductance as 1.23 uH although in your post you
mistakenly said Xl = 1.23 uH vs L = 1.23 uH.   170 milliohms is .170 Ohms  which is so small that
for practical purposes we can ignore it.  Using the formula for XL:

XL = (2*pi) * F * L  where  2*pi = 6.28     and    F=frequency   and   L=inductance   * means multiply

At  7.000 MHz

XL = (6.28) *( 7*10^6)*(1.23*10^-6) = 54.0708 Ohms  Call it 54 Ohms.

If the primary winding were just a coil and coupled to nothing its impedance, Z, would be 54 Ohms.
There is some tiny capacitance between turns of the wire but so small that for practical purposes
we can ignore it.

 Z = square root (XL^2 - XC^2)     Z=square root (54^2 - 0)   Z=square root (2923.651) = about 54 Ohms.
At 3.5 MHz Z  and XL would be about 27 Ohms.

(DC resistance and AC reactance are different. Go measure a 110 V. input transformer of some sort
with your ohmmeter. It will read a few Ohms. I=E/R so several amps. Why doesn't it trip a circuit
breaker? Because of XL, inductive reactance and same principle at RF frequencies).

But it is a transformer coupled into the parallel resonant circuit L2 and those two caps and the control
grid of V1a part of the 6U8 RF amp. That changes things but not seriously.

A receiver should receive signals with 3 feet of wire connected to its antenna input.
Impedance is important in a transmitter but much less important in a receiver.

150 mV is a huge signal to a receiver. Most every signal generator decreases in output under load.
If you simply want to put a signal into the receiver for test purposes try "loose coupling".  Connect
a short wire, a few inches long, to the receiver input. Connect a short wire to the signal generator
"hot" (center conductor) output. Insulated wires. Twist them together a few turns and you'll have
enough signal to drive that receiver crazy. Connect the grounds if you want to. Whatever works best.

Now, to see a small RF signal buried in power supply ripple. I'd use an oscilloscope. Set the input
coupling to AC mode (blocks DC B+ from the 'scope). The AC ripple will pass but so will your RF
signal. Turn the vertical sensitivity way up and adjust vertical centering so that you can see the
trace. Set the sweep speed to something very high, around the speed of the RF signal to be observed
if the 'scope can go that fast.  I'd expect that there would be very little AC ripple.

Not sure if you have built the receiver and it doesn't work or what. Maybe you could post more details.
If you are troubleshooting don't worry about input impedance, its just not that important right this
minute. I'd check the wiring throughout the receiver if it isn't working.

I probably have not properly answered your questions and some engineer is going to tear me a new one
for what I have commented Smiley

Just looked at your latest post on the other computer. Sounds like you know a heck of a lot about
what you are doing! If a filter makes the received signal sound "bassy" it is too narrow.

Anyway... was any of this helpful or did I totally miss?

Alan
WB6THE


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KA4POL
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Posts: 1959




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« Reply #4 on: October 27, 2011, 10:44:56 PM »

After this very detailed explanation I just want to add a tiny bit. The IG-5280 works into a 1k resistor from which you can tap your voltage. At full voltage you add a parallel resistance of about 50 Ohm. So it is not astonishing that your voltage shows the drop as experienced. However, some ┬ÁV at the input are close to reality and should work perfectly.
It is good to see that the old circuits still are getting attention, moreover that it is still possible to acquire the parts.
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VK2TIL
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« Reply #5 on: October 27, 2011, 11:00:06 PM »

"... was any of this helpful or did I totally miss? ...".

Well, I thought it was a very good reply!

'KEP mentioned the "impedance triangle" but didn't get the correct result.

An impedance expressed as Z = R +/- jX can be calculated by Pythagorean methods as an impedance triangle.

The end result is that the impedance equals the square root of the sum of the squares of the two components R and X.

R is small; R sqrd is also small so, in this simple calculation, may be ignored.

So, for practical purposes, Z = X as 'THE described.

And yes; 150 mV is a VERY big hit on a receiver front-end when normal receiver sensitivities are expressed in microvolts.

But the output impedance of the Heath is ill-defined; as is usual with these instruments, the output is through a pot so the impedance can be anything.

(Incidentally, I added a 50-ohm output amp and and a fixed attenuator to a couple of basic signal generators to establish a good 50-ohm output; it's worth doing if you like the generator and want to use it).

Voltage-divider action between the unknown Z of the Heath output pot and the 50-something ohms of the MM might have reduced the signal level but there should still have been plenty there.

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KA4POL
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« Reply #6 on: October 27, 2011, 11:54:10 PM »


Voltage-divider action between the unknown Z of the Heath output pot and the 50-something ohms of the MM might have reduced the signal level but there should still have been plenty there.



