Boost Converters - DC to DC Step Up Voltage Circuits

This electronics video tutorial provides a basic introduction into boost converters - circuits that can step up the voltage of DC sources such as batteries and solar cells. The electronic components needed are an inductor, a switch, a capacitor, a voltmeter, and a diode.

Light Emitting Diodes:

Zener Diodes:

Varactor Diodes - Voltage Capacitors:

Generate Electricity With Crystals!

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Simple Voltage Booster Circuit:

100,000V High Voltage Voltmeter:

Buck Converter Circuit:

Low Pass and High Pass Filters:

RC Band Pass Filters:

RLC Band Stop and Band Pass Filters:

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NPN and PNP Transistors:

Final Exams and Video Playlists:

Full-Length Videos and Worksheets:

[marthareddy]

Excellent teacher and educator. When he mentions that there is no free lunch, he wanted to educate the conservation of energy, very rarely others may have touched upon. 🙏🙏

[naamaratsabi]

I wish u were my teacher.. none of my proffesors explained material in such a simple, clear and gentle way!

[andre_c_almeida]

thank you so much, just by saying that the interruptor is there for shorting the circuit i finally understood this

[dinosaur]

You know how to make things soooo easy for others to understand...👍☺

[Zeioth]

Best explanation on the internet. Thank you so much.

[aymanehar]

thanks a lot, I was looking for this video for more than a week.

[Criiimesdowntime]

Oh hell yeah, I was wondering how you'd make thr mosfet do that. AC current, I love this channel ❤ learning all this as a hobby

[CAJO]

Thank you, I don't know if I learned more about eletronics how to teach, or how to communicate

[Flmask-xhlb]

good video!Explain clearly than anyone!thx

[hillaryaura]

Nycc, i think of coming up with a website where my clients will be able to see these tutorials, my views will be of benefit to you, thats what i can attribute to your hard work.thanks again 👍👍👌

[ChrisDIYerOklahoma]

Great video. I use those sealed marine grade 12v to 36v (5a) step up boosters to convert 12v battery power to power Bluetooth digital chip amplifiers in my speaker builds. They work flawlessly. Here's my question, why do my amplifiers (running at 36v) sound better (more power, fidelity) vs using their stock (included) AC/DC adapters that come with these little digital chip amplifiers? I don't have an audio dyno or distortion tester...but, I can "hear" and "see" the difference when the woofers on my speakers are pounding harder. Thanks.

[dramatalk]

1:52 why it is small current? even switch off whole current will flow through inductor L?

please reply

[nodriveasusephotos]

Would there still be an increased voltage output without the transistor? Perhaps using the most powerful inductor and highest useful frequency. Thanks Alan

[frankbaron]

simle but effective. you need a hi tolerent diode though

[david-tracy]

How is it determined how much energy flows to the capacitor & voltmeter ?

Thx

Great videos

[robanada]

Nice explanation on boost converters, thanks for making this video. I was confused about one part, though--at 7:38, you say that increasing the frequency applied to the mosfet gate will increase the dI/dt and hence V across L. I was thinking about how all of the (various few) buck/boost converter designs I've seen use PWM control, and not FM control. Wouldn't dI/dt depend on the rise time of the MOSFET rather than the switching frequency? Seems like a square wave would give you the minimum rise time and hence maximum dI/dt.

[madhub]

How the circuit will work/functional? and How the voltage and current waveforms at the output?, If we keep the load as a inductor or capacitor instead of resistor?. Please could you answer my question?.

[ahmedkchok]

it is possible to create a self current generating circuit using this?

[gorilladisco]

So, the switch has to be closed then opened again?
Do we have to close-open-close-open continuously, or do we just do it once?

*I don't have electronics knowledge 😶

[medghozlani]

I think the current will be increased due to the charge accumulated In the output capacitor and the inductor current, not as you have explained because no way to add the current of the inductor and the input source when the switch is off ( circuit ON)