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Registered Member #1792
Joined: Fri Oct 31 2008, 08:12PM
Location: University of California
Posts: 527
edit: I kind of skimmed before I posted and didn't realize that Bjorn pointed this out, and lightlinked said exactly what I did.
The problem is that you are using an N-channel device for high-side switching. When the FET voltage drop is low, the voltage drop on the load is the full 12V. Which means that the source voltage is 12V, and therefore even if the gate voltage is 12V, the gate-source voltage is zero so the device cannot be on. Instead it should operate in a partially on state, so the FET's voltage drop allows the gate-source voltage to be positive but not fully on.
Solutions: a high-side driver which allows gate voltage to exceed 12V, a P-channel FET for high side switching, or use existing N-channel device for low-side switching where source is grounded.
Registered Member #2008
Joined: Tue Mar 03 2009, 05:11AM
Location: USA, Frederick, MD
Posts: 118
Mattski wrote ...
edit: I kind of skimmed before I posted and didn't realize that Bjorn pointed this out, and lightlinked said exactly what I did.
The problem is that you are using an N-channel device for high-side switching. When the FET voltage drop is low, the voltage drop on the load is the full 12V. Which means that the source voltage is 12V, and therefore even if the gate voltage is 12V, the gate-source voltage is zero so the device cannot be on. Instead it should operate in a partially on state, so the FET's voltage drop allows the gate-source voltage to be positive but not fully on.
Solutions: a high-side driver which allows gate voltage to exceed 12V, a P-channel FET for high side switching, or use existing N-channel device for low-side switching where source is grounded.
Registered Member #2008
Joined: Tue Mar 03 2009, 05:11AM
Location: USA, Frederick, MD
Posts: 118
Mattski wrote ...
edit: I kind of skimmed before I posted and didn't realize that Bjorn pointed this out, and lightlinked said exactly what I did.
The problem is that you are using an N-channel device for high-side switching. When the FET voltage drop is low, the voltage drop on the load is the full 12V. Which means that the source voltage is 12V, and therefore even if the gate voltage is 12V, the gate-source voltage is zero so the device cannot be on. Instead it should operate in a partially on state, so the FET's voltage drop allows the gate-source voltage to be positive but not fully on.
Solutions: a high-side driver which allows gate voltage to exceed 12V, a P-channel FET for high side switching, or use existing N-channel device for low-side switching where source is grounded.
Registered Member #1792
Joined: Fri Oct 31 2008, 08:12PM
Location: University of California
Posts: 527
cavemen wrote ...
Mattski wrote ...
... Solutions: a high-side driver which allows gate voltage to exceed 12V, a P-channel FET for high side switching, or use existing N-channel device for low-side switching where source is grounded.
Can you draw a sketch of this circuit please.
The first circuit is more complex and there are many different ways to make a high side driver, google for it if you are interested. The second circuit I provide a link to. The third circuit Bjorn provided a sketch of, and you apparently got working. So I would go with the third one if I were you, R2 is your load in this case. And instead of the switch you drive the gate with your complementary darlingtons.
cavemen wrote ...
Can it be that the multisim model is wrong?
Not terribly likely, but you should be able to look up a Spice model from a device manufacturer and use that instead if you like.
Registered Member #2008
Joined: Tue Mar 03 2009, 05:11AM
Location: USA, Frederick, MD
Posts: 118
Mattski wrote ...
cavemen wrote ...
Mattski wrote ...
... Solutions: a high-side driver which allows gate voltage to exceed 12V, a P-channel FET for high side switching, or use existing N-channel device for low-side switching where source is grounded.
Can you draw a sketch of this circuit please.
The first circuit is more complex and there are many different ways to make a high side driver, google for it if you are interested. The second circuit I provide a link to. The third circuit Bjorn provided a sketch of, and you apparently got working. So I would go with the third one if I were you, R2 is your load in this case. And instead of the switch you drive the gate with your complementary darlingtons.
cavemen wrote ...
Can it be that the multisim model is wrong?
Not terribly likely, but you should be able to look up a Spice model from a device manufacturer and use that instead if you like.
Well, well, well the 10kohm resistor and the switch in the driver circuit are the darlingtons. One taps the gate to power, while the other one taps it to ground. They never bias at the same time or they burn. The source of signal drives the bases of two transistors. One - on other - off contioniously.
Do I put the signal source between the bases of the transistors and the high side of the circuit or between the bases of the transistors and the low side of the circuit?
The voltage drop on the MOSFET can never go below around 2 volts that it exhibited when I built the circuit that Bjorn set as an example? 2V * the current through the circuit branch gives me the powwer loss on the MOSFET Correct?
Well. Thank you fro help.
If I cannot squeeze that voltage drop out of MOSFET at 12 V I either have to use a different MOSFET or just live with it. As long as My design does the best it can.
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A "totem pole switch" I got problem with basic electronics understanding.