Anyway the proposed loose coupling by hooking up some wires is the best solution.
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VK2TIL
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« Reply #7 on: October 28, 2011, 12:22:47 AM »

Agree.
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W8JI
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« Reply #8 on: October 28, 2011, 04:37:35 AM »

Tom(?) KC9KEP always asks great questions!!!

I probably have not properly answered your questions and some engineer is going to tear me a new one
for what I have commented Smiley

The reactance of the inductor and capacitor, other than making the system resonant, have nothing to do with input impedance of the receiver. Once resonant, if we ignore resistive losses, impedance would be either zero or infinite depending on if it is a series or parallel connection.

The input impedance of a receiver is very complicated. It is determined by what loads the circuit to consume power and how the antenna port is coupled into those system losses. I'm afraid it is far beyond the scope of an eHam reply. :-)

It sounds like he has a classic link coupled input circuit, so this would apply if "load" is the antenna or generator:

http://www.w8ji.com/link_coupling.htm

KC9KEP (Tom??),

Your generator might have a very high output impedance, and be easily loaded. The receiver, unless it has a lot of loss on the input tank, also should have a very low input impedance.

Test your generator by connecting a 50 ohm resistor across the output. My bet is it drops level substantially. That would indicate the generator has a very high source impedance, which was normal for old cheap trouble-shooting generators designed for radio service work.  If it doesn't change much, it is a low impedance output generator.

If you really want a 50 ohm generator, the normal practice is to use at least a 10 dB 50-ohm attenuator pad on the generator output. This stabilizes the output impedance at 50 ohms. Some more expensive generators are designed to have a 50 ohm (or 75 ohm) output. They usually have built in attenuator pads.

The input impedance of the receiver is determined by turns (inductance) in the link, turns (inductance) in the tuned part of  the input coil, losses in the coil and capacitor, and by the resistive loading caused by circuitry in the receiver's input system. It just is not nearly as simple as XL or Xc = input resistance.....not even remotely close at all.

If there were no losses using power, and if the input was resonant, the input Z would be zero or infinite at resonance (depending on the input configuration). Why is that you say??? Because a pure reactance consumes no power, and thus has to look like an open or perfect short for waves! The input SWR, without power loss, would be infinite.

Since loss is the only thing that moderates the SWR, we have to know circuit losses ad how they are transformed in the path to the input port of the receiver. I guarantee you, calculating this will give you a real headache.  :-)

Most of my receivers have 20-100 ohms input impedance, but they are designed and intended to be in a 50 ohm system. For example my Drake R4C's, on 160 meters, have about 20 ohms input impedance. My Yaesu FT1000MK V is ~80 ohms on most bands.  My NC303 is about 1100 ohms on 160, 240 ohms on 40 meters, and 100 ohms on ten meters.

All work fine, and have more than enough sensitivity.

Some homebrew receiver from an old Handbook might be anything at all. It might be just a few ohms, and it might be a few thousand ohms. It probably is not worth worrying about.

73 Tom


« Last Edit: October 28, 2011, 05:43:19 AM by W8JI » Logged
KC9KEP
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« Reply #9 on: October 28, 2011, 07:03:23 AM »

Hello all,

Thank you for all your thought, time, and replies!  Very helpful.

I made a huge faux pas in my original assumption .. I used "inductance"
instead of "inductive reactance" in my formula!

I feel like Chris Farley on Saturday Night Live; "Stupid!  Stupid!".

But, I guess that there's even more to this impedance calculation than
meets the eye.

It's "one of those things" that I was interested in understanding more about.
(Maybe one of those fancy Agilent network analyzers would reveal much more
about the input impedance?)

re: tracing through the RF amp circuit:

The problem I was experiencing was that the miniscule RF signal is riding
upon a huge AC power supply ripple, about 2V p to p (upon ~130 VDC).

The receiver sports a "cheap" DC power supply, half wave rectification and "PI" type
filter using a resistor rather than a choke.  I have a 200 MHz Tenma scope
that refuses to trigger on the RF that is riding upon the power supply
ripple .. even with the vertical amp AC coupled and the horizontal sweep speed maxed out.

I was wondering if I just need a better scope (I bought mine for $100 used)
or if a different piece of test equipment would be more appropriate.

The radio is working quite well, but I am very much interested in learning more
about the invisible nuts and bolts of how these circuits are quantified and
designed.

Thanks again all!

73

--KC9KEP (Tom)
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WB6THE
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« Reply #10 on: October 28, 2011, 08:58:41 AM »

Your oscilloscope is more than good enough to look for those RF signals.
 
Poking around on RF circuit tube plates....
You would like to have a low-capacitance, high impedance probe. Any measuring instrument
affects the circuit on which it is used to some degree. The idea is to make the instrument
as "invisible" to the circuit as possible.

