4 ms·

Specs like "6A MOSFET" don't exist in isolation. The current rating of that MOSFET is due to the on resistance(Ron), the resistance between source and drain.

by mbell 14y ago

Specs like "6A MOSFET" don't exist in isolation. The current rating of that MOSFET is due to the on resistance(Ron), the resistance between source and drain. As current flows through the FET this resistance causes power to be dissipated as heat. The 6A number comes from how much heat the junction can handle without frying and is measured under a set of assumptions. Usually testing is done at room temperature (25C often used) with the best possible heat-sinking and ideal Vgs. Most MOSFETs of this type actually tie the internal thermal components to the drain pins. Its intended that the PCB designer use a large copper pour for the drain pads, this copper area acts as the heatsink for the chip, transferring heat away from the junction. In other words, the MOSFET can only handle 6A if ambient temperatures are kept at room temp (usually means fans on it or its wide open to a room), there is adequate heat-sinking provided by the PCB design, and Vgs is high (higher Vgs reduces Ron). Further its important to remember that FETs have a "run away" characteristic in that as junction temperature rises so does Ron, which creates more heat. So, there is a lot more to the functional rating of a MOSFET than the 6A on the datasheet. In reality you would never want to nor expect that FET to hold 6A. With just a cursory look at the PCB layout and previous experience with what the ambient temperatures tend to be on back-planes of server chassis loaded with many drives, 1A would be pushing it, so I'm not at all surprised that these things blew. Perhaps the most WTFery aspect of that PCB is the lack of any suitable capacitance near the SATA connectors. If you aren't using staggered spin up and many drives were spinning up at the same time the rail voltages can dip which in this setup causes Vgs to drop, increasing Ron at the worst possible time as mechanical drives draw a lot more current at spin up. This could trigger thermal run away.

