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Topics: BOOST CONVERTER HELPS on Power Supply
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Deepak Chikane
12-17-2013 03:12 AM


Dear all.

i have been confused in understanding the working of boost n buck converters....

i am trying to make a project for boost converter.
vin=2v dc,v out=14.4 vdc,i out=2amp,switch freq=100khz

i have been calculated the values for inductors & capacitors by uding some texas application notes..

basic thing is that, i have not deepely understand the concept of boost n buck ....

in boost regulator,
if mosfet is triggered by gate then equation will become vin-vboost (l)-rds(on)*current....

pls correct the values for me.. vin=2v, vboost=2v n rds on is negligible..

wher is voltage to boost to 14 voltage..??
pls help me with links/articles /good examples to understand the same
12-17-2013 05:50 AM
Top #2
12-17-2013 05:50 AM
It is the inductor that stores the energy and support the difference between input and output voltage. When the MOSFET is turned on, voltage across inductor is same as input voltage. When the switch is turned off energy stored in inductor is used to develop the voltage across inductor. In steady state this voltage is equal to the difference between output and input voltage. Remember that current through inductor cannot change instantaneously but voltage can change. Voltage across any inductor is developed due to rate of change of current through it. The relative polarity of voltage across inductor is positive when dI/dt is positive and voltage is negative dI/dt is negative. In boost converter, when you apply input the current through inductor increases storing the energy in it. When the switch is turned off energy stored in inductor starts flowing out of inductor and there by current starts decreasing. This makes the voltage across inductor negative. The magnitude of the voltage is governed by rate of change of current. Faster the rate of change of current higher the voltage will be. If off time is very small compared to ON time, the current decreases at faster rate (because in steady state, current must be same at the beginning and at the end of a cycle) therefore voltage across inductor is higher and net output is also higher than input.

You can find several references on boost converter. I think you are looking for simplified physical understanding, that I tried to give.

The operation of switch mode power supply can be explained and understood with clarity if you think from point of view of energy balance.

If you wish to understand, how the output build-up to desired steady state voltage, do transient simulation on ORCAD. You can easily see the build up of output voltage and voltage across inductor building up till a steady state is reached.

You can contact me by email if you wish for more help.
12-17-2013 08:16 AM
Top #3
Robert Piqué
12-17-2013 08:16 AM
According to the explanation of Nayan.

From another point of view, a boost is a "charge pump", so that, at steady state, the output capacitor is charged in proportion to the integral of the average value of the current injected by the inductor, to stabilize the output voltage at a constant value by the dissipativity of the output resistor (or load that feeds the boost!).
12-17-2013 10:59 AM
Top #4
Marty Brown
12-17-2013 10:59 AM
That is one heck of a power supply design. I would say almost impossible. For a two V input, the peak primary current would range from 45A (very continuous) to 78A (discontinuous). The primary AWG would be around #8 - 10 and the input source better be extremely low impedance (along with the interconnections). That these currents, a very low RDS(ON) MOSFET (or multiples in parallel) would be the only power switch choice. Any semiconductor switch with a junction (IGBTs, etc) would have a very large output switching voltage, comparable to the input voltage or higher.
This project is do-able, but its volume/weight/cost would be greater than 5x of a 28W flyback converter from a higher input voltage. Plus you would need an external power source for the AUX supply (to power the controller and gate drives). I would go back to whoever wrote the specification to see if there are any other higher input sources that can be used. This is a systems-level consideration, since some of the system design may need to be redesigned to accommodate this change.

Good luck Deepak. This is almost impossible for seasoned power engineers, but especially for someone who hasn't done this before.
12-17-2013 01:47 PM
Top #5
Edward Herbert
12-17-2013 01:47 PM
It may help to understand a boost converter if you realize that it is a buck converter run backwards. I am not sure why this design should be such a problem, most would not consider a 14.4 V to 2 V, 15 A buck converter to be that challenging. Except for the direction of current flow, it's the same. (Am I missing something?)

A multi-phase design might make sense, or a two stage design. Either would reduce the peak current.
12-17-2013 03:59 PM
Top #6
Ray Ridley
12-17-2013 03:59 PM
Marty - you might have missed on the input current calculation there. I come up with just 15 A on the input.

The challenge is to get the 7:1 step up, but that is just about doable.
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