High Side on Mosfet driver IC doesn't work

It helps to LABEL all pins in circuit diagrams.

Tom... :grinning: :coffee: :coffee: :australia:

@mroverecast you are getting excellent help from @TomGeorge but one thing I've read from you at least twice is:

You seem to think that disconnecting an input is the same as making it low, it is not. Disconnecting it is leaving it floating, it could be anything. You have to put defined, known voltages on inputs if you want defined, expected outputs. See Buttons and other electro-mechanical inputs (introduction) for a complete explanation.

I just looked again at what you have.
The only way you will get the globe to light is to put HIGH on Lin and Hin.
You need both MOSFETs to conduct to get a voltage across the lamp.
I assume this is how you have the lamp connected?
Have you checked the wiring configuration of the MOSFETS?
A question I should have asked first, link to data/specs of your MOSFETs?

Tom... Not enough coffee today... :grinning: :coffee: :coffee: :australia:
PS. I need to get back to work, not enough practice on holidays, start work tomorrow Yaaaaaayyyy. :+1: :+1: :+1:

sorry about the lack of information guys. I am trying my best. I am using IRFZ44N Mosfets here is the data sheet.

You have the MOSFET symbols upside down.
Correct the symbol orientation.
IRFZ44N I believe is not a logic level MOSFET.
That is 5V on the gate with respect to source will not fully turn on the MOSFET.

Vgs of 2 to 4V is the parameter that states at what gate voltage the MOSFET begins to conduct.

Have you tried Hin and Lin both HIGH to see if the load lights up?

Tom... :grinning: :+1: :coffee: :australia:


ive been having trouble understanding how some MOSFETS work. what exactly does logic level mean? Also I have tried using a G30N60A4 IGBT but that did not seem to work either. If HIN and LIN are both high the light illuminates. If LIN is High and HIN is Low the bulb still illuminates

You are providing 9V (VCC) to the IC, but they recommend 10-20V.

In this case, "logic level" is not an issue as I see it.

IRSDS17613-1.pdf (widen.net)


I have tried it with 17 volts and still nothing.

How are you providing HIN and LIN? With jumpers/wires to 5V or GND?

PE -
(I see that the inputs are pulled-down internally.)

Jumper wires are going from the arduino to both HIN and LIN.

Maybe leave the Arduino outputs out of it and work it with just a wire to 5V as needed.
(Are you clear on the all of the GNDs (Grounds) being connected together deal?)

thank you I will give that a shot. the ground from the 9V, arduino, and 14v are all connected.

It means they will fully turn on with 5V gate to source voltage. Many non-logic level MOSFETs are only just beginning to turn on with 5V, they are a long way from fully conducting.

You can see from this illustration that your 'S' & 'D' LABELS are swapped and the N-channel
mosfets are indeed BACKWARDS.

Three things I notice in your circuit:

  1. MOSFETs upside down - sources at the bottom please.

  2. Vs is not connected to the mid-point of the bridge, midpoint of the bridge isn't connected together - both are essential to circuit function - follow the datasheet (see the "typical connection" diagram).

  3. The capacitor values are completely crazy. 10mF and 1000mF? Or do you mean 10ยตF and 1000ยตF? Either way they are plain wrong.

The boostrap cap between Vb and Vs is usually of the order of 100--330nF and must be a MLCC ceramic cap placed close to the chip. Using a large value cap here may burn out the bootstrap diode (typically there would be a few ohms resistor in series with this diode for extra robustness).

The decoupling cap between Vcc and COM needs to be at least 10 times the value of the bootstrap cap (1-- 10ยตF is reasonable) - again this must be MLCC ceramic to have any decent effect, and be placed within a cm or so of the chip supply pin. This is high-current high-speed decoupling and electrolytics have far too much inductance to do anything useful at these speeds (the transitions are measured in nanoseconds).

To keep the charge pump working the low-side must be switched on regularly (at PWM speeds, 1kHz or faster), as the high side circuit requires this to recharge the bootstrap cap. This means you cannot park the circuit with the low side switch off, only with it on. So using this with PWM means the maximum duty cycle has to be less than 100% to ensure the low-side switch gets used to do the bootstrap. With 10kHz PWM you might arrange 0% -- 99% duty cycle, allowing 1ยตs every cycle to recharge that cap.

At least the bottom MOSFET has D and S right, only the image is reversed.

The high and low side drivers work independently. A load has to be connected (to GND) for the high-side driver to work. And you are right, Vs must be connected to the high side MOSFET.

In an H-bridge, which if you read the OP is the context, they are connected - using them separately has some extra caveats (like maybe needing free-wheel diodes).


The cap values weren't mentioned in the schematic, and I saw the values on another schematic someone made so I went with those. The mid point of the H bridge wasn't connected because the site in which I made the schematic didn't have the option to put a load there. I did not connect Vs to the midpoint because I thought it would short something out and it made the light bulb go dim. If you could make a schematic for me with all the values you mentioned I will order the parts and give it a shot

It should be connected to the high side MOSFET. For control of a load you can use the high side or low side switch, but both at a time does not make sense. Both sides together make sense only in a full H-bridge with 4 transistors.

That's a misunderstanding. A H-bridge consists of 4 transistors what the OP obviously does not want - see the code in #14. We still have a long way to go :frowning: