I found a few teardown videos for the Daly BMS and the JK-BMS. These BMS are rated for higher currents like 150A - 250A. Based on the teardown, the 16S Daly BMS uses JMSH0804AE N-channel MOSFET. According to the datasheet, JMSH0804AE has a Vds of 85V and a continuous Id of 139A, but if we check its safe operating area (SOA) then we see that the device is capable of handling 10A maximum at 80V for 100us. Daly has used twenty JMSH0804AE to handle the current.

My thoughts were that our desired output current must always be within the DC graph in the SOA. The way Daly has selected its MOSFET is perplexing. Moreover, I have only mentioned the Daly BMS, but the JK-BMS has also selected their MOSFETs similarly. Thus, I wish to understand the role of SOA in FET selection, and when can we ignore this rule.

  • \$\begingroup\$ A full circuit would be very useful. Even a circuit of the area under concern - source, load, voltages, drive, ... would be useful. || The comment by howardschlunder9754 at the top of the second video seems at least useful. And the results after 200A x 5 minutes suggest that results are about what would be expected. || Many will not want to watch videos to get information. What key points are made in them? What heat sink temperatures di they achieve? || "Off the cuff" with an infinite heatsink things should be "doable". \$\endgroup\$
    – Russell McMahon
    Commented Dec 27, 2023 at 8:13
  • \$\begingroup\$ I skipped through the videos. They don't show measurements of the switching signals or schematics. Can you please include schematics or relevant sections of the BMS datasheets? How did you select the 100µs ON time? The SOA basically describes how long the MOSFET can be turned on by given current and voltage. \$\endgroup\$ Commented Dec 27, 2023 at 8:15

2 Answers 2


These BMS are rated for higher currents like 150A - 250A

MOSFET conduction losses are RI², and a BMS is usually designed for continuous use at the rated current. So high currents like 250A mean you need ridiculously low RdsON to compensate, along with lots of copper, because every fraction of a milliohm counts.

Losses for paralleled MOSFETs.

  • Number of MOSFETs in parallel: N

  • All MOSFETs have the same RdsON

  • Total current: Itotal

Current per MOSFET: Itotal/N

Power per MOSFET: RdsON (Itotal/N)²

Total power loss: N RdsON (Itotal/N)² = (RdsON/N) Itotal²

As expected, the result is exactly the same as one bigger MOSFET with (RdsON/N) resistance carrying the total current.

Considering only one model of MOSFET, total loss is inverse proportional to the number of devices. Also the heat is spread over a large area, which makes it much easier to handle.

The table show total losses:

enter image description here

BMS subtleties:

Due to body diode, one MOSFET can only block current in the direction where the diode does not conduct. In the diode direction... if the FET is turned off, current will pass through the diode.

enter image description here

BMS is supposed to block current in both directions, so that needs two MOSFETs in series, back to back.

So if your BMS has 20 MOSFETs total, it has two groups of 10 parallel MOSFETs, and current goes through both.

Each group of 10 parallel FETs acts like a 10x lower RdsON FET, but there are two groups in series, so the total BMS resistance is only 5x lower than the RdsON of one FET.

You also don't want them to get too hot, because:

  • RdsON increases with temperature (+50% at 100°C) so more losses mean... even more losses

  • All the wasted heat comes from the battery, so it reduces runtime

  • The FETs are often next to Lithium cells, which are sensitive to temperature

  • Most BMS have no fan or other means of getting rid of lots of heat (although this one does)/

They could have used fewer FETs with lower RdsON, say 1mOhm. Maybe they got a good price on those, or they have a huge stock of them.

Another reason to use a large number of MOSFETs is to be able to switch off in overload condition (short circuit) without blowing a FET.

According to the datasheet, JMSH0804AE has a Vds of 85V and a continuous Id of 139A

It is rated for a continuous Id of 139A at Tc=25°C. This is a standard datasheet spec which is meant to allow you to compare MOSFETs: it combines RdsON and RthJC, ie how much heat it's going to dissipate (RI²) and how good it is at conducting this heat from the chip to the thermal interface part of the package (RthJC to the D²PAK tab).

It is a hypothetical number, the FET is not intended to be run at this current. With 3.6mOhm, it would dissipate 70W at 139A, and it's a SMD package, so to obtain Tc=25°C (on the tab of the package) you would need to solder it on a watercooled copper block. It's never going to happen in real use.

Besides, they also give a lower maximum current, limited by the package bondwires (see note 6 in the datasheet).

Regarding SOA graph:

enter image description here

When conducting DC current, it's on the blue line: Vds=RdsON*Id. It is limited by dissipation and how it is cooled.

