Hi, I thought I understood this but recent discussions left me wondering.
A Bird Wattmeter provides a voltage that is proportional to the forward power
minus the reflected power. This assumes the output impedance is known and
constant, usually 50+-j0. Assume a Bird has a 0-10vdc meter for indication. I
don't know what it is, but for discussion.
Input 100 watts into a 50 ohm load, and you get 70.7 volts, the Bird scales
this to 1 volt, and you get 10% deflection on your wattmeter, or 100 watts.
Input 1000 watts and you get 223.6 volts which the Bird scales to 3.16 volts,
or 316 watts. 316 does not equal 1000, so the scale on the meter has to have
V**2 relationship to indicate 1000 watts. So 1000 watts is 31.6% of full
scale. I thought all Ham wattmeters did this. Of course it is highly
dependent on the Z the wattmeter sees. Wattmeters that actually multiply V and
I are another subject, it is hard to keep them in line without causing some
insertion loss, is it not?
73 Gary N4AST

JGBOYLES wrote:
Hi, I thought I understood this but recent discussions left me wondering.
A Bird Wattmeter provides a voltage that is proportional to the forward power
minus the reflected power. This assumes the output impedance is known and
constant, usually 50+-j0. Assume a Bird has a 0-10vdc meter for indication. I
don't know what it is, but for discussion.
Input 100 watts into a 50 ohm load, and you get 70.7 volts, the Bird scales
this to 1 volt, and you get 10% deflection on your wattmeter, or 100 watts.
Input 1000 watts and you get 223.6 volts which the Bird scales to 3.16 volts,
or 316 watts. 316 does not equal 1000, so the scale on the meter has to have
V**2 relationship to indicate 1000 watts. So 1000 watts is 31.6% of full
scale. I thought all Ham wattmeters did this.

That isn't how any directional wattmeter like the Bird works. There's
information at:
http://www.ifwtech.co.uk/g3sek/in-pr...-of.htm#bruene

This is mostly about the Bruene bridge type (the ones with a toroid).
Those are easiest to understand, because you can see very clearly how
they take separate voltage and current samples from the line.

The original explanation came from the classic 1959 QST article by
Bruene (of the Collins company, who knows how his own bridge works) and
is backed by Walt Maxwell. I repeated it in my magazine column because
today's readers still need it... perhaps now more than ever!

My article goes on to show how exactly the same principles apply to the
Bird. Once you know about taking V and I samples, it becomes much easier
to see how the Bird's pickup loop is doing both at the same time -
taking the I sample by magnetic coupling and the V sample by capacitive
coupling.

It's also very easy to see that the V sample and the I sample are never
multiplied - only added and subtracted at RF, *before* the resulting RF
voltage goes into the detector diode.

Therefore the Bird does not directly sense RF power. As Walt confirmed
very recently, the power indication on the scale is obtained by external
calibration.

Of course it is highly
dependent on the Z the wattmeter sees.

Yes - but the same concepts tell you why the Bird indicates what it
does.

Wattmeters that actually multiply V and
I are another subject, it is hard to keep them in line without causing some
insertion loss, is it not?

And harder still to find one - I've never seen or heard of one that does
it that way for RF.


--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
Editor, 'The VHF/UHF DX Book'
http://www.ifwtech.co.uk/g3sek

Ian, G3SEK wrote:
"Therefore the Bird does not directly sense RF power."

Problem with that statement is that to the uninitiated it makes the
indication suspect. A voltmeter doesn`t measure volts directly. The
ordinary voltmeter measures current scaled by a multiplier resistor.
Wish you and Walt would knock off the nonsense.

A Bird senses all it has to sense and legitimately converts it to a
reasonable indication of watts. The Bird can do so thanks to Ohm`s law,
and that`s that.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
A Bird senses all it has to sense and legitimately converts it to a
reasonable indication of watts. The Bird can do so thanks to Ohm`s law,
and that`s that.

