Hi,
I want to incorporate a MOSFET module in my PCB and came across this schematic of a D4184 MOSFET module (PFA). However, I am a bit confused regarding the diode connected across the gate and source of the MOSFET in the attached schematic. The schematic does not reveal what diode this is and I'm not sure if it connected in the right way.
I did a bit of research and found out that this could be a zener diode that is being used to keep the voltage across the gate of the mosfet below a certain threshold. I'm not sure if that is the case. So here are my queries regarding this:
Is this a zener diode?
Under what circumstances can the gate voltage be too high to cause a breakdown?
How do I select this diode, given that the mosfet is D4184 and the pwm input of the optocoupler PC817 will be driven by a 3.3v microcontroller?
Shouldn't the 100 ohm resistor be in series with the diode (connected to the output of the optocoupler)?
Please let me know if any other information is needed.
Yes, the diode should be a zener say 12 volts. It keeps Vgs within specification.
The gate/gnd 4.7k resistor should be much higher to allow operation down to 6 volts.
The 100R resistor can be dropped.
The two 4.7k resistors form a potential divider approximately halving the gate voltage.
The Mosfet has an Rds(on) rated at 4.5volts. Half of 6 volts is not enough to ensure the Mosfet is fully conducting with a reasonably low effective resistance.
Edit.
What frequency is the PWM ? In the 500Hz to 1kHz range ?
Sorry, I don't understand. Which two 4.7k resistors are you talking about? I can see only one across the gate and source of the mosfet. If you are referring to the 4.7Kohm connected to the collector pin of the opt isolator, I'm curious to know why it has a 4.7kohm resistor in the first place? Can't we just ground that pin?
I wasn't considering the current limiting resistor on the power on indicator LED which also happens to be 4.7k. Yes, I didn't make that clear.
The network formed by the two other 4.7k resistors is potential or voltage divider between the power rails, that is between +36v and GND.
The 4.7k resistor between the (up to) 36v positive power rail and the opto-isolator is necessary to limit the current through the zener diode when the voltage exceeds the zener voltage (say 12 volts). Niether can be eliminated although, as I said, their resistances should be adjusted to permit operation of the circuit down to the 6volts which appears on the schematic.
The circuit should be OK with a PWM frequency of 1kHz (the typical "analogWrite()" frequency range).
ok, now I understand what you're saying. If the power side voltage is 6v, then the potential divider causes the gate voltage to become 3v which will not fully turn on the mosfet. So, in order to make the circuit useable with 6V, the gate-source resistor should be something like 12k so that the gate-source voltage is above 4.25v. Am I right?
Nevermind the power indication LED, I'm not using it in my circuit.
Also, if I use a 12V zener, then the upper voltage limit of the load should be less than 24V correct?
12k sounds about right. The other thing to bear in mind with PWM is that the mosfet should switch sharply on and off. There is a small capacitance at the gate ( under 2nF for the D4148) so the resistors should be low enough to ensure that this capacitance does not significant impact the switching time. However, if the resistor values are too low, then the voltage divider starts using a significant amount of power. Somehow you find a compromise.
The 12 volt zener will ensure that the full voltage range can be used, that is 6 volts through to 36 volts, yet a maximum voltage of 12 volts appears a the mosfet gate (Vgs in the data sheet).
Can you please explain why the 4.7K resistor was shifted to the right of the 100ohm resistor? Also, I'm eliminating the 100ohm resistor as per @6v6gt 's suggestion.
If you were outputing PWM signal to the MOSFET, the MOSFETs small gate/source capacitance would mean the some current would flow in the gate circuit.
The 100R is there in normal practice to limit the gate current and protect the driver circuit.
The 4K7 is sometimes moved so that it does not form a potential divider with the 100R at the gate, thus reducing the applied gate voltage.
In your case this the 100R is not needed as you are driving from the opto-coupler circuit the 4K7 in the pin 4 of the opto circuit is doing that.
Not sure I understand your statement fully.
AFAIK, when there is 3.3V across pin 1 and 2 of the optocoupler, pin 3 and 4 gets shorted which forms a potential divider and cuts the supply voltage in half across the gate (two 4.7k resistors). If I connect the gate of the mosfet with the emitter and connect the collector to ground, how does the gate cross the threshold voltage?
