# MOSFET High Side conducts when low side switches in H bridge circuit

I have designed an H-bridge circuit using IR4110 Mosfets.
Q1 and Q4 are OFF, Q2 and Q3 are conducting. Gate of Q1 is left floating. Q2 being high side, its gate is connected to Vcc + 12V. The gate of Q3 is connected to a PWM signal (12V square wave 15kHz, 70% duty cycle) for motor speed control through a 10 Ohm resistor. Gate of Q4 is grounded through 10ohm resistor. All MOSFETs have a 10k resistor between their gate and source, though it is not shown in the diagram.

Now, when gate of Q3 is high, it conducts, but at the same time Q1 also conducts for a brief moment, causing shoot-through and heating up Q1. When I hooked up the gate pin and source pin of Q1 to a DSO, it is observed that $$\ V_{gs} \$$ of Q1 goes above its threshold for a brief moment, turning it ON, which is undesired. The exact reason is unclear.

This never happens for IR4115 MOSFETs, which I am currently using without any such problems. Maybe it is because $$\ V_{gs} \$$ threshold of 4115 is higher than that of 4110.

Any explanation or suggestions for design modification will be highly appreciated.

• Try to unfloat the gate or add a dead-time on the PWM signals if possible. You're not using FET driver chip to generate the signals, is that correct? – Unknown123 Aug 28 '19 at 10:00
• I am not using FET driver chip, how do i unfloat the gate? By connecting it to ground? This will exceed the -Vgs rating of the MOSFET – Winner Aug 29 '19 at 9:39

My guess would be that whatever drives the gate of Q1 has some capacitance, tending to keep it from changing voltage suddenly. Thus, when Q3 is turned on, the source is pulled to ground while the gate is still idling at 12V momentarily. The gate gets pulled low, but only after you observe some shoot-through, and since this occurs on each PWM cycle, it will heat up a bit.

I'd say take that 10K from the base of Q1 to ground instead of the emitter, but I couldn't easily find a spec to determine whether it will take -12V Vgs. If not, a small capacitor between the gate and source would be my second preference.

• This must be it. The self-turn-on phenomenon due to parasitic capacitance. – Jeroen3 Aug 28 '19 at 13:21
• If i connect the 10k resistor from base(gate) of Q1 to ground instead of its emitter(source), it solves the shoot through problem, but then Q1 might not be able to take negative Vcc. For Vcc=12v this might not be a problem, but for higher Vcc, it exceeds the negative Vgs limit of the MOSFET. I will try reducing gate to source resistance to 1k or by replacing it by a capacitor – Winner Aug 29 '19 at 9:20
• I dont know if self turn on phenomenon is applicable here, but I will try to analyse it, thanks – Winner Aug 29 '19 at 9:42
• The phenomenon is the same as in the app note Jeroen3 referenced, the only difference is you're switching the power to the FET from the low side instead of the high. Unless you're driving the gate from a push-pull output, I'd suggest using a cap in parallel with the resistor rather than in place of it, so you still have a DC path to keep Q1 inactive. – Cristobol Polychronopolis Aug 29 '19 at 13:28
• @Winner How did you fix it in the end? – Unknown123 Aug 31 '19 at 14:36

The best design modification you could do is not to never leave the gate of a MOSFET floating. If you want it off and to remain off then either drive it low with something able to sink current continuously from the gate or decrease the fixed resistance to closer to 1k between gate and source. You could also add some more gate path capacitance close to the gate terminal to make the device 'harder' to turn on (i.e. increase the amount of charge needed to raise the voltage to the threshold voltage). Something around 10 nF is a place to start but note the more you add the more the device will switch a bit slower and so your switching losses will increase too.

As for an explanation, my initial guess would be current being injected into the gate via the gate-source capacitance and due to the gate being left floating it won't take a great number of pulses to raise the voltage on the gate to a point where the MOSFET will start to conduct. When the motor turns off the voltage at Vs rises, for that to happen you need to charge the output capacitance of both Q1 and Q2. Charging Cds of Q1 will also expose Cgs to a somewhat unknown voltage transient as one side is floating and depending on dv/dt of the transient you'll get more or less current injected into the gate. The IRF4110 has much larger Ciss (Ciss = Cgs + Cgd) and so one or both of them is larger and require a lower dv/dt to inject current through them.

Think of the voltage transient as a brief burst of voltage frequency spectrum, highier than the fundamental frequency, required to represent it. The lower the dv/dt the lower down the frequency spectrum the frequency content required to represent it will be. A larger capacitance ( from larger Ciss) will have a lower reactance at lower frequencies and therefore will behave more like a resistor to the lower frequencies and allow more of the energy in them through to charge the gate of the device up.

That's my best guess without having drawn out all the current loops and visualised the waveforms properly but the short version is to drive your gates low when you want them off, to decrease your fixed Rgs or to add some additional Cgs. All of those should fix the problem you're having.

• So, i will lower Rgs and if still shoot through persists, add additional Cgs...hope this solves the problem for IR4110s – Winner Aug 29 '19 at 12:11
• That should solve the problem but i would strongly urge the use of a gate driver that can hold the gate low with a low impedance path to the source potential of the power device. – hooskworks Aug 29 '19 at 17:13

I replaced 10k resistors between gate and source of all MOSFETs by 1k resistors. I also added a 0.01uF capacitor in parallel with all these 1k resistors, and the problem seems to have solved....no more shoot through. Vgs of Q1 always remains below its minimum threshold... MOSFET self turnon phenomenon as mentioned in the replies. This was my first post ever on an electronics forum, so thank you everyone for helping me out... :)