Maximum Output Current of Buck Converter - Page 1
Forum section: electronics / projects

author: martys
author_rank: Regular Contributor
timestamp: May 11, 2017, 01:29:30 am
content: I have a 12V 2.5A regulated wallwart switching power supply and need 2V at 6A.Is it possible for this power supply with some large value filter capacitor feeding a simple buck converter to deliver more than 2.5A?Although the wallwart may be able only to deliver 2.5A wouldn't the buck converter input capacitor make larger currents possible?Since 2V at 6A output is only 12W wouldn't the 12V 36W wallwart make this possible?

author: DBecker
author_rank: Frequent Contributor
timestamp: May 11, 2017, 02:16:54 am
content: The "extra" current comes from the freewheel diode, not a capacitor. The inductor stores energy while the switch is on, then continues to pull current from the diode with the switch off.

author: martys
author_rank: Regular Contributor
timestamp: May 11, 2017, 02:23:57 am
content: I didn't know you could pull current from a diode, I thought the catch diode could only act as a switch allowing output current to continue to flow when the pass switch transistor is turned off and current then flows continuously because of the stored inductor energy.During the off-time of the converter, the inductorpushes current through the catch diodeand at the same time this same current continues to deliver current to the output capacitor until the inductor is magnetically reset, thereby keeping the output voltage constant, and on/off intervals during each cycle of a fixed freq. convertor creates sawtooth ripple seen at the output, and the amount of ripple depends upon how large a value is chosen for the output capacitor.The switching transistor turns off by the control circuitry when the set current limit is reached through the series inductor.If the current limit of the converter is set at to 8-amp, then the input capacitor must supply this current in excess of the wallwart to reach this limit.If that is possible, that how do I calculate a value for the capacitor required, say if the inductor (lets say the inductor can be some uH value with a switching frequency of 200KHz) can be easily found that can handle/exceed this amount of current without saturating?

author: Leon23
author_rank: Contributor
timestamp: May 11, 2017, 08:19:53 am
content: Hi Martys,Perhaps you could add another DC/DC converter, switching from your input 12V to your desired 2V output with the current ratings you desire.Typically the buck converter current rating is limited with the switching elements (series transistor, typically P- or N ch mosfet), rectification stage (typically n-ch mosfet) or diode of non synchronous buck, inductor (both rms and Isat current values), and input/output capacitor rms current values.//Leon

author: martys
author_rank: Regular Contributor
timestamp: May 11, 2017, 10:47:56 am
content: I am choosing all components of my buck converter to be rated at >8A, but that is not the question I would like answered.The switching transistor turns off by the control circuitry when the set current limit is reached through the series inductor.If the current limit is 6-amp, then the input capacitor must supply this current in excess of the wallwart to reach this limit.If that is possible, that how do I calculate a value for the capacitor required, say if the inductor (lets say the inductor can be some uH value with a switching frequency of 200KHz) can be easily found that can handle/exceed this amount of current without saturating?

author: MagicSmoker
author_rank: Super Contributor
timestamp: May 11, 2017, 11:29:52 am
content: Quote from: martys on May 11, 2017, 10:47:56 amThere is considerable freedom in the choice of the values for all of the components in a buck converter.The input capacitor is sized on the RMS value of the ripple current vs. the desired value of the peak to peak ripple voltage. The equation used to calculate the input capacitance value in a buck converter (neglecting any contribution to ripple from ESR!) is:C > (Iout * (D * (1 - D)^0.5) / (Vripple * Fsw)Where D is the duty cycle of the switch (approximate 0.17 here), Vripple is the peak to peak input voltage ripple (1% of Vin is a good initial value), and Fsw is the switching frequency in Hz.Plugging the relevant numbers in gives a capacitance value of ~35uF, so using 47uF or more of either MLCC or polymer electrolytic capacitance should be fine.EDIT: forgot the square root operator in the numerator!

