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Signal source question

M

Mjolinor

Jan 1, 1970
0
Think maybe it's been too long since I had to do this sort of thing but I
can't get my head round it. Maybe a bit of basic help will sort me out.

I have a 50 ohm sweeping signal generator and a 50 ohm spectrum analyser.

Set the output of the sig gen to 0 dBm (1mW) and connect it to the spec an.
I get a flat trace at 0dBm on the spec an.

Does this mean that the sig gen internals are producing a fixed 2 mW, half
of which is being dissipated in the 50 internals of the sig gen and half in
the spec an?

If that is the case and a third 50 ohm device is connected in parallel with
the previous arrangement then the spectrum analyser trace will drop. At this
point is the sig gen still producing 2 mW and there will be 1/3 dissipated
by each of the devices or is the output of the sig gen unspecified because
it isn't driving what it was designed to drive.

I want to connect the sig gen / spec an arrangement to an unknown impedance
and use the spec an trace to give me an indication of the unknown impedance.
I crunched the numbers for an equation based on the above but I get no sense
out of it. The problem is, I think, somewhere around whether or not the sig
gen is a current source or voltage source, assuming either doesn't give
sensible results and I reckon the truth is that the sig gen is somewhere in
between maybe being constant power but even when I assume that I don't get
what I think I should.
 
J

John Larkin

Jan 1, 1970
0
Think maybe it's been too long since I had to do this sort of thing but I
can't get my head round it. Maybe a bit of basic help will sort me out.

I have a 50 ohm sweeping signal generator and a 50 ohm spectrum analyser.

Set the output of the sig gen to 0 dBm (1mW) and connect it to the spec an.
I get a flat trace at 0dBm on the spec an.

Does this mean that the sig gen internals are producing a fixed 2 mW, half
of which is being dissipated in the 50 internals of the sig gen and half in
the spec an?

If that is the case and a third 50 ohm device is connected in parallel with
the previous arrangement then the spectrum analyser trace will drop. At this
point is the sig gen still producing 2 mW and there will be 1/3 dissipated
by each of the devices or is the output of the sig gen unspecified because
it isn't driving what it was designed to drive.

I want to connect the sig gen / spec an arrangement to an unknown impedance
and use the spec an trace to give me an indication of the unknown impedance.
I crunched the numbers for an equation based on the above but I get no sense
out of it. The problem is, I think, somewhere around whether or not the sig
gen is a current source or voltage source, assuming either doesn't give
sensible results and I reckon the truth is that the sig gen is somewhere in
between maybe being constant power but even when I assume that I don't get
what I think I should.

It's easier to work in voltage.

Assume your sig-gen is an ideal sinewave source in series with an
internal 50 ohms.

Connect the gen to an external 50 ohm load, which it's advertised to
dump 1 mw into. So the voltage across the load is 0.2236 volts RMS
(which makes 1 mW into 50 r)

So we have a generator feeding a 50:50 ohm resistive voltage divider
giving us 0.2236 out. So the ideal generator is making 0.4472 volts
inside.

Now just treat situations like you propose as a simple voltage
divider, do the ordinary stuff, and once you know the voltages you can
convert any element's voltage back to power and dBm if you like.


0.4472v--------Rint50--------+-----------Vout
|
|
|
Rload
|
|
gnd


so it's just like a DC voltage divider.

John
 
M

Mjolinor

Jan 1, 1970
0
John Larkin said:
It's easier to work in voltage.

Assume your sig-gen is an ideal sinewave source in series with an
internal 50 ohms.

Connect the gen to an external 50 ohm load, which it's advertised to
dump 1 mw into. So the voltage across the load is 0.2236 volts RMS
(which makes 1 mW into 50 r)

So we have a generator feeding a 50:50 ohm resistive voltage divider
giving us 0.2236 out. So the ideal generator is making 0.4472 volts
inside.

Now just treat situations like you propose as a simple voltage
divider, do the ordinary stuff, and once you know the voltages you can
convert any element's voltage back to power and dBm if you like.


0.4472v--------Rint50--------+-----------Vout
|
|
|
Rload
|
|
gnd


so it's just like a DC voltage divider.

John

I did it that way too but still don't get sense out of it. It's probably my
spreadsheet equation that has brackets in the wrong place :)

Is it reasonable to treat it as fixed V, I thought that it was better to
treat it as a fixed power to be dissipated by total |Z| with variable V and
I from source, it got horribly messy and didn't work :). I suppose it does
depend on s[ecific design of sig gen though.
 
J

John Fields

Jan 1, 1970
0
I did it that way too but still don't get sense out of it. It's probably my
spreadsheet equation that has brackets in the wrong place :)

Is it reasonable to treat it as fixed V, I thought that it was better to
treat it as a fixed power to be dissipated by total |Z| with variable V and
I from source, it got horribly messy and didn't work :). I suppose it does
depend on s[ecific design of sig gen though.

---
It's better if you treat the internal source as a stiff voltage
source, since that's what it probably is. To find out for sure,
measure the unloaded output voltage from the generator with a
high-impedance device. It should be precisely twice what it is when
it's loaded with a proper load. If you wanted to, you could load it
with all kinds of non-inductive resistances, calculate what the output
voltage should be with that load, and then back into what the source
voltage should be for that load. It should stay constant. If it is,
and you do that for a few loads above and below the nominal 50 ohms
and calculate the power dissipated by the load resistance, you'll find
that it's at a maximum when the load is precisely 50 ohms.
 
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