Question about heat generation of a SMPS MOSFET - Page 1
Forum section: electronics / projects

author: WyverntekGameRepairs
author_rank: Regular Contributor
timestamp: December 13, 2019, 03:21:39 pm
content: I was taking apart my Sega Genesis when a question about switch-mode power supplies - regarding heat generation, to be specific - hit me.Here's my thought:When I'm working on an SMPS that I just unplugged (and tested for remnant voltages, never work on a power supply until the capacitors are fully discharged!), I notice that the transformer is warm, but in this power supply, the MOSFET is not. In my PS1's power supply, the MOSFET is not noticeably warm, even under operation. Usually, I see MOSFETs on heatsinks in other switch-mode power supplies that I work on. However, the MOSFET in this power supply has no heatsink at all. You see, the Inrush Current (IRC) of the transformer when it is initially turned onshouldbe shoving a lot of power through the MOSFET (or other switching device being used), thus heating the 'FET up a little. As the cycle repeats because the transformer is being switched a a few hundred kilohertz, the 'FET should end up settling at a consistent but high temperature (around 35°C, though this changes depending on ambient temperature).And of course, the transformer gets warm due to the small losses of energy in the transformer in the form of heat, and being used for a long period of time.In my PS1's power supply, I think the reason my MOSFET is not getting noticeably warm even after heavy use is because a damper is implemented to stop the IRC from shoving power through the controlling components. If this is the case, why doesn't everyone implement a similar damper, and if they do then why does it work better in some cases than others? Is there something I'm missing?I'm quite curious about this matter, and I appreciate your feedback on it. Don't be afraid to correct me if I'm wrong about something, as I'm eager to learn more wherever possible(Also, make sure your feedback doesn't look like an advertisement if you are referring me to any sort of external source or website!)Thanks!

author: MagicSmoker
author_rank: Super Contributor
timestamp: December 13, 2019, 06:01:29 pm
content: SMPS transformers don't have inrush, per se - that is something that really only applies to mains transformers when they are switched on randomly w/r/t the line voltage.An RC damper across the switch (or transformer) doesn't affect the peak current or power through the transformer, either; it just reduces the amplitude of ringing due to the series resonant network formed between the transformer leakage inductance and the switch output capacitance.SMPS transformers tend to get noticeably warm for two main reasons: 1) ferrite has a high thermal resistance, so a little heat results in a lot of temperature rise; 2) most ferrites exhibit minimum loss at 100C, so the core tends towards that temperature as long as total core and copper losses are reasonable (if losses are unreasonable - or ambient is, say, >60C - then once the core exceeds 110C or so it could go into "thermal runaway," which is exactly as bad as it sounds).Whether a semiconductor gets warm or not during normal operation very much depends on the current going through it, plus losses from switching. If the power rating of the semiconductor is sufficiently high relative to the current going through it then it might not get hot at all even at full load. A wise manufacturer will carefully weigh the cost of oversizing, say, the primary MOSFET to avoid needing a heatsink as the latter adds both to the BOM and labor costs.

author: WyverntekGameRepairs
author_rank: Regular Contributor
timestamp: December 13, 2019, 11:11:17 pm
content: Quote from: MagicSmoker on December 13, 2019, 06:01:29 pmOh, that is interesting. Could it be because of fewer turns on the primary? I have a feeling that more turns = larger IRC in an inductor such as a motor or transformer.Quote from: MagicSmoker on December 13, 2019, 06:01:29 pmOh, I think I understand now! So as demonstrated in a circuit with a capacitor and an inductor in parallel, when voltage is removed from the circuit, the capacitor pours its charges into the inductor, which resists change at first but begins conducting more as the magnetic fields generated by it collapse. The energy stored in the inductor is poured pack into the capacitor, which charges up and the cycle repeats. The voltage of course gradually decreases due to losses or any load attatched to it. This is the "ringing" you speak of, and it occurs with the transformer and the MOSFET, because of the capacitance of the MOSFET and the transformer's inductance properties (as it is basically an inductor). The damper absorbs that ringing, and the diode helps by clamping the voltage to a certain level. The resistor makes sure that the capacitor cannot charge up too quickly and cause more ringing. So basically a RDC network uses the resistor and capacitor for dampening, and the diode for clamping. And this is all to stop that ringing effect.Now that I think about it, this is pretty elementary stuff - it's just being employed in a big system that has many moving parts to it, and as such it appears to be complicated.Quote from: MagicSmoker on December 13, 2019, 06:01:29 pmHm. I'll need to consider this in my own power supply then. I think this would beextremelyimportant.Also, I've seen thermal runaway happen on an SMPS before, and the origin was the transformer. Turns out, it was a very cheap, crappy power supply from China (typical, I know), and there was no current draw monitoring that would stop the system drawing too much current and heating up the transformer. The SMPS was shorted out, and the transformer began drawing (i think it was) around 13 Amps. The transformer heated up so much it entered thermal runaway, and literally melted in seconds before finally going *bang* very volently. Yeah, that's someinsanelyscary shit.Quote from: MagicSmoker on December 13, 2019, 06:01:29 pmThis is interesting as well. Maybe you and Tim could help me figure out what I should use for my MOSFET? (seeing as you've already built and designed power supplies before, you probably know right away what would work the best).Anyway, I definitely agree. I think what is happening in my SMPS is that the MOSFET is rated for higher power applications, most likely specifically SMPS devices, so it is designed to handle high current and not need to dissipate so much heat or even need a heatsink.

