:Hi:
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:That is basically correct although I arrive at a resistance value somewhat empirically. I add a 10 ohm, 10 watt resistor in series with the diode, install the unit and check the value of B+. If it is high, I increase R. This may be trial and error but you may not exactly know the value of load current you are going to deal with. The resistor is sized from experience ( and I squared R plus some margin).
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:Lou
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::I'd like to make a solid state replacement for some 6X5 rectifier tubes that I have in a couple of signal generators that I have. I have a couple of questions and was wondering if anyone had any input. First the Diodes, depending on which tube you're using the regular, G or GT the PIV could be between 1250 and 1600 but the diodes (I'm using IN5408's) have a PIV of 1000V. Do I need 2 of these in series for each leg to get an effective PIV of 2000V or is that just overengineering in this case?
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::For the resistance to bring B+ down to normal. Do i just take the voltage drop (difference between normal B+ and the B+ with just diodes installed) and divide by the DC current which I think is about 750mA, to get the resistance. Then just I^2 R to get the wattage.
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