# Tag Info

8

You are misreading the graph. The horizontal axis is the drain to source voltage -- the voltage drop across the MOSFET along the high-current path. It is not the gate to source voltage -- the control voltage. You can think of the graph as showing the safe upper limit on the power lost due to the MOSFET's on resistance. It tells you that you should be ...

8

It's called a wettable flank to make the soldered joint easier to inspect. If part of the pad reaches the side, a correctly soldered joint with enough paste that reflows correctly will develop a little fillet on the exposed side. See figure 5 for "side wall plated": https://www.st.com/resource/en/technical_note/dm00298756-leadless-packages-with-...

7

You need to prevent back emfs from the washer motor destroying your MOSFET when it deactivates. The normal method is a "reverse" diode in parallel with the motor such as this (equally needed for a BJT as a MOSFET): - This suggestion is made assuming that the motor is never operated in reverse - if it is then a slightly different arrangement is ...

6

Not just SiC. All high-voltage MOSFETs. There's no point due to how they are or would be used. SiC is for high voltage and the max gate-source voltage is the limiting factor when using a PMOS high-side switch to simplify gate drive. 30V is pushing it, let alone 600V. So for high-side, high-voltage switches you need gate circuitry anyways no matter what you ...

5

Your output cap is reverse polarity. It's a PMOS so it's required state is inverted relative to the output of the IC. If it were an NMOS requires a high-side floating drive instead. MOSFETs require actively pulling it LO or shorting the gate to source to turn off. Simply disconnecting the drive signal leaves the gate-source capacitance charged thus leaving ...

5

Your problem is that you are using all N-channel MOSFETs. The high side ones are being driven by +12 so they act as source followers and drop several volts. Q2 and Q4 are the ones in question. In order to avoid that, either drive them from higher voltage than 12V and add circuitry to make sure the +/-20V Vgs(max) is never exceeded, or use P-channel MOSFETs ...

5

Safe Operating Area (SOA) limitations could apply during the switching operation. While it is switching the current from the inductive load continues to flow through the MOSFET as it tries to turn off, meanwhile as Vds increases, it is "fighting" the gate driver trying to suck gate charge out of the part via Miller capacitance. It may take several ...

5

Can SiC SoA rating be extrapolated for faster pulse widths? No it can't. I've recently fallen foul of this using a Genesic 1200 volt SiC MOSFET and, well, darn me, but the device failed and I had to eat humble pie and recognize that I had done what you might be thinking of doing. I don't make many mistakes and, the unit was a prototype (so no great loss) ...

4

The 100V rating is indicating the maximum drain-source blocking voltage that the MOSFET can block when turned off. The maximum current you can pass through the MOSFET is mainly a thermal design issue. You need to estimate the power dissipation of the MOSFET and make sure you can provide sufficient cooling using a heatsink. When using the MOSFET with a ...

3

Q2 and Q4 will get particularly hot because you are using them in "source follower" mode, analogous to emitter follower mode of a BJT: the drain is a fixed voltage (12V), and as the MOSFET allows current through, it increases the source voltage, which reduces the gate-to-source voltage - i.e. the MOSFET turns itself off as it turns on! It will find ...

3

Multiple paths exist to determine the turns ratio of a transformer in a flyback converter. One of them is to start from the $BV_{DSS}$ of the selected MOSFET and apply a derating factor for a safe operation. The derating factor is important to make sure that in any worst-case situation (open loop, short circuit, over-voltage situation etc.) the drain-...

3

Lower RDson often means higher gate charge/capacitance which slows down switching and increases switches losses. Things could get hotter depending on the balance between conduction and switching losses. So pay attention to both total gate charge and RDson when choosing.

3

To switch a load where the ground is common using an NMOS, I would need: simulate this circuit – Schematic created using CircuitLab Note how this circuit needs an extra battery to create a voltage that is higher than the voltage on the load. This is needed to apply a large $V_{gs}$ to the NMOS so that it will operate in linear mode and have a low on ...

3

There may be a number of reasons the mosfet failed. it got too hot. Assuming you drive the mosfet correctly, it has an on resistance (Rds on) of 0.077 Ohms. At 20A, the mosfet is wasting around 30W (P = I * I * R = 20 * 20 * 0.077). Without a heatsink the mosfet will get to around 62 (Rth ja) times 30W = 1860 degrees C. The mosfet will have expired well ...

3

LTspice gives me hundreds of watts? I hope it's not relevant. Unfortunately, it is relevant. You have shoot-through, i.e. current flowing through both transistors at the same time. simulate this circuit – Schematic created using CircuitLab Datasheet for BSZ520N15NS3 Datasheet for IRF9640 With the biasing resistors as in the schematic, significant ...