author: martys
author_rank: Regular Contributor
timestamp: May 11, 2017, 01:56:54 pm
content: Thanks MagicSmoker for a way to calculate the capacitance required at the input of the convertor.If wish someone could explain a little how this equation makes sense, see how the formula was derived from circuit analysis to better understand what I am doing.Very interesting, when I see a physicist analyze this kind of circuit, their approach starts talking volt-second integrals and coulombs etc being tossed around instead of volts and amps and farads.The formula given by MagicSmoker offers no explanation of its derivation at all, while the physicist's explanation requires me to get a master's degree to start to understand. At the same time my troglodyte brain figures it could cut and try different values and get something to work in minutes.Now's the remaining problem, since MLCC caps or polymer are expensive and MLCC have a negative coefficient of capacitance versus applied voltage,how could I calculate the values of achieving the required capacitance by using two types of capacitors to reduce the value of the MLCC capacitor to lower the cost of the convertor by choosing a lower cap. MLCC input capacitor and getting the rest of the required capacitance adding a much cheaper aluminum bulk capacitor in parallel.How much current does the 12V 36W wallwart actually need to supply to the convertor in question? Could I cut a corner here if I was gonna make a buncha these to save money?Is this supposed to be a different topic?? If so, then tell me and I will start a new topic.

author: Leon23
author_rank: Contributor
timestamp: May 11, 2017, 04:39:54 pm
content: I still don't understand what you're trying to achieve.You HAVE an existing 110/230 ---> 12.5V 2.5A wall pluggable SMPS. And now you need 2V 6A instead using the same HW? So your intention is to reduce the duty cycle of that specific converter and HW circuit?Or is the architechture so that you wish to have 110/230-->DC/DC -- 12.5V 2.5A ---> DC/DC -- 2.0V 6A?//Leon

author: martys
author_rank: Regular Contributor
timestamp: May 11, 2017, 04:46:37 pm
content: Use very common(low cost)120/240 VAC to 12VDC out wallwart as supply to buck down to 2V. Impossible(very expensive custom item?) to find 2V-8A wallwarts!

author: MagicSmoker
author_rank: Super Contributor
timestamp: May 11, 2017, 05:22:57 pm
content: Quote from: martys on May 11, 2017, 01:56:54 pmHmm... well, the formula I gave is really just the capacitor equation, which in its simplest form is: C = (I * dt) / dVIn this case, the numerator - along with Fsw in the denominator - gives the I and dt values, while the peak to peak ripple gives dV, of course.Quote from: martys on May 11, 2017, 01:56:54 pmI probably shouldn't waste my time looking up parts because who knows what you think is "too expensive" but Nichicon part number RNS1C470MDS1 is a 47uF/16V polymer capacitor with sufficient ripple current rating and costs $0.262 each at the 10 quantity...Otherwise, it is difficult to predict the division of ripple current among two vastly different (in value, ESR and ESL) capacitors numerically. Probably the simplest solution is to whip up a schematic in LTSpice with realistic values for the parasitics.Quote from: martys on May 11, 2017, 01:56:54 pmAs long as the efficiency of your buck converter exceeds 50% this wallwart is fine.

author: mbless
author_rank: Regular Contributor
timestamp: May 11, 2017, 07:01:57 pm
content: Quote from: martys on May 11, 2017, 01:56:54 pmA lot of questions you are asking can be answered by reading the data sheet for a buck converter, e.g.TI TPS54160. They typically walk through the theory of operation (at a high level) and show the calculations with an example. That should give you a starting point to learn the derivation of equations should you so desire. Do note that different manufacturers have different equations.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: May 11, 2017, 07:21:52 pm
content: I don't know why you're getting so hung up on the DC/DC converter's input capacitor...it's just a regular component, it has nothing to do with "supplying 6A from a 2.5A source", just pick an input capacitance suitable for whatever DC/DC converter you choose, the datasheet will have the necessary equations.You seem to be approaching switching DC/DC converters from "linear regulator land". Linear regulators maintain current. Current out = current in, and the reg burns off the voltage difference between input and output internally to regulate the output. Switching converters operate on a COMPLETELY different principle. They do not maintain current, they maintain power. Power in = power out + loss due to inefficiencies, at whatever voltage/current combination the system is designed for. You have a 30W source and a 12W load, which means as long as your converter as at least 40% efficient (plus some margin, call it 50%), you're fine. If you're putting out 2V and 6A, that's 12W. If the converter has an efficiency of say 80%, that's an additional 3W to run the converter, so it will be pulling 15W from the source. If the source is 12V, that's 1.25A, which is perfectly fine for your 2.5A supply.I recommend starting with TI's Webench tool. Put in your requirements, see what solutions it comes up with, see what values it chooses, etc. You can use one of those or not, but it'll give you an idea of what's required to make this work.

