MOSFET Circuit - I know this works...But Why?

[kringen]

I was hoping someone could help me explain why this circuit works when a 5V signal from the MCU is applied to the base of Q1.



I'm baffled why the MOSFET doesn't stay on all the time since it is connected directly to 12V.



Also, is Q2 necessary?  I was wondering if Q1 can be connected directly to the MOSFET gate eliminating the need for Q2?



Many thanks in advance!

[SeanB]

When Q1 switches on it is driven to saturation so typically the voltage drop across the CE junction is around 0.2 to 0.3V. This is less than the 0.6V required to keep Q2 on, so it switches off, and then R3 pulls the gate of M1 high and turns it on fully. You can dispenbse with Q1 if you are ok with having an inverted drive signal, with a low turning the M1 on and a high turning it off. If you drive the mosfet from 5V though you will find that it will not turn on very much, it needs over 10V to turn on to the point where it can switch 9A with minimal voltage drop across it. At 5V it is barely on, and will depending on the actual device either pass under 1A or will still be off.

[kringen]

Thanks for the explanation, Sean.

So, am I understanding this correctly?

Under normal circumstances, when there is no input from the MCU:

Q1 = OFF
Q2 = ON (And drawing 12V or 1.2ma)
M1 = OFF

And when Q1 is triggered:

Q1 = ON
Q2 = OFF
M1 = ON (And drawing 12V and full load)

So since Q2 is normally on, does that serve as a voltage divider to keep M1's gate closed?

Thanks again!

[SeanB]

Correct, it keeps the gatew of M1 below the threshold where it conducts. If you used a logic level mosfet for Q1 that would be capable of turning on fully with 4V then you can run it direct from the MCU pin.

Just note that both the circuit you first proposed and using the logic level FET are going to be limited in switching speed. The first one will be slow to switch on ( at least compared to your view, it will still switch on in under about 2 milliseconds, so do not expect this to work at more than 1kHz as a power switch) and might have issues with oscillation during switch on and off, as the gate capacitance charges and discharges. This can be stopped or reduced by using a gate resistor right by the gate  of around 100R, and or by using a ferrite bead on the gate or drain lead to damp the oscillation. MOSFETs are rather well known to oscillate at 100 MHz while switching a low frequency or doing DC linear work, they work but get hot.

[kringen]

Thanks again,

That is good to know about the switching speed.  I will be using PWM to control the motor speed so I will try to keep the frequency below 1kHz.  I have run into trouble in previous versions of this H-bridge trying to switch P-Channel FETS on and off too fast and they definitely got hot!  I'm using low-side (N-Channel) switching for the PWM signal in this version.

[SeanB]

The reason they get hot is because they are being switched too slowly. Each time you switch them you have to charge or discharge the gate capacitance, and this is quite high, normally over 1nF. To get a faster switching rate you either have to use a gate driver IC, or reduce the value of R3 in your diagram to something a lot lower, like 1k or so.That will charge the gate faster, and will allow a higher switching or PWM frequency. At high switching speeds you might find you have to have up to 1A of current flow in the driver, depending on the frequency and the particular devices used and the voltage. Can be a trap if you think only in the DC case where the gate is only allowing a small leakage current to flow.

[kringen]

So much to consider using these FETS!

I think at this point I'll experiment with changing R3 to 1K and different PWM frequencies.  A lower R3 will also mean losing more battery to ground when the PWM signal is LOW.