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Understanding The Gate Of A MOSFET
MOSFETs are remarkable devices which provide many benefits when driving various loads. The fact that they are voltage driven and that, when on, they have very low resistances make them the device of choice for many applications.
However, how the gate actually works is probably one of the least understood characteristics ...
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simulate this circuit – Schematic created using CircuitLab
Note 1: The input voltages are only \$V_{cc}\$ and \$V_\text{High Voltage}\$. You don't apply anything at the \$V_{BS}\$ node. It is only for representation.
Note 2: Notice that there are two different type of grounds. Those grounds must not be directly connected to each other.
You must drive ...
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If you selected this driver, which has huge output current (7A) then I presume you need this gate drive current to switch a very large FET very fast.
The gate resistor will only slow things down by reducing gate drive current, so its optimum value is zero ohms. Its maximum value depends on acceptable switching losses (switching slower causes more switching ...
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I've answered a question very similar to this here (How do I design correctly ground plane separation for Texas Instruments TPS63060 IC?), but I'll tweak a reply for you here.
IRF is asking you to keep those grounds "separate" in the sense that they do not want (as an example) 5A of current flowing through the output switches/stages to perturb the ground ...
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First off, the rules of the site do state not to ask for recommendations of products, so I will skip that bit. Just read the datasheets as everything will be explained in there. If there is something on a datasheet you do not understand, please post a separate question about it.
Now, on to your problem. From what I think you are trying to do, you may find ...
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Holy Carp! You're trying to do 10's of nsec switching on a solderless breadboard? And you don't have a flyback diode on your transformer?
If you're going to do this stuff, you've got to learn to respect fast switching and inductive parasitics. Go to a ground plane and make all your switching paths as short as possible. Also, put a 100 uF cap (tantalum for ...
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These transistors don't conduct unless Vbe>0.6V for the NPN, Vbe<-0.6V for the PNP. And as the bases and emitters are tied together, it is impossible for both these conditions to be true at the same time. So when one transistor is turned on, the other is turned off.
HOWEVER
if R2 is too low, the transistor being turned on will "saturate". And when ...
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Lose the diode. I can't even guess what you think it does, but it will prevent the FET from being turned off quickly.
The FET driver will drive the gate high quickly, which turns on the motor. However, when it tries to drive the gate low, the diode prevents it from doing so. That means the gate just floats. It probably slowly drifts low, running the FET ...
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This is indeed an interesting problem, because of the variation of effective load capacitance with the load resistance due to Mr. Miller, and your need to not overcompensate it.
I suspect a biased push-pull BJT output driver would work fine- maybe 4 small BJTs (2 connected as diodes) a couple bias resistors plus maybe a couple ohms each of emitter ...
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The trouble with using a high side P channel MOSFET driven from a signal that doesn't get close (less than 0.5 volts) to the high side voltage is that there is a decent probability that it will appear to be still active when you believe you have it turned off.
However, with some care you can put a zener diode in series with your 3.3 volt GPIO drive voltage ...
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which is best for fast switching do I need to use MOSFET driver IC such as ICL766 with IRFZ44n or is it better to use lower VGS MOSFET such as AO3400
You don't want fast switching to drive LED strips! If the FET switches in a couple tens of nanoseconds, you will send very high di/dt currents into your wires which will radiate electromagnetic interference ...
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Here is your circuit being discussed:
R3 and R5 should not be getting "hot".
Do the math. Even if Q1 were a perfect switch, there would only be 5 V across R3. (5 V)2(1 kΩ) = 25 mW. Unless this is a very tiny resistor, you wouldn't normally notice it getting warm. A 0805, for example, can usually dissipate about 150 mW safely in open air on a ...
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Figure 1. When Q1 is off both Q2 and Q3 are forward biased (green) and will turn on. The result will be shoot-through the two transistors (red).
If the supply voltage doesn't collapse (1) will be at about 4.3 V, (2) at 2.5 V and (3) at 0.7 V.
Ignoring R3 the current through Q2, R4, R5 and Q3 will be about \$ \frac {5 - 0.7=0.7}{2k} = 1.8 \ \text {mA} \$. I ...
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In a true totem pole configuration, shoot thru usually does occur for a very short time during switching.