author: Codebird
author_rank: Regular Contributor
timestamp: May 11, 2017, 07:22:25 pm
content: There is another factor limiting the current output of a buck converter: Inductor saturation. High current inductors are expensive, a designer isn't going to get one rated for too much more current than they expect to need. As soon as your peak current hits that saturation limit, everything stops working.

author: martys
author_rank: Regular Contributor
timestamp: May 11, 2017, 08:05:44 pm
content: Codebird: From the waveforms I see on my scope a buck converter doesn't stop working when the inductor saturates, but the current into the buck conv. increases exponentially and dissipation in the transistor switch element and resulting heating of the catch diode and inductorand input and output cap. cause efficiency to drop off rapidly beyond the sat. rated current.I can see that saturation occurs not instantly but over an interval of increasing on-time past the point that the rated current rating is reached.

author: martys
author_rank: Regular Contributor
timestamp: May 11, 2017, 08:18:50 pm
content: Thanks Suicidaleggrol, but I my designs at present are using MCU's as controllers so TI may not apply. At the same time, it is certainly a good idea to study their approach to further my understanding.If all buck switching supplies are essentially the same, except for the control element, switching transistor and maybe some mfg's data sheets for their buck converter chips are more right than others and so I wonder which engineer has best made the capacitor choice formula.Consider this: If a wallwart is rated at 2.5A then exceeding this current output will cause its output to drop and the wallwart will shutdown into a latching off state to prevent damage. This means that the input capacitor must hold up the voltage to prevent this by storing enough power to prevent the wallwart's output voltage droop as the on-time of the duty cycle causes the inductor current to reach its peak current value.

author: Leon23
author_rank: Contributor
timestamp: May 11, 2017, 09:23:06 pm
content: Quote from: martys on May 11, 2017, 04:46:37 pmAlright. As stated earlier you could just use a common buck filter topology to switch the 12V (up to 2.5A) to 2V 6A. Assuming no losses, you will draw 12W which is equal to 1A from the 12V source.From now on it's a standard buck inverter design. What's your design target. Is it an high volume Product? is it a single test item? Should it be short Circuit protected etc? Without proper information I can only give poor advice, but you should be able to use a standard buck inverter with ceramic input/output capacitors, a suitable inductor with higher Isat than your Irms+half the current ripple p-p, etc.//Leon

author: Codebird
author_rank: Regular Contributor
timestamp: May 11, 2017, 09:58:26 pm
content: If you're bucking down to 2V, you're probably going to want to look at a synchronous converter. Anything less is going to dissipate a good chunk of the power in the diode, which means inefficient supply and toasty diode.

author: martys
author_rank: Regular Contributor
timestamp: May 11, 2017, 10:11:31 pm
content: Thanks Codebird,I was thinking instead of two 5-amp schottky diodes in parallel and two 4 amp inductors in a parallel configuration.Two 3 to 4-amp inductors are cheaper and smaller than a single 8-amp inductor, as far as I can tell so far.I am well aware that PC motherboards use synchronous converters to supply as little as .8V to the CPU at very high currents(>30 amps), but the efficiency is not great by my guess(the inductors on the mother board are often too hot to touch) and the controlling IC has a much more complicated circuit to connect its many pins(12-18 pins?) and requires at least two power N-Chan Power MOSFETs and are made to work with having a 12V and a 5V and 3.3V supply. Much more expensive (requires a controller IC)and maybe 4-layer PCB to be stable and more PCB space in general.

