Current question in switchmode boost converter - Page 1
Forum section: electronics / beginners

author: galpi
author_rank: Newbie
timestamp: October 09, 2010, 02:18:13 am
content: Hi everyone,First post from a newbie here ...I'm making a switchmode boost converter using the MC34063 mentioned in blog #110. Input is 12V, output should be 24V with 350 mA of current. I figured out all the parts I needed using the formulas Dave goes over, and I end up with a peak current of 1.62A with feedback resistors (R1 and R2 in MC34063 specs) having values of 100 KOhm and 1820 KOhm. Dave mentions that you have to be careful when selecting your diode and inductor to make sure they can handle the peak current (here 1.62A), but what about the resistors ? Should I be concerned about peak current going through the resistors ? How do I figure out the necessary power rating required for those two resistors ?Thanks a lot folks,Pierre

author: TopherTheME
author_rank: Regular Contributor
timestamp: October 09, 2010, 02:48:02 am
content: There should be very little current going through the resistors so its not an issue (i = V/R).QuoteAre you sure this is correct? A value of 1.82M is a pretty big resistance. You may want to cut that down a bit to maybe a 10k and 180ishk resistor.

author: galpi
author_rank: Newbie
timestamp: October 09, 2010, 03:41:37 am
content: QuoteRight, yet somehow there is mention of a peak current Ipk of 1.62A, and with no load all of it has to go through the resistors for a ridiculous power consumption ... that's what I don't get.Thanks !Pierre

author: Rhythmtech
author_rank: Regular Contributor
timestamp: October 09, 2010, 04:27:29 am
content: Quote from: galpi on October 09, 2010, 03:41:37 amWe need to consider the plausibility of where the peak current is actually going. If 1.62A were going through a 100kohm resistor then you would have 162,000 V somewhere in your circuit, which is somewhat impossible. So the peak amps are finding their way to common or ground via another path. Perhaps the peak current is travelling through the load or the output filter cap or is maybe even inrush into the inductor. I would tend to think of it as the largest average current the diode will see over a short period of time until the regulator circuit settles for most cases.Maybe how transient responses move into steady state would help describe what I am trying to simplify better. This is just my untech way of taking a stab at it without an analytical approach, perhaps somebody else would like to detail out where the peak current comes from actually.

author: DJPhil
author_rank: Frequent Contributor
timestamp: October 09, 2010, 04:52:21 am
content: Buckle in for way more mediocre advice than you need, I've had too much coffee!You get the formula you need by taking the general instantaneous power formula:P = I * Vand subbing in Ohm's law . . .V = I * R subbed in for V OR I = V / R subbed in for Iso you get . . .(P = I * I * R) ->P = I2* VOR(P = V * V / R) ->P = V2/ RI remember a bit from the very beginning of Art of Electronics. One of the first half dozen exercises is to prove that you can't roast a 1/4 watt resistor over 1k with a 15V supply. If you flip it about, you can figure just about whatever you like.Bad advice that often works anyway:Resistors are fairly slow to fail (maybe whole seconds!) compared to other electronic parts (Google for 'silicon fuse') You often only need to worry aboutaveragepower, especially above audio frequencies, unless you're dealing with something fairly crazy. Carbon composition and carbon film (the cheap ones) tend to behave best under power surge, with the non-wirewound metal resistors tending to fry more quickly. It's all about thermal mass, so higher power resistors (that are naturally larger) have the mass to average out peaks better than tiny 1/8W SMD units. Remember that through hole resistors dissipate most of their heat through their leads./end bad advice (use as a rough guide only, some may consider this wreckless [or wrong, if so let me know!])Now to address the actual problem. The current that's hurling through the inductor and diode is large because their series resistance is low.Break out ohm's law for this one, just as Topher said:I = V / RI = 24V / 100KI = 240uA (yep, that's microamps)Let's flip it with the equations above, what's the lowest value resistor you could use at 24V and stay below 200mW (50mW safety margin, woohoo!)?P = V2 / R ninja flip! both sides times R then both sides divided by PR = (24V)2/ 0.2WR = 2880?Now that's assumingyou're shorting out the supply with this resistor, so don't assume you can't have a resistor in the circuit anywhere below that value. Consider R1 in the circuit from the datasheet. It's only 2.2K, but it's got a giant resistor restricting it's max current input.Whew. Then also, you don't need (also as Topher said) such large resistors. High value resistances are inherently noisy (small problem) and produce tiny currents which are susceptible to interference (much bigger problem). With an inductor nearby throwing an EMI party this could be trouble. You also don't want them too low or they'll be drawing power out of your power supply and dumping it to ground! The datasheet circuit is a good example.Update cause I'm way gabby and taking too long:Quote from: galpi on October 09, 2010, 03:41:37 amAh, I see the problem, you've got an incorrect concept of current. Current must flowthroughsomething. Ipk is actually Ipk(switch) to express what it's flowing through, and is the maximum current drawn across the switching circuit. What's the switching circuit? The Q1 and Q2 in the datasheet, plus any external transistor used to switch current to the inductor for energy storage. The inductor is always heaving and shoving current around (low internal resistance), and that's where you get your big peaks. The diode has a low internal resistance as well, hence Dave's warning. Co smooths out a lot of the mess, and the optional filter circuit at the end does as well.The output circuit itself is a voltage source, so it will attempt to deliver a fixed voltage until it can't keep up enough current (falls out of compliance or out of regulation, they say), after which the voltage will drop. In your design it's supposed to be 350mA, so that's the most you'd expect without the voltage dropping. With no load it's still flowing, but consider the resistances. You've got the resistive divider, the leakage of Co and the optional filter cap, flux losses in the inductors, and other such. The feedback network is the highest one there, but work out the current at 24V and you'll see that it's very, very small. The remainder is even smaller still. The balance of the output current doesn't exist, because there's nothing for it to flow through. If you work out the lowest resistance the supply can handle at 350mA (exercise for the reader, muahaha), then you'll actually see all 350mA flowing through that single resistor if you shunt the supply with it.Good grief, I'm gonna stop now, I need more coffee. Hope that helps.Edit: Included relevant datasheet schematic for reference.Edit: Included the correct relevant datasheet schematic (sheesh).

