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Why can a thick copper wire handle a large current? Because it has a low resistance. As long as you keep the resistance low (switch the MOSFET fully on, for example use Vgs = 10 V as in the datasheet of the IRL7833) then the MOSFET will not dissipate much power. Dissipated power $P$ is: $P = I^2 * R$ so if R is kept low enough the MOSFET can handle ...

29

The reason to use multiple MOSFETs is to lower power dissipation resulting in a cheaper design. Yes one MOSFET can handle the current but it will dissipate some power as it does have some resistance, typically 9 mohm for the IRFB3607. At 25 A that means 25 A * 9 m ohm = 225 mV drop At 25 A that means 25 A * 225 mV = 5.625 W of power dissipation A ...

27

There is no real inherent distinction between breaking one side of a loop or the other side- it's all in series so breaking the negative or the positive side of the supply keeps electrons from flowing. When you have an electronic switch and are turning off part of the circuitry with other circuitry it's easier to break the negative - called a low side ...

27

The blue line shows how the current flows when the switch is open and the motor is off: When the switch is open, the electrons can only flow by going through the LED. Very little current goes through the LED, so very little current can flow through the motor. The LED lights because it only needs a tiny bit of current. The motor doesn't turn because it ...

20

Acting as a closed or open switch is just an extension of it acting as an amplifier at its limits. Imagine you use your weak little fingers to push some buttons to control a massive flood gate. Anything in between full closed and fully opened is throttling the flow of water somehow, but when fully open or fully close it's just acting as a switch to block or ...

19

This circuit tends to be used quite often: -

19

Absolutely. This circuit takes advantages of the properties of a MOSFET to bidirectionally switch a signal between two different voltage levels.

19

When is a MOSFET more appropriate as a switch than a BJT? Answer: 1) a MOSFET is better than a BJT when: When you need really low power. MOSFETs are voltage-controlled. So, you can just charge their Gate once and now you have no more current draw, and they stay on. BJT transistors, on the other hand, are current-controlled, so to keep them on you have to ...

19

You need to re-draw the circuit to be more readable. Something like this: simulate this circuit – Schematic created using CircuitLab This has been simplified to match the diagram in the question. The motor needs the switch to complete the circuit. The LED, however has a return path to the negative side of the battery (GND), thus completing a circuit. ...

18

I actually have some relevant experience on this very subject. Many, many years ago I grabbed a bunch of PS2505 optoisolators which turned out to be PS2506s. No big deal right? It turns out the PS2506s are INCREDIBLY slow compared to the PS2505s. My friend and mentor, Don Shepherd, gave me this sage advice. Choose R1 so that about half of the available ...

18

Lots of problems with your circuit: T1 is used as a emitter follower. Its output will therefore be less than the 3.3 V input. Figure 700 mV for the B-E drop, and the maximum the gate of Q1 is driven to is 2.6 V Q1 is used as a source follower. Its output will thefefore be less than its input. Unlike a BJT, as T1 is, the G-S voltage is not so easily ...

18

You can leave the board design as-is, but make a short adapter on both ends of the cable, and make the actual cable either as a non-ribbon cable (micro coax, this will be the best), or use proper grounding between signal wires. Essentially you need to make a different cable to fit the IDC plugs (or whatever they selected as board-to-cable connector). ...

17

When your article says this (wrongly): - MOSFET in saturation region is preferred to make it work as a switch. It's because it's written by someone who thinks that the name of the equivalent section of the BJT's characteristic is 100% transferable to MOSFETs. To clear this up: - When a MOSFET is operated as an on-switch it works in the triode or ohmic ...

16

simulate this circuit – Schematic created using CircuitLab We know that for a BJT the collector current increases with the increase in base current. For example, a 0.01mA increase in base current has caused a 10mA increase in collector current. Now, let's assume, you have connected the collector and emitter through a copper wire (i.e. short-...

14

Many MOSFETs are characterized to allow overvoltage and the subsequent breakdown of drain to source provided the energy limits in avalanche are not exceeded. Provided you don't exceed these limits the device will not be damaged and will perform to the manufacturer's specification. For example, this is a section of an IRFP254PBF power FET datasheet. Both ...