Let's approach it a little differently. Assume a signal generator equipped
with a circulator load driving a lossless 50 ohm feedline. The load is
variable.

100W SGCL----50 ohm lossless coax--------variable load

Are there any locations up and down the line and/or any value of
load that will cause the Bird wattmeter not to read 100 watts
forward power? In the above example, I can show that the two
voltages that get added inside the Bird always yield a constant
value of voltage, i.e. and therefore a constant power indication.
--
73, Cecil, W5DXP

Cecil, W5DXP wrote:
"Are there any locations up and down the line and/or any value of load
that will cause the Bird wattmeter not to read 100 watts forward power?"

Cecil has eliminated a re-reflection with his circulator at the
generator.

Therefore, the Bird, being insensitive to power traveling in the
opposite direction to that sensed, should read 100 watts forward power
at any location in the line.

There must be enough transmission line to enforce Zo. The width of the
Bird case is likely insufficient line length.

Another caveat is that the Bird elements were designed to balance using
a 50-ohm load on the line, but were intended and have proved workable on
mismatched lines.

Another frequent question is, "How about a large reactance termination?"
Not to worry, again, given sufficient line. Terman and other authorities
report that this reactance only displaces SWR patterns on a line. That
is logical considering lengths of transmission lines as reactances
themselves. So as Cecil would likely be quick to point out, adding a
reactance is tantamount to adding a length of transmission line, and
vice versa.

SWR should have little effect on power determination with the Bird
wattmeter which is performed as : Forward power minus reflected power,
period! With the circulator, disposition of all power is clear. The
forward power must contain all load and reflected power which only makes
one round-trip.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
Ian, G3SEK wrote:
"Therefore the Bird does not directly sense RF power."

Problem with that statement is that to the uninitiated it makes the
indication suspect. A voltmeter doesn`t measure volts directly. The
ordinary voltmeter measures current scaled by a multiplier resistor.
Wish you and Walt would knock off the nonsense.

I wish you would credit "the uninitiated" (as you disdainfully call
them) with the intelligence to understand things when they're explained
to them.

A Bird senses all it has to sense and legitimately converts it to a
reasonable indication of watts. The Bird can do so thanks to Ohm`s law,
and that`s that.

No, that's just a half-truth. "The uninitiated" deserve better.

--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
Editor, 'The VHF/UHF DX Book'
http://www.ifwtech.co.uk/g3sek

Gary, N4AST wrote:
"So 1000 watts is 31.6% of full scale. I thought all Ham wattmeters did
that. Of course it is highly dependent on the Z the wattmeter sees."

The Bird has more than one scale on the same meter face. It`s possible
to scale them any way the designer wants. With the complications of
special circuits, the scale could be made linear. I for one wouldn`t
want it.

31.6 is the sq. rt. of 1000, approximately. Power is proportional to the
sq.of volts or amps. So, if you want to represent watts as a function of
transmission line volts and, or transmission line amps, you can print
the scale of a linear meter so that it advances as the sq. rt. of the
current. That is, current increases to 141.4% of its previous value, and
the watts indication doubles. That`s because 1.414X the volts multiplied
by 1.414X the amps equals 2X the watts.

In the Bird, the impedance is always supposed to be 50 ohms. That Zo
enforces a lock on the volts to amps ratio the Bird must work wiith. So,
all you need to know is the volts or amps in the incident wave and you
can calculate its power. Same for the reflected wave. The Bird very
effectively separates the incident wave from the reflected wave by
setting their samples exactly equal. In the forward wave the incident
volts and amps are exactly in-phase. In the reflected wave, they are
exactly out-of-phase. So, when set to measure power in the forward
direction, the two equal samples, voltage and current, add. The equal
reflected wave samples exactly cancel. Reverse the arrow on the Bird.
That reverses the polarity of the current sample. Now the forward
(incident) wave samples cancel, and the reverse (reflected) wave samples
add.

