tiny components with large current rating - Page 1
Forum section: electronics / beginners

author: Clear as mud
author_rank: Regular Contributor
timestamp: August 29, 2013, 05:10:20 am
content: I recently designed a stepper motor driver board using the Allegro A4989. It requires external MOSFETs. My driver is designed for a specific motor for the CNC machine I plan to build. The stepper motor takes 3.5 amps and has 2.4 mH inductance.I chose components by sorting by price, and I ended up with some really tiny FETs, only 1.6 mm square! Also, the current sense resistors are tiny, only 0603 (reverse) size - inch size .06 by .03, with the leads on the long edges.This is the MOSFET data sheet:http://www.vishay.com/docs/62715/sib456dk.pdf.It says they can do up to 5 or 6 amps, but my question is, how to be sure they can carry that much current in a real-world application? My controller limits the current at 3.5 amps, but that still seems like too much for such a tiny component. I laid out the board with large copper areas with solid connections to the drain and source pads. The traces carrying the main current are 4 mm wide, and by the MOSFETs I used filled copper regions of at least 50 square mm. If I have the board made with 2-oz copper, will that suffice for the 3.5 amps of current? I did the same thing with the current sense resistors - a filled copper area wider than 4 mm tapers down to the 1.6 mm wide pad of the resistor. The sense resistors are only 36 mOhm, so they are not dissipating much power, but I still worry about them being big enough to carry the current.This was the first design I have done that is this complicated (84 components total), and I think my main problem was not understanding the size difference between components until I started laying out the board. The Schottky diodes I chose to protect the MOSFETs from reverse voltage are a SMB footprint, and they are small (compared to my finger, for instance), but still huge compared to the MOSFETs themselves. I feel kind of silly using such a huge diode next to such a small MOSFET.Other questions:Should I add a ground plane? I usually do one, but I didn't on this board. I guess I was trying to keep all my electrons in a row.This design includes an on-board u78M05 voltage regulator (5 volts at 1/2 amp). External power will be 12 volts (regulated to 5 volts by the regulator), and 36 volts for driving the motor. I realized after I finished the design that I had used the 5-volt regulator output as a reference voltage, through a voltage divider made from two 1% resistors. I think maybe this was my worst error, since this will not be such a stable reference.I already bought enough components for six boards. I think the next version of this board will have some slightly larger current-carrying components, and a better voltage reference, but should I go ahead and build some of these for testing? I can't decide if this will be acceptable for driving the CNC machine I am making, or if I should redesign the board, and not build any of version 1.I attached part of a screenshot from KiCad, but it's not detailed enough to see the individual pins on the A4989. That's another very small component!

author: KJDS
author_rank: Super Contributor
timestamp: August 29, 2013, 07:24:57 am
content: You need to use Rds on and current to work out the power dissipated in the MOSFET, then the thermal specs to find the junction temperature rise.

author: Psi
author_rank: Super Contributor
timestamp: August 29, 2013, 08:34:25 am
content: As KJDS pointed out, to work out which fet to use for switching X amps you need the fets Rds on.You also need an idea of how your going to mount/heatsink it.The current rating given in a mosfet datasheet is the max safe continuous current before damage may occur.This current rating doesnottake into account the heat generated at all which is always the limiting factor.Take for example a mosfet i use in one of my projects.NVD5117It's in a D-Pak package and is rated for 60V and 61A with a Rdson of 16mR.If I actually put 61A through it the voltage across the fet would be 61*0.016 = 0.98V which would dissipate 0.98*61 = 59WIf I could cool/heatsink the mosfet well enough to maintain a safe temp while burning off 59W then it would run ok.In reality that's never going to happen. Even if i had a huge copper block with fins and a large fan on it, the package is just too small. I wouldn't be able to mount it to a heatsink well enough to get the heat away quick enough. The D-Pak package just doesn't have enough surface area.So, when spec'ing a mosfet to switch X amps the datasheet current rating is a rather useless value, except that it obviously has to be higher than your current.In my project i use that fet to switch <10A at ~14V which makes the fet dissipate about 1.6W. Small enough that i can have the fet soldered to some PCB copper for heatsinking.

author: cowana
author_rank: Frequent Contributor
timestamp: August 29, 2013, 08:47:54 am
content: For some rough back-of-the envelope maths, your FET has an Rds of 0.185R at Vgs=10v.With 5A, that gives you 4.6W of heating - a pretty big amount for such a small package. Junction to case is typically 7.5C/W - meaning that even if you could keep the outside at 30 degrees, the juction temperature will be 65 degrees.That's starting to get pretty warm - so it really comes down to how good your external heatsinking is at getting rid of heat.I'd be tempted to play it safe and use a FET with a slightly lower Rds(on).

