34

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

Can you change the ribbon cable, or insert an adapter to a higher pin-count cable? Consider what IDE/ATA did to increase bandwidth -- it was switched from a 40-wire cable to an 80-wire cable, with every other wire inside the cable tied to ground within the connector. A similar solution could apply here. Alternatively, can you reduce the slew rate? At 1 MHz, ...


18

This circuit tends to be used quite often: -


18

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


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

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 ...


15

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 ...


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 ...


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 ...


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

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 ...


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 ...


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 ...


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 ...


8

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 ...


8

If you don't have a fast flyback diode on the motor, the MOSFET will avalanche and it will overheat.


8

Either choose a better MOSFET or use a push-pull driver like this: - Notice that this chip uses identical MOSFETs in the output stage. Here's another using the FAN7842 from Fairchild: - You should also make sure there is enough deadtime between one turning off and the other turning on. Both devices can be used to drive single MOSFET outputs if needed. ...


8

The circuit you show should work. You probably don't have something connected correctly. For example, if the drain and source of the FET were flipped, you'd get exactly the symptom you see due to the body diode of the FET conducting. Check the FET datasheet and your connections carefully. Measure the gate voltage with a voltmeter, and verify that it ...


8

With a mechanical switch or relay? No, that will absolutely not work. Here's what a SATA cable looks like internally. The shiny bare wires at the top and bottom, and in the middle, are drain (ground) wires; the copper wires in between those are what carry the signal. As you can see in the picture, they're embedded in a dielectric material. This is to ...


7

The input is digital at 3.3V and should output 5V TTL levels. For this situation you very likely do not need any conversion circuit at all. Both 3.3 V and 5 V TTL logic switch with a threshold of about 0.8 V. Therefore no conversion circuit is needed to drive a 5 V TTL input with a 3.3 V logic signal. To be absolutely sure, check the minimum Vih of your 5 ...


7

Just for conduction losses, the power dissipation in the MOSFET could typically be \$I^2 \cdot R_{ds(on)}\$ or about 5.4W at Tj = 120°C, assuming 4.5V drive, which your 5V micro should be providing. At only 500Hz the switching losses should not be too bad even with a 100R gate resistor, but they can still add. You need a fairly large heat sink or a fan to ...


7

The best solution is to select the correct relay. The data sheet shows that the part is available with a 9V coil. The 9V relay coil only draws 40mA from your supply whilst the 5V coil will draw 72mA according to the data sheet. The easiest solution if you have to stay with the 5V relay is to simply put a resistor in series with the coil of the relay that ...


7

Start with this: - And then invert the output of the OR gate. Both gates should be schmitt trigger types. The RC time constant and the schmitt trigger high and low thresholds produce the deadband timing.


7

Have a look at the LM124 comparator. simulate this circuit – Schematic created using CircuitLab Figure 1. The LM124 can compare inputs down to 0 V. They feature an open-collector output suitable for driving the LED. How it works: The comparator, as the name suggests, switches its output based on the comparison of its inputs. If IN+ > IN- output ...


7

If the TxD output is high the diode is reverse biased, i.e. it blocks. The voltage at RxD is given by the 10k pull-up resistor connected to \$V_{DD}\$ (the lower positive supply voltage). If TxD output is low the diode is forward biased, i.e. it is conducting and the voltage at RxD is ca. \$0V + V_{forward}\$ (about 0.6V for normal diodes or 0.3V for ...


7

Contact bounce at 10x rated current with inductive lines = Arc + surge current forces to separate the filament. Since breaking an inductive current causes an arc, the switch is now at the filament, so the arc occurs at the switch during contact bounce during a cold filament turn on. This is the short explanation for the switch arc and failure during turn ...


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