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Understanding pre-made high-side drivers

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poofjunior

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Hi everyone,

I'm looking to switch a load with an n-channel mosfet in high-side configuration. I need a high-side driver circuit, and I've noticed that there are plenty of chips that do this already. A few are very similar, like the LT1910 and the IR2127. The load is only about 2[A] at a voltage from 12 to 20[v].

While reading the high-side driver datasheets, I've noticed that they both have a "current sensing capability" and a "fault" pin, but I'm not sure how these play into the circuit. I think they act as feedback so that the IC can shut down the driven mosfet if too much current flows through, but I'm not sure exactly how. Does anyone have any good insights as to how these play in?

Also, on the LT1910's datasheet, there are some hints to making such a current sense resistor with just a few inches of pcb trace. Is that sufficient to make a resistor that will act as the feedback resistor?

What if fault-detection isn't important, and I just want to saturate the mosfet fully on when the input signal is given. Is there a quick way to make this possible? I noticed one more chip that appears to do just that, the MIC5018. Could this be the drop-in chip to just turn on a mosfet in high-side configuration from a logic-level input?

Thanks, and I've attached the datasheet links below!

IR2127: http://www.irf.com/product-info/datasheets/data/irs2127pbf.pdf
LT1910: http://cds.linear.com/docs/en/datasheet/1910fa.pdf
MIC5018: **broken link removed**
 
It depends on your application. The IR part uses a bootstrap to make the gate voltage higher than the source voltage (necessary to turn on an NFET), but for it to work the input must always be switching. In other words if you want to just turn on the FET and leave it on it won't work.
The Mircel part is to low voltage.
The LT part is the goldie part :p It has what is called a charge pump so it can be on all the time and it's voltage is high enough. You don't need to use the over current if you don't want to.
 
thanks for the quick reply. To not use the overcurrent protection, do i just short the current sense connection displayed in the datasheet schematic, or should i just wire it to the supply pin through an arbitrary resistor since it has a 65mV offset from the supply pin?
 
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