dietrichepp 14y ago
> as junction temperature rises so does Ron, which creates more heat What's interesting about this is that bipolar transistor has the opposite property -- as it heats up, resistance decreases -- and yet this opposite property causes the same problems with thermal runaway. (Okay, when talking about bipolar transistors we don't really talk about "resistance", but if you know that then you probably know what I'm going to say anyway...) When hooked up to a load with a relatively constant voltage, there will also be a relatively constant voltage on the transistor. As the bipolar transistor heats up, the resistance decreases and more current flows through the transistor, and the transistor will dissipate power according to the law P=V^2/R. So once a transistor gets hot, it gets hotter until it blows. The question is, "what kind of load looks like a voltage source?" There's actually a quite common load -- any amplifier with parallel output transistors will look like it's driving a constant voltage load, from the perspective of one of the parallel transistors. Basically, hooking up bipolar transistors in parallel does NOT multiply the power rating as you would think, because thermal runaway might cause one of the transistors to dissipate all the power.
- deleted 14y ago
  [deleted]
rdl 14y ago
The thermal runaway is the "fun" part of the Tesla motor controller, right? It's got 3 MOSFETs, since there isn't a single one with enough power handling, and thus if one of them sucks a little more, it'll end up with more power, which will make it suck more, which will end up with more power...
- tlb 14y ago
  Bipolar transistors suffer from the thermal runaway of parallel devices you describe, but not MOSFETs. MOSFET on resistance increases with temperature, so with multiple devices in parallel the hottest one will flow the least current. This is an important effect even within one device, which is actually an array of thousands of junctions.
cyanoacry 14y ago
The FETs blew as soon as we turned on the case, oddly enough, and even if we had a single drive plugged in, that FET would blow by itself. This is even the case when we took the backplanes out and started testing them individually in free air in the server room, where ambient is more like 18C instead of 25. Is the design solution to use a FET with a lower Rds at 5V Vds for smaller heat dissipation? But more importantly, is there a standard current that most chassis are tested to? I would expect all SATA hot-swap bays would support all SATA drives on the market, since nobody ever gives power dissipation or consumption figures.
- mbell 14y ago
  > Is the design solution to use a FET with a lower Rds at 5V Vds for smaller heat dissipation? Honestly - Its to not cheap out on the design and use a proper hot-swap controller. Examples (many companies make these): http://www.ti.com/ww/en/analog/power_management/system-protection-hot-swap.html http://www.ti.com/ww/en/analog/power_management/system-prote... You can rig up something slightly better by designing a current sense circuit with feedback into the FET gate if your adventurous. In either case the goal is to limit and control inrush current during power on or disk insertion. Based on what you've said I doubt the steady state current of the disks is the problem at all. It just sounds like the 3TB drives have a higher inrush current during power up and its either high enough or lasts long enough to blow the FET junction.
  - cyanoacry 14y ago
    The funny thing is, I'm not sure that the hot-swap circuitry is even necessary in the first place due to the connector design. The SATA spec defines "pre-charge" voltage pins that are connected after ground, but before any of the other voltage pins are connected. The idea is that by inserting a small (10 ohm) resistor, you can limit inrush current to a tolerable value while capacitors charge and regulators start up, and then when the other pins connect a couple of milliseconds later, the drive gets the proper low-impedance power connection. Do you know if implementing pre-charge via connector mechanics obviates hot-swap protection circuitry? Supermicro seems to have hot-swap protection on their backplanes as well, but I haven't had a chance to closely inspect it.
mindslight 14y ago
Those failures don't really look like moderately too much power dissipation. I'd be worried about static at the drain or counterfeit FETs. I suppose the switcher in the drives could just be drawing several amps as it's input voltage decreased and maybe those FETs don't have the best thermal resistance, but sheesh.
- mindslight 14y ago
  Oh I see, 62 K/W junction-to-ambient for an SO-8. Whereas the D2PAKs I've had desolder themselves are 1.5 K/W junction-to-case.
gvb 14y ago
A related "gotcha" is not driving the gate hard enough. The 6A rating is also predicated on driving the gate hard enough to drive the on Rds to its minimum value. I would be suspicious of this, in conjunction with inadequate heatsinking (i.e. heavy copper pads under the FETs) especially if the FETs are being driven by 3.3v. Looking at the specs, the FET is rated at 50mOhm given 4.5v gate drive - very acceptable - but if the FET is driven by 3.3v, it will have much higher Rds (may be running in the linear region which would be very bad). Note that the gate threshold voltage is 1.5v typical but 3v max so driving with a 3.3v logic signal would be marginal in the worst case situation.
- cnvogel 14y ago
  In this application you typically use p-ch mosfets with gate pulled down to turn on, a resistor will provide pull-up of the gate to turn off. So Ugs is the same as the rail voltage (3.3, 5 and 12v).
ChuckMcM 14y ago
Came here to say essentially the same thing. In designing a 400A speed controller for my Battlebot I got to learn all about MOSFETs and their current ratings. In this case the MOSFET is rated with a max current of 5A @ 25 degrees C if the gate is driven with a 5v signal. It has an 80 mOhm resistance when fully enhanced. So at 5A that is 2Watts of power dissipation (p = i^2r) and since the case has a thermal constant of 62 degrees C/W that means the case will be at 149 degrees C (25 + 62 2W) which causes Rds(on) to double and that 'explodes' the MOSFET. If the device was successfully carrying .5A loads then then the case is only good for about 20mW of power dissipation (which looks reasonable given the package size). So gluing a copper heak sink to the FETs with at least 2 sq inches of surface area would probably keep them alive with a .75A load. If you combined that with about 150 CFM of air flow at sea level (225 CFM at 8,000 ft) you'd stay pretty solidly inside the 'not to be exploding' parts of the parameter graphs.

by mbell 14y ago

dietrichepp 14y ago

> as junction temperature rises so does Ron, which creates more heat What's interesting about this is that bipolar transistor has the opposite property -- as it heats up, resistance decreases -- and yet this opposite property causes the same problems with thermal runaway. (Okay, when talking about bipolar transistors we don't really talk about "resistance", but if you know that then you probably know what I'm going to say anyway...) When hooked up to a load with a relatively constant voltage, there will also be a relatively constant voltage on the transistor. As the bipolar transistor heats up, the resistance decreases and more current flows through the transistor, and the transistor will dissipate power according to the law P=V^2/R. So once a transistor gets hot, it gets hotter until it blows. The question is, "what kind of load looks like a voltage source?" There's actually a quite common load -- any amplifier with parallel output transistors will look like it's driving a constant voltage load, from the perspective of one of the parallel transistors. Basically, hooking up bipolar transistors in parallel does NOT multiply the power rating as you would think, because thermal runaway might cause one of the transistors to dissipate all the power.

deleted 14y ago
[deleted]

rdl 14y ago

The thermal runaway is the "fun" part of the Tesla motor controller, right? It's got 3 MOSFETs, since there isn't a single one with enough power handling, and thus if one of them sucks a little more, it'll end up with more power, which will make it suck more, which will end up with more power...