author: galpi
author_rank: Newbie
timestamp: October 09, 2010, 06:09:25 pm
content: Thanks all ... Sounds like such spikes in current shouldn't be a concern for my resistors. Works for me. I will change their values though for smaller ones, as recommended.Cheers,Pierre

author: jahonen
author_rank: Super Contributor
timestamp: October 09, 2010, 06:48:08 pm
content: 1.6 A peak current is high (too high for my taste) for this kind of converter. Datasheet specified maximum switch current is 1.5 A. I suggest using something else (like LM2585) for this purpose, to get decent efficiency. Besides, by using proper regulator with error amplifier and PWM-modulator you'll get smaller ripple as a bonusThere is a application circuit inLM2585 datasheet, page 20, with just the voltages you'll need.Also, make sure that your inductor does not saturate, otherwise there is a risk that switching device will be destroyed. Current rating of the inductor should be as high as your other constraints will allow, since reducing the DC resistance of the inductor will increase efficiency.Regards,Janne

author: Feanor
author_rank: Regular Contributor
timestamp: October 09, 2010, 11:19:22 pm
content: The bottom line is the two resistors R1 and R2 do not conduct any of that peak current, they are there simply to give feedback of the output voltage for the chip. The chip needs to know if the output voltage is too high, too low or just right so that it can adjust it's operation to control that voltage That is all that these resistors do.If the output voltage is high enough then the chip will stop switching until the output voltage drops again. (In reality it does not just stop switching or start switching it can vary its switching a bit instead of turning it on or off. But you get the idea.)So when you have no load the output voltage stays right up where it should be 24V in your case and the chip stops switching, so that peak current no longer exists in fact there is no current except for the leakage current in your feedback resistors which is wasted (So note that the higher these resistor values the lower your leakage current just make sure they are low enough to make things work alright).In theory this chip could have been designed so that we could connect the output straight back into the chip and it could read the voltage directly, some chips do this, but their voltages may not be adjustable. We use the "resistive divider" as feedback. As far as the chip is concerned 1.25V on pin 5 is just right, more than that is too high and less than that is too low. The voltage we get out of our resistive divider isVresdiv = (Voutput * R1)/(R1+R2)So if we know that Vresdiv should be 1.25V for our desired Voutput we can choose R1 and R2 to satisfy this formula.So what would happen if we did not feedback the output voltage? Lets say we just disconnected R2 or R2 somehow broke out of our circuit, say we soldered the resistor in badly. Then Pin 5 on our chip would be pulled down to ground always. So our chip would believe that the output voltage was always too low it would keep boosting the voltage higher and higher until either some protection kicked in or the chip was killed by high voltage on Pin 1. In fact using an external switch (Transistor, MOSFET, IGBT) and a high voltage inductor (car ignition coil) this chip could produce a very high voltage indeed thousands of volts if you like.*NB* you can not use the external switch schematic in the data sheet to produce high voltages

author: Rhythmtech
author_rank: Regular Contributor
timestamp: October 11, 2010, 02:08:23 pm
content: This is definitely getting into some good meaty explanations. Thanks AchMed99 and Feanor!

author: scrat
author_rank: Frequent Contributor
timestamp: October 12, 2010, 11:33:42 am
content: Quote from: Rhythmtech on October 11, 2010, 02:08:23 pm@gapiI'm glad to see the question about divider resistors has opened the door to feedback explanation.Feedback is one of the most useful concepts for understanding things. Once you catch it, you find that it's everywhere, from natural systems to the most complex control, to the way you can drive a car staying between the lines on the road.