Not exactly tube related but I intend to use this regulator for a tube preamp. When I use the supplied calculator and input 330 v as the desired output (I think the input will be around 350), the spreadsheet says Output voltage adjustment range is 475 high to 375 low. First question, how can that be if I specified 330v output? Second question, heat sink for Q1 say thermal resistance of 5.32. Since I'm a tube guy, not a SS guy, can anyone direct me to an appropriate part number (Mouser or DigiKey)?
The MOSFET is going to burn most of the excess energy. If you have 390 unregulated, and are aiming at 330 regulated out @80mA you need to vent 60V * 0.08A = 4.8Watts
The math works like this
T(junction) - T(ambient) = Power Dissipated * Thermal Resistances.
Where Thermal resistances are that of the junction to case, case to heat sink and heat sink itself.
To make it easy, just use DK part number HS276-ND! Get the mounting kit as well.
The math works like this
T(junction) - T(ambient) = Power Dissipated * Thermal Resistances.
Where Thermal resistances are that of the junction to case, case to heat sink and heat sink itself.
To make it easy, just use DK part number HS276-ND! Get the mounting kit as well.
Last edited:
Thanks, Jackinnj. I'm planning about 350 in and 330 out so 20v of compliance at around 30ma = 0.6 W. So a 1 watt heat sink should be sufficient?
With the 21st Century Maida circuit board, I deliver a spreadsheet for calculating the heat sink. This "calculator" takes into account mains voltage variation, mains frequency, capacitance of the reservoir cap, etc., thereby taking the burden of doing the math off of you.
Tom
Tom
Tom, thanks and yes I have the spread sheet. My original question was how to translate "thermal resistance" into a physical product. As I noted, I'm not used to working with SS stuff...
So, I then tired to figure the amount of watts because I've seen heat sinks specified by watts.
So, I then tired to figure the amount of watts because I've seen heat sinks specified by watts.