9/28/2023 0 Comments Cathode diodeThese all use active tabs that must be insulated with mica and silver oxide grease for electrical insulation, then you can earth ground the heatsink.īut worse, you have huge surge currents into a low ESR cap and huge dynamic range of DC for input tolerance and unspecified load tolerance. You won’t get 50Vdc out of this design as provided since the Vac peak voltage varies from 130 to 340Vp. At about $US0.25 per square inch it seems reasonable value for what it achieves. $US10c/in^2Īnd this magic material* achieves 0.013 degrees C / Watt per inch^2 - but does not mention electrical resistance - but is "probably" a usable insulator. This 3M thermal tape has a 2 mil / 0.05mm !!! version rated at 0.35C/in^2/Watt and 1500 V rated. Most materials designed for this application have a low enough degrees/Watt thermal resistance that while they should be accounted for, they are not major components in design decisions except in extremely high power applications. Use of a series "spreading resistor" to increase the conduction angle of the diodes may well be in order.Īs a "bonus", rectifiers with high current peaks can generate 'useful' amounts of EMI.Įach thermal material has a specification that can be used as part of the design calculation. When feeding rectified DC to a capacitor that is large enough that the voltage does not vary significantly across a power cycle means that the diodes will conduct only near the waverform peaks, and current may be in bursts of much (or very much) higher than average current and you need to calculate accordingly. If you have an AC 4 diode bridge then duty cycle is 50% or less and heating is half or less what is expected.ĭissipation can easily be much higher than expected if significant output capacitance and consequently low ripple voltage occurs. It may not quite glow in the dark, but it also will not last very long at all.Ī 10 C/W heatsink will give you 10 x 6 = 60C rise. Pd x Rth_JA = 6W x 50 C/W = 300 degrees C temperature rise. ![]() SO at 6 Watts dissipation with no heatsink temperature rise is Ryh-JC = 1.25 C/W (both leads) = junction to case (leads) Use of the two diodes per package in parallel will reduce the net wattage dissipated - but see below re peak currents. If insulating washers are used on all diodes then a shared heatsink can be used.ĭata sheet says Vf forward voltage at 8A is 0.85V typical.Īt 7A then dissipation = I x V = 7 x 0.85 ~= 6 Watts. As above, use of a suitable insulating washer is recommended. If they used separate heatsinks they could be electrically connected to the package tabs but this would be 'unwise'. This is not a "problem" if sufficient allowance is made for this, but it would be more usual to ground the heatsink and to use low thermal resistance insulating "washers" between the packages and heatsink.ĭ23 & D24 Cathodes and therefore package tabs need to be electrically isolated from any other electrical connection. Heatsink grounding is usual but may depend on application.Īs the packages have the Cathode connected to tab, in the circuit shown D25 & D26 could share an electrical connection to a common heatsink - but it would then be at Vout_positive = 55 VDC. ![]() Thermally conductive electrically insulating rubber pads are easy to use and well priced. ![]() There are numerous insulating thermal mounting systems available for this package. Heat sink isolation of the diodes is required OR individual heatsinks that "float" could be used, but this is less common. If insulation is not used, should the heatsink be connected to the rectifying diode's thermal tab terminal (cathode) through the PCB, through the heatsink's mounting screws? Do I need to use insulation to isolate the diode's thermal tab and the heatsink from each other? If yes, should I connect the heatsink to Ground?
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