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Message ID: 13     Entry time: Wed Oct 7 17:58:20 2009
Author: Stefan Ritt 
Subject: VDD switch off speed 

It turned out that the VDD switch off speed plays some important role. On our VME board, we have a linear regulator, then a 4.7 uF capacitor, then the DRS4 chip (DVDD and AVDD). When switching off the VME power, it takes quite some time to discharge the 4.7 uF capacitor, since the DRS4 chip goes into a high impedance mode if VDD < ~1V. This gives following VDD trace:

no_res.png 

Rising edge is power on, falling edge is power off. Note the horizontal time scale of 2 s/div. So to get below 0.3 V or so, it takes up to 30 seconds. If the power is switched back on when AVDD is above 0.3V, the DRS4 chip can get into a weird state, where probably many domino waves are started and the chip draws an enormous amount of current. Typically the linear regulator limits the current, so the 2.5V drops to ~1.5V, and the board is not working. If people are aware of this and always wait >30sec. before turning the power on again, this is fine, but people might forget.

So the solution is to put a resistor (typically 100 Ohm to 1 kOhm) parallel to the 4.7 uF capacitor in order to have some resistive current load of a few mA. The discharge then looks like this:

100ohm.png

Note the horizontal scale of 10ms/div. So after 30 ms AVDD is discharged and powering on the chip again does not do any harm. The same should be done to DVDD.

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