3

Devices like that are meant to be used with a heatsink. The junction to ambient thermal resistance will mainly depend upon that heatsink. You will need to calculate the thermal resistance of the combination. The device manufacturer doesn't have that information to put it into the datasheet. Low power device datasheets can declare a device to ambient thermal ...

3

I believe the process of producing them is more complicated but I am not sure. They do have a higher on-resistance than N-channel MOSFETs this is disadvantageous due to higher power losses. So therefore most people prefer N-channel MOSFETS. Also, they are enabled when applying a negative voltage to the gate compared to the source (Vgs), N-channels can simply ...

2

what is the general guideline and in what theoretical principles is it grounded to choose a MOSFET rating for a load? First check voltage rating, with some margin. For 54V supply voltage, a 60V FET is too tight, a 80V FET would work if you are sure there won't be voltage spikes due to inductive load switching or layout inductance, and a 100V FET would be ...

2

@Andyaka Vgs = 5.0V Look at the data sheet and explain to me how you would expect this MOSFET to be suitable for 20 amps with a Vgs of only 5 volts: - The 5 volt Vgs trace is 2nd from the bottom and a typical device cannot even reach 20 amps with Vgs = 5 volts. what is the general guideline and in what theoretical principles is it grounded to choose a ...

2

NTE does not make parts. The buy normal parts, add their own number then raise the price. Normal parts have a correct detailed datasheet, NTE parts do not. You copied the words wrong. NTE says Breakdown Voltage but you said Cutoff Voltage that is completely different. The breakdown voltage is the voltage that they have breakdown (you do not want breakdown). ...

2

Yes it is possible. However, for mains voltage they're almost never used because it is just impractical as you need much more than just "a MOSFET" to be able to switch a mains powered device. A MOSFET can only switch DC so for mains AC you also need a bridge rectifier. MOSFETs tend to be more sensitive to voltage spikes which are often present on ...

2

So after taking your suggestions, I tried doing Rnmos and Roh in parallel, and the rest in series, and it fits the examples given. Thanks guys!

2

It is two resistance in parallel. Summary (Rnmos*Roh)/(Rnmos+Roh). Document has an example of calculation.

2

Yes, there's a way, with a few more components. Let's talk about your design for a moment. Note that for the load switches to work, the Vgs must enough above the threshold voltage to turn them on fully. Your circuit doesn't do that. While the N-FET gate is at 5V, the drain will never go higher than 5V-Vgs. If the drain goes above that, the FET turns off. So ...

2

I crudely overlaid the two photos you showed to get a better picture of the overall wiring. The transistor appears to be wired correctly. You've got the gate being pulled to ground, driven by a GPIO pin. The drain goes out to your load. The source goes to ground. The most likely cause of failure is that you mixed up the red wires and connected the red wire ...

2

You have a source follower, not high-side switch. MOSFETs are NOT controlled by the gate voltage. They are controlled by the voltage DIFFERENCE between gate and source. How is your MOSFET supposed to know what the voltage at GND is when none of its pins are connected to GND? Yet, you are applying a gate voltage relative to GND. What happens to the voltage at ...

2

IR2113 is a bootstrapped high and low side (half H-bridge) driver. Your MOSFETs are both low-side. Consider using low-side drivers instead. These are basically inexpensive high current buffers, to push enough current into the gate to switch quickly. But it may still work. Here's the internal schematic of the driver: The top driver is normally bootstrapped, ...

2

the inpust H1, H2, L1, L2, VS1 and VS2 are unconnected but that can burn the mosfets? YES. A floating gate means that there is a high chance that the MOSFETs are neither fully on nor fully off. They are probably "somewhere in the middle" meaning that they act like resistors. So current can flow and the MOSFETs heat up and destroy themselves. NEVER ...

2

No, this won't work because you won't be able to apply a high enough gate voltage to turn off the high-side P MOSFETs. You need to bring these gates to 24 V to turn them off. You should be looking for something called an "H-bridge". You should be able find one device that contains all of the logic you need, and works with your desired voltages. ...

2

Having shortened and used larger gauge wire from driver to gate, and most importantly changing my probing technique (x10, shortened ground path) I was able to get a time that was more in line with my calculated estimate (~160ns with a gate resistor of 24Ohms). Probe ground was way too long; I attached my ground to a more suitable location. Below is the ...

Only top voted, non community-wiki answers of a minimum length are eligible