In this thread we will discuss what are the switching losses, Causes of it. We will also derive an equation for Switching losses for Resistive Loads.



In Power Electronics we mostly work on switch mode power supplies due to their better efficiency and smaller size, they have very less power losses in terms of heat as compared to conventional power supplies but still their are other losses which need to get removed.

What do we Need Switching in Power Supplies?

In uncontrolled Rectifiers output cannot be controlled by any means but in Switch mode Power supplies output can be controlled by Switching the thyristors at different duty Cycle, Switching is not always only used in Power Supplies it has various different Applications.


What are the Switching Losses?

When device is turned off, for instance lets say MOSFET was conducting and suddenly it is turned off by removing Gate Voltage, So MOSFET internal Resistance will be maximum and Voltage drop across MOSFET will be equal to Applied Voltage and Current will be zero, Ideally this should Happen in no time but Practically Every Device take some time to turn off itself and this time Toff Depends upon Nature of device. During this device Turning off period Toff voltage and current levels are not zero which we wanted to be zero, so this unwanted voltage current levels are actually Power Losses because we have turned off device yet device is not completely turned off. 


Turning ON and OFF Profile for Resistive Load

Same scenario goes for Turning ON the device but in this case Current rises and voltage drop accross device is decreased.

Now we want to calculate the Switching loss so we have to calculate it separately for both Turning ON and Turning OFF.


Device Turning OFF Profile
In this Graph Power Losses can be calculated by integrating V x I in the Period Toff




 Eq..1


Woff= Power Loss during Time Period T off.
V= voltage drop accross Device
i= current flowing through device

We cannot two different variables in single integral, so we have to transform these variables into one variable 't' .

we will use straight line Linear Equation 


y = mx + b,
                                                                                     


where m= slope 
 b= constant or offset
                                                                    
m=\frac{y_2-y_1}{x_2-x_1}.

Now Applying Straight line equation for Voltage




Now as you can see in Picture at Point Y2 Value is Vbus, at Y1 Voltage is Zero. 
x2-x1 = Toff 
b=0 Offset
x= t     because x axis is time 
y=V      because we have considered Voltage value on Y-axis

so Straight line equation would be


                                   Vbus-0 x t + 0
                             V=   ---------
                                      Toff

                             V= Vbus x t / Toff


Now Applying for Current I:



Note: In Current's case the offset 'b' will not be Zero it will be Imax, because current was initially Imax and after Toff became Zero so initial Offset was I max

Y= i       Because straight line is of variable I 
x=t        because time is plotted on x-axis

                                   (0 - Imax ) x t + Imax
                               I=  ------------
                                        Toff

                                I= Imax (1 - t/ Toff  )


Now Put these results in Eq no 1 and Integrate Woff by yourself, Its easy still if you need help regarding any step ask Below in comments



As you can see above we have integrated V x I through time Toff. Note that Imax, Vbus and Toff are constants in the equation, the only variable in equation is 't'.

The above Equation [J] is the expression for Switching losses in Resistive Loads, please note that it is expression for Turning off losses, Turning ON loss will have same expression but it will have Ton instead.

For Example Question solved Click below



More on Switching Losses , how to reduce them and their Practical Scenario Based Problems, Designing of RCD Snubbers will be posted Soon.                                                        

Post a Comment

  1. Noman Ashraf i am very happy to see your blog........And best wishes for your informative work.....And keep doing this kind of activities......i really appreciated that....and you explained it very very easiest way for seekers...nice :)....

    ReplyDelete
  2. noman can u plz upload the solution of numericals of power

    ReplyDelete
    Replies
    1. Now i have uploaded the Solution 1 check here

      http://www.perpendicularstuff.com/2015/10/switching-losses-calculation1.html

      Delete

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