The AM Forum

Here is a tip I thought I would pass along. I have used this approach to change the range of an RF ammeter (originally 15 amps full scale!) to the range I needed (now 5 amps full scale). Since the RF ammeter uses a thermocouple sensor, the reading on the face of the meter is a very non-linear function of the r.f. current passing through the meter. Therefore, it can't be scaled using the methods we use with regular AC or DC ammeters.

What I did was to make a current transformer using the same FB-43-1020 ferrite cores that many of us use to build transformers for Class E rigs. [I use these for many things around the shack/house, including making rf transformers, and for RFI filters on audio cables (i.e. they slip right over most phono plugs and most phone plugs)]

In the above case, I wanted to make the current in the meter 3 times the current in the output lead of my homebrew linear amplifier. That way, the meter would read 9 amps when the current in the output lead was 3 amps [3 amps x 3 amps x 50 ohms = 450 watts].

I made the current transformer using four (4) cores, with 1 turn of #14 insulated wire in the primary and 3 turns of #14 insulated wire  in the secondary. The meter is across the secondary. The primary is in series with the amplifier's output lead. Thus the presence of the transformer does not change the impedance of the output port of the amplifier.

Since the meter has a very low series impedance at r.f. frequencies, it looks like (essentially) a short circuit across the secondary of the current transformer. Thus the presence of the current transformer adds a negligible equivalent resistance (much less than 50 ohms) in series with the output lead. [Actually, the added resistance is 3 x 3 = 9 times the very low series resistance of the meter.]

It works perfectly. The reading appears accurate (comparing I* x I* x 50 ohms to the reading on a Bird power meter), and there is very little heating of the cores on old buzzard transmissions on 75 meters and 160 meters... running the linear amplifier at legal limit power. [I* is what the meter reads, divided by 3]

In retrospect, if I had to do this again, I would try using 6 turns in the primary and 2 turns in the secondary (same 3:1 ratio) to reduce the magnetic fields in the transformer. This would probably allow me to use fewer cores, for the same power passing through the transformer [or to reduce the small amount of heating I am getting with the current design, if I still used four (4) cores]

I just acquired a nice 500 mA RF ammeter on Ebay. I plan to use a 1:10 current transformer (1 turn on the primary and 10 turns on the secondary) to turn this into a 5A r.f. power meter.

Best regards
Stu

Would not a 1:10 transformer make that a 50ma RF meter?
Think you want a 10:1 ratio?

But a cool idea... next, clamp on RF ammeters?  ;D

                 _-_-bear

Bear

The turns are correct. With 1 turn on the primary, the current in the 10-turn secondary will be 1/10th the current in the primary. [ # primary turns x primary current = # secondary turns x secondary current].

It would, indeed, be interesting to try using a split, clamp-on ferrite bead to couple the RF ammeter to a wire with r.f. current flowing through it.

Best regards
Stu

Ah! <sound of small lightbulb flashing as the filament blows...>

The power remains the same, so 10x the voltage means the current is reduced proportionally.

               _-_-bear

Bear

There you go! ;)

I just hooked up the 500mA meter with the 1:10 transformer... and it works like a charm (reads 250 mA with 2.5A flowing of r.f. current in the primary).

Best regards
Stu

now just make it directional and...

           _-_-bear

Hello Stu,

Am I am a bit confused?  I believe the original configuration must be a 3:1 step down turns ratio, with a 3 turn primary in series with the amplifier's output, and the one turn winding in parallel with the meter, rather than as defined in your quoted text  above.  A step down would result in less voltage, more current, to cause the meter to read 9 amps when only 3 amps flow from the transmitter.  Is this correct?  Thanks for clarification.  BTW, I really appreciate the info that you and Bear share on these technical topics.  Much more worthwhile than reading that Qxx rag.

73, Rick

Rick

You are correct:

3 turns in the primary (in series with the 50 ohm output of the linear amplifier)
1 turn in the secondary (across the meter)

Therefore 3 amps of current in the output of the linear yields 9 amps of current through the meter.

Thanks for the correction.

Best regards
Stu

It works, I use it to measure current in vertical wires and radials.   
                                       Ian VK3KRI