Now I actually do understand.So a long story short, I have a transistor BS170 that acts as if always on when using 12 power supply (for led strip) but acts appropriately if connected to arduino 5v pin. I mis-interpretted because when I imagine a driver, I don't see the final load, all I see is the transistor and pull-up, that to me IS the driver and so I took the parts of the driver itself as what switch and load referred to.īut anyway, thank you both for clearing it up. That kind of detail is never covered just the terms "switch" and "load". Like in this example, the SWITCH is the mosfet, where as I thought it meant the transistor, the load is the final load whereas I thought it meant the pullup resistor. ![]() but because I don't know the definition of "high side" or "low side" and/or mis-interpret it, I'm buggered because no where there is a real unambiguous definition. ![]() I can tell you on a physical quantum scale why a mosfet works, I understand doping etc. Everyone automatically assumes the beginner is going to understand every bit of jargon and terminology. This is the problem in this field for beginners. I'll stick to programming, it makes more sense. To be non-inductive (not wire-wound) and right next to the gates.įorget it, the terminology is too ambiguous and counter-intuitive for me. If you do add gate resistors because of paralleled devices, the resistor should Many common MOSFET drivers have an output resistance of a few ohms to a fewĭozen ohms and typically need no resistors if driving only a single MOSFET and EMI Switch-on, as a safeguard against shoot-through). (typically the diode conducts to discharge, causing a faster switch-off than You can add a resistor and a diode in order to make switching asymmetrical High frequency EMI from the switching circuit, its basically a compromiseīetween switching efficiency and EMI reduction. You can add a resistor to slow down switching time in order to reduce Resistors also have the role of allowingĭevices with different plateau voltages to switch in unison (plateau and threshold Same driver, because there is a differential oscillation mode in this configuration You normally have to add gate resistors if driving two or more MOSFETs from the Outputs enough current to switch your MOSFET as fast as you want. Gate resistors are not normally required when usingĪ MOSFET driver driving one MOSFET. Those resistors can also help prevent ringing or oscillation. So without current limiting somewhere (in the driver or external to it), the current can spike very high. ![]() More correctly stated, the Gate takes a large amount of charge. The resistors to the Gates of the MOSFETs are there to limit current, because the Gate of a MOSFET has a very high "capacitance". High side: On the "high" (upper) end of the load. I'm still however a little confused, now I understand the resistors are there to protect from these high currents, but again, wouldn't they mess up the driver's function? The gate would then have to charge and discharge through those resistors and introduce a delay making it no better than using the transistor circuit above? Also if the spiking current is high wouldn't those resistors burn out if they were regular 0.25W ones? Is the term is relative to the device it's driving? If I used the same circuit to drive an N-Channel would it then become an inverting high side? I still don't get why it's non-inverting though, as the logic is being reversed by the transistor. So High Side would be clearer to understand if they said the LOAD is on the High Side rather than the SWITCH being on the high side as the switch by definition is actually on BOTH, the load can only be on one. It's one thing to understand circuits and another to know the terminology.I guess, it seems very counter-intuitive to me. I thought low-side meant the switch (Q2) was on the low (ground) side of the load and inverted because the gate of the mosfet is at an inverted logic to the base of the transistor. I do understand how the circuit works, I'm just not up on the jargon.
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