Electronic – Reduce MOSFET switch off time

mosfetswitching

I am using this switching circuit to switch a lamp. I am controlling a rising slew rate at Vout (from 0-Vcc) that is adjustable by setting R2 with a poteniometer. It's not necessary for my project, but I would like to control the switch-off time for the MOSFET. Here is a my output at R6 with a 3.3v input.

As far as I understand it, Vout is staying high after Vin has gone low because the FET will stay on until Vgs is below the threshold voltage, so it makes sense that Vout stays high until Vgs goes below the threshold, but why is the slope after that point so much shallower than the rising Vout slow?

How could I reduce the fall-time of the voltage at Vout? This post has shed some light on the issue, but I haven't been able to figure out a good solution yet. I'm somewhat of a beginner to electronics, so any help is really appreciated.

EDIT:

Simulation with Vg

Best Answer

A very effective way of reducing the MOSFET turn-off time if you are driving it by a single pull-up (or pull-down, as it is in the case of a P-channel MOSFET) is to use an active pull-down (pull-up in our case circuit). This can be realized simply as follows:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

The circuit works according to the following points:

When the current mirror turns on, the \$I_{C1}\$ current flows through \$R_1\$, \$D_1\$ and the MOSFET gate-source capacity \$C_{GS}\$, charging it by increasing the gate charge \$Q_G\$ up to its turn-on value. \$Q_3\$ is off since it is a NPN BJT and its \$V_{BE}\$ is less than zero during this switching phase.
When the current mirror turns off, the current stops to flow through \$R_1\$ and and the voltage of the cathode of \$D_1\$ starts to rise, while its anode voltage is kept constant by the (large) gate charge stored in the gate capacitance. When \$V_{BE_{Q_3}}=-V_{AK}\approx 0.6\mathrm{V}\$, \$Q_3\$ turns on and discharges \$C_{GS}\$ quickly though \$I_{C3}\$, turning off \$M_1\$.

NOTES

This circuit does not require too low values of the pull-up resistor, and this is good since this lowers the static power dissipation. On the other and it requires a diode and a BJT.
This circuit is an adaption to a p-channel MOSFET of a circuit for n-channel MOSFET I saw in an old (1979 circa) SPRAGUE manual: once upon a time, the famous capacitor manufacturer manufactured also ICs, particularly a switching power supply controller, and I saw this circuit in the application examples of its data sheet.

Related Solutions

Switch 15V source with a MOS by digital 0/5V (dc)

You have an N channel MOSFET connected as a source follower. A source follower (just like an emitter follower for BJTs) can't have as it's output a voltage any higher than what is put on the gate. In fact, for a MOSFET, the source voltage will be a couple of volts less than what is on the gate - this is why you see 3V. The act of connecting your load to the source ensures that you cannot achieve full-turn-on unless the gate rises higher than 17 or 18 volts from ground. But all is not lost....

Ask yourself - what turns the FET on - answer - the gate to source voltage has to be a couple of volts or even higher on some fets. This can only mean that the fet is partially turned on.

You ought to consider grounding the source and putting the train lights in the drain. Like this: -

enter image description here

Rd is where you'd connect the lights. Now with the mosfet gate at 5V with respect to grounded source the fet turns on much better.

Another way is to use a P ch MOSFET driven from a BJT: -

enter image description here

Because the P channel FET's source is up to the positive supply, pulling the gate down to GND ensures it turns on to the lowest resistance. Don't run this circuit any higher than about 15 volts though because the maximum FET gate voltage might be limited to 15 volts - read the data sheets - a lot of mosfets are OK for +/-20 volts gate-source but some are only rated at max 12 volts.

Electronic – Mechanical relay replacement for mosfet bias

Option 2: Replace the analog circuit with a microcontroller

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Without external input, pull-down R1 firmly applies ground to Vgate through Q2.

(Existing)R2 is large-ish since FET gate does not need large current.

R3 is current protection for the LED inside the opto. A 380ohm resistor with a 1.2V drop across the LED yields about 10mA.

When uC_PowerEnable becomes high, Q2's gate is biased at DCIN, allowing Vgate to rise.

The recommended capacitor across the ATTiny power supply pins can be up-sized to ensure the Tiny is the last thing that turns off. Without signal from the uC, R1 resumes its original role of 'permanently' grounding Vgate. In this setup, the uC can shut down Vout for any reason(s) and can be reprogrammed, instead of redesigned and/or resoldered.

SetPin(uC_PowerEnable, VinPlus15Good && VinMinus15Good); //Shut down if either input rail fails

Best Answer

Related Solutions

Switch 15V source with a MOS by digital 0/5V (dc)

Electronic – Mechanical relay replacement for mosfet bias

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