Hot answers tagged

9

If you add V1 and V2 you get 2.7V. Unless there's some other source of resistance in the circuit, that's the battery voltage under about 0.534A load. That's not much load for the 18650, unless it's severely aged or fairly completely discharged. Measure the voltage directly across the battery terminals while the element is active; if it reads near 2.7V try ...


5

You have a massive power MOSFET with a lot of capacitance. If you reduce the size of the MOSFET you can go faster, or change the type to a different technology than silicon. A cheap one that should show a lot (like 5 or 10:1) improvement is the FDD1600N10ALZ. They're cheap enough, but you can run some simulations to see what the predicted performance is. ...


4

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 ...


2

It seems to me like you could just use an emitter follower circuit. If you do this with the NPN transistor then the output voltage would just be Vout=Vd-0.6 If you need a current then just scale the output resistor accordingly. If this answer isn't enough then you need to restate the problem in a more complete way.


2

This MOSFET is only specified for operation at Vgs = 4.5V. The RdsOn curve in the datasheet represents "typical" values but there is quite a lot of dispersion. If you want to be sure that the FET RdsON is below, say 10-20 mOhms at Vgs=3V then you need to have this specified in the datasheet. The voltage across the 5ohm heater is 2.67V. 30mV across the ...


2

There are a few problems with your layout , I can imagine from your resonance. The 10:1 probe 10pF to 15pF causes Drain resonance with your wire jumper inductance from Vcc low ESR 5V Cap to Drain @ 10nH /cm. Solution add 50 OHm series near drain for a test pin The inductance in the power to drain increases the resonant gain which can be reduced by using at ...


2

Basically, in a FET, lower RdsON means a bigger chip which means higher Qg, higher capacitance, ie a slower FET and higher gate drive current if required to switch fast. RdsOn i sonly relevant if you use the FET as a switch, ie fully turned on. Since your circuit uses the FET in linear mode, RdsON is not a relevant selection criteria for your FET. There ...


1

Ignoring the ringing (Which might just be poor scope probe technique) you see basically straight line rise and fall, which says to me that the mosfet capacitance is introducing a slew rate limit. Pick a mosfet with much smaller gate capacitance's or lower the 600R resistor (or both). Or go with a BJT in place of the mosfet, will need a little base ...


Only top voted, non community-wiki answers of a minimum length are eligible