However, what you have is not a totem pole configuration. You have two emitter followers back to back. In this case, you will not get shoot thru. For each transistor to be on, the base has to be one junction drop towards the collector voltage from ...
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Here is another approach that uses an N-MOSFET in a level shifter configuration that does not invert the control signal polarity.
simulate this circuit – Schematic created using CircuitLab
You need to chose MOSFETs with under 1V gate thresholds, M1 needs to be low Rds-ON. M2 can be a small signal device.
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To protect against possible future edits, here is the circuit we're talking about:
Major problem are:
There are no component designators in the schematic, making the circuit hard to talk about.
The 1 mF cap makes absolutely no sense. You have a double emitter follower driving the FET gate. This provides substantially more current capability than the ...
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On your layout the decoupling cap (finger-painted red) is very far away from the chip, there is a 5R resistor in series (purple) and the current loop area (highlighted in yellow) is quite large.
This large loop area adds inductance to the power supply impedance as seen from the chip. Wet finger in the wind, about 10-20 nH. With a di/dt around 1A per 1-2ns ...
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They got destroyed due to inductive kickback. The TVS diode is not a correct choice, you should use a schottky diode at that position.
When the MOSFETs were switched off, the motor induced a high voltage, that was clamped with TVS, but these devices are meant to surge a short pulse, not to dump entire motor energy, so it has blown, consequently MOSFETs also ...
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A combination of the IRF840's reverse transfer capacitance (120pF), the dv/dt of the drain voltage and the rather weak driver (MCP1402) is my best guess.
Starters, read the data sheet on the driver - it says on page 3 that the "latch-up protection withstand reverse current" is typically greater than 0.5 amps - that is a clue to why that device might be ...
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Outcome Report
Okay, the short story is: adding a discrete buffer worked! That said, I don't think I'll design my circuit this way, rather I'll go with the recommendation of @Spehro and @WhatRoughBeast and just use an op amp with higher current output capability, basically having the buffer stage built right into the op amp.
Here's the circuit I used. ...
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Sometimes...
Assuming the point of interest is power MOSFETS and not small signal MOSFETS and silicon (as oppose to SiC, GaN)
The first characteristic to check is the output voltage. For power devices they should be 0V to 12-15V (acpl-312T) to cater for gate thresholds around 4V (as well as being able to drive to -15V if miller turn-on is a concern). As ...
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There is nothing wrong with driving a FET like this if there is no need to PWM.
What I would add is
Add a series gate resistor ( ~100R)
consider adding a 15V zener gate-source just incase there are some transients. You do not want to burn the gate region out
NOTE: you might want something bigger than just a Zener on the gate. TVS maybe. its dependent ...
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That is a pretty weird way of "driving" LEDs.
OP's schematic:
A few issues:
essentially you have an "equivalent" resistor, shared by the LEDs (if the LEDs have any imbalance - which they have - the current draw will be uneven)
power draw - when the LEDs are "off", the FET will be drawing the current (even more than the amount ...
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This is no easy task to do by simply re-using bits of old motherboards. It's a problem that I feel requires a properly designed solution. I can't imagine that the laser diodes are cheap and because of this I would recommend a properly designed circuit board with a design solution that is tailored for this application. I'm initially thinking of maybe the ...
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No. The datasheet tells you what the bare part can do, already assuming very fast gate transistions. The test conditions are usually listed in the datasheet. Read it.
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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. ...
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Your BJTs are in a follower configuration. This means that they can provide current gain, but not voltage gain. In fact the emitters will be a diode drop BELOW the base for positive going signals. If you got to 6V on the gate you must have had around 6.7V out of your signal generator.
The BJT Wiki page has links to the 3 common forms of amplifier which ...
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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 ...
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There's a continuous base current path for both Q2 and Q3 when Q1 does not conduct. You can easily expect continuous 200mA current through them from your +5V. That's about 1W dissipation which well can make also other parts behind short wires hot than Q2, Q3.
Thermal runaway can short your +5V source.
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You should look at the I-V curves specified for the device.
The first plot refers to IRF520. As you can see, you need at least 4.5V on the gate in order to get a current just a bit higher than 1A. With 3.3V the current could be even lower than 1A. (I can't tell from the graph, but I'm quite sure this device isn't going to work with your setup).
In the case ...
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