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Myth: manufactures conspire to put internal diodes in discrete components so only IC designers can do neat things with 4-terminal MOSFETs. Truth: 4-terminal MOSFETs aren't very useful. Any P-N junction is a diode (among other ways to make diodes). A MOSFET has two of them, right here: That big chunk of P-doped silicon is the body or the substrate. ...

58

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 ...

51

In a low-side switch, shown on the left, the load is between the power rail and the N-channel MOSFET doing the switching. In a high-side switch, shown on the right, the load is between ground and the P-channel MOSFET doing the switching. The low-side switches are convenient for driving LEDs, relays, motors etc. because you can generally driven them ...

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The 10k$\Omega$ resistor is there to pull down the gate when the input is floating (thus avoiding an undefined/uncontrolled gate voltage). On the other hand, the 100$\Omega$ resistor is there to limit gate charging/discharging current (due to the presence of gate capacitance) and to prevent oscillations. But, as you have already detected, both resistors ...

48

In short: You have linear control of the 'speed' by applying a PWM signal, now the frequency of that signal has to be high enough so that your DC Motor only passes the DC component of the PWM signal, which is just the average. Think of the motor as a low pass filter. If you look the transfer function or relationship angular speed to voltage, this is what you ...

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Your problem is the gate drive voltage. If you look at the datasheet for the STP16NF06, you'll see that the 0.08 Ω Rdson only applies for Vgs = 10 V, and you're driving it with only (a bit under) 5 V, so the resistance is much higher. Specifically, we can look at Figure 6 (Transfer Characteristics), which shows the behavior as Vgs varies. At Vgs = 4.75 V ...

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It is generally a good idea to include a gate resistor to avoid ringing. Ringing (parasitic oscillation) is caused by the gate capacitance in series with the connecting wire's inductance and can cause the transistor to dissipate excessive power because it doesn't turn on quickly enough and hence the current through drain/source in combination with the ...

35

It's for two reasons. Well, actually just for one, but with two factors. A MOSFET can conduct in both directions when turned on, as it truly is just a resistive channel that is opened or closed. (Just like a tap, it's open with a tiny resistance, closed with huge resistance or a small gradation in between.) But, a MOSFET also has what is called a body ...

35

You're mixing up implementation technology with colloquial terms for functionality. CMOS - Complementary Metal Oxide Semiconductor - is a method of making logic and related circuitry using both N-Channel and P-Channel field effect transistors. One of its defining characteristics is extremely low static power consumption - power is almost only used when ...

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 ...

33

Compare the actions of a P and N channel MOSFET in your circuit. (I've left the junction transistor in to aid comparison.) The PIC output does not like being connected to 12V so the transistor acts as a buffer or level switch. Any output from the PIC greater than 0.6V (ish) will turn the transistor ON. P CHANNEL MOSFET. (Load connected between Drain and ...

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MOSFETs are a bit unusual, in that if you connect several of them in parallel, they share the load quite well. Essentially, when you turn on the transistor, each one will have a slightly different on-resistance and a slightly different current. The ones carrying more current will heat up more, and increase their on-resistance. That then redistributes the ...

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Your description is correct: given that $V_{GS}>V_T$, if we apply a Drain-to-Source voltage of magnitude $V_{SAT}=V_{GS}-V_{T}$ or higher, the channel will pinch-off. I'll try to explain what happens there. I'm assuming n-type MOSFET in the examples, but the explanations also hold for p-type MOSFET (with some adjustments, of course). The reason for ...

32

simulate this circuit – Schematic created using CircuitLab Figure 1 a, b and c. Since the circuit is not isolated the bottom line of your circuit moves around with the mains voltage. On positive half-cycles (b) the bottom of M2 would normally be held at about 0.7 V above neutral voltage. Since this is connected to mains earth - that's 0.7 V above ...

31

Vgs is just the voltage from gate to source (with the red lead of the multimeter on the gate and the black one on the source). Everything else is from context. The Absolute Maximum Vgs is the maximum voltage you should ever subject the MOSFET to under any conditions (stay well away). Usually the actual breakdown is quite a bit different (borrowing from ...

