SMPS Maximum MOSFET Qg Calculation - Page 1
Forum section: electronics / projects

author: SparkFly
author_rank: Regular Contributor
timestamp: October 11, 2017, 02:59:42 pm
content: Hi all,I'm looking into SMPS design, specifically the trade off between frequency and Qg for use with a LT3757. I'm a bit confused, however, by the procedure taken by LT in determining values for a graph (figure 2, page 10 in datasheethttp://cds.linear.com/docs/en/datasheet/3757Afe.pdf). Look two paragraphs above it and it mentions that figure 2 was derived from the Idrivemax equation (which I understand) as well as the INTVcc vs input graph (page 6). However, I'm lost as to how they've used the graph here.I've calculated a Idrivemax of 0.113A, which should equal the product of the switching frequency and Qg, but as the INTVcc graph has some effect my values for max Qg are about twice that in figure 2.My apologies if what I've said is a bit tangled, if you need me to explain anything clearer please just ask - I'm very new to SMPS design. :]Regards,SF

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 11, 2017, 06:25:57 pm
content: You used the equation wrong then..?Offhand, the INTVcc curve shows 60mA at 20V, and the Qg curve shows 200nC at 20V, 300kHz. 200nC * 300kHz = 60mA, so it seems to be consistent.What are you trying to do?Tim

author: SparkFly
author_rank: Regular Contributor
timestamp: October 11, 2017, 07:11:55 pm
content: Many thanks Tim.I am almost certain I am doing something wrong. :]My understanding is, if I plug the values (Tj = 125C, Ta=70C, VIN=10V, theta=43C/W, Iq=0.0016A) into the eq. I should get the drive current. From there it should be divide by the switching freq to get the Qg? For VIN=10V, with f = 300kHz, I get 420nC, which is very different to the graph. Though when I go to 20V, I get ~200nC, which matches nicely. Am I missing something obvious here?Regards.

author: ocset
author_rank: Super Contributor
timestamp: October 11, 2017, 07:24:32 pm
content: Page 4 to about 10 of this tell you about gate drive calcshttps://pdfs.semanticscholar.org/8a69/693a73f96a64a4148893fb2b50b219176370.pdf?_ga=2.188046251.201822431.1507749723-1963478271.1507749723Design And Application GuideFor High Speed MOSFET Gate Drive CircuitsBy Laszlo Balogh..dont forget when you switch on theres also the current neeeded to discharge the drain gate capacitance. (as well as to charge up the GS capacitance)

author: Leo Bodnar
author_rank: Frequent Contributor
timestamp: October 11, 2017, 07:37:12 pm
content: Your current is limited by both - allowed power dissipation formula and internal current limiter.In your case dissipated power suggests 126.3mA but internal current limit is only guaranteed to provide 78mA to charge the gate.78mA/300kHz = 260nCLeo

author: SparkFly
author_rank: Regular Contributor
timestamp: October 11, 2017, 07:57:06 pm
content: Thanks both for your answers, treez that looks like a great resource, many thanks.Leo, thanks for the clarification. The equation calculates the limit based on it's ability to dissipate power and the graph is for the internal current limit. I think the wording had me a little confused when it say "based on...".Regards,

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 11, 2017, 08:15:16 pm
content: I see, you're calculating the maximum current based on power alone.Keep in mind what the equation is doing:- Difference in temperature: this leads to how much power is flowing out of the device.- Divide by Rth: this converts temp rise into watts. Bingo.- Divide by Vin: this converts watts to amps (DC).- Amps are coulombs per second, so you multiply by frequency (1/s) to get charge. Aha!But this is an equation based on power alone, and no other constraints. Note that amps and volts are inversely proportional, at the same power. If Vin = 1V (an absurd value, you can already guess the chip won't work down there), you would get over an ampere!What you can actually draw from the chip, however, is given by the V(I) curve mentioned above. Note that it isnota hyperbola (i.e., V and I inversely proportional), but a straight linear line: this means they used a foldback current limiting circuit. It's easy to design, and keeps power from being catastrophic, but doesn't protect the chip against continuous abuse.So the current limit curve is what youcandraw, while the equation is telling you what youshoulddraw (or less). The true value you can design with is: min(Ilim, Idrive(max)).Tim