author: jbb
author_rank: Super Contributor
timestamp: May 12, 2017, 06:11:43 am
content: Hi MartysI'm afraid that power electronics does involve a fair few equations. In fact, a colleague of mine once said that power converters are like machines which a) embody calculus and b) are 20 microseconds away from disaster. The Wikipedia article actually seems to be surprisingly good.The synchronous rectification argument does boil down to complexity vs. efficiency (with a sideways shimmy into circuit behaviour at low currents). An important question: what are you supplying with this circuit? If it can feed energy back into your buck converter, that complicates matters.Let's think about your circuit operation. Please note all calculations are approximate because going into full detail is just awful algebraically.Vo = 2V, Io = 6A, Pout = 12WThe power stage will go from 12V -> 2V. This means that the duty cycle D of the control MOSFET will 2V / 12V = 17% (approximately!). D = Vout / VinThe power inductor will have an average current flow (plus ripple) equal to the output current, in this case 6A. IL(avg) = IoutThe inductor current is sourced from the main switch when it's on, i.e. D, which is 17% of the time. Isw(avg) = D * IL(avg) = 1.0AThe inductor current is source from the rectifier when the main switch is off , i.e. (1-D) which is 83% of the time. Irect(avg) = (1-D) * IL(avg) = 5.0ALet's think about power losses in therectifier. (Specifically conduction losses. Rectifier switching losses shouldn't be a big deal at this voltage level.)I'm guessing a reasonable Schottky will have a forward voltage drop of 0.4 V when hot. This would give an average dissipation of 0.4V * 5A = 2W. This is fine for a TO-220 with a wee heatsink but will make your life difficult if you want a surface mount e.g. D2Pk.This puts a maximum bound on your efficiency of Pout / (Ploss + Pout) = 12/(12+2) = 85%. Plus you'll loose some in the main switch. Expect to end up in the 75% - 80% efficiency range.If you put 2 diodes in parallel, they probably won't share current evenly because the hotter one will have a lower voltage drop (and inductive effects of the circuit layout will also have a surprisingly large impact) and the froward voltage won't go down much anyway.If you use a synchronous rectifier, there will be added complexity. But let's think about losses. Assuming a 20mOhm MOSFET, the voltage drop will be around 0.02 R * 6A = 0.12V and losses around 0.7W(*). This should be within the thermal capacity of a surface mount MOSFET and improves the maximum efficiency bound to 12/(12+0.7) = 94%.So, from a thermal and power budget viewpoint, a simple Schottky rectifier may be quite acceptable - especially if you don't need your 6A output for a long time.On inductors:If you parallel inductors, they may not carry current evenly.The inductors on a PC motherboard will be carefully selected to be as small as possible without getting too hot. Operating at 80 degrees is probably OK for reliability but definitely too hot for fingers!If you're doing a small number of units, the time spent making parallel inductors play nice might be worth more than just forking out a bit more for one bigger one. Also, what switching frequency are you thinking of? Many modern switching supplies go at or above 500 kHzA word of warning about MOSFET resistance Rds_on: the datasheets all give values at 25 deg C. Once the junction warms up to 100 - 125 deg C, Rds_on typically doubles. Remember that the people writing the datasheets want to sell you stuff.*You really should use RMS currents for MOSFET losses rather than averages, but I didn't want to throw an ugly great derivation in.

author: martys
author_rank: Regular Contributor
timestamp: May 12, 2017, 11:32:24 am
content: Jbb, thanks so much for your help.The 2V output is supplying power to a passive load.I have now deadbug breadboarded my circuit on a solid copper DS FR4 PCB blank using 1 TO-252 12-amp common cathode Schottky rectifier soldered to 2 cm^2 FR4 DS PCB and it runs reasonably hot to the touch at 6-amps out.Since my circuit requires 6A and I have two 4-amp inductors through hole mount with .1 ohm copper loss, they share the current quite well in parallel(they seem to be equally warm) and have a lower use of PCB area compared to a 8-A rated inductor.A 30-amp N-Chan Power MOSFET with a RDSon of .008 ohm also runs only warm when mounted on a 2 cm^2 of FR4 DS PCB stock.Simplicity rules, efficiency 60-70%, parts total will be 14-20 discrete components depending on MLCC and Al bulk capacitor combinations.

