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Common heatsink for IRD510's
#ubitx
This may have been answered already. The mounting tab on the IRF510 is internally connected to the DRAIN which is the output of the transistor (connected to the output transformer). Therefore would it not be hot both with 12V and also with RF; and if mounted on a common heatsink, wouldn't they be capacitively coupled together through the insulating material? Would this be a problem on the higher bands, like 10m?
I guess the way to check would be to measure power out on 10m with individual heatsinks, and then with a common heatsink. The dielectric properties of different insulators (mica, silicone, ceramic etc) could also affect things. 73? AL |
Not a problem.
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Impedance from drain to ground is quite low, my guess is the push-pull output transformer gives it 12.5 ohms because the center of the primary is effectively at ground.? (Some push pull amps power the finals through a center-tap which is at AC ground, instead of using the extra chokes at L8, L9). How much not a problem? Check this out:?? Capacitances around the IRF510 gate do mean it requires more drive at 30mhz than some of the other FET's.? But I've never heard of the drain-to-ground capacitances being a problem.? Allison recommends extremely short source leads to a good ground plane, and keeping the gate far far from the drain to avoid oscillations.? She clips off the drain lead and solders any drain connections directly to the tab. Jerry, KE7ER On Fri, Mar 23, 2018 at 11:43 am, Al Duncan VE3RRD wrote:
This may have been answered already. The mounting tab on the IRF510 is internally connected to the DRAIN which is the output of the transistor (connected to the output transformer). Therefore would it not be hot both with 12V and also with RF; and if mounted on a common heatsink, wouldn't they be capacitively coupled together through the insulating material? Would this be a problem on the higher bands, like 10m? |
That was my gut feeling, figured I'd best back it up somehow.
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Let's assume the IRF510's tab is 1/2" by 1/2", the insulating material is 0.003" thick, and the relatively permitivity of that insulator is 3.0 To convert to meters, divide inches by 39.37 So area = (0.5/39.37)^2 = 0.0001613 square meters, separation is 0.003/39.37= 0.0000762 meters. Capacitance =? k * e0 * area / separation =? 3.0 * 8.854e-12 * 0.0001613 / 0.0000762 =? 56e-12 Farads ? ?? Impedance of 56 pF at 30 MHz is? ? 1/(2*pi*Hz*Farads) = 1/(2*3.14*30e6*56e-12) = 95 ohms ? ?? So the 12.5 ohm impedance presented by the output transformer should swamp any effect that the 95 ohm capacitive impedance from IRF510 tab to ground. Jerry, KE7ER On Fri, Mar 23, 2018 at 12:21 pm, Jerry Gaffke wrote:
Not a problem. |
¿ªÔÆÌåÓýThe issue with not using the chokes is that now you are putting DC through the core in addition to the RF and adding to the saturation of the core from the DC component.? That¡¯s why you can get away with smaller cores for the RF portion if you route the DC through the chokes instead. ? ? Dr. William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ PJ4/K9HZ VP5/K9HZ PJ2/K9HZ ? Owner - Operator Big Signal Ranch ¨C K9ZC Staunton, Illinois ? Owner ¨C Operator Villa Grand Piton ¨C J68HZ Soufriere, St. Lucia W.I. Rent it: Like us on Facebook! ? Moderator ¨C North American QRO Yahoo Group. ? email:? bill@... ? ? From: [email protected] [mailto:[email protected]] On Behalf Of Jerry Gaffke via Groups.Io ? Not a problem.
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