Zo only permits wave propagation in which volts and amps are exactly
in-phase or 180-degrees out-of-phase. The arithmetic is simple and Zo is
the friend of the Bird wattmeter.

Best regards, Richard Harrison, KB5WZI

31.6 is the sq. rt. of 1000, approximately. Power is proportional to the
sq.of volts or amps. So, if you want to represent watts as a function of
transmission line volts and, or transmission line amps, you can print
the scale of a linear meter so that it advances as the sq. rt. of the
current. That is, current increases to 141.4% of its previous value, and
the watts indication doubles. That`s because 1.414X the volts multiplied
by 1.414X the amps equals 2X the watts.
Richard, That is exactly what I was trying to say, one can measure the
power absorbed by a known resistive load by sampling V or I, rectify, scale and
display on a log scale meter, or is it square law? Whichever one it is, it
ain't linear.

73 Gary N4AST

All this time, I thought it was a Byrd wattmeter. I mean, if we are
talking details here, then let's get it right. Right?

Dave
KZ1O

p.s. And before you give me s**t about this, look below where he says
"70.7 volts". Down to three digits of accuracy, but the spelling is
wrong (in fact, a copyright violation) two words later.

Come on, do the Math (and Spelling).

JGBOYLES wrote:

Hi, I thought I understood this but recent discussions left me wondering.
A Bird Wattmeter provides a voltage that is proportional to the forward power
minus the reflected power. This assumes the output impedance is known and
constant, usually 50+-j0. Assume a Bird has a 0-10vdc meter for indication. I
don't know what it is, but for discussion.
Input 100 watts into a 50 ohm load, and you get 70.7 volts, the Bird scales
this to 1 volt, and you get 10% deflection on your wattmeter, or 100 watts.
Input 1000 watts and you get 223.6 volts which the Bird scales to 3.16 volts,
or 316 watts. 316 does not equal 1000, so the scale on the meter has to have
V**2 relationship to indicate 1000 watts. So 1000 watts is 31.6% of full
scale. I thought all Ham wattmeters did this. Of course it is highly
dependent on the Z the wattmeter sees. Wattmeters that actually multiply V and
I are another subject, it is hard to keep them in line without causing some
insertion loss, is it not?
73 Gary N4AST


--

Please yank that last "t" from my email address.
It's "net", not "nett". You know how to do that,
but the spammers won't.

The output of the element is a function of both voltage and current at the
point it is measured. The element sums 2 voltages, one that represents the
voltage at the point of test and the other that represents the current. When
SWR is 1:1 these 2 voltages are equal so that the voltage output is actually
twice that of what is contributed by voltage alone at the point of test when
measuring forward power. This is why it would be zero when measuring
reverse( the voltage representing the current adds negatively). In the
hundred watt example you used the voltage out to the meter would be 141.4v
not 70.7v

"JGBOYLES" wrote in message
...
Hi, I thought I understood this but recent discussions left me wondering.
A Bird Wattmeter provides a voltage that is proportional to the forward
power
minus the reflected power. This assumes the output impedance is known and
constant, usually 50+-j0. Assume a Bird has a 0-10vdc meter for
indication. I
don't know what it is, but for discussion.
Input 100 watts into a 50 ohm load, and you get 70.7 volts, the Bird
scales
this to 1 volt, and you get 10% deflection on your wattmeter, or 100
watts.
Input 1000 watts and you get 223.6 volts which the Bird scales to 3.16
volts,
or 316 watts. 316 does not equal 1000, so the scale on the meter has to
have
V**2 relationship to indicate 1000 watts. So 1000 watts is 31.6% of full
scale. I thought all Ham wattmeters did this. Of course it is highly
dependent on the Z the wattmeter sees. Wattmeters that actually multiply
V and
I are another subject, it is hard to keep them in line without causing
some
insertion loss, is it not?
73 Gary N4AST