Oops! N-channel Mosfet on sinking output

Question

I have designed a simple circuit to drive LEDs using a MAX6964 I2C LED driver/GPO chip. Unfortunately, I somehow missed missed that the outputs are open-drain current-sinking outputs. This is not working because my circuit uses an N-channel mosfet in a low side configuration. From my research on this site, I found a few articles on N-channel and P-channel mosfets but none seem to address my specific question.

From my readings on this site, the N-channel gate needs a positive voltage to switch it. Again, unfortunately the output of my Max6964 is a current sink not a current source. I have already made some pc boards with the design that includes surface mount parts. I am trying to "salvage" these pcbs with the smallest possible change of components and/or trace cutting & rewiring. Here are my thoughts:

1. Simply replace the the N-channel mosfet with a P-channel mosfet. Is that correct? My hunch is this will not work.

2. Leave in the N-channel mosfet and add a pull up resistor (value?) between the gate and the supply (VAA) work? This would invert the logic for turn on the LED but I can live with that.

3. Since the Max6964 can sink 50mA on each port, remove the N-channel mosfet, jumper the gates to the sources, and increase the value of resistor R6 to some higher value (~100 ohms) to limit the current. I am trying to maximize the intensity of the LEDs so I am not sure this option is the best.

Here is my circuit:

Here is the simplified Max6964 output port from the data sheet:

[![Max6964 simplified output port][2]][2]

Thanks a bunch in advance for any feedback!

[IMPORTANT EDIT] VAA is +5V PWM signal. So although I liked the pull up idea, I don't think it will work.

[EDIT 2016-04-20 proposed:]

Answer 1

Go with the pullup, if you want to keep your MOSFET driver.

Make sure you don’t sink too much current into the pin by checking that $$V_{aa}/R < 50\text{ mA}.$$

If you don’t need super fast switching, a resistor in the order of a few kOhm (such as 4.7k) will do fine.

Answer 2

1. Use a P-channel mosfet. You have the opposite problem with this, you need a pull-up to turn off the mosfet. Also, depending on the voltage of VAA this may not work even with a pull-up.

2. Add a pull-up from Gate to VAA. This would be my choice. Since the MAX6964 uses PWM you will need a value of around 4k7 to rapidly charge the gate capacitance..

3. Bypass the mosfet. This will limit your options on the amount of current you can drive through the LED.

Answer 3

Since the '6964 can already do current control just tie the drain pad to the gate pad, omit the MOSFET entirely, and fix it in software.

Answer 4

^{simulate this circuit – Schematic created using CircuitLab}

The P/N of BJT and LED in schematics have no meaning. You should use your LED. As for BJT, it must reliably handle 100 mA load current (very easy), have transient frequency above 300 MHz for fast switching (easy) and preferably have low VCEsat (saturation voltage between Collector and Emitter).

You must put a resistor between MAX6964 output and Node "From LED driver". It must provide about 3-5 mA current at your minimum VAA-1V. For example: VAA = 5V, then the resistor is 1 kOhm.