The IPS1031 is a Low-Side MOSFET Driver that drives its own MOSFET. I am planning on incorporating it into my design, but most of its spec sheet is total gibberish to me. I am driving two of them off of a single PIC output which can sink
3.7-5V @ 25mA. The IPS1031 datasheet:
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Attached is a schem I used for its simulations purposes. Through the use of "OR" diodes, only two IPS1031s are on at any single time; Q2 is saturated when either Q1 or Q3 is saturated. S1 and S2 signafy two different PIC Outputs. Schematic is located at:
LMAO, I forgot to mention what it was that I needed help on. I was concerned about the required voltage to turn on the mosfet, and if in combination with the diode and the pic output, if it would even see this voltage.
Hi, Jed. The 1N5282 is a controlled forward voltage silicon diode, and will have a Vf of 0.55 to 0.6V at 1mA forward current. The IR Low Side Switch IC requires a minimum of 4.5V for guaranteed operation. If your power supply is 5.0V, and your PIC output was exactly 5.0V, that means the IN terminal will see 4.45V to 4.4V -- not quite enough for guaranteed operation. And there will be a voltage drop of a tenth of a volt or so at the PIC output with 1mA output.
Of course, 4.4V or even 4.3V should be enough. But if you want to be sure, you might want to replace the 1N5282s with schottky diodes. That way, your diode "Wired-OR" configuration will be guaranteed to work. Schottkys have a typical forward voltage of less than 0.3V at low forward current.
You might want to use standard transistors, though -- the PIC output pin can certainly source the current, and a 2N3904 will cost a lot less than your low side switch ICs (view in fixed font or M$ Notepad):
With 2N3904s or other standard NPN small signal switching transistors and shottky diodes for D, this will work well for any 12V relay with a coil requiring less than 100mA.
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