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13

According to datasheet: But you're not driving the gate with 10V, rather it's close to 6-7V so RdsON will be higher. Note this is a guaranteed maximum value. Actual value may be lower. As Andy says in his answer, if the FET follows the "typical" characteristics graph, its RdsON will probably be around 0.7 ohms, but notice the text in the corner of ...


6

So what is it then? This is the correct graph: - Taken from this wiki page. Saturation refers to the channel being saturated and, as you said, no matter what \$V_{DS}\$ you apply, current remains constant. It is sometimes also referred to as the active region (not to be confused with the MOSFET being activated or ON). The linear (triode) or ohmic region is ...


5

I've builded this simple circuit that has the purpose to swiching on ad off a 12 V / 25 W lamp That's a load current of about 2 amps (25 watts divided by 12 volts). The IRF730 MOSFET just isn't very suitable for a 2 amp load: - Whether the gate control voltage is 6 volts or 10 volts you will get about 2.5 watts to 3 watts dissipated in the MOSFET and that ...


4

Capacitor: 40uF, DC Power Source: 100V, DC Switching time: ~us (originally nanoseconds) Current in a capacitor can be calculated from the derivative vs time of voltage across it: \$ i = C dv/dt \$ With dt=1µs, dv=100V, C=40µF we got i=4000 A. This is quite impractical. If the charger does not use inductive energy storage, instead simply connecting a ...


4

On the IRF730's datasheet I can read that it can handle more than 7 A Never trust the current ratings given on a MOSFET datasheet. They are obtained under artificial lab conditions you will never achieve in practice and are essentially useless. You can use them as a metric to non-rigorously compare the relative current capability of MOSFETs, but that is ...


4

this circuit needs the mosfets to switch once every 5 minutes (not pwm, that's why I didn't put a driver), does 12V reach all gates? – Ricardo Casimiro 4 mins ago Yes all the current is drawn by the 10k resistor to 11.7V with Vce(sat) on the opto. or > 1mA This works well with 4mA input or more. If using 3.3V input (3.3V-1.2)/(220+25 cmos) = 8.5mA If ...


3

Two cascaded common emitter BJT amplifiers will have many more times the voltage and current gain of a single BJT amplifier. Because of this the threshold of the LED turning on and off is much more singular in nature as you adjust the potentiometer.


3

It's a P-channel MOSFET. But it's in backwards, so it's basically acting as a diode due to the body diode. That's ok, as I will explain below. Assume that the battery is one LiPo cell (3.7 ~ 4.1V max.) With the V_USB off, then the FET body diode is forward biased. The FET is also turned on (fun fact: current can flow in either direction in a FET.) Now turn ...


3

There are a lot of improvements that could be made to this design. Fundamentals that need to be taught Vgs gate voltage for low resistance needs to be at least 250% time the worst-case threshold Vt ( Vgs(th) ) for classic FET types from 2 to 4V. This means 10V min. for Vgs The subthreshold types <=1V only need >= 200% times the threshold voltage. ...


3

Capacitor: 40uF, DC Power Source: 100V, DC Switching time: ~ns Source Output impedance: ~50Ohm The charge/discharge rate should be in the nanosecond range To charge a 40 μF capacitor from a source with 50 Ω impedance cannot be done in nano seconds. The time taken to charge the capacitor to approximately 63.2% of 100 volts will be: - $$\text{50 Ω } \times \...


2

Your answer is approximation. The following is a graph of output characteristics for an arbitrary 2N7000 MOSFET:- Assume a fixed gate drive \$ V_{GS} \$ of 5V and ignore all the other curves. So it's fixed. Consider the MOSFET then as an opaque component with only two legs, Drain and Source. As \$V_{DS}\$ (voltage across component) changes, \$I_D\$ (...


2

A 12V 50 W high beam+low beam bulb in series will limit current to 2.4A as the Cap charges up. dt= 33F * 12V / 2.4A = 165 seconds. The bulb will draw more current as it dims below visibility and speed up a bit.


1

Your wire is not defined on the schematic. Typically Delays are made for Half Bridges to delay turn-on to avoid short circuit or shoot -thru and depend on load inductance and FET Capacitance which I modelled here for a power FET with Ciss. The Turn-ON delay is circled and OFF is assumed to be x ns with low diode resistance. You must define more details on ...


1

If I pull up the IN pin to 3.3V or 5V will my MOSFET turn on, and if I give a logic 0 signal to IN pin will the MOSFET turn off. Yes. Applies to all three gate drivers mentioned. None of them have inverted inputs or outputs. You can look for the timing diagrams in the datasheets to see for yourself. That said, there’s a catch. None of the gate drivers you ...


1

to turn the mosfet on you need to put the gate voltage to be 3.3V above the source voltage. This means that if you want the battery voltage on the source you need more voltage than the battery has on the gate. these drivers all use capacitors to make the high voltage by adding VCC to it, but it only makes a pulse of charge, so if you're not occasionally ...


1

In Fig 2.6(c), it's not negative ions that are under the gate, but negative carriers (electrons) that are mobile and free to move in their plane. Underneath that there is an insulating depletion region (fixed negative charge from the p-type material where holes have been repelled from), and underneath that there is conductive p-type material. The depletion ...


1

Is this an acceptable way to drive a mosfet? Quite possibly not acceptable - you have to ensure that the gate drive voltage can fully turn on the MOSFET and, because the MOSFET source is expected to reach the same voltage as the +14.5 volt supply, the gate drive voltage needs to be several volts above that (anything between 4 volts and 10 volts higher). So ...


1

The circuit works fine as a high side power switch with bidirectional isolation when off. For example, if you use it with a battery, it can prevent the battery from both charging and discharging. But it cannot turn on when both A and B nodes are near or below ground. The reason is that the sources will be pulled down to ground and if A and B are both also ...


1

I've used this Electrical forum for the first time and I'm very impressed for the real high quality of all replays. I'm a teacher, as I said, and I've very appreciated all explanations supported by diagrams and datasheets. But let me go to my results (for now only simultated on Multisim). Setting Vds to 10 V, changes very little because the Pd is about 2,80 ...


1

Use a P-channel MOSFET eg. AO3401A to cut Vcc. Low to the gate = ON. simulate this circuit – Schematic created using CircuitLab Add a 1K resistor series with the TRIG input and make sure that you drive TRIG low before shutting off the Vcc and do not drive it high until after Vcc has been turned on. Also do not drive ECHO high as an output (it should ...


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