author: poorchava
author_rank: Super Contributor
timestamp: August 29, 2013, 08:54:00 am
content: You need to pay attention to gate charge of the mosfet and voltage rating. Small mosfets with large current rating are typically for low voltage and have very high gate capacitance, which means that they can be easily damaged by inductive voltage spikes and that internal drivers in A4989 chip will have hard time switching the mosfets fast enough to minimize power losses during transitions.What transistors did you exactly choose?

author: Kevin.D
author_rank: Frequent Contributor
timestamp: August 29, 2013, 09:30:48 am
content: I think current rating's given by the manufacturers data sheet are just about grabbing headline specs and little about reality . I think they must freeze them close to 0 K and when they do the current specs and publish those as max current . I mean look atsome of the headline specs on a typical device in a to220 package , irf2804 - max current 200 A .Your not going to get near 200A of course because of package limitations .they do though state another another spec below this (75 A Package limitations) ,so there goes your headline spec .Though even that at 75 A for a to220 is being over optimistic I think .Would you put 75 A through a 1mm * 0.5mm thick pin ? ,I dont think I would put more than 20 A through it , if it was a piece of wire that thick it would be rated at ~10 amps . When choosing power transistors its usually always the package style that will set your limits and makes the other current specs look like manufacturer's fantasy.

author: Psi
author_rank: Super Contributor
timestamp: August 29, 2013, 09:39:23 am
content: The nice thing about mosfets is most can handle a large amount of current for a short time. This makes it easy to protect them from most types of short circuit, even in software. Usually the resistance of the cabling in the system is sufficient to keep the short circuit current well under the mosfets rated peak current or even its continuous current.With P channel fets you can do it with minimal external components by having a sense resistor in the load to ground and use the ADC or analog compactor to keep an eye on the voltage on the top side of the resistor.If it jumps up beyond a limit you can stop driving the fet.You can even make the sense resistor out of pcb tracks since the tolerance isn't critical to detect a overload.

author: Zero999
author_rank: Super Contributor
timestamp: August 29, 2013, 12:20:22 pm
content: What's the duty cycle? You don't normally apply the current to a stepper motor for that long.If you don't want to replace them, you could parallel them, there's enough room on the PCB, but I assume you don't have have enough for all the boards so would have to buy more anyway.

author: tszaboo
author_rank: Super Contributor
timestamp: August 29, 2013, 12:36:46 pm
content: I have 3mmx3mm MOSFETS handling 10A of current continuously, fully tested, sold, working 24/7. It is all about enough cooling, wide copper traces, calculated temperatures, and proper soldering. Be aware, that the switching losses could be actually higher than the conduction losses, depending on frequency. Altough the traces could heat up, and if you fail to solder the MOSFET correctly, they will loose the magic smoke.In this particular case, I would assume that the board temperature is 70 degrees (it will get hot, trust me), Rds is 0.185 Ohm, the current is 3,5A, than for continous drive, the dissipation is 2.26W, RthJC is 9.5 K/W so the temperature is only 90 degrees. So yes, it might work properly.If you are not comfortable with this size, use some bigger ones. Power SO8 is hand solder-able, without hot air blower. Also, I would suggest to increase the size of the shunt to 1206.

author: Clear as mud
author_rank: Regular Contributor
timestamp: September 03, 2013, 12:22:58 am
content: Thank you for all the suggestions. It looks like it will actually be OK to build this board with these MOSFETs, but in the future, I will always keep in mind to check the actual power dissipation, not the manufacturer's current specification. I think I probably won't build this board exactly this way, but I think I might be able to use the same FETs for a similar board.You are right to say that the MOSFETs would not be handling 3.5 amps continuously. For my 3.5 amp current limit, the average current per H-bridge is only 2.1 to 2.5 amps, depending on which microstepping mode is selected. And, in each H-bridge only two of the four MOSFETs are conducting at any time, so the actual average current through each FET is only between 1 and 1.25 amps. So, using a similar calculation as NANDBlog posted, I get .185 Ohm x (1.23 A)^2 = .28 W power dissipated per MOSFET on average. Times the 9.5 degrees per watt, it only runs 2.5 degrees above the board temperature. And that is for continuous stepper motor operation. In reality, it may not run for very long without a break.I would like some clarification on why you chose to use the RthJC together with an assumed temperature of 70 C? I think the 70 degrees is an educated guess of the PCB temperature, and the RthJC can be used because I said I will have a solid connection from the drain to the board copper. Is that correct? I suppose that alternatively, if I didn't have a solid connection, I could use the RthJA together with the highest ambient temperature expected to be encountered. But that would be higher, and as long as I am careful about soldering solid connections, I can use the RthJC.I think I understand the steady-state dissipation, but I am not so sure about switching losses. I picked a maximum speed of 3300 RPM for my stepper motors, then sized the gate resistors so that the di/dt can not change much faster than my chosen maximum speed. The A4989 datasheet says "External series gate resistors can be used to control the slew rate seen at the gate, thereby controlling di/dt and dv/dt at the motor terminals." It seemed like a good idea to me, to limit the slew rate and thereby drive the motor with something better approximating a sine wave. So, I calculated a minimum step time (91 us for a 200-step/rev motor at 3300 RPM), and divided the motor drive voltage by the step time to get my maximum necessary dV/dt at the output. From there, I think I did something wrong. I took the same dV/dt and multiplied it by the input capacitance given in the MOSFET datasheet, then divided 10 by the result (thinking the A4989 drives the gate at 10 volts above the source). That doesn't look correct at all, now that I try to explain it.How would I correctly do the calculation to find the proper value for a series gate resistor to limit my motor di/dt to the maximum required for my chosen maximum RPM?And then, how would I calculate switching losses in the MOSFET, given the series gate resistors? Would it be different from the way to calculate it if I did not have an external gate resistor?

