Circuit Snippet

I am designing a circuit to drive the brakes on a DC motor. The brakes are a 19.9Ω coil that is disabled when energized (think wheelchair motor style brakes). The brakes work on 24VDC.

I have found myself in a unique situation where the brakes are switched on and off from the main microcontroller (5V logic), but I want to be able to disable the brakes if certain conditions aren't met on the auxiliary microcontroller (also 5V logic). The auxiliary microcontroller is set up as a type of safety interlock device, with the idea of being able to disable power to the motor controller in the event of an emergency. This is done by the switching of a solenoid (separate system from what I am trying to ask about with this post). Sure this task could be moved to the main microcontroller, but I like the idea of keeping the systems separate.

My idea is to use an AND Gate to control two individual MOSFETS to control two individual brakes IF both microcontrollers have the pins set to HIGH. I feel like I've gone around in circles, trying to find the answer to this/implement this design in an intelligent fashion...

My question is in multiple parts:

  • Can an AND Gate drive two MOSFETs in the configuration below? (If not, what changes should be made?)

  • Have I chosen the correct resistors for the MOSFET gates?

  • Are pull-down resistors needed on the input side of the AND gate?


AND Gate - https://www.mouser.com/datasheet/2/916/74LVC1G08-1319751.pdf

MOSFETs - https://www.mouser.com/datasheet/2/916/PMV40UN2-1320360.pdf

Microcontroller - Arduino Yun Rev2

Aux Microcontroller - Arduino Nano

Brake - Inductive load 24V 19.9Ω ~1.2A

This is my first post so please be gentle and thank you in advance for any input!

  • \$\begingroup\$ There isn't really any need for the resistors. That is a (somewhat poor) method used to bias BJTs. Also, we will probably need to see more of the circuit in order to help. How is it connected to the motor? Is there an H-bridge? \$\endgroup\$ Oct 18, 2019 at 2:47
  • \$\begingroup\$ Caleb, thanks for the speedy reply. The motor controller is handling all motor related functions and this circuit doesn't control the motor itself. Sorry for the confusion there. The function of this circuit is to switch an inductive load so no need for an H-bridge here. So you would recommend removing the 100 ohm gate resistor or the 10K resistor? Happy to leave off components if they aren't needed! \$\endgroup\$
    – Rybitski
    Oct 18, 2019 at 3:15
  • \$\begingroup\$ Both. Neither of them should necessary in order to drive a MOSFET with a low enough threshold voltage. However, I don't think the MOSFETs you chose are going to be able to withstand the conditions you specified. You will most likely need something in a larger package with a thermal tab. \$\endgroup\$ Oct 18, 2019 at 3:58

3 Answers 3


Let's break it down:

  1. Can the AND gate drive two MOSFETs?

As a rule of thumb, any device can drive any FET as long as A) it can exceed the gate-source leakage current (100nA in your case, no problemo) and B) get the voltage over the threshold (I'd call 2V safely on, looking at page 7 of the FET datasheet). The only question is how long does it take?

To a driver, a FET pretty much looks like a capacitor -- table 7 says this is about 635pF. 100Ω of gate resistance makes your R*C time constant 100*635e-9 = 64us. Your capacitor voltage rises to 63% of the source voltage (63%*5V ~ 3.15V, more than enough to turn your FET on) within one time constant, so it takes less than a heartbeat to turn the FET on -- assuming your driver can maintain a constant voltage...

But can it? Per Table 5 in the chip datasheet, the AND gate can drive 50mA at a dead short to ground. At turn-on a capacitor looks like a dead short, so you're really only dealing with your 100Ω caps in parallel, for 50Ω to ground. At 5V this is 100mA, which is a more than your chip can supply. Is this a problem though? No, the chip will happily supply 50mA until your FETs turn on, which will still happen in less than 1ms.

  1. Do the gate resistors and MOSFET pulldowns work?

There are two main reasons gate resistors are used: one is to prevent ringing that causes EMI and the other is to slow down ("slew") device turn-on. Since you probably don't really care about either of these it is safe to leave them off, but you can leave them on if you like. 100Ω or less is a good value, you can't really go wrong in that range.

The pull downs aren't really necessary: your AND gate operates in "push-pull", meaning that it will "push" the output high when it is a 1 and "pull" it to ground when it is a 0. On some devices it will only pull in one direction, meaning you need a pull up/down to counter it when the device is floating, but that isn't the case here.

Of course, having them there won't hurt anything and 10k is a good value if you like them. Sometimes devices connected to FETs can leak current and cause odd behavior if they are in weird states (think a microcontroller that has internal pulls that get disconnected when it is in power-saving mode) but once again, that isn't something you really have to worry about here.

  1. Are pulldowns needed for AND gate input?

For the same reason as above, you will be fine without them.

In short, the design should be good to go. You could switch the FET pull-downs to pull-ups for safety, in which case your device would brake itself if it loses 5V (assuming brake voltage is still present) -- not sure if this is desirable or not You could do the same at the AND gate input if there's ever a chance the microcontroller gets disconnected.

Just a note - if possible, schematics should be drawn with the voltage rails pointing up. Flip your +5V (or flip pins 4 and 5 on the part) and flip your clamping diodes with the brake input lines.


You dont need R8, R9, R10 and R11... if you want a current limiting resistor for the gate of each mosfet you can keep R10 and R11 in your design, depending on how fast you want this mosfet to turn on you might want a lower value for them, I think 100 ohms is okay for this application.

Second issue, I would add a driver IC between your gate and your MOSFETs, logic gates are not meant to provide driving currents, even though MOSFETs require small currents for their turn on(few mAs or so) I would add a proper driver IC that takes the signal and can drive some current to the gate of your switch. a 5V dual MOSFET driver IC or something alike will do the trick here proving current for both switches.

Third, seeing you current and voltages it seems your MOSFET will be okay, but see what are the tradeoffs of getting a bigger one and having more leeway on that end


Series gate resistors serve two purposes. They reduce the gate current drawn when switching and protect the driver against transients from the load beyond the output connector.

When the AND gate output switches from low to high, it is instantaneously presented with a short-circuit, in theory anyway, by the FET gate capacitances of 2 x 635 pF. These will stress the upper driving transistor in the AND gate. This may affect the long-term reliability of the part. Although modern logic gate outputs are tougher than those of decades ago, it is better to put a series gate resistor to limit the current drawn.

From the datasheet, the AND gate can source/sink up to +/-50 mA and an output loss of approx. 0.6 V can be used i.e. output driving 50 mA gives 4.4 V or less (ignoring 5 V supply accuracy), output sinking 50 mA gives 0.6 V or more. This suggest a series resistor of 4.4/0.05 = 88R. So I would use 100R, which you have. When the AND gate switches, the FETs gates will be fully charged or discharged within 4.6RC = 584 ns. You could get a more accurate FET switching time using the actual transfer characteristics but I imagine this time is fine anyway.

When a large transient appears on the FET drain when switched-off, its drain-gate capacitance Crss can cause the transient to pull the AND gate output excessively high or low and damage it. (I have seen this in a couple of other people's designs that drove medium loads in noisy environments. My adding series resistors removed this fault.)

Pull-up or pull-down resistors bias the FETs to a preferred state when undriven. The AND gate has a push-pull output and always drives. So the pull-down resistors have no purpose and should be removed.

So: (1) yes, the AND gate can drive the two FETs; (2) keep R10 and R11 as they are; (3) remove R8 and R9.


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