@ witsend
I tried to amend my last post as there were calculation errors. I have therefore reposted below: -
OK witsend, lets use your method. Firstly the 120V supply will drop 88V across R1 and 32V across the coil (based on the coil resistance being rounded to 8R). Taking R1, P= Vsq / R = 7744 / 22 = 352W * 8% = 28.1W. The mosfet volt drop can be ignored as its RDS (Drain to Source resistance is very low). Taking R2, the inductor is the generator and therefore the source of the output voltage, akin to the supply battery. This is very important to understand because it explains the increased heat in R2. The full voltage stored in the inductor will be developed across R2 and the diode. The diode forward voltage drop is less than 1 volt, so can be ignored for this exercise. The inductor will charge to over 120V, so assuming 120V across 22R, P= 120V sq / 22 = 14400W / 22 = 654.5* 8% = 51.6W. In practice, much less power will be available from the inductor because of switching losses, so the real power dissipated in the resistor will be a lot less.
The input resistor is cooler than the output resistor because the input power has to be shared between R1 and the charging inductor. This is not the case at the output, because the inductor has already recieved its energy from the input and can therefore deliver virtually all its stored energy to R2.
Hoppy
I tried to amend my last post as there were calculation errors. I have therefore reposted below: -
OK witsend, lets use your method. Firstly the 120V supply will drop 88V across R1 and 32V across the coil (based on the coil resistance being rounded to 8R). Taking R1, P= Vsq / R = 7744 / 22 = 352W * 8% = 28.1W. The mosfet volt drop can be ignored as its RDS (Drain to Source resistance is very low). Taking R2, the inductor is the generator and therefore the source of the output voltage, akin to the supply battery. This is very important to understand because it explains the increased heat in R2. The full voltage stored in the inductor will be developed across R2 and the diode. The diode forward voltage drop is less than 1 volt, so can be ignored for this exercise. The inductor will charge to over 120V, so assuming 120V across 22R, P= 120V sq / 22 = 14400W / 22 = 654.5* 8% = 51.6W. In practice, much less power will be available from the inductor because of switching losses, so the real power dissipated in the resistor will be a lot less.
The input resistor is cooler than the output resistor because the input power has to be shared between R1 and the charging inductor. This is not the case at the output, because the inductor has already recieved its energy from the input and can therefore deliver virtually all its stored energy to R2.
Hoppy
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