LED matrix - MOSFET driver leakage?

Question

I have 4 common-anode RGB LEDs in a 4x3 matrix. I have a BSS138 on each of the red, green and blue cathode lines and a BSS84 on each of the anode pins. The power supply is 3.3v, which is close enough to Vf for each LED (red is slightly lower, and there is a 5Ω series resistor just before the red MOSFET drain to compensate).

When the matrix is not being driven at all, all the LEDs are dark. When I perform rastering, I get ghosting. The ghosting is quite low, so I could probably live with it, but I'd like to know what's causing it.

The gates are driven by a microcontroller and the gate waveforms are clean and square. When I look at the drains of the MOSFETs, their on-state is correct (either 0v for the cathode lines or 3.3v for the anodes), but the off-state voltage of the anode MOSFETs is like a downward ramp - like discharging a capacitor.

If I add a ~50k or so pull-down resistor to the anode MOSFET drain the problem goes away and the on-state drain waveform becomes square, like you'd expect.

The rastering duty cycle is 25% - that is, I actually have a huge dead-band between each anode activation period. The raster rate is one period every 125 microseconds. So the full raster is 16 periods - LED 1, then 3 off, LED 2, then 3 off etc (they're very bright LEDs).

Is a pull-down on the drain something to be expected to be necessary in this situation?

EDIT: Complete schematic is at https://hackaday.io/project/166766-Симон in the Files section.

Answer 1

As the voltage across a LED drops down through the forward operating voltage the current through the LED drops non linearly and very quickly. If self and stray capacitance is maintaining even a small current the LED can remain slightly lit.

How bad the problem is depends on the operating parameters of the LED.

Using a pull down resistor is an acceptable way of dropping the current through the LED very quickly the fact that these resistors are quite large in your case demonstrates that the LEDs will still produce some light at very low currents.

Answer 2

The problem is on the input capacitance of the MOSFET. The switching of the cathode is using the capacitance between gate and drain to feed the current to the LED. You could use a N-channel MOSFET in the high side to avoid this problem.