CMOS NAND Circuits Draw Current? Lab Notes Explained

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CMOS NAND circuits are designed to draw minimal current, primarily during steady-state conditions, where leakage current is typically negligible. However, when inputs are driven to mid-scale voltages, both NMOS and PMOS transistors can partially turn on, leading to significant current flow. This situation occurs when the inputs are not at clear logic levels, resulting in a continuous current draw. The confusion arises from the term "non-negligible," which refers to current levels that exceed expected leakage during these transitional states. Understanding the operational conditions of the circuit clarifies why current draw can appear contradictory.
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In my lab notes, it says that the connection of CMOS NAND circuits will draw non-negligible current...but I thought the point of the CMOS was because it uses such little current, at least that's what I wrote in my notes..I don't get it, what's going on?
 
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Yes, the CMOS implementation draws little current. Your lab notes are wrong ?!
 
I don't think they are wrong as it says so explicitly, and it's done in a way that acknowledges that this is unusual..this is the circuit if it makes a difference [PLAIN]http://img10.imageshack.us/img10/8317/lab7w.jpg.

I just don't get this "contradiction".
 
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Hm, I'm not sure what they consider non-negligible, it just says non-negligible...I guess any more current than it theoretically should. I don't know the explanation for this non-negligible current though, it shows us how the CMOS NAND is built and tells us the reason has to do with that, but I don't know the explanation.
 
Hey mmmboh,

The idea is that CMOS circuits use virtually no current when they're in steady-state. If you have some network of NANDs and all of the inputs are constant, not changing in time, then the CMOS circuit uses very little current. The only current it uses is called "leakage," and it's negligible in most situations. When the inputs change, though, the gate capacitance of the transistors has to be charged or discharged, and that does consume current.

The circuit they gave you is kind of a trick. They're driving the input of the NAND gates to mid-scale, halfway between the two supplies. This is a no man's land, somewhere between logic-0 and logic-1. This voltage is adequate to partially turn on both the NMOS and PMOS devices in the gates, so a continuous (and possibly large) current will flow.

Normally, when a CMOS gate changes state, its output changes so rapidly that the voltages are only in the no man's land for a short time. Except during these brief transitions, the voltage is stable, very close to either supply. Because the transitions are so rapid, the total power loss is very small.

- Warren
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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