Roger, For measuring the output power take a 50 Ohm dummy load (50, not 55 Ohms or so) and measure the peak voltage with an oscilloscope. Type this into the excel file and calculate the output power
The IRF510 gate voltage is set just below 3V with a resistive pot. The drive is not a square wave though, so I suppose this could be part of the issue? Hello Roger, Try higher gate voltage (> 8V) and
Don't even try biassing the FET with a pot. You must use a squarewave drive- preferrable from a proper MOSFET driver like an ICL7667 or TC4426 type, or at a push a stack of paralleled 74C drivers
Hi Roger, 40 Vpp is just normal if the power supply has about 12V! You can assume about 3,5...4 times the supply voltage for u(DS), if well matched! Most programs will give you just an imagination on
Hi Stefan, I thought the power was the RMS voltage squared (i.e. peak voltage X 0.707) divided by the resistance. Chris, G4AYT. -- Original Message -- From: [email protected] Stefan Schä
My 137.5kHz WSPR TX transverter is now working fine, so I'll be on-air with WSPR tonight (Sept 7th) from 5.30pm - 11pm UK time. After my success on QRSS3 with G3XIZ I'm hopeful he'll now decode my WS
Take a look at G3NYK's Class E design page http://www.alan.melia.btinternet.co.uk/classepa.htm Don't even try biassing the FET with a pot. You must use a squarewave drive- preferrable from a prope
Hello Roger, Maybe you can send us a circuit with its values. I have built class E PA for 2200/160/80/40/30/17/10m and have some experience. If you are using a IRF510 and stay in the QRP range (< 10
Hi Roger, 40 Vpp is just normal if the power supply has about 12V! You can assume about 3,5...4 times the supply voltage for u(DS), if well matched! Most programs will give you just an imagination o
Not sure if anyone else has had this problem, but I was using the small 16mm 3C90 cores (cream colour) as the PA bifilar output transformer and I had all sorts of issues with the darn thing overheati
Hi Chris, Yes, right. But my _expression_ is the same. 0,707 = 1/sqrt(2), so U^2/R = P =(û/sqrt(2))^2/R =û^2/2/R =û^2/2*R. Measuring 100 Vp at 50 Ohm is 100 W :-) 73, Stefan Am 09.0
Hello Roger, Maybe you can send us a circuit with its values. I have built class E PA for 2200/160/80/40/30/17/10m and have some experience. If you are using a IRF510 and stay in the QRP range (< 10
