New answers tagged

0

I used the solution I gave in the question with: C1 = 100 nF (min 90.7 nF calculated with Crss = 1.7 nF, Vds = 64 V and Vgsth = 1.2 V) R1 = 47 k (value not critical, BoM reuse) D1 = STPS1L30 Also be warned the inductive voltage spike at turn-off is actually a significant problem contrary to what I thought, you must do something about it if your application ...


1

Your MOSFET driver 1EDI has excellent description ! "The 1EDI EiceDRIVER™ Compact is a general purpose MOSFET gate driver. .... The separated rail-to-rail driver outputs simplify gate resistor selection, save an external high current bypass diode and enhance dV/dt control." So You can control On and Off time simply by selecting two resistors: Ron from Out+...


-1

Why not simply increase the switch-on resistance by adding R3? simulate this circuit – Schematic created using CircuitLab


0

LTC7001 with internal charge pump will do this 100% duty cycle high side MOSFET driving well. LTC7001 is a static high side MOSFET driver which can make MOSFET on for infinite time. The LTspice simulation results show this feature comparing LTC4446 with no such capability.


2

Higher operating voltages translate to higher stepping rates (and higher peak RPM, higher torque at higher RPMs). If you operate the motor from 3.12V, you will get the rated torque at 0 RPM (aka holding force), but performance at any speed will be poor. Since the TMC2130 driver is a constant current PWM driver, higher voltages (up to the rated voltage of ...


1

I simulated your circuit using a similar FET with a transient suppressor added, switching a 'short circuit' drawing 1000 amps from a 50 V battery with 10 nH of wiring inductance. It 'worked' in that it had no effect on the switch-off time or power dissipated during the transition. However in this scenario the transition is slowed down by Miller effect, so (...


0

Besides your suggestion, there is something else you can do in order to reduce the unwanted gate voltage rise due to the parasitic leakage current. Consider the following circuit: The mosfet is permanently grounded through the resistor R1 and a square wave with a slew rate of 100n is applied to its drain. How it works: the leakage current is mostly a ...


2

To turn off that FET, the gate driver Vout needs to be up at 24 volts. This circuit does not provide that.


0

According to its datasheet, the MP6540 has 'automatic' synchronous rectification:- When both the HS-FET and LS-FET are turned off and the voltage on an Sx output pin is driven below ground, the LS-FET is turned on until the current flowing through it reaches near zero or until the HS-FET is commanded to turn on. Similarly, if Sx rises above VIN, the ...


3

Q1 and Q2 are a pair driving an inductor to form a buck (step-down) DC-DC converter. There are two phases of operation: Q1 on: current flows into the inductor L and load, meanwhile the inductor stores energy (builds flux) Q2 on: inductor L energy released, current flows into load as the flux collapses The key is, neither Q1 or Q2 are kept on long enough ...


2

What do you think is the job of the bq24650, here ;)? It switches Q1 (and Q2) on and off in a manner that employs L as the energy storage of a switch-mode power supply (a term that you should research; explaining SMPS here would definitely be out of scope). So, Q1 will never be switched on for long enough for current that you don't want to (effectively) ...


2

Is it possible to only use high side part of IR2103 mosfet driver ic You can: - Use the bottom only The top only Both as half a H bridge Not use both simultaneously (and independently) because of the shoot through protection prevents this (thanks to @StainlessSteelRat for pointing this out): - The lower side MOSFET connection is a high impedance with a ...


0

If I recall rightly, the triode-mode drain current in a long channel ( > 1 micron length) FET is Id = K/2 * W/L * (Vgs - Vt)^2 where K is Mu * Cox and where MU is the mobility.


Top 50 recent answers are included