IRFZ48 N and FAN7392 MOSFET Switch

Question

I want to make a MOSFET switch to drive a 12 V 300 W heat bed. Instead of buying one, I decided to make myself one. I used IRF48N MOSFET and FAN7392 Mosfet driver to make circuit and when tested it, it worked(I tested it by heating a 40 W hotend). I later had to solder few male header pins to provide connections for FAN to cool the MOSFET & Heat sink. Then when I tried to heat the hotend, it's not working(May be I damaged it when soldering).

IRF48N MOSFET data sheet
FAN7392 data sheet

This is my current circuit

I think I damaged the FAN7392. With input from Arduino, LO pin should give 12 V to turn the MOSFET ON, but now its not. I don't know how it damaged, but I'm little skeptical of my circuit.

One guy suggested me to

1. add capacitor to the supply pins(when checked the datasheet of FAN7391, it says that, my mistake)

2. Add a Zener voltage protection to MOSFET gate

3. Don't leave any pins left floating

Based on this I made following modifications

Not tried the circuit yet(need to source components).

Q1. What should be the value of capacitor? 1 micro-farad ? Is it always necessary even if the supply to driver is a regulated one?

Q2. 12 V zener for protection

Q3. Should I use pull-up or pull down for unused pins?

Q4. Do I need to make any additional modifications?

Q5. Which graph or table to give more importance when designing Switch? I mean required gate voltage to create channel for specific current. I came up with 12 V for Gate from Rds ON value and Vds vs ID characteristics

Answer 1

If you have damaged the circuit then there's nothing in your original circuit other than the adding of the 1 uF capacitor between Vdd and Vss that might be seen as problematic. 1 uF ceramic is the value recommended in the data sheet. Pull-up/down resistors on unused logic pins are not required because internally, the FAN7392 has pull-down resistors.

You don't need a capacitor on Vcc because you are not using the top driver and a capacior is required when using the top driver because of the bootstrap capacitor. Not applicable in your case.

If your Vcc power supply is 12 volts then you don't need a zener on the gate.

Apart from the above, I suspect layout/breadboard issues.

Answer 2

All the pins you have N/C can probably be left that way.
HIN & SD (NOT SIN) have internal pulldowns - see data sheet.

You have probably destroyed the driver
or The FET is gate-source short (as happens :-) ) or even dgs short.

Remove the 100 Ohm (giving your components names is a very good idea). Test resistance of gate to ground. Should be 10k. If 0 Ohms or << 10k FET is dead.

With 100 Ohms still removed, toggle LIN (input).
LO should follow LIN.
If not, driver IC is dead.

Either replace FET and/or driver OR, as this is presumably a low switching rate circuit, you can build a simple cheap driver.

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Discrete driver:

^{simulate this circuit – Schematic created using CircuitLab}

Vin high = Q1 on = Q2 on = FET on.
D1 zener voltage rating should be ABOVE Vdd = 12V so zener will not conduct due to supply alone.
D1 removes voltage spikes from Millar capacitance coupling from drain if load is inductive. [[ALL loads are inductive :-) !]].
D3 is a small Schottky diode mounted as close to FET bs pins as reasonably possible. On +ve drive cycles Schottky sleeps. If thgere are gate oscillations (as happen) schottky conducts on negative cycles and steals energy from oscillation.
Mount D1 as close to FET bs pins after D3 has been positioned.

R1 probably wants to be more like 10 Ohms but I have used your existing value. R1 mainly serves to damp drive oscillations (which happen).
R4 may want to be more like 1k depending on switching cycle rate. A heat bed element does not need a fast switching rate due to the large thermal time constant. A 1 second rate would usually be VERY adequate and even much longer is liable to be fine.
R4 affects the FET turn off time due to RC filter formed with gate capacitance - usually in the 1 -10 nF range. (IN this case about 2 nF. T=RC = 10k x 2 nF =~~~ 20 uS so FET switches off over maybe 50-100 uS. Dissipation can be "high" over this period (maybe 100W+ surge) but as long as not done very often is probably OK. Smaller R4 helps switch off speed and does little harm. (This is of much less concern with your driver IC due to active internal high/low drive.

Make R4 say 1K.
Make R1 10 Ohm.

Happy 3D printing - or is it T-shirts?

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Faster switching:

For faster switching th FET gate capacitance must be charged and (especially) discharged more rapidly than the above circuit allows.

The following circuit is from Olin Lathrop and appears elsewhere on the SE EE site. This is a high side P FET driver - add the extra stage as in my above cct to get NFET low side - but I've posted this more for the Q14 + Q15 emitter followers which give high current gate drive. That pair of transistors can be added directly to my circuit. Olin's circuit has a ~= 200 nS switching speed.

Carefully work through how Q2 R14 R15 work - it's 'trickier' than at first apparent.

Answer 3

Your circuit looks very similar to a motor speed control we make at work for an electric scooter. I don't understand the need for a mosfet driver. We use a HPWM output from a PIC microcontroller that produces a 5V square wave and drives the mosfet gate directly. Like your circuit, the other side of the load (motor) goes to B+ 12V. Being a motor with transients, we use a fast recovery rectifier instead of a 1N device. The only other difference is that we use an irLz FET. They require less gate voltage to turn on since we use them in many 3V products