30

Here's the datasheet that should be linked from your question. I shouldn't have to look for it. Each mosfet should handle 32 Amps That's with $V_{GS}=10$V You set $V_{GS}$ to $5V×\frac{R_2}{R_1+R_2}=4.54V$, you really want as much voltage here as you can (5V seems to be your maximum). If I were you I would change $R_1$ to 10~50Ω and $... 29 In a word: Efficiency. You can use a PMOS transistor to drive a logic output high (e.g. VDD) when the input is low (e.g. GND). However, you can't use that same PMOS transistor to drive a logic output low when the input is high. When you drive the input high in your PMOS inverter, it turns off, leaving the output effectively high-impedance, which is not ... 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 ... 29 It is an amplifier because you put a wiggly signal in, and you get a bigger wiggly signal out. The fact that the output wiggles down when the input wiggles up is a trivial detail, that -- if it's a problem at all -- can be solved in any number of ways (not least of which is following one stage by another, because two negative gains, when multiplied, result ... 29 MOSFET load drivers are much, much larger than the FETs used on a CPU chip: they have to be to deal with the voltage and current they’re designed to drive. As a consequence of their size they have large gate capacitance and so take longer to change their gate voltage from ‘on’ to ‘off’ state. Also, to achieve low Rds(on) appropriate for a driver, they need ... 28 Yes it does conduct in either direction. Due to the body diode, most discrete MOSFETs cannot block in the reverse direction, but the channel will conduct in either direction when the gate is biased "on". If you want to conduct and block in both directions you need two MOSFETs in series. MOSFETs used as near-perfect rectifiers are usually used in the ... 28 The MOSFET behaves like a resistor when switched ON (i.e. when Vgs is large enough; check the data sheet). Look in the data sheet for the value of this resistor. It's called Rds(on). It may be a very small resistance, much less than an Ohm. Once you know the resistance, you can calculate the voltage drop, based on the current flowing. 25 From the driver to the gates, the wires are ~15cm. Does this cause rining? Almost certainly, and it's a fair bet that this is destroying your MOSFETs, by one or more of these mechanisms: exceeding$V_{G(max)}$even for the briefest instant exceeding$V_{DS(max)}$simple overheating due to slow switching and unintended conduction #3 should be pretty ... 25 Vgs is the gate to source voltage. In the datasheet you'll find an absolute term Vgss this is the maximum voltage that can be applied between the gate and the source. Beyond this, you risk damaging the mosfet. An N channel mosfet is essentially a P type sandwiched between two N type regions. Party time. You are hosting a party and inviting all the ... 25 Yes, power MOSFETs have a parasitic diode called Body Diode. As a result of this diode, a single MOSFET can work only as a unidirectional switch. A single MOSFET can't switch-off the opposite direction, because the diode conducts independent of the gate. The body diode is usually fairly slow to turn on. I advise against using it as a the only flyback ... 25 As Andy says VGS(th), i.e. threshold gate-source voltage corresponds to a low current, when the MOSFET barely turns on and Rds is still high. From a user/shopping perspective what you want to look for is guaranteed (and low) Rds(on) for a given VGS that you plan to use in your application. Alas you did not link to any datasheets or name any specific parts ... 24 "Why is the Miller Plateau longer for bigger$V_{\text{ds}}$? " The short answer is that Miller Plateau width scales with the area under the curve for$C_{\text{gd}}$. But why? What does the Miller Plateau show? The Miller effect exists because there is effective capacitance between the drain and gate of the FET ($C_ {\text {gd}}\\$), the so ...

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The gate-source threshold voltage is the voltage that is required to conduct (usually) 100 uA of current into the drain. Different MOSFETs have different definitions and some devices define the threshold voltage at up to 1 mA drain current. It's a fairly useful comparitive indicator of how a certain device might operate when given a proper logic level ...

24

The intrinsic body diode is the p-n junction between the body and the drain. In a discrete (standalone) MOSFET, the source and body are usually tied together for convenience to make a three-pin package. This means there's a diode between the source and drain: If the source voltage is always lower than the drain voltage, the diode stays off, and everything ...

23

The part you are trying to get the heat out of is roughly in the centre of the black plastic part and mounted against the lead frame. Distorting the lead frame (the metal part) by overtightening will result in poor heat transfer and could even damage the bonding of the die to the lead frame or the die itself if it bends the soft metal leadframe, even ...

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