New Open Source Antispark Switch

Oh, they’ll handle that no problem, then.

Actually, since the drain and source pins are contiguous on opposite sides (aside from the gate pin) simple rectangular pads should work with the right spacing. If not, they fit perfectly in the same footprint as the NTMFS5C628NL, which I believe is “DFN5 5x6mm” or “SO-8FL”. What you can do in easyEDA is place the NTMFS5C628NL and simply solder the TPW1R306Pl,L1Q in its place.

I know this because before I switched out the NTMFS5C628NL in my FSESC 4.20, I double checked the package drawings and soldered in the TPW1R306PL,L1Q. The FSESC 4.20 works better than ever xD.

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That’s a valuable tip, I was going to do the package from scratch

@b264 @Pedrodemio here’s the firmware for the attiny85


Design and testing tip, while adjusting fet gate timer, do all your tests with a single fet. That single fet must be able to take the inrush, to wich will be greatly reduced by the gate cap. While testing at the max voltage, use batteries, not a power supply.

Adding // fets will improve the reliability and reduce the Rdson


Thanks Blasto, will see to it

I didn’t have much to work but was thinking, how does the LTC7004 current limiter is better than the Vedder implementation? Would’t we be also operating in the linear region as @b264 explained?

If yes, one way I thought to overcome that is to set the ramping time to a few seconds, but have another MOSFET with a current limiting resistor inline that turns immediately and a few seconds later it’s turned off by 555 timer, the majority of the current would pass by the limited current

Or maybe go back to my initial idea of using two 555 timers, one that turns the current limited MOSFET immediately and turn it off after a few seconds and then another 555 turn all the mains MOSFETs on after a few seconds

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I don’t think there is a problem operating in the linear region as long as you design the circuit correctly to not exceed the maximum ratings for the FET. Every product I have designed that uses a FET to switch power simply ramps the FET slowly enough to not exceed ratings. Don’t use 555 timers, they will make the problem worse, just use a resistor and capacitor on the gate and get the values right to not exceed the current rating of the FET.

Quickest way is to just do a spice simulation. Let me know if you need help with this.

Nice, would gladly accept some help, I have an idea how to do it but trust more in someone else

The LTC7004 datasheet has some formulas to calculate the rise time but I doesn’t take the current into account

On the spice we would just put a transistor being turned on by an RC network while powering a capacitance equal or greater than both VESC’s combined and watching how high the current gets?

My only suggestion for whole this project is not to put mosfet into linear mode they burn instantly doesn’t matter how many 2,3,4 you will put in parallel inrush currents are thousands of amps for few tens of ms you can check your mosfet SOA and will see that these inrush will kill it instantly.



That part I understood, but wouldn’t the soft turn on of the LTC7004 with an adequate rise time make the current low enough? Or since an empty capacitor is almost a short, no matter how long the ramp up is the current would still be big?

LTC7004 by slowing down opening will put Mosfet into linear mode

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Sorry can you help educate a noob? A resistor with a timer would be an effective current limiting precharge circuit for the capacitors, no?

what i don’t like about LTC7004 is the external Vcc that is required (15V MAX)

if you are able to work with that, i’m sure you can it can be done

did a little sim for fun, not sure how accurate it is


like @Kug3lis said, you need to look at the SOA curve of your chosen fet


same circuit but removing the gate cap, you can see the monster current spike, insta blow


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There should be formulas for calculating inrush current, if not, assuming you know the rise time, the inrush current will be (Load Capacitance * System Voltage ) / Rise-Time. For example, system voltage of 40V, load capacitance of 3,000uF, Rise-Time of 100ms, inrush current will be 3,000uF * 40V / 100ms = 1.2A.

The purpose of the RC timing circuit is to limit this inrush current from the hundreds of Amperes to something more manageable such as a couple amperes or even less than an ampere. In this case, since the rise time is significant, the circuit just needs to not exceed the maximum power dissipation of the MOSFET, which in this case is 170W.

From the above example, the peak power dissipation will be 50W (40V * 1.2A).

It has a very low quiescent current, less than 1mA, something like the ZXTR2112F-7 provides a 12V supply without issue.

In order to implement automatic turnoff, bi-directional current sensing and some kind of timing element is required. This is most easily realized with a current sense amp and a microcontroller. It can be done using a current sense amp and window comparator with timing circuit, but that is really annoying to design.

Unfortunately all of the LTC700x series chips with current sensing only have unidirectional current sensing, not sure how it would behave with negative load current. But, if the on-chip current sense amplifier recovers without a problem from negative saturation, then using the LTC7003 in tandem with a MCU can implement automatic turnoff.


Well specialist :smiley: You know better and u have design and replaced design n times :slight_smile: sot block mosfets burn which are rated 4kA impulse currents :smiley:

Mosfets in linear mode at low voltage with high resistance burn like popcorns do even 200, 1000ms Mosfet is not a resistor :smiley:


I have not worked with such high power circuits before, but I have destroyed my fair share of power MOSFETs (mainly from avalanche breakdown), and I have destroyed power MOSFETs in a low side switch which did not implement precharge.

I currently have two switches designed with the LTC7004 working in an electric mountainboard and electric scooter for a few months now, they switch 40V with a load capacitance of 1,200uF - 2,000uF. I have tested them up to 42V at a load capacitance of up to 6,000uF. They have a rise-time of 100ms (might be longer because I didn’t size my capacitors correctly) and use two IPB014N06NATMA1.

If they fail next week, then you have proven your point. There is a load capacitance at which the switches will fail, but that’s why designs have to keep things like those in mind, what is the highest load capacitance and system voltage this device will encounter and can it survive in those conditions?

In this case, longer rise-times will be required to handle large load capacitances and supply voltages.


This doesn’t work, MOSFETs don’t turn on like that, they turn on suddenly at the threshold voltage. Similarly the gate doesn’t rise linearly.

Are you guys discussing a LTC7004 because you need a high side driver? Why not put the FET on the low side?

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The RC timing circuit as described in the LTC7004 datasheet FORCES the gate voltage (referenced to ground of the system!) of the MOSFETs to rise linearly, and the source voltage trails by 2-4V (voltage at which miller region is encountered). The MOSFET is driven in the miller (linear) region, it passes the necessary current through it to have the source voltage rise at the same rate the RC timing network charges.

Because of the way capacitors work, in order for the voltage across them to rise linearly, a constant current must be passed through them.

Basically the MOSFET is being driven to act as a constant current source until the load capacitance is charged to supply voltage, and because of the way the miller region works, the rate of voltage rise ends up being linear.

This is exactly what happened in Blasto’s LTSpice simulation. Constant current through MOSFET (at non-constant drain-source voltage) and linearly decreasing drain-source voltage.

The issue with low side switching is since everything is referenced to ground, it runs the risk of accidentally shorting something to ground, the device being switched ends up floating at some significant voltage above ground.

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Ahh ok, sorry didn’t realise it was a fancy driver.


That’s all starting to make more sense, thanks for the great discussion so far guys

The MOSFET has to be capable of operating like a resistor, otherwise it wouldn’t be possible to build any current source with them