author: tszaboo
author_rank: Super Contributor
timestamp: September 03, 2013, 05:54:33 pm
content: 70 degrees is a very good calculation point for worst case scenarios. The PCB is 70 degrees, the components are usually warmer than that. A lot of commercial rated components are only rated for 85 degrees. Also, by calculating with this worst case, even if the temperature is not reached, and the outside temperature increases, you have more headroom. For very simple calculations for power electronics, I usually start with this. And hope for the best. Sometimes to divine interventionThe switching losses are not easy task to calculate. Normally you will have a gate resistance less than 50 Ohm. You need to take the gate charge, the drive voltage, the driver output characteristics, frequency into account, and they modify each other's parameters with some nonlinear way. Usually it is easier to just measure it than calculate. DMM to the mosfet driver power supply and done.

author: Clear as mud
author_rank: Regular Contributor
timestamp: September 03, 2013, 10:59:19 pm
content: OK, so I should start with gate resistors of 47 ohms and see what the output motor drive waveform looks like, to see if the di/dt is limited the way I want. Then, to measure the FET switching losses, I should make an open point in the trace supplying the motor-voltage power supply pin on the A4989 driver chip and put a DMM across it, measure RMS current into that pin and divide by 8 because of the 8 MOSFETs it is driving. OK, thank you for the plan. It's very empirical, but I guess that is what works!

author: sub
author_rank: Regular Contributor
timestamp: September 07, 2013, 01:21:24 pm
content: Quote from: Kevin.D on August 29, 2013, 09:30:48 amAs I understand it, that is the current that the bond wires (or however you move current around in a discrete device) can handle. That's why it's a current (rather than power) rating.

author: Kjelt
author_rank: Super Contributor
timestamp: September 07, 2013, 01:40:10 pm
content: If I look at the same datasheet it states under Maximum Ratings (page1) that Continuous Drain Current (TJ = 150 °C) TA = 70 °C is only 2.2 Amps and that this is valid if conditions b and c are met.Conditions:b. Surface mounted on 1" x 1" FR4 board.c. t = 5 s.So how I read this, is that if you place this component on its own on a 1" x 1" copperplane for cooling it can handle 2,2 Amps for 5 seconds. Or how do you read this?

author: Clear as mud
author_rank: Regular Contributor
timestamp: September 07, 2013, 07:40:10 pm
content: Yes, that does seem to be what it is saying. TA probably won't be so high, I was thinking the ambient temperature would remain more around 40 maximum. I think you are correct, the data sheet says that even if you allow 1 square inch of copper for cooling, at an ambient temperature of 70 degrees, the junction temperature will get too hot in only five seconds at 2.2 amps. Or 2.7 amps at 25 degrees ambient.They seem to be saying that the device is not meant for average currents of that magnitude, only pulses of short duration. In this case, I would be better off not building a stepper motor driver with these MOSFETs, and I should choose a different (larger, and ideally with lower RDS-on) type instead.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: September 07, 2013, 09:32:43 pm
content: ..and don't forget the option of using heavy copper (2oz)

author: peter.mitchell
author_rank: Super Contributor
timestamp: September 08, 2013, 07:37:39 am
content: Yeah, also, something big people don't realize that even the slightest amount of airflow can make a pretty huge difference, even just some strategically placed vent holes and vertical PCB mounting.