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Paragraph 2 of the description says: FAN3100 drivers incorporate MillerDriveTM architecture for the final output stage. This bipolar-MOSFET combination provides high peak current during the Miller plateau stage of the MOSFET turn-on / turn-off process to minimize switching loss, while providing rail-to-rail voltage swing and reverse current capability. ...


20

TL;DR Use BJTs for linear operation, not FETs Most FETs are not rated for safe operating area (SOA) at DC. Bipolar junction transistors (BJT) are. If you examine the SOA graph for any FET, you'll find a set of curves for pulses of duration 1 µs, 10 µs, 1 ms, etc., but rarely any curve for DC. You can try to extrapolate to 'near DC' if you ...


19

The answer is at the end, but, just in case you are not familiar with the concept of MOS capacitor, I'll do a quick review. MOS Capacitor: The Gate of MOSFET transistor is essentially a capacitor. When you apply any voltage to this capacitor, it responds by accumulating an electrical charge: The charge accumulated on the Gate electrode is useless, but the ...


18

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 ...


16

One good reason is to have this resistor to keep the gate low if the MCU pin is in high impedance state (e.g. during reset or after reset until the port is initialized). (Otherwise during high impedance state it could act as an antenna and pick up some voltage that turns it on)


14

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 ...


13

Unfortunately modern power MOSFETs fail when operated in the linear region at high power dissipations. MOSFETs are safe to use in the linear mode as long as the drain current decreases with increasing temperature. Most MOSFETs have a crossover below which they can experience thermal runaway and above which they don't. For very "good", low Rds(on) low Vth ...


12

If you look at the full schematic, it is easily apparent why. The MC34063 is configured in the circuit in such a way that it drives the MOSFET gate with an open emitter output: The chip can drive the gate high by turning on the NPN transistor between the pins 1 (switch collector) and 2 (switch emitter), but the chip itself has no way of pulling the gate low. ...


12

During normal operation there is no need for the resistor. However you may want it to put the FET into a known state during power up and reset. Otherwise on power up before the MCU starts to drive the pin the FET could turn on. This could cause glitches on the output or worst case (and this is very unlikely) depending upon what other current surges happen ...


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

The only control you have over the resistance of the FET is the gate-source voltage. You need to slow down the change of that voltage. The most common way of doing that is an RC filter at the gate. Put a resistor between your drive source and the device gate, and the gate's parasitic capacitance will form an RC filter. The bigger the resistor, the slower the ...


11

The CMOS and BJT output stages are combined to from one stage, the manufacturer calls this a "MillerDrive(tm)". Why they do this is explained in the datasheet: My guess is that they want to achieve a certain (output drive) performance that cannot be achieved by only using CMOS transistors or only using the NPNs with the manufacturing process that they're ...


10

Not enough Miller time? Just extend it. Spehro has the right approach here. I am going to ride his coat tails and expand on the idea a little, because it is such a good idea for this kind of thing. \$C_{\text{dg}}\$ is special in a FET because it provides negative feedback to the gate. Part of what that means is that it also gets multiplied by the ...


10

The reasons are different depending on if it is a BJT or a MOSFET. The effect, though, is the same - it's there to reduce current and protect the IO pins on the controller that are driving the transistors. On the BJT the base -> emitter junction is essentially like a diode, and without a resistor would be like a near short circuit. The resistor stops the ...


10

You might consider adding an attenuator probe to your PCB consisting of a single series resistor. The 50-ohm coax cable to your oscilloscope can be any length - the oscilloscope input at the far end of this cable must be 50-ohm-terminated, not left to its default 1MEG simulate this circuit – Schematic created using CircuitLab The series resistor Rs ...


9

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. ...


9

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 ...


9

Maybe. The MC34152 datasheet pp.8 on shows a series Rg to damp oscillations, and reverse-bias Schottky diodes for catching negative ringing spikes at the driver. Wouldn't hurt to have these in your layout. You could stuff zero-ohm for Rg if you don't need damping, and no-stuff the diodes if you find the ringing isn't too bad. Have one resistor/diode per FET, ...


9

Notice how the top NPN can only make the output reach VDD-0.7 V, I assume it is the job of the mosfet to take care of the last 0.7 V. It looks as if the BJT's are doing most of the grunt work and the mosfets are taking care of making the output reach VDD and a strong GND. I could be wrong though.


9

The P-Channel MOSFET needs to switch up to 20V. And that's the catch, when making the MCU output high impedance, you could have 20 V at that MCU's output. You would be violating the maximum ratings of the MCU. But actually you would not get to 20 V at that output as the ESD protection diodes in the MCU will pull down that voltage (to roughly Vdd + one ...


8

You can add a series resistor to the gate. That's often done to slow rise-fall times in order to reduce EMI or prevent excessive overshoot. Obviously this increases switching losses (but not conduction losses), so there is a trade-off. As well as causing the switching to slow, it will also add a delay time, so keep that in mind if there is a chance of cross-...


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

MOSFET switching Let's first approximately model MOSFET inductive load switching behaviour, far from a complete study on the topic I'll just recall main points and later on focus on the voltage slope period. simulate this circuit – Schematic created using CircuitLab Turn-on phase can be roughly divided into 4 sub-periods here depicted. Delay time (\$...


7

TL;DR: I was driving my MOSFET gate at the supply voltage I was switching/ so I didn't maintain a high enough VGS. I'd failed to consider that Vs is not 0V once the MOSFET is on. Per the accepted answer the simplest solution to that is putting the MOSFET on the low (ground) side so that VS ~= 0, making VGS easy to keep high. See corrected (but warning, ...


7

The Gate Threshold Voltage VGS(th) refers to the voltage below which the FET stops conducting completely, defined by some manufacturers as the gate voltage at which the drain current crosses the threshold of 250μA. It is not a maximum gate voltage (and nor is it the minimum gate voltage at which the FET can be used as a switch for useful amounts of current). ...


6

The reason why the drivers have two signals for the high-side and low-side MOSFETS (or other power switches) is because usually you want to first switch both sides OFF for a short period of time and then to switch the needed MOSFET ON. Usually, the switching OFF is slower than switching ON. This way if you switch the input signals in the same time, there ...


6

I notice that you do not have a single decoupling capacitor in the schematic. Each time the MOSFET switches on or off, there can be some very high transient currents. If you don't have any decoupling capacitors, those transient currents will work against the resistance and inductance of the power rails and cause all sorts of problems like noise and resetting ...


6

A MOSFET has a very high gate resistance because the gate is physically insulated from the other terminals on the device. This means that under DC conditions it doesn't draw any current, making you think it wouldn't need much to drive the gate. However, the gate has a large surface area over the drain-source channel and this introduces quite a lot of ...


6

suitable IGBT gate driver And the key to your question is "suitable". The short answer is yes you can. The IGBT combines an isolated-gate FET for the control input and a bipolar power transistor as a switch in a single device (wikipedia). Your question already contains the appropriate considerations, "threshold, plateau, and turn on voltage ratings,...


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