Switching MOSFETs for a buck converter

I'm confused about how to make the MOSFET Q1 act as an ideal switch in this buck circuit:

(schematic 1)


I need to have the voltage at N2 equal to the voltage at N1 when the transistor is turned on.
I've tried using a frequency generator in a setup similar to that in schematic 2. At N2 I ended up getting a a pulse more like V2 (voltage of the frequency generator) - Veb. I tried some BJT's and MOSFETS for Q1, different resistor values, different setups like schematic 1, different pulse frequencies, and never really got anywhere.

If I get a pulse from a frequency generator or microntroller- for experimental purposes- then how do I set up the pulse at the base/ gate to make the transistor work as an ideal switch?

(schematic 2)

Hi,

The pulse amplitude must be higher than the desired output because that is a voltage follower circuit.
The pulse amplitude with the MOSFET may have to be as much as 10v higher than the output to get it to switch to a low resistance, although it depends on the specific MOSFET used.

MrAl said:

Hi,

The pulse amplitude must be higher than the desired output because that is a voltage follower circuit.
The pulse amplitude with the MOSFET may have to be as much as 10v higher than the output to get it to switch to a low resistance, although it depends on the specific MOSFET used.

Click to expand...

Thanks. I tried a simulation with a BJT and it worked like you said:



However, the same setup with a MOSFET didn't work (it didn't work when I upped V2 to 20 V either).



For switching converters MOSFETs are recommended instead of BJTs. Why doesn't the MOSFET circuit operate the same as the BJT circuit?

Got it. Thanks! The pulse is running at 5V, vin is 12 v

VR ≈ Vin*duty cycle

you need to select a model for the mosfet.

Typically a PMOSFET is used for a buck circuit with the source terminal connected to the positive input voltage and the gate voltage driven between the input voltage and ground.

If you want to use an NFET, the gate must be driven ~10V above its source. Since its source is biased to the converter's output voltage, then the gate must be driven to Vout+10V. That is why it is better to use a PFET, because it turns on fully when the gate is driven ~10V more negative than its source, which is usually connected to the converter's input voltage.

MikeMl said:

If you want to use an NFET, the gate must be driven ~10V above its source. Since its source is biased to the converter's output voltage, then the gate must be driven to Vout+10V. That is why it is better to use a PFET, because it turns on fully when the gate is driven ~10V more negative than its source, which is usually connected to the converter's input voltage.

Click to expand...

Sweet, thanks