DC-DC buck Converter Tutorial  - Page 1
Forum section: electronics / beginners

author: Bryan
author_rank: Frequent Contributor
timestamp: August 25, 2014, 09:55:36 am
content: Very good video on a buck switched mode power supply.http://youtu.be/CEhBN5_fO5o?list=UUosnWgi3eorc1klEQ8pIgJQ

author: jGalt
author_rank: Contributor
timestamp: August 26, 2014, 05:42:06 am
content: This brings up a question of mine; the basic premise of the DC-DC buck converter that he outlines is that an LC lowpass filter will average out a step input, resulting in a DC output of a lower value. My question is why an LC lowpass filter? What makes an LC lowpass better than an RC lowpass filter, for this application?

author: Kremmen
author_rank: Super Contributor
timestamp: August 26, 2014, 06:16:03 am
content: Because the inductor is an energy storage device whereas the resistor is an energy dissipation device. With RC filter you will always turn part of the transmitted power into heat but with an LC filter you don't (excluding the unavoidable losses).

author: jGalt
author_rank: Contributor
timestamp: August 26, 2014, 06:29:02 am
content: Of course, thank you.

author: fubar.gr
author_rank: Unknown
timestamp: August 26, 2014, 09:39:38 am
content: Also an LC lowpass filter has a steeper cut-off than a RC filter.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 26, 2014, 02:48:49 pm
content: Actually, a better question is:Why not an RC or LR filter?The RC filter drops DC, which is bad. Or, if you omit an explicit R at all, the Rds(on) of the switches comes into play, which is even worse. If Fsw << 1 / (Rds(on) * C), the switches have to charge and discharge the C fully, on each switching transition -- and probably, smoke shortly follows! And, needless to say, no actual filtering is accomplished, you still get a square wave output. If Fsw is much greater instead, then the output ripple voltage might be low enough to be useful, but the transistor(s) are burning up in just a few cycles now, dissipating orders of magnitude more power than a basic linear regulator would.So, RC sucks. Scratch that.But what about LR? That's totally a thing.In fact, you can think of the LC filter as an LR filter, with extra C bypass to help it along. The L works, because the transistor(s) are happy switching into a load that doesn't draw huge current spikes. The load is happy...ier, because it's not seeing a raw square wave, compared to the RC case. But a pure resistive load sees ripple voltage = ripple current * R, which is probably still a lot. To get a smooth output, you need quite a lot of L, to get the ripple down. But that's expensive.For loads where ripple voltage doesn't matter, this can be advantageous -- high power LED drivers, for example. Here's one:http://seventransistorlabs.com/Images/LED_Light2.pngThe FJPF13009 is the switching part of the buck converter, and the UF5404 is the flyback diode. The series 1.8mH choke feeds a bunch of LEDs, with a current sense resistor at the bottom. As long as current doesn't drop to zero, the LEDs stay fully lit. (The light intensity will always flicker at the switching frequency anyway, whether current drops to zero or not, so this doesn't even matter, really. As long as it's fast enough not to see, it's not a problem.)Why not do this for most circuits? Because most circuits are more sensitive to ripple voltage, and most circuits require a low impedance at high frequencies -- bypass caps. Because the cap soaks up the extra ripple current, you can use a smaller and cheaper inductor, too.Tim

author: Pedram
author_rank: Regular Contributor
timestamp: August 26, 2014, 03:46:10 pm
content: What about a cap with a diode ?HighVoltage-->Switch--->Diode--->LargeCap---->LowVoltage(Unregulated?)| || ||------------<-------FeedBack<----|PWM Controller

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 26, 2014, 11:18:57 pm
content: Not sure how you envision the switch, diode and cap working together...There is such a thing as a switched capacitor converter, but it works by charging one or more capacitors to a given voltage, and rearranging them so that the voltage is physically moved to another set of pins. The voltage can be increased or decreased by a fixed ratio, inverted, etc. An arbitrary ratio cannot be made with high efficiency, because you can only stack capacitors in series or parallel to achieve the voltage conversion.A switched capacitor converter (also called a charge pump) operates at high efficiency if the ripple voltage is small, so that the voltage drop across switch resistance is small.A charge pump built with bidirectional switches (usually CMOS) is also bidirectional, in that it implements a voltage ratio, period, regardless of the direction of average current flow. Thus, a "step-up" device also serves as step-down, as long as the control logic is able to start up.There are "regulated" devices, but these work by skipping cycles, or dropping voltage across higher resistance switches, etc. The efficiency is low.Tim

author: Harshadb13
author_rank: Contributor
timestamp: March 21, 2019, 03:36:58 pm
content: Someone please help me with this.I have a doubt on conservation of charge in a buck converter.Even in a practical buck converter with non 100 efficiency the input current is less than the output. So more charge flows out than it comes in.. so how does this satisfy charge conservation

author: Kasper
author_rank: Frequent Contributor
timestamp: March 21, 2019, 04:06:35 pm
content: Quote from: Harshadb13 on March 21, 2019, 03:36:58 pmEfficiency and conservation are based on power, not current.Power = voltage * currentPowerIn * efficiency = PowerOutVoltageIn * CurrentIn * efficiency = VoltageOut * CurrentOutFor a buck, VoltageIn is higher than VoltageOut so CurrentIn can be lower than CurrentOut while still having similar PowerIn and PowerOut.

author: David Hess
author_rank: Super Contributor
timestamp: March 22, 2019, 12:03:09 am
content: Quote from: fubar.gr on August 26, 2014, 09:39:38 amThe 12dB per octave roll-off is nice but it comes with 180 degrees of phase shift which complicates the frequency compensation. Current mode control removes the phase lag of the inductor making it a popular control scheme.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 22, 2019, 10:44:40 am
content: Quote from: Harshadb13 on March 21, 2019, 03:36:58 pmThe difference flows through the ground connection, of course.This is particularly marked in inverting converters (buck-boost, Ćuk), where the ground current is the sum of source and load current magnitudes.Tim

author: Harshadb13
author_rank: Contributor
timestamp: March 20, 2021, 11:44:50 am
content: Okay thanks a lot!Extremely sorry for the late reply

author: Harshadb13
author_rank: Contributor
timestamp: March 20, 2021, 11:45:53 am
content: Yes, very good point. Your mention of inverting buck really forced me to think hardOkay thanks a lot!Extremely sorry for the late reply