13

When the lamps fails, it is often the case that the delicate filament collapes on itself, causing a short circuit. This causes a momentary peak of current. So much current that the short almost immediately blows itself open circuit again, due to a teeny explosion. The lamp usually fails when switched on because the resistance of most materials, including ...

13

This is what your circuit looks like: simulate this circuit – Schematic created using CircuitLab When you close the DIPSW the LED will be shorted thus the current will flow through the DIPSW contacts instead of the LED. So it will turn off. If you want to turn the LED on when the DIPSW is closed then you should hook up the circuit according to this: ...

12

I'm pretty sure the transistor is getting switched on all the way The TIP120 is a Darlington transistor and NEVER can get switched "all the way". It will always "drop" at least 0.5 volts across it and reduce the voltage across your solenoid. If you think about the picture above, for the right hand transistor to be "on" there must be about 0.7 volts at its ...

12

The signals are 1 MHz data signals with no ground wire separating them. This is pretty slow, so first check if there are source termination resistors on the driving side. If there are resistors, you can increase their value to lower the slew rate. If there are no source termination resistors, then whatever is driving this cable is going to push ...

12

The only mistake is using the NFET as a high side switch when it should be a low side switch with Vs=0V then with Vgs>=10V you pull down the load cathode and series R from the supply with the drain. So transistors used as switches (FETs and BJT’s) are always inverting. Vgs is chosen from the specs or as a rule of thumb Vgs>2.5 x Vt(max) the threshold of ...

12

You mention an IO pin; this is how you could do it with one IO pin if you are using a microcontroller (assuming one that uses 5V logic): simulate this circuit – Schematic created using CircuitLab When the IO pin is 5V, D2 will light up; when the IO pin is 0V, D1 will light up. Note that you will have to calculate the actual resistor values; different ...

11

Your LEDs may or may not work exactly as you expect, depending on what you expect. The change in brightness when operating the switch will vary depending on whether you have 1 or 9 LEDs lit. With 1 LED, the change will be small, with 9 LEDs, the change will be large. With the switch closed, any number of LEDs will have the same brightness. With the switch ...

11

That's what a relay is. In both cases you have a set of contacts carrying current. In both cases there is a mechanism for moving the switch: in a relay, this is done by a magnetic coil. All techniques available for switch construction can also be applied to relay contacts. The lifetime issue is only visible on relays because they generally get switched a ...

11

It's not as complex as you imagine. Let's look at a slightly re-drawn, but equivalent circuit diagram: simulate this circuit – Schematic created using CircuitLab There's no need for the switch, since we can just assume the switch is active and connecting the $+10\:\text{V}$ power rail to resistor $R_1$. That, plus separating the power supply rails ...

10

Most circuits that control switches are ground-referenced. It is therefore easier for them to drive a ground-based switch than a supply-based switch. The former is termed low side switching and the latter high side switching. Both methods are perfectly valid. Like most things in engineering, it's a tradeoff. No, this is not a indication of the education ...

10

The IRL540 has a parasitic capacitor between drain and source so, when you "turn off" the MOSFET, that capacitor ($C_{OSS}$) takes time to charge via the 1 kohm drain pull-up resistor. To add a little misery/complication, $C_{OSS}$ changes with $V_{DS}$ as per this graph: - So, when you initially turn-off the gate voltage ($V_{GS}$), \\$V_{...

9

You are asking for help adding an LED indicator, but there are some other problems with your circuit. The Pi can become unstable when it is given less than 5.0 volts, and the 1N4007 diodes have a 0.8 or 0.9 volt drop. One solution is to use Schottky diodes such as the 1N5820 which have only a 0.3 volt drop. An alternative to diodes that would provide better ...

9

I am answering my own question because I think it fits better than updating the question. Feel free to comment I update the post accordingly if it does not suit the EE.SE guidelines. As @dim suggested, I have contacted the manufacturer and I have received an answer. The IC inside the Adam-4055 driving Digital Outputs is a ULN2803 (specs, info). The reason ...

9

After the fact, you have a few choices: Use Schmitt trigger input receivers use shielded foil ribbon cable Edit: @Duskwolf has the best solution: I forgot all about the 80 wire cables ( senior's moment ) terminate with 470 pF as a starting value terminate with cable impedance 110-120 Ohms to ground terminate with driver impedance ~ 50 Ohms to Vcc/2 ...

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