author: David Hess
author_rank: Super Contributor
timestamp: December 14, 2019, 02:46:20 am
content: Quote from: WyverntekGameRepairs on December 13, 2019, 11:11:17 pmUsually the power switching transistor is selected to somewhat balance switching and conductive losses however reducing loss in the power switching transistor may eliminate the need for heat sinking.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 14, 2019, 03:57:24 am
content: The more important thing is the environment is completely different: a mains transformer is connected to a low impedance, constant frequency, voltage source. The transformer is designed to saturate not much higher than nominal voltage (or lower than nominal frequency). There is little or no inrush at the instant power is applied to the transformer; the inrush surge occurs after some delay (i.e., at least 8ms, but more likely closer to 20ms).In a typical (current mode) SMPS, the supply voltage or frequency is effectively variable, or the source impedance is effectively higher. To see why this is, we must cover several relations:* Voltage in RMS, is the root of the mean of the square of the instantaneous voltage. That is, for a signal v(t), take v(t)^2, then average it over some period of time; then take the square root of the result. In calculus, we can write:\[ V_{RMS} = \sqrt{ \frac{ \int_0^T v(t)^2 dt}{T} } \]The integral sign means to add up many tiny slices (of width \$dt\$) of the expression within, over the range specified by the indices (i.e., 0 to T; for a periodic signal like a sine or square wave, we can set T equal to the period).* The RMS of a "pulsed DC" wave (i.e., a load driven by a switch to a DC source) is therefore \$V \sqrt{D}\$, where D is the duty cycle of the wave.* In a current-mode controller, for a given current setpoint, and for an inductive load, the on-time is set by the inductance:\[ V = L \frac{\Delta I}{\Delta t} \]In DCM, current starts at zero so \$\Delta I\$ is also \$I_{pk}\$. So the on time \$\Delta t\$ is proportional to the set current and the inductance.Which means, as the load impedance (taking inductance as approximately impedance) decreases, the switch duty decreases, so the load voltage decreases. In other words, the source impedance is significant.We can't simply measure the source impedance, because the circuit is very nonlinear (there's a comparator setting switch time), and it's time-domain not AC-steady-state. We have to be very deliberate in defining what we mean by "impedance" in this case. Following the above definitions, we could indeed end up with an unambiguous proportion having units of ohms; it wouldn't be terribly useful though, so it will suffice to say that wecan, and that it more-or-less follows the usual rules of impedance, at least within the limits of our definitions.In short, it doesn't get inrush because it's not a low impedance mains source.And we can do a similar exercise for the variable-frequency case, where applicable (e.g., the MC34063 and ilk).Regarding power dissipation, it is usually divided into two aspects: conduction loss and switching loss. Roughly speaking, we can calculate conduction loss as Irms^2 * Rds(on) (Irms is calculated same as above, incidentally!), and switching loss as 0.5 * Vpk * Ipk * t_f * Fsw (draw the V and I waveforms during switching, to see a geometric proof of why this might be so; hint: what figure has area "one-half base times height"?). A lot of subtle dynamics is going on with modern transistors (mainly because of their highly nonlinear Coss vs. Vds relationship) which tends to dissipate less switching loss than this would suggest, although it can be increased in some situations instead (hard switching).Keep drawing circuits and waveforms, and running simulations; this won't (can't?) come instantly, but through careful study you can understand -- and prove -- everything that's going on here.Tim