Applying the oscilloscope to the control grid of the local oscillator tube or to the control
grid of the BFO tube might introduce enough loading that the circuits cease to oscillate
or oscillator frequency changes even though the probe presents a high impedance.

The plates of the tubes are much less susceptable to instrument loading.

Some means of blocking the 60 Hz ripple from the half-wave supply would be desireable.
A picture would be worth the next thousand words but here goes:

Suppose you constructed a simple circuit consisting of a small value series input capacitor of,
say 100 pF  and let's call it C1 . From the right-hand  side of C1 to ground place approximately a
100 uH choke and call it L1.  Where C1 right-side and L1 connect, place another 100 pF series capacitor.
The circuit looks like a "T" .  You have constructed a high-pass filter.  For those capacitors be
sure that they have a couple of hundred volt ratings.

Values are not at all very critical. The idea is that the capacitors block B+ DC and easily pass RF
and act as a huge impedance to the 60 Hz ripple.

So far you have left-side of C1 unconnected and right-side of C2 unconnected.
Between right-side of C1 and left-side C2 the 100 uH top wire is connected and the
bottom wire connects to ground. The filter is connected between the probe and the 'scope.
Left-side C1 to probe connector center. Right-side C2 to scope input. And all ground-sides
connected to each other.

Set the 'scope for AC coupling. At slow sweep speeds you may be able to see a
"carrier" wave. If there is enough RF present and the vertical gain is high enough
and sweep speed is high enough you *might* be able to get the sweep trigger to
"stop and lock" an RF sine wave.

Observations might still be difficult or impossible because:
    Typical signal from the antenna to receiver input might be only a microvolt or less.
    The gain of tube V1A might be (wildly guessing) 100. That is still only .0001 Volt of RF.
    The mixer's output may likewise be very small. Its job is to mix, not amplify.
    
I'd expect a great big signal at the local oscillator and BFO oscillator plates.

==> The grids of amplifier and mixer tubes might have far less power supply ripple
       and the coupling from the previous stage doesn't pass 60 Hz so you might not
       even need a high-pass filter. <==

The cathode side of CR2 is audio so the high-pass filter would need to be removed
to get a 'scope reading. Plate of the audio amp should have all sorts of signal as
should the headphone output.

Loose couple your signal generator to the receiver antenna terminal and there will be
lots of RF to try to observe.

If you are receiving a signal then rest assured the RF is there and if you cannot
observe it on the 'scope it is because it is just too difficult to resolve even on a very
expensive instrument. Your oscilloscope is, again, very well "qualified" to attempt
the observations.

I see one thing in the Mighty Midget schematic that asks me a question:
Most audio amplifiers have a "bypass capacitor". In a common emitter transistor
amplifier, parallel with the emitter resistor and in tube common cathode circuits, parallel with the
cathode resistor. The capacitor value is chosen to have a reactance of about 1/10th
the value of the resistor at the lowest audio frequency to be passed. In the Mighty Midget
something like 8 or 10 uF with a voltage rating of 50 Volts or better. The capacitor "filters"
the voltage drop across the resistor to maintain a constant grid bias so that audio gain
is not decreased. See how RF amp and IF amp have an appropriate value bypass capacitor?
Mixer doesn't have one because the local oscillator output doesn't want to be shunted
to ground.  Something to try if you were so inclined. I'm not saying that the circuit is
incorrect. Just wonder why the cap isn't used.

Also, why feed the audio output through headphones via that .01 capacitor. Seems to
me that an output transformer could be used to drive a speaker. There are 10 ways
to do things and maybe this is the way the designer did it.

Anyway, there's another essay for ya'

Alan
WB6THE
« Last Edit: October 28, 2011, 09:34:58 AM by WB6THE » Logged
KC9KEP
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« Reply #11 on: October 28, 2011, 12:28:25 PM »

Alan,

Thanks for all the information and ideas!

I will try the high-pass filter idea tonight as well
as loosely coupling the signal generator to the antenna
input.

Yes, the local oscillator and BFO outputs can be easily displayed
on the scope by connecting the scope probe/ground to a loop
of enameled wire, consisting of about 5 turns or so.  These
induced signals can even be "read" by my cheap-o frequency
counter.  (I can calculate frequency using the scope (1/period),
but the freq-counter is easier and a bit more accurate :-)  )

And, yes .. I can drive a small (3" X 5") speaker from the
secondary of the output transformer, which is what I have
been doing (in addition to the headphones connection).

And, I see what you mean about the "omission" of any
bypass capacitor on that audio amp.  That capacitor
also sets up the AC-Load-Line, no?  And, maybe changes
the frequency response to add more "lows" to the audio
output?

Seems simple enough to sub one in to see what happens.

And thanks again, all.  This is the best part about home brewing ..
learning!

