Buck Boost Converter -- how they step up the voltage - Page 1
Forum section: electronics / beginners

author: wael
author_rank: Contributor
timestamp: May 28, 2016, 08:29:59 am
content: Hi,Could anyone explain how the buck boost converter steps up the input voltage, for the boost converter it's clear that the inductor is in series with the voltage source so they add up and the voltage is stepped up, but in buck boost converter they are in parallel, how does the duty cycle affect the voltage to be stepped up, my first assumption they would have the same voltage (Vin, Vout). , Where did the extra voltage came from ?Regards,Wael

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 28, 2016, 08:32:40 am
content: They are equal [in magnitude] at D = 0.5.Draw the inductor waveform for D != 0.5. If the "on" half is +Vin for D of the cycle, what must the 'off' voltage be, for (1-D) of the cycle?Tim

author: ebclr
author_rank: Super Contributor
timestamp: May 28, 2016, 08:42:22 am
content: Take a look on this video, it's a explanation of what os going on

author: wael
author_rank: Contributor
timestamp: May 28, 2016, 08:46:29 am
content: Quote from: ebclr on May 28, 2016, 08:42:22 amYes, i understand the explanation for the mentioned circuit, but not for the circuit diagram posted in the op, they are different circuit diagrams.

author: wael
author_rank: Contributor
timestamp: May 28, 2016, 08:48:52 am
content: Quote from: T3sl4co1l on May 28, 2016, 08:32:40 amCould you please explain it in a more simple way

author: Ammar
author_rank: Regular Contributor
timestamp: May 28, 2016, 09:31:29 am
content: https://en.wikipedia.org/wiki/Buck%E2%80%93boost_converterIt is a little hard to answer this question since power converters are a broad topic. We did a whole subject on them at Uni in fourth year. We did SPICE simulations, built circuits, measured waveforms etc.I can try to provide some intuition though. Think of an inductor as having a high voltage across it when the current through it suddenly changes (v(t) = Ldi/dt). If the switch is in the off state then the inductor current "sees" the load through the action of the diode. If the duty cycle is say, 80%, the inductor current will "see" the load for 20% of the time, or a short amount of time. The "spikier" this current spiking is, the higher the voltage.I hope that helps. I can see why you might be confused. Since everything appears to be in parallel with the source, it may be tempting to think that everything sees the source voltage. However, notice that the diode forces the inductor to conduct during the on state, while not passing current to the load. In the off state, the inductor's voltage actually adds to the load, due to the reversed polarity of the output, due to the diode.

author: ebclr
author_rank: Super Contributor
timestamp: May 28, 2016, 10:47:39 am
content: Induct or likes to maintain the current stable, if you open the circuit they will generate a voltage on the opposite signal to try to maintain the voltage, your diode will get that and generate the negative voltage, this phenomenon is the same that need to put a diode on a relay

author: wael
author_rank: Contributor
timestamp: May 28, 2016, 11:13:08 am
content: Quote from: Ammar on May 28, 2016, 09:31:29 amBut why this doesn't happen with the buck converter , in the off state they (buck boost and buck converter) have the same circuit/current path.

author: ebclr
author_rank: Super Contributor
timestamp: May 28, 2016, 11:34:04 am
content: They key point is to undestand that the inductor generates two voltages , one on the turn on an another on the turn OFF, the diode is to get only the one you want, the turn one or the turn off one