On Mon, 24 Nov 2008 23:46:32 -0500, Joseph Smith joe@settoplinux.org wrote:
On Tue, 25 Nov 2008 03:36:34 +0100, Peter Stuge peter@stuge.se wrote:
Joseph Smith wrote:
A series of 1K ohms resistors 250mW (1/4W) on data lines D0 - D6 convert the signals from 5V to 3.3v needed by the flash chip.
This sentence and design is not so good.
Ok, how about this: A series of 1K ohms resistors 250mW (1/4W) on data lines D0 ??? D6 convert the signals from 5V to 3.3v needed by the flash chip.
This is still a bad claim. A resistor limits current and nothing else. Ignoring that will harm the rest of the circuit.
Any excess voltage is shunted by diodes inside the 74HC244 to its 3.3V power rail.
Umm, it could just break the IC. Please have a look at the data sheet again to confirm my fears. Going beyond absolute maximum ratings is not a good idea. Please avoid such recommendations.
Hmm. I just hooked everything up and tested the voltages across the circuit. D0,D1, D2, D4, D5, and D6 are 4.13V all the way across. D3 was 4.37V all the way across. /Error, /SELIN, /PE, and /Ack were a solid
3.3V.
The original author also put a 1k resistor across the 3.3V power rail to help to dissipate the current. Do you think that will help? I could
always
replace the 1k resistors with schottky diodes to lower the voltage 0.6V, but what about the bi-directional data lines? Would I have to put 2 schottky diodes togethor in both directions???
Also the Voltage on the LPC and FWH chips data lines should be + or -
0.5V
of 3.3V :-(
Ahh, I think the 1k resistor across the 3.3V power rail acts as a potential divider, I will add the 1k resistor across the 3.3V power rail, test and report back.