73

--KC9KEP

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KC9KEP
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« Reply #12 on: October 28, 2011, 01:37:28 PM »

Woops .. did I say I have a 200 MHz scope?  It's just 20 MHz. Tenma Model 72-320.

I'm working on my radio and not sleeping enough!

73

--KC9KEP
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WB6THE
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« Reply #13 on: October 28, 2011, 07:03:40 PM »

I think that tube load lines involve finding the center of the Eg-Ip curve to set the
class A or B or C quiesent operating point based on plate load (Eg-Ip = grid voltage vs plate current).
Know what... I like tubes but it has been years since I last touched one, and I do mean YEARS Smiley

I was a high school guy in 1966 or so when I began learning serious electronics and I kind of grew
up with both tubes and transistors. Old-school tube techs who couldn't adapt were dropping out of
the field like flies. I never knew there was a problem. Parts were plentiful and there was still WW2
surplus to be had for pocket change. It was IC's that for a while screwed me up. But... but... but
what's inside. Who cares. I do. It doesn't matter. Stuff inside. Look... apply power, signal in, the
innards do this and that and signal out. It doesn't matter, you can't fix an IC... you replace it!
I graduated high school with a Second Class Radiotelephone License and an Advanced Class
Amateur Radio License in 1968. Joined the Navy, spent 6 years as a steam turbine engineroom
mechanic, then switched to aviation electronics. Later got First Class Radiotelephone with Radar and
made Extra Class Amateur Radio. Saw the world and retired. Then the FCC told me I don't need NO
license anyway but here's a GRC good for life and never bother us again. But getting WAY off subject!
I had to work for those licenses and the high school diploma was "free". I valued my FCC tickets more
than that silly diploma. There was no money for college and I wasn't smart enough at the time. Those
FCC tickets were my "degree". And I learned most of what I know on my own from books. Now, I have
taught EE engineers things and THEY have taught me a LOT. Engineers and techs learn differently. Now,
about a receiver...

I said "wow" when I read 200 MHz in your post. But 20 MHz is still quite adaquate for what you are
doing. I once had a great big Tektronix 'scope that was designed for color television work. Don't
remember the model number but it was full of tubes! It was made for working on the old NTSC
analog color TV system of late demise. I had that puppy displaying all 8 cycles of the 3.54xxx
reference color burst looking for fleas at 270 degrees of the 5th cycle.  There weren't
any but it would have displayed them. Darn thing would stop and lock a single cycle of RF from
my 2M rig.

I  have two 'scopes. A Tektronix 454 dual trace 100 Mhz classical CRT instrument and a Tektronix
THS720 LCD display 'scope. The 454 was old when I got it and now its older. Years ago when they
were a great big deal I had a dead 5.25 inch floppy drive that I traded for it. I'm serious, I really did.
I found the LCD 'scope at a yard sale. Just sitting there. I was lovingly touching and fondling it. The
owner asked me if I know what it was (well it DOES look like a TV set). I assured him that I very well
DID know. I'd been wanting a digital LCD o'scope for a long time. I knew that I couldn't afford it but I
asked the price. $160.00 said the owner. O-M-G as I whipped out my weekly allowance. Owner knew
very well what he had, no longer used or wanted it and just wanted a bit of cash. Had the original receipt
for something like $2500 bucks. Saw the same 'scope at a hamfest asking price $1600.

Anyway, back to your receiver. Without that audio bypass cap you would actually have better fidelity
but much lower gain in a common emitter/common cathode circuit. Now, you spoke of signals sounding
too "bassy". The IF filter is too tight (too narrow bandwidth). The high frequency component of voice or
music are outside the IF filter passband. Well, this is primarily a CW/SSB receiver anyway. But are
you aware that the frequency of xtals is NOT exactly what is printed on them?  Depends on the
capacitance of the crystal socket and other components connected to the crystals. You can experiment
with variable series/parallel small value trimmer caps and small series inductances to "pull" the
crystals slightly +/- the IF frequency. To shift the IF passband. That may improve the bassy quality. I doubt that the quality has much to do with a cathode bypass capacitor.
 
Keep homebrewing because that is an excellent way to learn a lot! And let us know what you're up to.
Post more questions and we'll enjoy trying to answer them.

Alan
WB6THE



« Last Edit: October 28, 2011, 09:19:22 PM by WB6THE » Logged
G3RZP
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« Reply #14 on: October 29, 2011, 03:05:29 AM »

Tom said:

>The input impedance of the receiver is determined by turns (inductance) in the link, turns (inductance) in the tuned part of  the input coil, losses in the coil and capacitor, and by the resistive loading caused by circuitry in the receiver's input system.<

Also the coupling factor between the link and the main winding.

As the man said, it is not just simple XL and R!
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