Measuring complex output impedance
Lord Snooty
complex output impedance of a signal source, like an amplifier? I'm interested
in the MHz domain if that's relevant. I would want to model the source as
having a voltage source connected in series with one real impedance R and one
imaginary reactance jX.
Andrew
Reg Edwards
associated with the output amplifier, and knowing the circuit and operating
conditions CALCULATE what the output impedance R+jX must be.
It is certain to change with frequency. What frequency do you have in mind?
Put yourself in the position of the designer and work backwards.
A very imprecise method would be to switch it on, pass a largish, linear
signal through it, tune out the output reactance with a variable inductance
or variable capacitor, either in series or parallel, as observed with an
output voltmeter.
Then, with the reactance still tuned out, load the amplifier with known
values of resistance, measuring the output voltage as you do it. And do a
few calculations.
Frankly, whatever problem you've got, there must be another and easier way
of solving it.
----
Reg, G4FGQ
================================
"Lord Snooty" wrote
Tom Bruhns
Do you know for sure that your source is linear and time-invariant?
If it's not, then the linear impedance value doesn't have much
meaning. If it is, at least over some range of loads, you can use a
"load-pull" measurement technique. Do a google search. Basically,
you load the source with an impedance near the load it was designed
for, and measure the output; then "pull" the load a little and see
how things change. Because you are taking the delta between two large
values, you'll need to be very careful to get accurate results.
You'll need to do this for multiple loads (or a variation on the
theme) to resolve the resistive and reactive parts of the source
impedance. Obviously, the operating parameters, including frequency,
of the amplifer will affect the results.
Cheers,
Tom
Lord Snooty
The problem I've got is this:
In order to match a load (r + jx) to a source (R + jX) by designing a matching
network, you need to know all four values. It's insufficient to know only z =
sqrt(r^2+x^2) and Z = sqrt(R^2 + X^2), at least for a two-component matcher.
For this reason, I need to accurately model output impedance.
73 de G3UHD
"Reg Edwards" <g4fgq...@ZZZbtinternet.com> wrote in message
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Lord Snooty
"Tom Bruhns" <k7...@aol.com> wrote in message
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Assume we have an output impedance that does not significantly vary over our
load variations, and we only make measurements at one frequency. Assume I feed
the amp with a constant voltage input. That's good enough for me.
What kind of load do I use? A pure R? a pure C? an R and C in series? (I know
that jX is >0, so we don't use L in the load). If the latter, which do I vary?
just R? just C? both?
Can you write down the equation that you think solves the problem?
Something like
R = f(dr, dx,dV)
X = g(dr, dx,dV)
where
f and g are some functions I don't know but would like to
dr = change in load resitive component
dx = change in load reactive component
dV = change in output voltage
Best,
Andrew
Tom Bruhns
reactive part first, by putting a variable capacitor in series with a
resistive load. If the amplifier is designed for a 50 ohm load, then
use a 50 ohm resistive load. Use a variable capacitor which is
capable of resonating the expected inductive source impedance at the
operating frequency, preferably around the middle of the capacitor's
tuning range. Power things up, monitoring the voltage (or current or
power) across (in) the resistor. Adjust the capacitor for maximum
voltage. This is likely going to be a slowly-varying function,
assuming that the inductive reactance of the source is fairly small.
(If it's large, the tuning will be sharp...). When you've done that,
the capacitance and frequency will tell you the inductive reactance of
the source: XL = XC = 1/(2*pi*f*C). That fits into Zsource =
Rsource+J*Xsource; XL=Xsource.
Then you have to determine the Rsource part. If you leave the
capacitor in there, and now consider it part of the source, you've
tuned out the source reactance and you know that (to a good
approximation, at least) the source is resistive. That makes this
second part easier:
Using two non-reactive resistors (or one resistor with a second one
added in shunt or in series to get a second resistance), R1 and R2,
preferably about + and - 10% from the load you have designed for or
expect to be using with the generator, then:
o Connect R1; measure the voltage across R1. Call it e1.
o Connect R2; measure the voltage across R2. Call it e2.
Then since eload = Rload*egen/(Rload+Rgen), it's just two equations in
two unknowns. I believe if you work it out, you'll find that
Rsource = (e2-e1)*R1*R2/(e1*R2-e2*R1)
For example, if Rsource is 50 ohms, and R1 is 45 ohms and R2 is 55
ohms, and egen is 2 volts, then e1 = .9473684 and e2 - 1.047619 (but
you won't be able to measure them that accurately!) Those values in
the above equation yield Rsource=50.000 ohms, so perhaps I did the
math right for you. (Where do I send the bill??)
I leave it as an exercise for you to figure out what errors you get if
you don't read the correct voltages, or don't know the load resistors
accurately, or you don't have the reactance cancelled exactly, or the
source changes as you change the load (but maybe if the source changes
smoothly and quickly, it doesn't really matter; the effect is the same
as if it was linear, time-invariant).
Cheers,
Tom
"Lord Snooty" <bo...@dog.com> wrote in message news:<5SZpc.3399$H_3....@newsread1.news.pas.earthlink.net>...
Lord Snooty
This is excellent feedback. I was doing the first part right (cancelling the
reactance), but was using min SWR instead of voltage to figure out the second
part (the determination of R). I checked your equation and it's correct. The
only possible fly in the ointment as I see it is that "egen", for a real
system (in my case, the output transistor in an RF power amp at 1 to 20 MHz)
does not actually vary between the two R measurements. In this case, it is
prudent to keep the variance in R about nominal as small as possible just in
case, but of course large enough to swamp measurement inaccuracy.
Thanks a bunch!
Andrew
Lord Snooty
I had some nasty problems with this technique.
1. My measurements of V1,V2 lead to the inescapable conclusion that the model
fails, because the calculated value of R comes out negative.
Let us assume that we set (r2 > r1) and we obtain (V2 > V1), which is
predicted from the model, and is also the case for my measurements. Under
these conditions, a negative value of R can only be obtained from the
equation if V2/V1 > r2/r1, which in my case is true.
I undertook a full (and rather exhaustive and tedious!) calculation of the
expression for R when the value of (x) was not set correctly to -X, thinking
that perhaps this was the cause of the discrepancy. It turns out in this case
that, if the calculated value of R is negative, it has nothing to do with the
setting of (x), and depends ONLY on the condition V2/V1 > r2/r1. Since we know
that the actual value of R cannot be negative, this implies a failure of the
model.
How then can the model fail? Since we are maintaining frequency constant, any
collection of resistances and reactances, however complicated, can be modelled
as (R + jX), so it cannot be that. The only assumption left to question is the
constancy of V0, and this is what the failure must be.
This leaves me with more questions than answers, because the way forward is
now completely unclear.
2. I should also mention another, less serious, problem I had, and that is
with the determination of X. The value of (x) I determine from measurement
would be expected to be constant at a given frequency. It is in my case not
so. The derived value of X appears to depend on
a) the power level setting of my amplifier
b) the value of (r) I set in the circuit when determining X.
Quite probably this second problem relates to the first problem's
identification of the failure of the model, and can probably be subsumed under
that category.
Best,
Andrew