tlb 14y ago
Bipolar transistors suffer from the thermal runaway of parallel devices you describe, but not MOSFETs. MOSFET on resistance increases with temperature, so with multiple devices in parallel the hottest one will flow the least current. This is an important effect even within one device, which is actually an array of thousands of junctions.

cyanoacry 14y ago

The FETs blew as soon as we turned on the case, oddly enough, and even if we had a single drive plugged in, that FET would blow by itself. This is even the case when we took the backplanes out and started testing them individually in free air in the server room, where ambient is more like 18C instead of 25. Is the design solution to use a FET with a lower Rds at 5V Vds for smaller heat dissipation? But more importantly, is there a standard current that most chassis are tested to? I would expect all SATA hot-swap bays would support all SATA drives on the market, since nobody ever gives power dissipation or consumption figures.

mbell 14y ago
> Is the design solution to use a FET with a lower Rds at 5V Vds for smaller heat dissipation? Honestly - Its to not cheap out on the design and use a proper hot-swap controller. Examples (many companies make these): http://www.ti.com/ww/en/analog/power_management/system-protection-hot-swap.html http://www.ti.com/ww/en/analog/power_management/system-prote... You can rig up something slightly better by designing a current sense circuit with feedback into the FET gate if your adventurous. In either case the goal is to limit and control inrush current during power on or disk insertion. Based on what you've said I doubt the steady state current of the disks is the problem at all. It just sounds like the 3TB drives have a higher inrush current during power up and its either high enough or lasts long enough to blow the FET junction.
- cyanoacry 14y ago
  The funny thing is, I'm not sure that the hot-swap circuitry is even necessary in the first place due to the connector design. The SATA spec defines "pre-charge" voltage pins that are connected after ground, but before any of the other voltage pins are connected. The idea is that by inserting a small (10 ohm) resistor, you can limit inrush current to a tolerable value while capacitors charge and regulators start up, and then when the other pins connect a couple of milliseconds later, the drive gets the proper low-impedance power connection. Do you know if implementing pre-charge via connector mechanics obviates hot-swap protection circuitry? Supermicro seems to have hot-swap protection on their backplanes as well, but I haven't had a chance to closely inspect it.

mindslight 14y ago

Those failures don't really look like moderately too much power dissipation. I'd be worried about static at the drain or counterfeit FETs. I suppose the switcher in the drives could just be drawing several amps as it's input voltage decreased and maybe those FETs don't have the best thermal resistance, but sheesh.

mindslight 14y ago
Oh I see, 62 K/W junction-to-ambient for an SO-8. Whereas the D2PAKs I've had desolder themselves are 1.5 K/W junction-to-case.

gvb 14y ago

A related "gotcha" is not driving the gate hard enough. The 6A rating is also predicated on driving the gate hard enough to drive the on Rds to its minimum value. I would be suspicious of this, in conjunction with inadequate heatsinking (i.e. heavy copper pads under the FETs) especially if the FETs are being driven by 3.3v. Looking at the specs, the FET is rated at 50mOhm given 4.5v gate drive - very acceptable - but if the FET is driven by 3.3v, it will have much higher Rds (may be running in the linear region which would be very bad). Note that the gate threshold voltage is 1.5v typical but 3v max so driving with a 3.3v logic signal would be marginal in the worst case situation.

cnvogel 14y ago
In this application you typically use p-ch mosfets with gate pulled down to turn on, a resistor will provide pull-up of the gate to turn off. So Ugs is the same as the rail voltage (3.3, 5 and 12v).

ChuckMcM 14y ago

Came here to say essentially the same thing. In designing a 400A speed controller for my Battlebot I got to learn all about MOSFETs and their current ratings. In this case the MOSFET is rated with a max current of 5A @ 25 degrees C if the gate is driven with a 5v signal. It has an 80 mOhm resistance when fully enhanced. So at 5A that is 2Watts of power dissipation (p = i^2r) and since the case has a thermal constant of 62 degrees C/W that means the case will be at 149 degrees C (25 + 62 2W) which causes Rds(on) to double and that 'explodes' the MOSFET. If the device was successfully carrying .5A loads then then the case is only good for about 20mW of power dissipation (which looks reasonable given the package size). So gluing a copper heak sink to the FETs with at least 2 sq inches of surface area would probably keep them alive with a .75A load. If you combined that with about 150 CFM of air flow at sea level (225 CFM at 8,000 ft) you'd stay pretty solidly inside the 'not to be exploding' parts of the parameter graphs.