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0

I think what you are suggesting is that you use a MOSFET as the contacts of the relay, allowing current to flow in either direction. That is commonly done in VLSI design using MOSFETs. The key terms include "pass transistor" and "pass gate" or "transmission gate". You might also find some optoisolated MOSFETs that will do what ...


0

The FQP47P06 is a P-Channel mosfet, which means that it will have a negative threshold voltage. You can use it depending on your application. Which in this case would be like this: simulate this circuit – Schematic created using CircuitLab Or if you would like to use a N-Cahnnel Mosfet you could look for FDD8447L-F085 or similar from aliexpress. It ...


5

In this specific case the way you can tell that the second one is a better choice is that Rds(on) is specified down to 4V, but in the first one, Rds(on) is specified only at 10V. In power switching applications I usually try to stick to MOSFET's that have Rds(on) specified at the Vgs I will be using (or lower). Vgs(th) is only marginally useful in figuring ...


0

I suspect not. The problem with an n-channel MOSFET is that the gate needs to be a threshold voltage (Vt) above the terminals in order to turn on. So, your solar panel puts 12V on the source and your battery leaves 12V on the drain and your gate is a maximum of 12V. The transistor in that case is off. You would need to put 12V+Vt on the gate to turn the ...


1

That's why most battery and solar controllers have positive common and switch the negative. If you are not that much constrained, try changing to common positive. In order to switch positive side with N-channel MOS, you need some kind of charge pump driver that will drive the gate with a voltage positive than the source and the drain. See here: Charge Pump-...


1

so my question is can I use n-channel mosfet to pass positive voltage from source to drain? An N channel device as shown in your diagram will pass current from source (the positive input voltage) to drain (the 1N5822) quite naturally because an N channel device has an internal parasitic diode that always allows this: - The symbol for the N channel device ...


2

As a hint, what effects in a MOSFET cause the channel current to depend on the drain-source voltage, even in saturation mode?


2

If you have two indipendent power busses you need 4 bilateral switches. Eeach switch needs either two MOSFETs or (highiy deprecated due much higher losses) one MOSFET and a bridge rectifier. In both cases you also need four indipendente floating gate drivers. simulate this circuit – Schematic created using CircuitLab So it all just sums up to 8 ...


0

Try Snubber circuit. Snubber circuit eliminates or reduces voltage and current ringing. See this document for more details.


3

is a high voltage MOSFET appropriate for this application, keeping the requirement of low leakage current and fast switching in mind? Is there a better component? (I have purchased a Vishay IRFPG50 (1000v, 6.1A)) The types of components that are appropriate depend on what you need your switching to look like. If you just need to turn on quickly and drain ...


4

It is guaranteed to pass less than or equal to 250uA with 1V on the gate and drain. It is guaranteed to pass more than or equal to 250uA with 2.5V on the gate and drain. 250uA with 2.5V across it is about 10K equivalent which is not very much 'ON'. If you want it to be guaranteed to pass a useful amount of current greater than 250uA (like, say 50mA with a ...


-2

The correct way to read this is: ABOVE 2.5V the device is guaranteed to be ON BELOW 1V the device is guaranteed to be OFF Between? do not design for this case IF you are after deterministic behaviour as a ON/OFF device


-1

The question I did not see but was expected to be answered too: what size of heatsink is needed then. This link may be useful for answering that (It boils down to your cooling situation and temperature difference) https://celsiainc.com/resources/calculators/heat-sink-size-calculator/


1

Issues I see with this design: MAX4172 input voltage range issue. The supply of your MAX4172 is + 5 V while the sense inputs are at +12 V. That is not going to work, you will have to supply the MAX4172 with at least 12 V also. The circuit around Q1 and Q2 will probably work fine. However I would use a simpler circuit, I'd use one NMOS to make M2 switch on. ...


-1

TRIAC, that is a component that is used to control higher voltage AC. https://circuitdigest.com/microcontroller-projects/arduino-ac-light-dimmer-using-triac But for motor control have a look at https://create.arduino.cc/projecthub/saulius-bandzevicius/arduino-based-universal-ac-motor-speed-controller-a4ceaf that describes the creators journey into motor ...


-1

There are several ways, and which you use would reflect whether you have any stuff already lying around, or what you can get cheaply. a) Drop the voltage with a transformer. You may have a low power transformer with a split 110/220 V primary. If so, hook up the fan to the mid point. If that's too slow, put the secondary in series with the mid point to lift ...


0

Can you please let us know which mosfet you are using its number, you can control the fan speed using RC circuit also like we have in the fan regulator other cheap way is no need of arduino just use your fan regulator it will do your work and save your money also. But if you still want to use arduino to control the speed of the fan you can use pwm of the ...


