# Gate drive resistor size and/or gate driver IC

I am working on a design for a 16-channel LED driver. At the moment the basic design is based on three main components:

The circuit needs to initially drive a 110 V LED light with 250 mA LED strings, but should to be adjustable to drive sets of other LEDs e.g Ledengin LZ7

simulate this circuit – Schematic created using CircuitLab

Without further components I think this design should work in principle with M1 getting an RDS(on) of ~30 mΩ (other similar MOSFETs had a larger RDS(on)). The RDS(on) can be reduced further if I can supply a higher voltage (14.1 mΩ at 6 V and 12.6 mΩ at 10 V).

(This combination of R2, M1, Q1, R2, and D1..n are repeated for each channel though R1 will be varied dependent on different D1..n on each channel.)

The PCA9685 can supply a limited current of 10 mA per channel and if I'm reading it right 25 mA total, and it can operate PWM at up to 1526 Hz.

Simulation at 1526 Hz PWM (assuming I did it right) shows if I have R2 = 10 kΩ I get an inrush current of 500 μA which should be no issue.

If I reduce R2 to 1 kΩ I get 5 mA which is likely too large to operate more than 4 channels at once, and 100 Ω gives 50 mA which is too large for a single channel and the PCA9685.

How do I determine the best value for R2?

Would the inefficiency resulting from a large R2 make it worth putting in a gate driver e.g a MAX620EWN+ (would need 4) or a cheaper alternative?

Or is there another cost-effective way to make this more efficient?

Simulation model and output (the New macro are a PLW7070GA Spice model)

• If you only need 250mA of current, then you are really over spec'ing your RDSon. Also, your FET is only rated for 100V, not 110V+margin.
– W5VO
Commented Jun 12, 2018 at 21:50
• correct me if im wrong but isn't the voltage across the MOSFET (Source - LED Forward Voltage) Commented Jun 13, 2018 at 13:33
• so for a 110V supply as long as the voltage drop across the LED array is >10V+margin this wouldn't be an issue ? and would ideally choose a voltage source to minimize the voltage across the mosfet to me more efficient Commented Jun 13, 2018 at 13:39
• When the MOSFET is off, the current through the MOSFET and LED chain will be close to 0A. That makes the voltage drop across the LED almost 0V, and the rest is across the MOSFET. Picking a bigger MOSFET (lower RDSon) means you have larger gate capacitance, slower turn on/off, and more power dissipation from switching. Unless you're using a switch-mode driver for controlling LED current, "efficiency" is irrelevant - you will burn the difference between the forward voltage of the LED and your power supply somewhere.
– W5VO
Commented Jun 13, 2018 at 17:29
• that's interesting as the simulation of the circuit in TINA-TI doesn't show that simulating a set of PLW7070GA leds and CSD19534KCS, i only get a ~0-30V swing across the mosfet is there something wrong with the simulation? (ive added simulation info to post) Commented Jun 13, 2018 at 18:42