author: MagicSmoker
author_rank: Super Contributor
timestamp: May 12, 2017, 12:02:38 pm
content: Quote from: martys on May 11, 2017, 08:18:50 pmThere are quite a few equations for finding the amount of capacitance and ripple current rating of the buck input capacitor, but the more complicated ones I have seen are attempts to find a "more accurate" answer to a question that is itself very fuzzy (are the load current, input and output voltages always the same?) and for parts with some of the worst tolerances in electronics: capacitors...For example, the portion of the equation I gave earlier that finds the RMS AC current in the input capacitor - Io * sqrt(D * (1 - D)) - assumes that the current waveform is a rectangular pulse with a height equal to the output current. It ignores the contribution from the peak to peak variation in the inductor current which turns the flat top of the rectangular pulse into a ramp. You can find the contribution from the inductor ripple by using the equation below:IAC= IOUT* sqrt(D * (1 - D + (r²/12)))Where r is the peak-to-peak variation in the inductor current (ie - above and below the average, or load, current). A common recommendation is to choose the inductance so that r is 0.4 (ie - +/-20%), as this tends to optimize the size/cost of all the components.If r is 0.4 then the expression r²/12 gives a value 0.0133. Plugging that into the more complex equation will result in an increase in the AC current seen by the capacitor of a whopping 20mA:Simplified version: 6 * 0.3756 = 2.25ArmsComplex version: 6 * 0.3786 = 2.27ArmsWhat if r = 2, which means the inductor current ramps up from 0 to 2IOUTevery cycle (aka "boundary conduction mode)? Well, r²/12 now equals 0.333, so the complex version gives a ripple current of Iout * 0.4471 = 2.67A, or 0.4A more. That's an 18% increase, which seems reasonably significant except it took a quintupling of the peak-to-peak ripple to get there.Quote from: martys on May 11, 2017, 08:18:50 pmUt-oh... the underlined sentence implies you expect the capacitor *inside* the wallwart to also act as the buck's input capacitor... that's a bad idea because the buck input capacitor needs to be as close as possible to the buck switch. You basically have to pretend the output capacitor in the wallwart doesn't even exist if there is more than a few inches of separation.

author: martys
author_rank: Regular Contributor
timestamp: May 12, 2017, 03:27:29 pm
content: Thanks MagicSmoker, I donno the cap. uF inside the wallwart which has approx a meter of cable to plug into my device, but am only considering what combo of caps I place on my PCB nearest to the converter's power MOSFET on my PCB.I can quickly approximate the value of the wallwart's output cap by only loading the output with a 1k resistor and measure the time constant during discharge and solve for C. I can easily see the exact discharge curve on my storage scope.Since the converter MOSFET ontime applies a constant voltage to the inductor the current curve is a linear ramp until the peak current is reached and falls instantly to zero. This is a linear sawtooth ramp, certainly not a rectangular current pulse shape.

author: MagicSmoker
author_rank: Super Contributor
timestamp: May 12, 2017, 06:18:10 pm
content: Quote from: martys on May 12, 2017, 03:27:29 pmThe current supplied by the input capacitor to the MOSFET is only a sawtooth ramp that starts from 0 each cycleifthe inductance and/or the load current is low! Otherwise it is rectangular with a ramp on top (often called a "ramp on a step"). The attached picture shows a buck converter in LTSpice that steps down 12V to 2V and with the inductor chosen so that the peak to peak ripple is about 40% of the average output current. The green trace is the output voltage while the blue trace is the current delivered by the capacitor via the switch, M1.

author: Leon23
author_rank: Contributor
timestamp: May 12, 2017, 06:55:49 pm
content: The simulation looks good. Perhaps the output (and input) capacitance values are a bit luxurious. Depends on what switching frequency is desired. Current ripple in inductor looks good (perhaps a bit high). Possible to increase the inductance value?Again don't know whats the target. Low cost/high volume, performance, okay input/output voltage ripple, size requirements, emi performance etc.Thanks for the simulation,//Leon

author: MagicSmoker
author_rank: Super Contributor
timestamp: May 12, 2017, 07:46:39 pm
content: Quote from: Leon23 on May 12, 2017, 06:55:49 pmOh, I just quickly sketched something in LTSpice to show what the capacitor current waveform looked like because the OP was attempting to argue with me about it; nothing in the circuit was really optimized. Attached is the LTSpice simulation with the values/component choices tweaked a bit.The current ripple in the inductor was pretty close to ideal before. You don't want to reduce ripple current too much because that results in a physically large (and costly) inductor, as well as slows down the transient response. The capacitor values were, however, very luxurious for a 100kHz switching frequency