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5

The resistor is there to control the gate charge current and through that the MOSFET off-to-on time. It's important for the whole bunch of reasons. My personal favorite is that MOSFET going on too fast creates current through the miller capacitance (D to G) and makes itself oscillating (sometimes kills the driver and other circuits, if totally out of control)...


3

Yes you can. When Vds changes, drain-gate capacitance Vdg communicates some current to the gate. Then you have a potential divider formed by Cdg, and the parallel combination of Cgs and the source impedance of the gate drive circuit. simulate this circuit – Schematic created using CircuitLab Note that this is an AC effect, not observable at DC. More ...


3

First of all, I'm sure you ment Vds >= Vgs - Vth for a MOSFET in saturation. Vds is defined as the potential difference between drain and source, Vgs as the potential difference between gate and source. simulate this circuit – Schematic created using CircuitLab By shorting gate and drain, they share the same potential. Therefore, Vgs = Vds. That ...


2

Assuming Vdd is large enough, it is no different from driving Vgs with a separate voltage high enough to completely turn on the MOSFET so it conducts as much as it can and behaves as a switch (ie. a very low resistance) across the source drain.


2

C_GD2 is called the "bootstrap" capacitor. When the MOSFET is off it charges up to VCC<->Source voltage trough bootstrap diode D_GD1. When the MOSFET has to turn on the chip will connect VB to HO, effectively putting the charged capacitor in parallel with the MOSFET gate-source. As the MOSFET turns on the source begins to rise towards V_BAT(x)...


2

After three days of tinkering, I found out that ST is offering two products with a very similar name. I've accidentally purchased STUSB4500LQTR instead of STUSB4500QTR, which does not support USB PD, and has the voltage fixed to 5V. Unfortunately, apart from the name, also the pinout and I2C programming interface are similar. It allows you to set the voltage ...


1

The heat sink doesn't just dissipate heat. When you connect both devices to the same heat sink, the heat sink will quite effectively carry heat from one device to the other. So, even when one of the devices isn't dissipating much (if any) power, if the other device is dissipating power, it's going to heat up both devices. For an extreme example, consider two ...


1

My comment about using a MOSFET as a variable resistor was a general suggestion (before the presence of the schematic), now it became nearly meaningless after seeing the schematic. It appears that your aim is to adjust the output voltage via the Arduino. Then you have these options: As the schematic is a Boost Converter, you can simply remove the IC, place ...


1

I have found a solution: The circuit was behaving as if there was a 25nF output capacitance. I don't know where this was coming from. The losses were all occurring in the MOSFETs since no other components were getting warm. Despite what I said the current draw was proportional to DC Bus voltage and the switching frequency (sorry for misinformation). I ...


1

In principle, active elements are "one-way devices" that are controlled from the input side (gate, base, grid); they are not reversible, such as a resistor that can act as both a voltage-to-current and current-to-voltage converter. But there is a way of connection in which it is possible they to be controlled from the output side (drain, collector, ...


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