SOA for switching depends on the load. With an inductive load, for example in a buck converter, as the FET turns off, current will remain pretty constant as Vds rises. Current only falls when Vds allows it to flow through the other MOSFET or diode. So you get a I-V trajectory like the one drawn in purple. In a non-inductive load you may get an easier I-V trajectory like the one in red. In both cases, during switching, there is a small time window where dissipation is very high (high current * high Vds). The SOA graph gives information about the maximum allowed switching time the FET can withstand.

If you want it to be able to switch off in case of an output short, the maximum current allowed in the SOA graph is important. Here it's around 3-400A for 80V, so with 20 of these FETs in parallel you could (in theory) switch off 6000-8000A. More than that, and you can expect explosive disassembly. If the battery pack runs at 67V and a short occurs, there's not much resistance in the circuit to limit current, maybe a few mOhms, so these extreme current numbers are what you'd expect to find.

Circuit inductance does limit the rate of rise of the current though, so it is important to have a fast current sense comparator and gate driver to turn off the FETs as quickly as possible before current keeps rising. This should always be implemented in hardware, never as software, there's no time for interrupt latency.

Further reading:

Power MOSFETs in linear mode: application of SOA for linear mode (not for switching).

Understanding power MOSFET data sheet parameters

Failure signature of electrical overstress on power MOSFETs

  • \$\begingroup\$ If I (extremely) broadly summarize, I can operate this FET at 30V and (continuous) 2A (10ms line), given I can dissipate 60W of heat from the FET, am I right? \$\endgroup\$
    – Momobear
    Commented Dec 27, 2023 at 12:47
  • \$\begingroup\$ "Continuous" means you have to use the "DC" line in the SOA plot, so current would be limited to 1A. When you see a dual slope in the SOA (like this one) it means the MOSFET is subject to Spirito instability: in linear mode, Vgs threshold goes down at high temperature, so hotter areas of the chip have a lower threshold, which means they pass more current and get even hotter. These hotspots can overheat and destroy the chip. It's similar to second breakdown in BJTs. If you want to use a FET in linear mode, to make an amp or a current source, it's better to pick an older, higher RdsON model \$\endgroup\$
    – bobflux
    Commented Dec 27, 2023 at 12:53
  • \$\begingroup\$ as these are less vulnerable to this problem. See here: nexperia.com/applications/interactive-app-notes/… Note this only applies in linear mode. If you want to use it as a switch, to switch 30V @2A, then when the FET is open, Vgs=30V but Id=0, and when it is conducting Vgs is tiny and Id is 2A. In this case there is no problem. \$\endgroup\$
    – bobflux
    Commented Dec 27, 2023 at 12:54
  • 1
    \$\begingroup\$ The BMS has a lot of mosfets in parallel so current is shared between them. If the MOSFET doesn't have DC SOA specified then you shouldn't use it in linear mode. As a switch, it's fine, because RdsON increases with temperature, so there are no hotspots. \$\endgroup\$
    – bobflux
    Commented Dec 27, 2023 at 14:44
  • 1
    \$\begingroup\$ @Momobear Just in case it wasn't clear, when Bobflux said " ... When you see a dual slope in the SOA (like this one) ..." he meant "when you see SOA line(s) with a kink/change of slope in them .." . || In such cases the SOA curves MUST be followed with great care for linear operation. \$\endgroup\$
    – Russell McMahon
    Commented Dec 28, 2023 at 9:48

You can’t ignore any of the specifications. They’re stated for a reason.

As soon as you pass current through the MOSFET, it is going to heat up due to losses associated with mainly RdsOn. There will also be a voltage drop across DS. You need to ensure that you have adequate heatsinking and that you stay on the safe side of the SOA line.

Why did the BMS designer use 20 devices in parallel? Price probably. They might be using zillions of them in other products and get them for 5c each. The large board area allows for heavy tracks and some heatsinking and since it is not required to switch fast, the combined gate capacitance is not an issue.

If price was not an issue but you wanted small size, high speed and/or light weight, then one would most likely choose differently. That’s what engineers are paid to do.

  • \$\begingroup\$ Even if the device is working, as an engineer should we implement this kind of solution? Won't IGBTs be better? \$\endgroup\$
    – Momobear
    Commented Dec 27, 2023 at 13:15
  • \$\begingroup\$ The charging and discharging FETs are supposedly split if equal. Therefore, each condition gets ten FETs. At 250A and 67V, the power dissipation will be 1.67KW. Let's take into consideration the diode in the FET, then the power dissipation will be 167W per FET, this is still too much power. \$\endgroup\$
    – Momobear
    Commented Dec 27, 2023 at 13:34
  • \$\begingroup\$ What would make an IGBT ‘better’ in this application? I’d suggest that the forward voltage drop might make it a poor choice. Don’t confuse the system power with the power dissipated in each device. Assuming 10 devices, each device sees 25A. Factor in the RdsOn. What does each device dissipate? \$\endgroup\$
    – Kartman
    Commented Dec 27, 2023 at 22:11

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