So, I understand of the NMOS transistor work's here (at least I believe I do). Electrons are flowing towards the capacitor plate from the drain here. So, I don't get how the plate connected to the drain is positively charged instead of being negatively charged. Can someone explain?
-
$\begingroup$ I don't know if this is the best link for this but bartleby.com/subject/engineering/electrical-engineering/… the answer is that the current is the opposite of the flow of electrons. At least, that's what I think is at play here. $\endgroup$– Alex KJan 30 at 1:05
-
$\begingroup$ hmm... So, why does a battery not do the opposite $\endgroup$– ChemDudeJan 30 at 1:23
-
$\begingroup$ Doesn't the linked source say that the electrons come out of the - terminal (the short side in schematics)? $\endgroup$– Alex KJan 30 at 1:25
-
$\begingroup$ The electrons come out the drain from my understanding. So, I don't understand why the plate is positively charged. $\endgroup$– ChemDudeJan 30 at 6:11
-
$\begingroup$ Would Electrical Engineering be a better home for this question? $\endgroup$– Qmechanic ♦Jan 30 at 7:59
1 Answer
The transistor is symmetrical, and used as a switch. Pull the gate positive, and the N channel goes to a conductive state. The voltage on the top of the capacitor thus duplicates the voltage on the bit line to the left. The voltage at the bottom of the capacitor is held constant by its connection to the substrate (its bottom plate is the chip substrate).
If, during this operation, the bit line is forced high or low, the corresponding voltage is stored on the capacitor (write operation). If it's not forced, the voltage on the capacitor influences the voltage on the bit line, and this may be sensed (read operation).
Did you notice that I didn't use the word "electron"? Tracking electrons is often more confusing than illuminating. Pay attention to voltages and currents. The only place it might be useful to think about electrons is in understanding why a positive voltage on the gate turns on the N-channel transistor.
-
$\begingroup$ Thanks this makes sense. However, I want to understand it from the electrons. The electrons flow from the source to the drain to the capacitor so shouldn't that plate be -ve charge? I'm trying to understand what I'm missing. $\endgroup$– ChemDudeJan 30 at 2:26
-
$\begingroup$ @ChemDude I'm not sure what you mean. When you use a FET as a bidirectional switch, current can flow either way through it. $\endgroup$ Jan 30 at 13:20
-
$\begingroup$ @ChemDude I think part of your conceptual problem is that you haven't drawn a complete circuit. Electric current flows in circles: that's why we call electrical networks "circuits". Often it's electrons, but it can be ions, holes, and most importantly here, displacement current. Where's the rest of your circuit? $\endgroup$ Jan 30 at 15:54
-
$\begingroup$ @ChemDude Usually in DRAM, the substrate is P-type, so the current that closes the circuit between the missing part on the left side of your diagram and the capacitor is actually holes, not electrons. $\endgroup$ Jan 30 at 15:58