I think the fact that you do not have a load on the output tank, is what is
limiting the power output. Power tubes require a particular load Z to
deliver 'rated' power. Right now all I see is a parallel LC tank which will
be high Z. You quote a voltage of 50V "in the tank coil". How are you
measuring this?
Can you monitor plate current? If so, that times the applied plate voltage
will give you your DC input power. Power out will always be less than
this amount.
Also, are you sure your HV supply has the current capacity for the 200-400W
you want to get out? (This may mean something like 300-600W, or more, DC
input power).
Are you taking steps to prevent radiation of the generated RF into the wider
environment? The level of power you are talking can reek havoc with other
licensed users. The FCC could pay you a visit.
Finally, BE CAREFUL! These DC voltages can kill, and you have not lived
until you have gotten an RF burn from 100W, much less 300-400W.

Bob WB0POQ
wrote in message
oups.com...
Hi,

I am trying to build a tube power oscillator running at 25MHz from a
Russian tube (GI6B), and although I have it running, I can't seem to
ramp up the power to anywhere near the level I want out of it.

Hi Bob thanks for the attention,

Re the load, I am using a soft iron 'poker' which I bring close to /
into the tank coil to provide a load which I can easily tell when
power is delivered into. So I fire the circuit up with the scope
connected to a certain point, then increase the load by bringing the
poker closer to or into the coil. The iron looks like a very close to
resistance load, the Z varies with the proximity. So I cat test all
loads from high Z to a few ohms. If the iron gets hot, I am delivering
a decent amount of power - got some decent power with a circuit like
this running at 7 MHz with the iron getting very hot.

While doing this I monitor the voltage on the plate or tank coil with
a scope. So the 50V is the best peak to peak voltage across the tank
coil.

Loading the circuit more reduces the pp voltage in the tank. And since
the tank Z is about 200R, p=e(squared)/r so p=1.6 watts in the tank
circuit.

Plate current is around 100mA but this is mostly dumped into the anode
as heat.

PSU is half amp MOT plus variac, bridge rectifier and 10uF cap. Shunts
are removed from MOT. Plenty of power available, about 1kW.

As for shielding, when running at full power the whole thing will be
screened including the load. Till then though, the tops off although I
dont think that much will radiate from the exposed leads even when the
power gets up to decent levels.

As for RF burns, yes I am rather nervous about that too... I am used
to working around up to about 20kv but this RF angle is a new thing,
so I am being cautious. Painful experience is the best way to learn
your lesson and I have had it :-(

Anyway thanks for the suggestions.



On Jul 25, 10:09 pm, "Bob Liesenfeld" wrote:
I think the fact that you do not have a load on the output tank, is what is
limiting the power output. Power tubes require a particular load Z to
deliver 'rated' power. Right now all I see is a parallel LC tank which will
be high Z. You quote a voltage of 50V "in the tank coil". How are you
measuring this?
Can you monitor plate current? If so, that times the applied plate voltage
will give you your DC input power. Power out will always be less than
this amount.
Also, are you sure your HV supply has the current capacity for the 200-400W
you want to get out? (This may mean something like 300-600W, or more, DC
input power).
Are you taking steps to prevent radiation of the generated RF into the wider
environment? The level of power you are talking can reek havoc with other
licensed users. The FCC could pay you a visit.
Finally, BE CAREFUL! These DC voltages can kill, and you have not lived
until you have gotten an RF burn from 100W, much less 300-400W.

Loading the circuit more reduces the pp voltage in the tank. And since
the tank Z is about 200R, p=e(squared)/r so p=1.6 watts in the tank
circuit.

Where are you getting the "200R" value from? (And what is 'R'?)
Another thought is to change the L/C ratio which should give a different
loaded Q with a given poker position. This may provide more power out.

On Wed, 25 Jul 2007 21:22:56 -0500, Bob Liesenfeld wrote:
Loading the circuit more reduces the pp voltage in the tank. And since
the tank Z is about 200R, p=e(squared)/r so p=1.6 watts in the tank
circuit.

Where are you getting the "200R" value from? (And what is 'R'?)

200R is British for 200 ohms.

(which does seem a bit low for a proper match, for 1kv plate voltage and
~300w RF out shouldn't the tank Z be closer to 3,000 ohms?)

(or is my sleep deprivation showing?)

Well im not completley certain - heres some of my maths.

Far as I have read, for Class C operations,

Load Z should = Plate voltage/2 * Plate current

And since I am aiming for plate current of 300mA or so (for power of
300 watts at 1kV) then thats

1000/2*0.3 = about 1500 ohms (R).

Then tank Z should = load Z / Q

For a Q of 7.5, which is slightly below average I think, thats
1500/7.5 which is 200R tank Z.

Then via Xl = 2piFL, Xl is 2 * 3.14 * 25M * 1.3u which = 204R
and
via Xc = 1/2piFC, Xc is 1 / ( 2 * 3.14 * 25M * 30p) which = 212R

So the tuned circuit is going to buzz at a slightly higher F than
25MHz but only slightly, at a Z of about 208R. I think stray values
are larger than the difference here so this is not worth worrying
about.

Thats my understanding of the maths having studied it over the last
few weeks and I am pretty sure its in the right ballpark, although the
Z could be a bit high or a bit low for the tube conditions. But since
the thing buzzes at 25MHz the F is right, and if the Z is a bit off
this just means the Q is a bit off and I understand this is not a
critical parameter for power.

Also to back this up, the tank Z and tank coil and tank cap values are
about the same as you see in published designs for this tube (GI6B),
this power level, and this frequency (eg http://www.nd2x.net/yu1aw/1xgi7-6m-schem.GIF
- its a GI7B but apparently they are almost identical). Ive compared
with a few circuits for these tubes and they are all similar.

Only thing of course is that they all use grounded grid, which is not
anything like the grounded cathode I am using.

Of course if I have my tank Z wrong, this would surely cause power
problems. But as I say, I cant see anything wrong with these values,
but maybe thats my lack of knowledge...

Thanks for the help so far...

On Jul 26, 4:35 am, Doug Smith W9WI wrote:
On Wed, 25 Jul 2007 21:22:56 -0500, Bob Liesenfeld wrote:
Loading the circuit more reduces the pp voltage in the tank. And since
the tank Z is about 200R, p=e(squared)/r so p=1.6 watts in the tank
circuit.

Where are you getting the "200R" value from? (And what is 'R'?)

200R is British for 200 ohms.

(which does seem a bit low for a proper match, for 1kv plate voltage and
~300w RF out shouldn't the tank Z be closer to 3,000 ohms?)

(or is my sleep deprivation showing?)