JedOs86 Posted August 11, 2006 Report Share Posted August 11, 2006 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: http://www.irf.com/product-info/datasheets/data/ips1031.pdfAttached 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. Quote Link to comment Share on other sites More sharing options...
JedOs86 Posted August 12, 2006 Author Report Share Posted August 12, 2006 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. Quote Link to comment Share on other sites More sharing options...
Theatronics Posted August 12, 2006 Report Share Posted August 12, 2006 The part is basicly a LOGIC level MosFet.You can treat it like a on/off switch. Quote Link to comment Share on other sites More sharing options...
JedOs86 Posted August 12, 2006 Author Report Share Posted August 12, 2006 Thanks Theatronics, that clears up most of the issues. But with the combination of the diode voltage drop, and the pic output which probobly wont be anything more than 4.5 volts, will the MOSFET still function? a voltage drop of .7 with 4.5 volts is about 3.8 volts. The MOSFET can't hardly function at this? This was my intended question, becuase I was told that a PIC output could handle a dozen these particular MOSFETs. Quote Link to comment Share on other sites More sharing options...
Theatronics Posted August 12, 2006 Report Share Posted August 12, 2006 Ick, I see what you mean.The Min Switch on voltage is 4.5v Quote Link to comment Share on other sites More sharing options...
JedOs86 Posted August 12, 2006 Author Report Share Posted August 12, 2006 Theatronics thanks for the diagram, I really appreciated that you took time to help me on this! I have decided to go with a design that I had in mind, which was earily similar to what you drew up for me. I'm just glad that you came up with a similar design , it shows that this might work afterall ;) . I might post the new schematic up later. Quote Link to comment Share on other sites More sharing options...
JedOs86 Posted August 15, 2006 Author Report Share Posted August 15, 2006 Mike:Taking a second look at the schematic you offerd, arent the transistors that are attached to the collectors always on? Shouldnt those second NPNs actually be PNPs? Quote Link to comment Share on other sites More sharing options...
Theatronics Posted August 15, 2006 Report Share Posted August 15, 2006 see, that’s the problem with using a CAD program to 'Sketch' something.I should have just drawn out the thing by hand and left it moreambiguous.I am not sure. I did say that the values needed to be 'fiddled' with.I built a setup like that once but it was a while ago. the 10Kohm should show+5v when the transistor is in cutoff.When the NPN goes into sat the current through the 10K will be determined by (beta) times Ib. If the 2.2k is seeing 5v then calc would be (Roughly)(5v - .7 = 4.3v)(4.3v / 2.2k = 1.95ma)(Given a beta of 120: 1.95ma * 120 = 234ma)So the current through the 10k should be 234ma ?well, the Vcc is 5v and 234ma * 10K = 2340v. Because that is not possiblethe Transistor must be in saturation. As a result the entire Vcc must appear across the 10K. As a result the Vbe at the second NPN would see near 0 Volts. Even if it saw above .7 volts due to junction voltages within the first transistor, that should not be enough to cause the second transistor to go into saturation and pull the output low.If the first transistor is in sat. and there is a bias on the base of the second stage of lets say 0.7V. That 0.7v wouldn’t overcome the Vbe knee and therefore the second transistor should stay in shutoff. If there were a Bias on the base then a simple voltage divider (A second 2.2k to ground) would cut the bias in half and put us back in control.When the first transistor goes back into shutoff, the entire 5V is expressed on the base of the second transistor, through 12.2k ohms of course. 12.2K ohms @ 5v yields 0.41ma.0.41ma * 120 = 49ma. 49ma though the second 10K resistor is 491 volts. Again the transistor is in sat so the whole voltage applied appears across the second 10K and the Collector appears to be at 0V. (Or darn close)Unless I am overlooking something, I think I got it right. Or at least close enough.It was just an example of the concept. If it were me I'd use a driver chip to shape the signal, A quad Op-amp would work. But because he had already mentioned a transistor style set up I went with that as an example.-Mike Quote Link to comment Share on other sites More sharing options...
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