-1

Just install a 10 ohm resistor in series, such as from Dale. In gold_colored metal case with 2 mounting tabs to some metal sheet, to remove heat. Or try 1.0uF capacitor in series. At 60Hz, that has impedance of 250 ohms and will allow about an amp at 250 volts.


6

The resistor is there to control the gate charge current and through that the MOSFET off-to-on time. It's important for the whole bunch of reasons. My personal favorite is that MOSFET going on too fast creates current through the miller capacitance (D to G) and makes itself oscillating (sometimes kills the driver and other circuits, if totally out of control)...


0

Here's a way, without the mosfet. The sensor, capable of switching 200 mA, can easily drive the relay's 12.6 mA coil. The 'NC' contact of the relay is used to turn off the pump.


0

I suggest using a logic level n-MOSFET for driving the high side MOSFET. Notice the freewheeling diode which is mandatory when driving an inductive load like a motor. simulate this circuit – Schematic created using CircuitLab


1

The heat sink doesn't just dissipate heat. When you connect both devices to the same heat sink, the heat sink will quite effectively carry heat from one device to the other. So, even when one of the devices isn't dissipating much (if any) power, if the other device is dissipating power, it's going to heat up both devices. For an extreme example, consider two ...


0

Surely it must depend upon the details of the heatsink. I can think of en example where the volume is identical. Envisage a prismatic heat-sink with one device in each half. The heat-sink would require a certain volume and the devices would attain attain a certain junction temperature. Now split the heat-sink between the two-devices with an infinitesimal cut ...


2

C_GD2 is called the "bootstrap" capacitor. When the MOSFET is off it charges up to VCC<->Source voltage trough bootstrap diode D_GD1. When the MOSFET has to turn on the chip will connect VB to HO, effectively putting the charged capacitor in parallel with the MOSFET gate-source. As the MOSFET turns on the source begins to rise towards V_BAT(x)...


1

I have found a solution: The circuit was behaving as if there was a 25nF output capacitance. I don't know where this was coming from. The losses were all occurring in the MOSFETs since no other components were getting warm. Despite what I said the current draw was proportional to DC Bus voltage and the switching frequency (sorry for misinformation). I ...


1

My comment about using a MOSFET as a variable resistor was a general suggestion (before the presence of the schematic), now it became nearly meaningless after seeing the schematic. It appears that your aim is to adjust the output voltage via the Arduino. Then you have these options: As the schematic is a Boost Converter, you can simply remove the IC, place ...


3

First of all, I'm sure you ment Vds >= Vgs - Vth for a MOSFET in saturation. Vds is defined as the potential difference between drain and source, Vgs as the potential difference between gate and source. simulate this circuit – Schematic created using CircuitLab By shorting gate and drain, they share the same potential. Therefore, Vgs = Vds. That ...


2

Assuming Vdd is large enough, it is no different from driving Vgs with a separate voltage high enough to completely turn on the MOSFET so it conducts as much as it can and behaves as a switch (ie. a very low resistance) across the source drain.


0

This is a high-side, single-channel gate driver IC with 100V blocking and level-shifting capability from <2.2V to 10V, while non-inverting. This IC is just like the impedance of CD4000 logic except rated for 100V. Do not assume that any high side logic FET will work well with your reactive load. That would create a large negative spike on shutoff ...


1

In principle, active elements are "one-way devices" that are controlled from the input side (gate, base, grid); they are not reversible, such as a resistor that can act as both a voltage-to-current and current-to-voltage converter. But there is a way of connection in which it is possible they to be controlled from the output side (drain, collector, ...


2

After three days of tinkering, I found out that ST is offering two products with a very similar name. I've accidentally purchased STUSB4500LQTR instead of STUSB4500QTR, which does not support USB PD, and has the voltage fixed to 5V. Unfortunately, apart from the name, also the pinout and I2C programming interface are similar. It allows you to set the voltage ...


3

Yes you can. When Vds changes, drain-gate capacitance Vdg communicates some current to the gate. Then you have a potential divider formed by Cdg, and the parallel combination of Cgs and the source impedance of the gate drive circuit. simulate this circuit – Schematic created using CircuitLab Note that this is an AC effect, not observable at DC. More ...


0

simulate this circuit – Schematic created using CircuitLab There's a couple issues that I see. One being that your Mosfet is only rated for 13V higher than your rail. With that amount of current plus an inductive load, you are likely to see voltage spikes well above that. The second being that your mosfet is going to turn on very slow due to the large ...


0

The keyword here is Yield. Manufacturer produces X number of ICs, tests them, and finds out that Y are good. Yield is Y/X. That's one of the biggest issues with new processes, or when making more complex ICs (adding cores for instance). If the yield is not good, that means your manufacturing cost per functional unit is higher (because you're throwing away a ...


