Loss calculation in a switching Regulator rms current vs Avg Current

Thread Starter

hoyyoth

Joined Mar 21, 2020
74
Dear Team

I was going through this video which explains the losses in a switching regulator(Basic Buck Converter). In this the author is calculating only resistive losses and diode losses.

For calculating resistive losses he is considering RMS current through the resistor and for calculating diode losses average current is taken to consideration.May I know why is it so.

How to decide when to consider average current and when to consider the average current.

Screenshot 2021-06-01 at 10.58.29 PM.png
Regards
HARI
 

crutschow

Joined Mar 14, 2008
27,213
If the voltage drop is an ohmic resistance (linear with current) then you use RMS current, since the power proportional to the square of the current (RMS is the square-root of the sum of the squares).
If the voltage drop is essentially constant (such as a diode) then you use average current, since the power is proportional to the current.

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Thread Starter

hoyyoth

Joined Mar 21, 2020
74
If the voltage drop is an ohmic resistance (linear with current) then you use RMS current, since the power proportional to the square of the current (RMS is the square-root of the sum of the squares).
If the voltage drop is essentially constant (such as a diode) then you use average current, since the power is proportional to the current.

Edits in Italics:
Hi Curt,

Thank you.

If we have a constant voltage source and need to find the power across it always use average current.Correct me If I am wrong.

For any linear device we can use RMS current.If I need to find the loss due to MOSFET Rds then I need to use RMS current.

Hope my understanding is correct.

Regards
HAri
 

vanderghast

Joined Jun 14, 2018
55
The general loss in a MOSFET is more complicated.
For low frequency, you are right, when the MOSFET is blocking, then the current is zero and the power, still current times voltage, is zero. And when the MOSFET is conducting, it is the difference of voltage between the drain and the source, generally called the voltage at the MOSFET, times IC which matters. So, you are rigth using the avarage.
But at high frequency, when you often switches from passing to blocking, your MOSFET come from (low voltage, high current) to (high voltage, low current), in a non zero time. If you draw, on the same graph, versus time (horizontally), both the current and the voltage (vertically), that makes an X. And the product of the two is NOT zero. Each times you switches ON to OFF or OFF to ON, that relatively small loss is repeated. And that may become the MAJOR source of loss if you do it 1000 times, 10 000 times, or 1 000 000 times a second. It is dependant of the switching time, and generally, the datasheet supplies a graph for it, versus the frequency (and the duty cycle, if they add both "types" of loss).
 
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