Saturday, May 6, 2017

Who ya gonna call?

Grid Driver Configuration
I decided to dig deeper into the ghost segments appearing on the VFD.

The Toshiba RN4604 I used in the anode and grid drivers contain two pre-biased bipolar transistors, one PNP and one NPN; the bias resistors are all nominally 47K ohms. Driving the CTRL input to +3.3V turns on Q1B, which drives Q1A into saturation and pulls the grid to +30V. With the CTRL input at ground Q1B and Q1A are off, and R2 pulls the VFD grid to ground. The ends of the filament swing ±2V around a point +6.8V above ground, so an inactive grid should be at least -4.8V with respect to the nearest filament segment.

I connected my oscilloscope to the VFD grid leads for the active and ghosting digits. When the active digit's grid went to +30V with respect to ground I saw the ghosting digit's grid rise to about +10V. That's about +3V wrt to the average cathode voltage, which attracts enough electrons to cause the active anode in that digit to fluoresce too. Although this is not a truly stable state and tended to dissipate over time, it appeared the pull-down resistor R2 wasn't doing its job.

Just to be sure I looked for any indication that the ghosting digit's grid driver was turning on spuriously, but I couldn't find any. The circuit is sensitive enough to turn on momentarily if you touch an undriven CTRL input, but wiring CTRL to ground didn't change the behavior. Scoping the input to the PNP transistor showed nary a wiggle even as the grid lead voltage rose.

I expected some leakage current on an inactive grid, but the datasheets I saw for ICs that had built-in pull-downs used 125K ohm resistors so I wasn't expecting much current on an inactive grid. I tried placing a second 100K ohm resistor atop the existing 100K pull-down resistor, giving an effective 50K ohm pull-down, and although that reduced the brightness of the ghost segment it didn't eliminate it. Given 10V across 100K ohms, the leakage current on the ghosting digit's grid must be about 100 uA. Hanging a through-hole 10K resistor to ground on the lead killed the ghost. That's probably excessively low, but I haven't been able to find any information on this VFD that would give an expected inactive grid current.

My best guess at this point is that there is some capacitive coupling between the grids within the tube, and it causes the adjacent grid to rise. The condition isn't stable because the charge eventually bleeds off through the pull-down resistor. But I don't really understand the mechanism at work here.

I don't have a lot of 0603 SMD resistors laying around so I decided to replace the 100K pull-downs on the grids with 10K resistors. This will cost me an extra 3mA on the active grid, but there's only one grid active at any time. Active anodes draw less than 600uA each, while there are always 2 to 9 anodes lit per active digit, so I decided to leave their pull-downs as 100K resistors.

Next I'll work on getting the FPGA to drive this thing.

No comments:

Post a Comment