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| Saturating a MOSFET with a constant current sink driver |
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| Oleenick:
So for a high side driver a PCH MOSFET should be used? Sorry to keep you explaning things, but before I jump to conclusions, why should this be a PCH? Also as noted before I should flip the MOSFET to have the +5V on top of the diagram :-+ |
| Wimberleytech:
--- Quote from: Oleenick on January 07, 2020, 02:42:47 am ---So for a high side driver a PCH MOSFET should be used? Sorry to keep you explaning things, but before I jump to conclusions, why should this be a PCH? Also as noted before I should flip the MOSFET to have the +5V on top of the diagram :-+ (Attachment Link) --- End quote --- Well, you are driving the anode side of the LEDs toward 5V. If all you have is 5 volts, an NCH device cannot drive the LEDs to a high enough voltage. A PCH can. |
| Oleenick:
--- Quote from: Wimberleytech on January 07, 2020, 03:11:51 am --- --- Quote from: Oleenick on January 07, 2020, 02:42:47 am ---So for a high side driver a PCH MOSFET should be used? Sorry to keep you explaning things, but before I jump to conclusions, why should this be a PCH? Also as noted before I should flip the MOSFET to have the +5V on top of the diagram :-+ (Attachment Link) --- End quote --- Well, you are driving the anode side of the LEDs toward 5V. If all you have is 5 volts, an NCH device cannot drive the LEDs to a high enough voltage. A PCH can. --- End quote --- Ahh i see. I've read over everything again and I understand why I'd need a PCH device. Because an NCH device is only useful for driving a load at a higher voltage than the gate (logic level) can supply. And because the Vgs is the voltage across gate and source. So if I have 5V on an NCH gate and the source is at 0V: Vgs = 5V-0V = 5V (NCH MOSFET on) ... 5V then goes from drain to source, Vgs = 5V-5V = 0V (NCH MOSFET off): So an NCH will never stay on. Whereas with a PCH MOSFET, Vgs = 0V-5V = -5V (PCH MOSFET on greater than -2.2V (SiSS05DN)) ... 5V then goes from source to drain, Vgs 0V-5V = -5V the PCH MOSFET is still on. Does my understanding make sense? Thanks again. |
| Wimberleytech:
--- Quote from: Oleenick on January 07, 2020, 04:39:34 am --- Because an NCH device is only useful for driving a load at a higher voltage than the gate (logic level) can supply. --- End quote --- Not sure what you are trying to say here...I think you are still a little confused. --- Quote --- So if I have 5V on an NCH gate and the source is at 0V: Vgs = 5V-0V = 5V (NCH MOSFET on) --- End quote --- Correct --- Quote --- ... 5V then goes from drain to source --- End quote --- No. First of all voltage does not go anywhere...it is "across." If the NCH transistor is on, it will conduct as much current as it can based on its physical properties and external circuit elements. --- Quote ---Vgs = 5V-5V = 0V (NCH MOSFET off): So an NCH will never stay on. --- End quote --- Correct --- Quote ---Whereas with a PCH MOSFET, Vgs = 0V-5V = -5V (PCH MOSFET on greater than -2.2V (SiSS05DN)) ... 5V then goes from source to drain, Vgs 0V-5V = -5V the PCH MOSFET is still on. --- End quote --- Again voltage does not go anywhere. Current will flow --- Quote --- Does my understanding make sense? Thanks again. --- End quote --- You are getting closer. |
| Oleenick:
--- Quote from: Wimberleytech on January 07, 2020, 01:36:30 pm --- --- Quote from: Oleenick on January 07, 2020, 04:39:34 am --- Because an NCH device is only useful for driving a load at a higher voltage than the gate (logic level) can supply. --- End quote --- Not sure what you are trying to say here...I think you are still a little confused. --- End quote --- I was trying to say that I an NCH device is only useful if another, higher voltage is availible. For example, switching 10V with 5V signal, would that have a Vgs (off) = 0V-0V = 0V and a Vgs (on) of 5V-10V = -5V?. Is it right that I have that -10V on the end there? I'm assuming that when the gate is at 5V the source is at 10V. --- Quote --- --- Quote --- So if I have 5V on an NCH gate and the source is at 0V: Vgs = 5V-0V = 5V (NCH MOSFET on) --- End quote --- Correct --- Quote --- ... 5V then goes from drain to source --- End quote --- No. First of all voltage does not go anywhere...it is "across." If the NCH transistor is on, it will conduct as much current as it can based on its physical properties and external circuit elements. --- Quote ---Vgs = 5V-5V = 0V (NCH MOSFET off): So an NCH will never stay on. --- End quote --- Correct --- Quote ---Whereas with a PCH MOSFET, Vgs = 0V-5V = -5V (PCH MOSFET on greater than -2.2V (SiSS05DN)) ... 5V then goes from source to drain, Vgs 0V-5V = -5V the PCH MOSFET is still on. --- End quote --- Again voltage does not go anywhere. Current will flow --- Quote --- Does my understanding make sense? Thanks again. --- End quote --- You are getting closer. --- End quote --- So apart from my irresponsible description of voltage (thanks I'll make sure to correct it in future), its just that I assumed that source will be at 5V after saturing the MOSFET given the load. In a previous case, does that mean if the MOSFET was on and 10A of current started flowing from drain to source, the voltage across drain and source would be ~3V? Thank you for continuing to help. |
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