Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Register Log in

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

Status
Not open for further replies.
7

777funk

Newbie level 6
Joined
May 21, 2012
Messages
12
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,392
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?
 

**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".
 
enjunear said:
**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".
Click to expand...

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

Q2 is NPN, not PNP. What enjunear wrote is correct.
 
  • Like
Reactions: asking

    A

    asking

    Points: 2
    Helpful Answer Positive Rating
enjunear said:
**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".
Click to expand...

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:
 

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

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top