Electronics > Beginners
Low voltage drop transistor (or other device)
permal:
--- Quote from: oPossum on December 02, 2018, 08:33:32 pm ---If you replace the NPN with a N channel MOSFET, you will have greater voltage drop. You have to bodge in a P channel device as T3sl4co1l says.
--- End quote ---
Wait, what? Why greater voltage drop? The on-resistance on for example IRLML2402TR is just 0.25-0.35 \$\Omega\$. At 100mA, that's just 3.5mV, compared to the current ~0.7V.
Is it that the 3.3V I can put on the gate isn't enough to open the Mosfet fully? (how do I tell
IanB:
--- Quote from: permal on December 02, 2018, 08:42:10 pm ---Wait, what? Why greater voltage drop?
--- End quote ---
Because you have to put the transistor on the correct side of the load to obtain good switching properties.
For example, an NPN BJT has to go on the low side of the load in order to be a good switch. If you put it on the high side it won't work efficiently.
Similarly, a PNP BJT should go on the high side of the load.
By similar logic, the same is true of MOSFETs. According to the type of FET you need to put it on the correct side of the load. If you put it on the wrong side it won't switch efficiently.
HB9EVI:
If you want to switch high side with a N-FET, you need a considerably higher Vgs than the Vdd of 3,3V.
Use a P-FET. For 100mA the higher Rdson of P-FETs doesn't really matter; that's an issue if you switch higher currents
permal:
--- Quote from: IanB on December 02, 2018, 08:48:06 pm ---
--- Quote from: permal on December 02, 2018, 08:42:10 pm ---Wait, what? Why greater voltage drop?
--- End quote ---
Because you have to put the transistor on the correct side of the load to obtain good switching properties.
For example, an NPN BJT has to go on the low side of the load in order to be a good switch. If you put it on the high side it won't work efficiently.
Similarly, a PNP BJT should go on the high side of the load.
By similar logic, the same is true of MOSFETs. According to the type of FET you need to put it on the correct side of the load. If you put it on the wrong side it won't switch efficiently.
--- End quote ---
Ok. So, the rule of thumb is that Ibe/Igd should never have to pass the load to reach ground for the transistor/mosfet to be an effective switch. Is that correctly understood?
--- Quote from: HB9EVI on December 02, 2018, 08:58:28 pm ---If you want to switch high side with a N-FET, you need a considerably higher Vgs than the Vdd of 3,3V.
Use a P-FET. For 100mA the higher Rdson of P-FETs doesn't really matter; that's an issue if you switch higher currents
--- End quote ---
That'll invert the control signal though, like T3sl4co1l pointed out. I'll have to think of how that affects the rest of the circuitry.
Thanks for all the input so far.
HB9EVI:
That inverts the control signal, but there are many ways to handle that.
otherwise, if you go with the N-FET, you can take a bootstrapped FET driver to solve the problem
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