# MOSFETs: Gate-to-Source resistor and Gate resistor value calculation [duplicate]

I want to control several Mosfets through a microcontroller. I have a low-power application in mind, therefore the power consumption should be minimized.

1. When do I need a Gate-Source-Resistor, and when not? How to calculate the value for minimum power consumption?
2. Regarding the gate resistor, Does it make a difference in terms of power consumption, if I choose a 100 ohms resistor vs. a 1k ohm resistor?

Added schematic (Dx-pins: digital pins from microcontroller):

• 1. When you need to know you turn off when the power goes away and you start from zero each time. 2. P=D*U^2/R where D is your duty cycle. I go for 10 kohm as standard unless power consumption is critical. – winny Jan 30 '18 at 16:13
• electronics.stackexchange.com/a/343286/139766 – Trevor_G Jan 30 '18 at 16:22
• Believe it or not, it would take several paragraphs to cover everything you need to consider. To simplify a bit, can you let us know more about your application? What are you switching with the FET's? What voltage will be applied to the gates? Are you doing PMOS on the high side or NMOS on the low side? Are you using high-speed PWM control or just more like a load switch? – mkeith Jan 30 '18 at 17:11
• @mkeith I'm using N-Channel Mosfets. A voltage of 3.3V will be applied to the gates. The mosfets are used to switch various things on and off, e.g. a small pump or a buzzer. PWM will be applied to the buzzer. The pump will only be switched on and off. Oh and also one P-Channel MOSFET to switch a step-up converter (used for pump and buzzer) on and off. – Henry Jan 30 '18 at 17:24
• You will put the N-channels on the low side, right? In other words, source connected to ground, and drain connected between load and ground? I think you should add a schematic to your question. You can edit the question, then click on the built-in schematic editor (one of the buttons at the top after you start editing). – mkeith Jan 30 '18 at 17:47

Assumptions based on comments: The three NPN transistor symbols actually represent NMOS devices. The IO voltage used to drive the gates is 3.3V. The PNP transistor symbol actually represents a PMOS device.

That is it for assumptions.

The three low side switches will probably be OK. There are two things I would double-check to make sure.

First, make sure the specific NMOS you use is fully turned on at 3.3V. Look for Rds(on) to be specified to an acceptably low level at Vgs of 3V or 2.7V. This should be no problem to find. For this kind of thing, look at Rds(on) not Vgs(th). Because you need to know that 3.3V will fully turn on the transistor. Vgs(th) is specified at a very low current.

Second, make sure that the IO signals which control the gates of these switches are at a well defined on or off voltage at any time that the load voltage is present. Sometimes IO pins toggle at reset or during boot up, looking for hardware that is not present or something like that. Sometimes they may default to inputs with weak pullups prior to when your application code takes control of the processor. I have been burned by this before. But as long as they do not get driven high unexpectedly during bootup, I don't think there is any need to add gate-to-source resistors. And anyway, the resistors would only help the weak pullup case. If the output is driven hard to a high state, a pulldown will not help you.

For the high-side switch to the regulator, you MIGHT get away with only driving the gate up to 3.3V if the highest voltage on RAW is 3.7V. However, if RAW is a Lithium Ion or Lithium polymer battery, it may be as high as 4.2V when the battery is fully charged. So in that case, the best bet is to add a gate-to-source resistor of, say, 100k, and a small NMOS (e.g., a BSS138) to pull down the gate when needed. The NMOS will be controlled by the GPIO. High means regulator on and low means regulator off. See schematic below.

simulate this circuit – Schematic created using CircuitLab

The BSS138 will work fine to turn on M1. PMOS M1 needs to have an acceptable Rds(on) at Vgs of 2.7V or 3V (just like the NMOS). This circuit consumes no power when the regulator is off. When the regulator is on, the only wasted power is in the 100k. You can probably go to 1M if you can't tolerate the 100k. But if the boost regulator is enabled, you will probably not be worried about 30 uA or 40 uA in the 100k resistor.

It is up to you. The danger with very high resistances is that maybe the PMOS will start to turn on just a little if the NMOS drain-to-source leakage creeps up. This is more likely to be a problem at very high temperatures. Check the datasheet for the NMOS. Remember, with a 1M pullup, 1uA = 1V. The PMOS may not turn on at 1V, but it probably will turn on more than you want at 2V.

One last thing. Do the NMOS devices need series gate resistors? Probably not. But for the buzzer, since you plan to PWM it, I would keep the gate resistor in there in case you determine that it is needed. If it is not needed, you can use a 0 Ohm jumper instead of a resistor, and maybe eliminate it in a future board revision.

• Thanks, excellent answer, contains everything I wanted to know. For the NMOS I'm using aosmd.com/pdfs/datasheet/AO3400A.pdf and for the PMOS aosmd.com/pdfs/datasheet/AO3401A.pdf - there is a graph in the datasheet which shows Vgs in relation to Id(A). Can I use this graph do determine if the nmos is fully turned on? – Henry Jan 31 '18 at 8:54
• There's one thing I still don't understand though, and that is why I need the BSS138 to turn on the PMOS? (and yes, RAW is a li-ion battery) – Henry Jan 31 '18 at 10:52
• The reason you need the NMOS is because the NMOS drain will not conduct even if it is pulled up to 4.2V (as long as the gate is near 0V). But the IO pin from the processor, if it is pulled up to 4.2V, WILL conduct. This is because there is, in just about every micro, a protection diode from each IO to VCC. So if the IO pin was directly connected to the FET gate, since VCC is only 3.3V, the diode will become forward biased, causing a voltage to appear across the gate-source resistor. This is uncontrolled current leakage, and it may turn on the PMOS. – mkeith Jan 31 '18 at 15:29
1. Gate to Source resistor, assuming it's Gate to Ground as Source = Ground, can be, in your case, anything between 100K and 600K. Consumption is negligeable no matter what value you choose for this resistor as long as it's above 100K.
2. Gate resistor: between 140 ohms and 200 ohms. 100 ohms is also good, but from what I read 140 offers complete safety for the IC and the mosfet. This resistor protects both. But too high resistance, like 1K will cause, in your case, too much voltage drop and too slow rise time for the PWM application. Also here, the exact value has no effect on consumption.
3. MOSFET gate consumption will increase with PWM frequency, especialy high ones.
4. Make sure your MOSFET gate maximum treshold voltage is not more than 3V. If the output is 3.3V and the max Vgs(th) is 4V, you will have some may-or-may-not work issues. Many MOSFETs have max Vgs(th) of 4V.
• Thanks for the answer! Do I need Gate-to-Source resistors? If so, why? – Henry Jan 30 '18 at 23:06
• The Gate to Source is in fact Gate to Ground. I prefer to call it like that because it's goal is to connect the Gate to the ground. You always need it because without it, you can't make sure that the gate will be always at zero volt due to parasitic interference, electrostatics and so on. Without it your MOSFET may turn on unpredictably and sometimes may stay on all the time. 600K seems a lot but is already enough to sink this parasitic power to the ground. Lower values are more effective. But too low values (less than 100K) will cause an important voltage drop. – Fredled Jan 30 '18 at 23:24
• I will use an arduino to control the mosfet, in this case I shouldn't need a gate to source resistor should I? Because the Arduino pulls the pins to Ground if you Set them to LOW – Henry Jan 30 '18 at 23:40
• Maybe not (I'm not expert in Arduino). But it doesn't cost anything to add an extra pull to ground in case the arduino is disconected or fail to pull to ground for some reason, and the rest of the circuit is on. – Fredled Jan 30 '18 at 23:55