NPN BJT to drive N-channel MOSFET into saturation. How would I do this?

[777funk]

MOSFET's are pretty simple to operate as a switch. However it seems like to get the lowest Drain Source resistance the Gate voltage needs to be quite a bit higher than what most logic circuits can supply (in the range of 12V for Vgs). How would I do this using an NPN transistor between a Microcontroller and a MOSFET driver?

 

[D.A.(Tony)Stewart]

Yes perhaps you should choose another part. 30,000 to choose from https://www.digikey.ca/product-sear...nductor-products/fets-single/1376381?k=mosfet

 

[enjunear]

**broken link removed**

The MCU connects to PORT-C2. The problem with this is that when the MCU is not driving a voltage to the output pin, it gets pulled up to +12V by the 10k resistor. This means that if left unconnected (or the MCU goes crazy/dies) the MOSFET defaults to the "ON" state.

You can make this circuit work with a default "off" condition by using a P-channel MOSFET.
**broken link removed**

With that circuit setup, a high gate voltage (approximately = Vsupply) makes the MOSFET be in cutoff. When you turn on the NPN from the MCU, it pulls the gate voltage down, which makes a P-channel FET start to conduct and turn "on".

 

[asking]

**broken link removed**

The MCU connects to PORT-C2. The problem with this is that when the MCU is not driving a voltage to the output pin, it gets pulled up to +12V by the 10k resistor. This means that if left unconnected (or the MCU goes crazy/dies) the MOSFET defaults to the "ON" state.

You can make this circuit work with a default "off" condition by using a P-channel MOSFET.
**broken link removed**

With that circuit setup, a high gate voltage (approximately = Vsupply) makes the MOSFET be in cutoff. When you turn on the NPN from the MCU, it pulls the gate voltage down, which makes a P-channel FET start to conduct and turn "on".

PNP Transistor Q2, Requires Negative Signal to start conduction....... as its PNP so Negative signal is required to drive the transistor.

 

[godfreyl]

Q2 is NPN, not PNP. What enjunear wrote is correct.

 

[asking]

Q2 is NPN, not PNP. What enjunear wrote is correct.

Yes, you are right. I mistaken...

 

[monkeyfist]

**broken link removed**

The MCU connects to PORT-C2. The problem with this is that when the MCU is not driving a voltage to the output pin, it gets pulled up to +12V by the 10k resistor. This means that if left unconnected (or the MCU goes crazy/dies) the MOSFET defaults to the "ON" state.

You can make this circuit work with a default "off" condition by using a P-channel MOSFET.
**broken link removed**

With that circuit setup, a high gate voltage (approximately = Vsupply) makes the MOSFET be in cutoff. When you turn on the NPN from the MCU, it pulls the gate voltage down, which makes a P-channel FET start to conduct and turn "on".

Apologies for opening an old thread, but this is the only search result relevant to driving an N channel MOSFET with an NPN transistor. Simply adding a 15k pullup to the base of Q2 makes the MOSFET normally off, although it does increase power consumption and is definitely not an optimal design. However, the only things I had lying around were NPN so... yay not waiting for shipping

Thanks for the thread, very useful today:thumbsup:

 

[mjsh401]

Which mode must be selected for BJT transistor? Active or Saturation?