TC RF AMP Meters

[Steve W8TOW]

OK, a lack of my knowledge...How do you connect a TC RF ammeter
to open wire line?

I have several RF ammeters that are "TC" Type...Do you run TC wire
from the meter to a leg of the open-wire and loosely couple it to the
feeders?
73
8tow

[W7TFO]

A bit of semantics at play here.

All RF ammeters are TC, or thermocouple types unless they are ancient hotwire jobs.  We are not talking about them.

There are remote thermocouple types, with the meter separate from the TC for ease of installation, and self-contained types.

Regardless of the style, you put it (usually them) in series with the OWL at any convenient point.

Since OWL is traditionally a balanced affair, one should use one in each side to make sure it is running in a balanced state with equal current in each leg.

73DG

[AB2EZ]

Traditional thermocouple RF ammeters, of the type you see on eBay, work by having current (RF or otherwise) pass through a thin wire that generates heat along its length (I**2 x R).

In one design concept, the thin heating wire is attached/welded (roughly at its midpoint) to a pair of thin wires made of dissimilar materials. The dissimilar materials form a junction at the point where they attach to the thin heating wire. Via the thermoelectric effect, this hot junction has an observable DC voltage across it that can be used to produce a current in an attached meter. Note that the other end of each of the two thin wires... that form the junction that is attached to the midpoint of the thin heating wire... also forms a junction at the point where it connects to a copper wire that goes to the meter coil... but since those additional two junctions are at ambient temperature, the net result of using this configuration, with three junctions in series, is that a current flows through the meter (around the circuit) when the heated junction is at a higher temperature than the other two junctions. [See the attached 3-page patent, describing the use of a thin-walled heating tube instead of a thin heating wire... filed September 11, 1936]

Two ordinary wires attach to the thin heating wire inside the meter via the terminals on the back of the RF ammeter. Depending upon the value of current that the ammeter is designed to read, there will be a current divider inside the meter to shunt some or most of the current around the thin heating wire.

To measure current in one of the two wires of the balanced line, you will need to do one of two things:

a. Cut one of the two balanced wires, and place the RF ammeter in series (being careful not to make the round trip path (to the meter- through the meter- and back to the balanced line) more than about 1/16 of a wavelength at the frequency of operation

or

b. Attach a broadband RF current transformer. E.g. a ferrite core with one side of the balanced line passing through that core; and a separate wire passing through the same core, and looping around through the meter. Note: since this is a current transformer, with a 1 turn primary winding and a 1 turn secondary winding, the current passing through the meter will be the same as the current in the side of the balanced line that passes through the core. If you use N turns on the "winding" that passes through the meter (instead of 1 turn), then the current passing through the meter will be 1/N x the current passing through side of the balanced line that passes through the core. Warning: if you don't attach the meter (or alternatively, a short circuit) across this "winding", then you will produce a large voltage across the ends of the open winding... and the core will get very hot.

Note: traditional RF ammeters that use a thermocouple are very easy to burn out. If the actual current passing through the RF ammeter exceeds the peak current on the RF ammeter scale, by even a small percentage, for even a short period of time, the thin heating wire inside the meter will fuse... and the RF ammeter will become a paperweight.

Stu

[Steve W8TOW]

Stu, please clarify:
" Cut one of the two balanced wires, and place the RF ammeter in series (being careful not to make the round trip path (to the meter- through the meter- and back to the balanced line) more than about 1/16 of a wavelength at the frequency of operation"


OWL _______________ +Ammeter    - Ammeter _____________ OWL
OWL _________________________________________________ OWL

73 Steve

[AB2EZ]

Steve

See the attached. The distance around the red loop (from the OWL to the meter, through the meter, and back to the OWL) should be no more than 1/16 of a wavelength at the frequency of operation. [1/16 wavelength corresponds to 22.5 degrees of phase shift]. Less is better

Note:

In typical applications, the SWR on the OWL is high. This means that, at a given frequency and a given position along the OWL, the rms current flowing through each OWL wire might be much larger than (or much smaller than): the square root of [the net forward power / the nominal impedance of the OWL]

For that reason, it would likely be very difficult to use a traditional thermocouple RF ammeter, in this application, to measure the current. I.e. the rms current will either be too small to measure with the RF ammeter or large enough to fuse the thin heater wire inside the RF ammeter.

[Steve W8TOW]

In most amateur applications, OWL is right into the shack (in my case, at least) and the
ATU is hanging on the wall...
The antenna is a 80 CF with 600 ohm feeders and I use it on 20-80m...
The length of the "red" lines in your example will only be about 2-4"...
much less than 1/16 wave...at 20-80m...this correct thinking?

[AB2EZ]

Steve

Yes... but there is still the problem of the small range of currents that the meter can accommodate (between too small to measure, and burnout), and the large range of currents that will exist at the point where you insert the meter.

Stu

[Steve W8TOW]

And obviously, this will change on each band of operation, right?

[AB2EZ]

Steve


Yes...

The SWR will be different on each band, and the locations of the current maxima and current minima will be different on each band.

If the SWR on the OWL at a given frequency is N:1, then the current on each wire of the OWL at the location of a current maximum will be N times as big as the current on each wire at the location of a current minimum.

For a typical thermocouple RF ammeter designed (as an example) to read full scale at 5A of RF current ... The needle will only be at about 4% of full scale with 1A of RF current; while 7.5A of RF current will quickly fuse the heating wire inside the meter.

