How is this capacitance affecting other parts of the circuit?

Question

I'm attempting to drive a matrix of LEDs. Columns. Rows. It flickers too much for my liking, so I tried smoothing the LEDs with capacitors, but I'm getting very odd behaviour.

Here is a minimal schematic of my test setup:

The A1-A3 and C1-C3 lines are tristated when not in use. They don't go high/low, they go active/tristate. I figured this would allow me to parallel a capacitor across each LED without them affecting each other, but apparently not.

If I close the A1/C1 switch, I would expect the capacitor to affect the operation of that LED and no others.

Test sequence:

1. Activate A1/C1. Wait.
2. Deactivate A1/C1.
3. Activate A3/C2. Wait.
4. Deactivate A3/C2.
5. Go to step 1.

I expected D1 to light, then fade out whilst D6 was lit. I expected the other LEDs to remain dark.

What actually happens is that D3 and D4 flicker noticeably as well. Given that the control lines are tristated when not in use, I'm at a loss to understand how the stored power is getting from the capacitor to these other LEDs.

It occurred to me that maybe some LEDs are passing reverse-bias current. LEDs aren't perfect diodes, but given that D1 is conducting forwards, I wouldn't expect that to be a factor.

Answer 1

I think td127's answer is correct. After you've turned off A1 and C1, and turned on A3 and C2, current takes the following path:

(The image has a text description of the path as alt-text. The pentagons indicate that current goes from the A3 pin of J1 to the A3 lead of R3.)

The microcontroller and the capacitor are effectively in series, so their voltages add. If the microcontroller is putting out 3.3 V and the capacitor is still charged up to 3.3 V, that adds up to 6.6 V. The two LEDs are also effectively in series, and 6.6 V is enough to light up the two LEDs in series.

Another way to see this is to apply Kirchhoff's voltage law to the circuit. If \$V_{uc}\$ is the voltage between A3 and C2 (produced by the microcontroller), \$V_{cap}\$ is the voltage across the capacitor, and \$V_{D3}\$ and \$V_{D4}\$ are the voltages across the diode-and-resistor pairs, we find that we must have

$$V_{uc} + V_{cap} = V_{D3} + V_{D4},$$

so D3 and D4 will light up if the capacitor hasn't discharged enough by the time that the microcontroller puts out voltage on A3 and C2.

This happens regardless of whether or not the clamping diodes are conducting. And I don't think that the clamping diodes will conduct, because D3 and D4 will probably have approximately the same voltage across them, and that voltage will be no greater than \$V_{uc}\$. If one of them does conduct, that will decrease the voltage across one of the LEDs, and thereby increase the voltage across the other one. It looks like there's no way for both clamping diodes to conduct at the same time.

Answer 2

Here’s my hypothesis:

When the capacitor is charged it holds A1 high and C1 low (for a while) even after they are tristated.

The capacitor will start discharging through D1 when A1/C1 is tristated, but as soon as A3/C2 is activated (assuming capacitor still has some charge) the C2 will provide a parallel discharge path from A1 through D4.

Likewise, when A3 is brought high then there is a residual discharge path through D3 to the still-somewhat charged capacitor. These parallel discharge paths manifest as flicker.

To reduce flicker you must run at sufficiently high update rate. I wouldn’t waste any time in tri-state – move on to the next row/column as soon as possible.