Bill, I think you missed part of the discussion:we were not talking about tuned circuits. We were talking about an amplifier with no tank circuit being coupled directly (no transmission line) to an a
It is quite normal for an amplifier to show lower efficiency when it is run at lower drive and output levels. As the drive is reduced, the class of operation moves in the direction of class A and we
Where are you guys finding this item? When I search for that number I get a not found message. 73, Gerald K5GW In a message dated 4/22/2010 2:18:09 P.M. Central Daylight Time, keepwalking188@yahoo.co
While it is the plate voltage swing that generates the rf power, it is customarily accepted that the k factor takes care of that issue. Plate V times plate I divided by k factor gives the complete an
Yes, I agree (E / I ) / K. Mental lapse before, sorry. 73, Gerald K5GW In a message dated 5/1/2010 12:51:14 P.M. Central Daylight Time, k0wa@swbell.net writes: I think it is Plate V divided by (Plate
Hi Peter, Since you brought the conduction angle into the discussion, the Pi/2 number equates to 1.57. The customary shortcut is use of 1.8 for the K factor and I am assuming the difference between 1
Cam, there are voltage drops in the commercial power lines just as in your house wiring. the drop is related to the resistance times current. The farther you live from the power source, the more resi
Bill, with plate current that low, the plate load impedance is pretty high. To match a higher impedance requires less coupling C. Less coupling C means you need more tune C to keep the plate circuit
Vic and all, some rough calcs at 100 mHz shows close agreement with your results. Assuming the 12 uH number is a parallel equivalent, converting that to a series form results in 100 +j1.33 or 2.1 nH.
There was a write up in QST back in the early 70's about parasitic suppression in the cathode circuit. Maybe you can track it down and read all about it. I remember it because I was building up an am
K2AW, of Silicon Alley fame, once supplied 50W zeners. Wonder if he still has them? 73, Gerald K5GW In a message dated 8/11/2010 9:42:22 A.M. Central Daylight Time, dhallam@knology.net writes: It's g
Or, use two connectors, one for B plus and another for B minus. 73 de K5GW In a message dated 9/10/2010 5:32:38 A.M. Central Daylight Time, rezycle.bin@gmail.com writes: OK, thanks. Will consider a i
That is why the 500 or 1000 ohm resistors and/or back to back diodes are important. If the B minus and B plus connections are made via grounded SHV connectors with coaxial cables, there would be no d
In my view, a diode pair reverse connected from B minus to ground is the preferred combination. Back to back would look like an open circuit until one or both fail shorted. The inclusion of a series
Dave and all, the last SHV connectors I purchased were from Pasterrnak, the high priced specialist folks. They have SHV cable and mating chassis connectors for RG213/U coax. II wanted the security of
Jim, you are 100% correct about a single diode being sufficient for B minus surge protection. I guess we habitually say/use back to back or back to front or reversed connected pairs just in case we s
Hi Jim, I could not find any ratings for the P{asternak connectors. When I received them, I was a bit dubious of them handling enough voltage. The dielectric of the center conductor of the coax has t
Scott, Jim and all, it certainly is not as simple as grounded grid B minus protection but there are a couple of options. There are some hefty surge limiting devices available. One series was made by
Sorry Dave, it is still baffling! 73, Gerald K5GW In a message dated 9/27/2010 4:37:20 A.M. Central Daylight Time, mausoptik@btinternet.com writes: Hi Alek - yes, I think that for the "real radio tha
To test for emission, it would seem that the grid voltage would need to be zero or even positive to simulate the effect of the peak driving voltage. If the emission is normal this would cause a very