Stu

[Steve W8TOW]

So, when setting this up with my RF ammeter 0-7 A, if I start out
with pretty low level RF Power, say a Ranger at 40 watts, I can
evaluate the response of the ammeter for each band of operation.
Create a scale of reference with various power levels, checking both
wires of the Feeders of the OWL (to confirm equal currents)...
Then, if nothing changes on the antenna, I can begin to increase the
RF levels up a a safe point, staying below the 7 AMP max....
Seem reasonable steps?
thank Stu for the conversation!

[AB2EZ]

Steve

You're welcome!

Good luck!

But I still think this is the wrong instrument for checking the balance of the currents in the two OWL wires. For all of the reasons described in my prior posts... the actual readings ( v. comparing the readings obtained from each of the two wires) will be of limited use... and the likelihood of inadvertently fusing the meter's heater wire when changing frequencies or adjusting the tuner is too high.

Best regards
Stu

[Opcom]

What is a good way to check the balance of the currents regardless of whether it's 500ma or 10A?

[AB2EZ]

"What is a good way to check the balance of the currents regardless of whether it's 500ma or 10A?"

http://amfone.net/Amforum/index.php?topic=34072.msg263405#msg263405

[K5UJ]

Unless you have a way to measure and/or calculate the Z of the line on a certain frequency at the point where the meters are inserted, the current reading won't mean much assuming you are trying to calculate power.  they can give a good idea of current getting transferred to the load of course and if you have two matched meters one on each side you can check balance.

If you want to measure power, it is best to have the rig on a length of line terminated with a 600 ohm noninductive purely resistive load and insert the meter into the line.  With that you can calculate the current for a given amount of power out and knowing the scale of the meter(s) know what to expect.  Let's say you have a rig capable of 300 watts.   the current to the load for that power would be about 710 ma.  square root of power divided by resistance.  If you have a 0 to 3 A meter you're okay.   If the load is unknown and you have a rig with continuously variable output like most modern plastic radios, you can use it to start off by feeding small amounts of RF to see what happens. 

I don't know the range of your meter so I can't say much more. 

[Steve W8TOW]

One approach to this analysis I've successfully done is to couple each channel of a
dual trace scope to the individual feeders of the OWL...and tx a sig (carrier)

The 2 sine waves should be 180 deg out of phase on the scope!

[YB1AHY]

Hi All, very interesting discussion

I just bought TC amps from ebay arrived a week ago with 4 amp max range.

I m trying to understand this discussion, and i try to express my understanding, because i m also worry if i cooked TC really quick.

My understanding: if we know the power from transmitter, for example 50 watts (50 ohm) mean 1 amp in TC because the formula is P=I2R 1x1x50=50 watts. In this case we don't need to worry about to insert TC in wherever position in the feeder line. If accidentally put in position where the lowest impedance occurs, the max current will be only 1 amps.

Is my understanding correct?

Best regards

Agus / yb0djh

[AB2EZ]

Agus

Hi!

Your understanding is, unfortunately, not correct.

You can think of what is happening in terms of two waves traveling in opposite directions along the transmission line. Every electrical half wavelength of distance along the transmission line, each wave changes phase by 180 degrees. Since they are traveling in opposite directions, the phase difference between these waves changes by 360 degrees for each electrical half wavelength of distance along the transmission line (i.e. the phase of the reflected wave changes by +180 degrees, and the phase of the forward wave changes by -180 degrees)

The current in each conductor, at any point along the transmission line, is the sum of the currents associated with each of these two waves. At points where they are in phase, the currents add with the same sign. At points where they are out of phase, the currents add with opposite signs.

In a situation where there is a reflected wave from the antenna (i.e. where the reflected wave's amplitude is not zero), the total current in each conductor will be higher at points where the forward and reflected waves add in phase. The current will be lower at points where the forward and reflected waves add out of phase.

This leads to the definition of the standing wave ratio. I.e. the SWR is the ratio of the current at a point of maximum current on each conductor to the current at a point of minimum current on each conductor.

Bottom line... unless the SWR is 1, the RF ammeter will experience a larger current at some points along the transmission line than it will experience at other points along the transmission line.

If the transmission line has an impedance of Z ohms, and the net forward power is P watts, then the maximum current that the RF ammeter will experience (i.e. at a point where the forward and reflected waves add in phase) is I(max) = the square root of [(P/Z) x the SWR]. The minimum current that the ammeter will experience (i.e. at a point where the forward and reflected waves add out of phase) is I(min) = the square root of [(P/Z) / (the SWR)]. As expected, the ratio I(max)/I(min) = the SWR

Example:

The net forward power is 100 watts. Z= 400 ohms. the SWR is 4

I(max) = the square root of [(100 watts/400 ohms) x 4] = 1 ampere

I(min) = the square root of [(100 watts/400 ohms) / 4 ] = 0.25 amperes

I max / I min = 4 = the SWR

The current in the forward wave is [I(max) + I(min)] / 2 = 0.625 amperes

The current in the reflected wave is [I(max) - I(min)] / 2 = 0.375 amperes

The forward power is 0.625 amperes x 0.625 amperes x 400 ohms =156.25 watts

The reflected power is 0.375 amperes x 0.375 amperes x 400 ohms = 56.25 watts

The net forward power = the forward power - the reflected power = 156.25 watts - 56.25 watts = 100 watts

Stu

[YB1AHY]

Bottom line... unless the SWR is 1, the RF ammeter will experience a larger current at some points along the transmission line than it will experience at other points along the transmission line.

Stu

I forgot to put in my assumption that SWR is 1. i agree.

So, if SWR is not 1, it seems like (in some position in feeder) TC looks load "like" short circuited means TC will burn or looks "like" open circuited. is it correct?

Thanks for your explanation Stu

73s Agus.