1

I would use a bipolar transistor with its collector to the right terminal of the on/off switch. Emitter to ground. And a resistor to the base that is driven by an output from your MCU. Set the signal high to turn power off. Ensure that the transistor does not get turned on during the power-up and initialization of the MCU.


4

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 ...


10

We don't "botch" them by the time they are sold. We've botched them long before you ever saw an IC with those features. I was working on 14nm SOI in 2012, and the single biggest reason that things don't get out into the wild is yield, but this does not explicitly mean that the transistor did not work. I was making FPGAs because it allowed me to ...


2

There are several reasons for using a gate resistor, like: Damping ringing between wire inductance and gate capacitance; Slowing down the switching transition and preventing spikes (will also help with EMC); Reducing the gate drive current spike to a level the microcontroller can safely handle. If any of these apply for your application, you must use a ...


6

First, when you manufacture such a circuit there must be a decent chance of botching some of these. How do the plants make sure that they all work? By putting additional logic on the chips. For example scan chains. These basically connect all flip flops on the chip into one huge shift register, allowing you to program any value (using an Automatic Test ...


1

MOSFETs are notably ESD sensitive on account of the high impedance of the insulated gate. The gate is insulated from the body of the transistor by a layer of silicon dioxide, which forms a capacitor between the gate metallisation and the body of the MOSFET. This input capacitance is called Ciss. https://techweb.rohm.com/knowledge/si/s-si/03-s-si/4873#:~:text=...


6

there must be a decent chance of botching some of these. They're not made separately; the photolithographic process reproduces an entire layer at once, and layers are built up gradually across the wafer. A huge amount of effort goes into ensuring alignment and linearity across the wafer, as well as the use of very high purity materials. That said, there is ...


1

The simple answer is: To include spare parts on the silicon that can replace the faulty part. A search for yield and fault tolerance will give you an answer. Yield = How many faulty vs functional is the factory producing. To improve the yield its a common practice to add spare parts or redundant routes to come around the fact that the silicon has ...


0

I've found the source for the squeaking noises. It was caused by power supply noise propagating through the voltage divider and making the mosfet 'flicker' when approaching it's cut off voltage. Adding a capacitor to the gate pin of the transistor solved the issue. I hade a 10u cap laying around and that's what I used. Just leaving it here in case somebody ...


0

Although you have resolved the problem, but since the actual cause is still unknown, I will suggest the following; Assuming a defective FET is ruled out, it is very possible that the problem was coming from the LATCH circuit. Considering that your pull-up resistor for Q4 is 100K, any leakage current in the LATCH circuit will result in a Vgs that can keep the ...


2

Briefly connecting both motors to the same terminal (either works, yours chose positive) shorts out the motor. This brakes it, which reduces current surges while reversing. What the MOSFET is for is anybody's guess, unless you know what is driving the gate. It could provide soft starting, or an electronic disconnect in the event of overload, overheating, etc....


2

Notice the fine print on DMP3056 datasheet about how RthJA of 91°C/W is measured... Note 5: Device mounted on FR-4 substrate PC board, 2oz copper, with thermal vias to bottom layer 1inch square copper plate. The other FET specs 131°C/W but... (a) For a device surface mounted on 25mm x 25mm FR4 PCB with high coverage of single sided 1oz copper, in still ...


1

Your circuit is working as intended. It is not an oscillation. Your inductance, L4, is very large so it takes a long time for current to build up, longer than your MOSFET actually conducts for. Every time you the MOSFET blocks, the current circulates around the inductor through D1 and you turn on the MOSFET again before the magnetic field in the inductor has ...


-4

Basically the reasons of that are: Use resistors the lower wattage than the mosfet needs. Short circuit in the circuit, making the mosfet die. Bad cooler or no disipation of temperature in the mosfet, it includes no disipator, no fan, no resistor. Overvoltage it can exceed the temperature, and make the mosfet burns like a FIRE!, very harmfull. Bad use of ...


0

This is how I see it. When Vin = 0 the device is turned off and Vout = VDD (This point is important). So when Vin = small voltage = Vgs; Vout is finally something less than VDD. Now as you mentioned; Vout > Vin - Vth is the condition for to work in saturation. So when you are starting, Vin is low and naturally it can't cross the Vout as it's close to VDD, ...


1

This will be easier if you add reference designators to all components. 2a) 150 mW (or 167 mW) is not a lot, although it would help to know the part number or at least the package type of the internal MOSFET to determine its power handling abilities. 1a) Look into an emitter follower circuit. This will give a more stable output voltage. Example: Increase ...


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