Home |
Search |
Today's Posts |
#3
![]() |
|||
|
|||
![]()
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. |
Thread Tools | Search this Thread |
Display Modes | |
|
|
![]() |
||||
Thread | Forum | |||
FA: Vintage 5 tube working exc. | Swap | |||
R-390A: feeding oscillator tube heaters in DC, w/o regulator tube? | Boatanchors | |||
FA: COLLINS R-390A TUBE Set, S-METER & Crystal Oscillator modules | Equipment | |||
FA: COLLINS R-390A TUBE Set, S-METER & Crystal Oscillator modules | CB | |||
FA: COLLINS R-390A TUBE Set, S-METER & Crystal Oscillator modules | Shortwave |