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At 12:45 AM 4/22/96 -0400, you wrote: >First of all, bear in mind that the memory socket is designed for 3.3V circuits >Secondly, for a 5V IC to work, it REQUIRES 5V!!!!! >Third, according to Ohm's law, V=IR, where V=voltage, I=current, R=resistance. >So for the 5V memory board to work, the thinkpad MUST output a 5V to the 5V >module, or else, there will not be enough voltage for the module to distinguish >0 and 1. (I think anything > ~4.2 is considered 1. Don't quote me on this. I >am not too sure). So the current output is now 5/R instead of 3.3/R. So the >IBM tech is probably right...... actually, it's the other way around. logic 0 is ranging from 0 to something like 0.7 volts (it really depends on the circuit) but the threshold for logic 1 is actually quite low, definitely less than 3.3 bear in mind that the 5volts is a rating on the memory module. it's clearer for notebook manufacturers that way. i guess they could quote it in watts (power consumption) but i don't think that any notebook manufacturers ever made notebooks with currents that could fry the memory chips, and it was easier for the public to follow (never mind about confusing!) if they just labelled the voltage rating. so, the chips don't NEED 5volts, but they will draw more current given less voltage, and (this part i'm not sure) that being the case, circuitry via the power supply to the memory module are likely to be fried too... had this problem with a robot last week. we put in another sensor without changing the resistor. it was added in parallel so we thought the voltage drop would be the same, but we forgot that more current was drawn, and so we blew the motherboard :( sam