# Full Bridge VSI problem

The above problem was asked in GATE(Graduate Aptitude Test in Engg) 2015 in India. I have no idea where to start so I can't show my attempt. But through google, I got solution by 2 different coaching institutes

First one - 56.72

Second one - 49.5

IIT(Indian Institute of Technology), the governing body of exam says that the answer will be in the range 60-64V

• If you really "don't know where to start" you better crack open your text books and start reading. :) – Tyler Jan 18 '18 at 13:37
• @Tyler your suggestion helped, was able to figure it out on my own :) – Nikhil Kashyap Jan 19 '18 at 18:34

Both the solutions are crap !!

VR = 0.7 * VDC = 0.7 * 100 = 70V ........Case of full bridge inverter

where VR is peak value of fundamental component

Vo = VR * [(-JXc||R)/(-jXc||R + jXl)] .......Voltage divider

On solving we get peak of fundamental component of Vo = 62.76V

• The mod. index definition is not declared by law, it can be different in different books. Generally it is one of the measurig numbers for how usefully the DC is splitted to 50kHz pulses when compared to what is got by splitting the available DC directly to 50Hz square pulses in the used switch configuration. One definition: Mod index = the amplitude of the resulted 50Hz component/the amplitude of the 50Hz component in 50Hz square wave. In your case the denominator would be 100V * 4/Pi = 400V/Pi = 127V and the resulted 50Hz component = 89V, not 70V. I haven't your book for comparison. – user287001 Jan 19 '18 at 10:27
• @user287001 this solution gives result in the range 60-64V which is ok with the official answer key released after the exam. There might be different definitions for mod index but I think this is what the examiners wanted to make things simple for students. – Nikhil Kashyap Jan 19 '18 at 10:49
• @user287001 please take a look at this problem electronics.stackexchange.com/questions/351084/… – Nikhil Kashyap Jan 20 '18 at 13:06