-4
\$\begingroup\$

When I use LTSpice to create a MOSFET, it comes with a very high resistance when closed and a very low resistance when open. My guess is that the creators of LTSpice just took two values that are some kind of average for the values that most occur for transistors in the industry.

I guess I assumed that those creators would also use a time delay that is some kind of average of time delay for most such transistors. A certain amount of time passes between the time that one of the inputs changes and the time that the output changes. It may be a very small amount of time, but I'm sure it's nonzero.

However, my attempts to discover what that transistor delay is using LTSpice have met with failure. I've got a schematic where a transistor's inputs change and I've taken the time interval looked at from tens of nanoseconds down to hundredths of femtoseconds, and I still don't see any difference between the time the input changed and the time the output changed in response. This leads me to believe that the creators of LTSpice very possibly programmed that transition to be instantaneous. Is that true? If it's not true, how small is the delay between input change and output change for a typical transistor generated by LTSpice?

\$\endgroup\$
  • \$\begingroup\$ The creator (Mike) of LTspice did nothing like you suggested, and you would do well to investigate more thoroughly before making accusations such as this. LTspice is quite capable of simulating the complex behavior of MOSFETs but the user must choose an appropriate model. I suggest that you convince us that you know what you are doing by providing your schematics and netlist. \$\endgroup\$ – Elliot Alderson Jan 14 at 0:22
  • \$\begingroup\$ Rise time(10~90%) =0.35/f-3dB , group delayTgd=dΦ/dt , which one? or propagation delay in conductors \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 14 at 0:31
  • \$\begingroup\$ You are asking about the time it takes for the output to react at all to any change at all in the input? Not the time it takes for the output to complete its transition after the input has completed its transition? Because you seem to describe both and then use the two interchangeably. \$\endgroup\$ – DKNguyen Jan 14 at 0:46
  • \$\begingroup\$ Please post you LTSpice schematic. There many people here that are very familiar with LTSpice and might be able to spot an error in it. \$\endgroup\$ – Aaron Jan 14 at 1:42
  • \$\begingroup\$ When in doubt, doubt your assumptions 1st \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 14 at 2:24
4
\$\begingroup\$

LTSpice lets you specify numerous parameters for its MOSFET model. For each mosfet, you can specify which model it should follow, and you can then add a ".MODEL" statement to provide the parameters for that model. For example if your circuit contains (for some reason) a 2N7000 and an IRF540, you would include .MODEL statements providing the parameters for those two types, and then for each NMOS in your circuit specify which type it should be.

If you want to see which models are built in to LTSpice, after you place a MOSFET on your schematic, right-click it, then choose "Pick New MOSFET". You'll get a list of a couple hundred models based on real world devices to choose from.

Generally you shouldn't rely on the default model (the one you get when you don't specify a model yourself) to reflect any real world device at all.

As for turn-on delay, if you drove the gate of the MOSFET with an ideal voltage source, then there's no reason to expect there would be a turn-on delay. Turn-on delay happens when the gate driver is current limited, or the gate is driven through a resistive connection. Of course there's a parasitic gate resistance on any MOSFET but this defaults to zero. Again you shouldn't expect the default model to correspond to any real-world device at all.

|improve this answer|||||
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.