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I have chosen the SiHK045N60E MOSFET for my circuit, but it has 4 different pins (standard MOSFETs always have 3: gate, source, drain). In this case there is an extra pin called 'driver source'.

My circuit is a synchronous buck converter, so I have one high-side and one low-side MOSFET, and I do not know where should I connect the driver source pin in both situations.

enter image description here

I found this picture, so I guess I should connect it to ground (negative component of the PWM):

enter image description here

This is my circuit (with 3 pins):

enter image description here

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    \$\begingroup\$ I think it's just another source pin, but routed independently from the source on the die to the pins on the package. \$\endgroup\$
    – Oskar Skog
    Commented Oct 14, 2022 at 6:32
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    \$\begingroup\$ eepower.com/industry-articles/… \$\endgroup\$
    – Oskar Skog
    Commented Oct 14, 2022 at 6:33
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    \$\begingroup\$ You can't connect the high-side driver source to ground. Show your proposed circuit using 3 pin mosfets please. \$\endgroup\$
    – Andy aka
    Commented Oct 14, 2022 at 6:55
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    \$\begingroup\$ An extra source pin for the MOSFET driver is also called Kelvin source, see electronics.stackexchange.com/questions/505030/… \$\endgroup\$ Commented Oct 14, 2022 at 8:15
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    \$\begingroup\$ High side "driver source" connected to SW? \$\endgroup\$
    – Antonio51
    Commented Oct 14, 2022 at 8:17

2 Answers 2

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The device is capable of switching 480 volts at 17 amps with a fall time of 14 nanoseconds. There is an internal inductance (shown as Lsource on your equivalent circuit diagram) in the current path from the die inside the part out to the pins. The instantaneous voltage (labeled VLsource) on this inductance is proportional to the rate of change of the current (V = L * di/dt). Even a very small inductance will produce a voltage spike on this current path, since di/dt can be 17amps/14e-9sec. This voltage can be significant compared the gate-source threshold voltage, and cause the part to turn "off" if the gate drive voltage is referenced to the current-carrying source pins. The part can turn "on" and "off" more than once on the edge of each switch, increasing switching time and reducing efficiency.

The part has provided a second, separate path between the source on the die and the package, which would not be in the high current path, and thus would not exhibit the same induced voltage from the high current. In your circuit, you could connect this lead on Q1 to your SW signal, and connect the other source leads to LF. Then your switch would be more closely referenced to the voltage on the source die, since your bootstrap charging current should be significantly lower than the current you are switching. If you are not switching high currents, you probably don't need to worry with this.

Good luck!

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    \$\begingroup\$ to illustrate: 17 amps / 14e-9 seconds is more than 1 gigaamp per second. That means every nanohenry of inductance creates more than a volt. \$\endgroup\$ Commented Oct 14, 2022 at 15:04
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Your circuit shows a glimpse of DC/DC controller similar to LM5146-Q1, so it seems quite natural to search TI's online documents for design examples where this type controllers are driving external power transistors having the Kelvin source pin. If the search gives no sensible results, the examination of designs with controllers for GaN FET transistors is useful, because GaN FET transistors also exploit Kelvin connection.

GaN FET power transistors in the circuit Simplified Typical Application, datasheet LMG1210 half-bridge MOSFET and GaN FET driver, page 1

LMG1210-typical-application

have a Kelvin connection pin, although the connection is shown only for a low-side transistor. But the document TIDA-01634, Multi-MHz GaN Power Stage Reference Design for High-Speed DC/DC Converters shows both high-side and low-side power transistors with Kelvin connection pins connected to the gate driver IC.

TIDA-01634

The low-side transistor's Kelvin pin is connected to VSS, low-side ground return. There is no pin in LM5146 specific for low-side ground return; I think you can continue using PGND for this purpose, only take care of PCB wiring recommendations as given in the manufacturers' documents for low-loss switching designs. The high-side transistor's Kelvin pin is connected to HS, a switch node and high-side ground return, which is a functional analog of LM5146's pin SW: it is used to reference the internal high-side gate driver input against a true source voltage of the external high-side transistor having the Kelvin source pin. Notice that the circuit shows no direct connection between the high-side-transistor-power-source/low-side-transistor-drain node AND the controller.

However different might be the two controllers (LM5146-Q1 and LMG1210) in their implementation, the comparison of their functions supports the decision to connect the driver source pin of high-side power transistor to SW. To provide the advantage of lowering switching losses with the help of a Kelvin source pin, the internal gate driver of the controller should be given the true gate-source voltage not distorted by induced voltage from drain-source current. Therefore, the power source pin(s) of the high-side external transistor and the controller's pin SW of your original 3-pin design should be disconnected (red circle in the drawing).

updated For4pins

But, in the LM5146-Q1 controller, the voltage from SW pin is also used in the adaptive deadtime delay subcircuit. How disconnecting this pin from external transistor high-current paths may affect switching losses, cannot be predicted without knowing the details of the controller internal circuitry.

adaptive-deadtime-delay

LM5146-Q1 might be designed lacking capability to benefit from driving Kelvin-pin-enabled transistors. Having no direct instructions from TI documents, one can only test, measure, and analyze design improvements in order to come up with a better solution (if it is ever possible). Or you can ask the advice of TI engineers on e2e.ti.com forums.

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