Mosfet gate voltage offset questions - Page 1
Forum section: electronics / beginners

author: engelsheidt
author_rank: Newbie
timestamp: December 06, 2020, 10:01:16 pm
content: Hello!A little about me , i haven't formally studied electronics, i do not own or have access to an oscilloscope.I have been researching about MOSFETs to use an IRLZ44N as a switch in a basic boost converter, on the IRLZ44N datasheet Vgs(th) = 1v to 2v, as i undertand this Vgs(th) is the V needed for the MOSFET to START conducting, in order to fully turn on the MOSFET higher V is needed , for the IRLZ44N 5V at the gate is enough (i think) . On the LTSpice file MOSFET_offset.asc i am using Q2 as an emitter follower to reduce the 12v of V1 to 6v, then Q1 is configured as a level shifter and there's a voltage offset in the node between R6 and R3 , the voltage source pin3 represents an arduino UNO pin 3 with a PWM frequency of around 31kHz and 50% duty cycle, my thinking is : since Vgs(th) = 1v to 2v , if a square wave instead of reaching 0V has a minimum voltage of lets say 0.8v and is used to drive an IRLZ44N , will this allow the gate to have something similar to a preload? Will this result in a faster gate charging (since it has capacitance)? Will the MOSFET get hotter if the gate V is not reaching 0v?Thank you!

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 07, 2020, 07:00:46 am
content: Elevated Vgs(off) will probably turn on slightly faster, sure. More importantly, it'll turn off that much slower: the critical point is discharging Cgd, which due to Miller effect occurs as a plateau near Vgs(th) (usually higher, due to load current, but can also be lower due to common source inductance). So you're spending whole, whatever, hundreds of nanoseconds, sitting around Vgs=2-4V, trying to discharge to 0.8V and that voltage drop just doesn't draw much current. Whereas if it were pulling down to 0V, or even negative, the voltage drop would be much higher, and so current flow as well.Why not make the emitter follower work for you? Move Q1 collector to Q2 base, now Q1 acts as an inverting switch with a pull-up to 6V. R3, R6 and C1 are deleted, R7 connects to Q2 emitter directly. Finally, add a diode across Q2, E to B, so Q1 can sink gate current when it turns on. A ~1A schottky will do (anything from BAT54 to 1N5819, B140, etc.). This can pull down to about 0.5V and up to 5.3V, with a few hundred mA in either direction.If you don't need fast switching (only a few kHz, or just generic on-off loads), or full load capacity (up to a few amperes), you can also wire the MCU directly to it (with a modest gate resistor, 100 ohms or so). Actually with a 5V MCU this will give pretty much full capacity (Vgs(on) ~ 5V) but it'll most likely still work even with 3.3V MCUs.Another pure-logic option: use a 74HC04 and wire a few gates in parallel. Use same supply as MCU (preferably 5V; can also use a separate 5V supply, with 3.3V MCU, with HCT type), and a modest gate resistor to the transistor (say 100/N ohms, for N gates in parallel). The full chip can deliver a couple 100mA total, with less DC current consumption and in about the same space as a discrete (BJTs and resistors) solution.Tim

author: David Hess
author_rank: Super Contributor
timestamp: December 07, 2020, 07:50:51 pm
content: Quote from: engelsheidt on December 06, 2020, 10:01:16 pmYes, that allows faster switching by lowering the amount of charge that has to be transferred in or out of the gate, however the particular way you are doing it increases the impedance at the gate limiting the current available for moving charge.What your idea cannot do is change the amount of charge which needs to be moved during the Miller plateau when the drain voltage is changing which is when switching is actually taking place so the advantages of precharging the gate are small and not related to actually switching faster.I agree with T3sl4co1l about changing the level shifter configuration to the one he describes which will take better advantage of the low output impedance available from an emitter follower without a high current passive load resistor.QuoteNo, it will not unless switching is taking longer.

author: engelsheidt
author_rank: Newbie
timestamp: December 08, 2020, 01:02:13 am
content: Quote from: T3sl4co1l on December 07, 2020, 07:00:46 amI really appreciate all the examples you gave me! I've modified my original circuit to match one of your designs, for future reference i attach the circuit in .pdf and .asc (LTspice XVII).