I'm making a mosfet circuit controlled by PWM from the arduino to run a windhield washer pump. Using the recommended circuit in the IR2101 datasheet (just the high side - I only have one FET), I can get the FET to turn on. However, even at very low PWM values, the gate capacitor continues to charge up and the voltage/RPM on the motor increases until it's pinned to the rail.
Even turning off the input on the Arduino does nothing, nor does pulling it low. If I pull the gate low, the MOSFET instantly shuts off, but I cannot get it to run with even the highest of pull down resistance (my voltmeter between the gate and ground is enough to prevent the driver from running).
Am I doing something wrong? There's nothing wrong with the MOSFET - it is an unmarked chinese mosfet but it does work fine if I pin the gate by hand.
Here's the datasheet with the recommended circuit - I'm using everything except Lin and Lout - the load is grounded on the other side.
packocrayons:
I'm making a mosfet circuit controlled by PWM from the arduino to run a windhield washer pump. Using the recommended circuit in the IR2101 datasheet (just the high side - I only have one FET),
You can't run a high-low MOSFET driver just on the high side, it cannot bootstrap.
Furthermore you cannot use the high side continuously, you have to PWM so that the low
side is on regularly to recharge the bootstrap.
And I suspect from your description you are erroneously using a p-channel MOSFET - both
high and low MOSFETs must be n-channel.
It is an N-channel mosfet, it was actually paired with the driver on a brushless motor controller.
I don't understand how the load will run if I only turn on the high side, won't the low side be floating? If I have to turn on both the high and low side, what's the point of having two mosfets in the first place?
Maybe I'm misunderstanding something here - it's been a while since I've done theory and of course university doesn't teach anything practical.
Yes, you don't understand how the bootstrapping works. The bootstrap capacitor only charges when the lower
MOSFET is on, so you have to regularly have the lower MOSFET on (every few ms), to ensure the bootstrap
cap is always charged. You can either have the lower MOSFET permanently on (for braking perhaps), or use
PWM with a duty cycle < 100% so that the lower MOSFET is regularly on (even if for only a few percent of
the time).
The high side part of the driver is powered by the bootstrap capacitor, and this floats up and down
with the output, like a tiny 12V battery.
There are a few driver chips with a charge pump as well (such as the HIP4081A) which allows indefinite
high-side operation, but most don't.
Okay, that makes sense. It's been years but seeing "charge pump" brought a lot back.
I guess it's not as relevant but how does the low side charge the bootstrap? Turning it on would take some power (to charge the gate), and after that as far as I can see it's no more useful than just having a dead short between load and ground.
The boostrap capacitor high side has a diode from the +12V rail, so if the low side of the cap is connected to
ground, the cap gets topped up through the diode (very fast - actually its a wise idea to add a few ohms of
resistance in series with the diode to limit the peak current).
Its good to scale the capacitances correctly. If the MOSFET equivalent gate capacitance is 1nF, then
the bootstrap cap should be 10 to 20nF, and the decoupling cap on the gate driver chip at least 10 times that
again, ie an order of magnitude or more between each.
The MOSFET equivalant capacitance is just the gate total charge divided by the 12V gate drive voltage...
So for instance you often see total gate charge listed in nano-coulombs, if its 10nC, then the capacitance
is 0.83nF, so 10nF bootstrap cap is fine, and at least 100nF decoupling on the chip (all ceramic). You can
use more decoupling on the chip to reduce the noise spikes coming from it to the rail, I tend to use 10uF
as a matter of course.
Don't fall into the trap of adding resistors between gate and source on the MOSFETs, that will discharge the
bootstrap cap rapidly and cause trouble - the gate driver chip takes on the job of ensuring the gate isn't
floating.