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  • if we look at the datasheet I linked to, Vgs(th) is the threshold voltage at which the MOSFET starts to conduct current (0.00025A). This threshold is between 0.6V and 1.5V, at room temperature. This is good information in the way that if we have 1.5V on the gate we should be able to run at least 0.00025A in the MOSFET and can expect that the drain voltage will be 1.5V or below. But most of the time our loads are larger than 0.00025A, are they not?
    Now, the threshold voltage is not useful information if we want to control a real life load. If the load is for example 2A we will need a lot more gate voltage if we want the transistor to turn on. If we look at figure 3 we could be fooled to expect that we only need 2V gate voltage. But don't forget that this curve is showing only typical behavior and with 5V across the MOSFET. totally useless information, because 2A * 5V = 10W. the MOSFET would expire in a puff of smoke if we tried that. The only section in the data sheet that gives us any solid promise is the Rds(on) value at 2.5V gate voltage. If we apply 2.5V at the gate and the load current is 2A we know that the resistance will be max 0.065 ohm, if the transistor is at room temperature, because the data sheet says so. the power dissipation then is 0.065 * 2^2 = 0.26W, and we can be sure that the transistor can handle this heat dissipation. the temperature rise will be about 34 degrees.

    So, just to be clear, this .045 and .065 for the different voltages is meant to give us a ratio? Does this ratio only specify the required voltages applied to the gate in relation to the current on the load? As in, assuming you had .00000000000001 amps on the load, but wanted to apply the maximum 20V on the Drain-Source, it would only take at a minimum .6V on the Gate since that's the beginning threshold for the transistor to turn on?

  • So, just to be clear, this .045 and .065 for the different voltages is meant to give us a ratio? Does this ratio only specify the required voltages applied to the gate in relation to the current on the load? As in, assuming you had .00000000000001 amps on the load, but wanted to apply the maximum 20V on the Drain-Source, it would only take at a minimum .6V on the Gate since that's the beginning threshold for the transistor to turn on?>

    the 0.065 ohm Rds (on) is the resistance between drain and source, when the MOSFET has the specified voltage on the gate. This doesn't depend on source to drain voltage - as soon as the MOSFET is turned on, Vds is becomes tiny, since the drain and source are practically connected.

    For even very small loads, yes, you'd need to get the gate up to around Vgs(th) for it to start to conduct. However, note that leakage current - current that will flow even when the transistor is off - is not quite zero; it's specified in the datasheet, and is generally very, very small, but probably larger than your example very small load.

    I've attached two screenshots of the datasheet that confuse me with what I'm being told about Rds(on) and this maximum peak current... The Rds says the transistor can only handle 15A at 10V but the maximum specs say it can handle 61A at 10V... what does it mean! LOL>

    First one specifies Rds(on) at the specified gate voltage and current.

    Second one specifies maximum continuous current at that gate voltage - at the higher current, the resistance may be a bit higher (see the charts)

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