Is it possible to amplify weak lows and weak highs?

The operation of a FET depends on the type and external components. For more clues goggle "fet transconductance graphs"

Jan Coombs

is low (say 0 volts), then the output is high impedance; and that if that gate is high (say 5 volts), then the voltage at the drain depends on the vo ltage at the source; if the voltage at the source is low, then the voltage at the drain is low; but if the voltage at the source is high (say 5 volts) , then the voltage at the drain is a weak high (say 3.5 volts). For a P-cha nnel MOSFET, if the gate is high, then the output is high impedance; and if that gate is low, then the voltage at the drain again depends on the volta ge at the source; if the voltage at the source is high (5 volts), then the voltage at the drain is high (5 volts); but if the voltage at the source is low (0 volts), then the voltage at the drain is a weak low (say 1.5 volts) . Does this sound about right?

This is a description of a MOSFET used as a switch. So the switch is open or the switch is closed. To turn the switch on requires the gate to have a bias relative to the source. That's why when on, changes in the source vo ltage will not be fully reflected in the drain. If the source changes enou gh that the gate to source bias is reduced too much, the switch is turned o ff. This effect is mitigated in a real application by using both P and N c hannel devices in parallel, so that at all voltages one or the other is ful ly on. Of course, this requires the gate drives to be complementary.

ible to design a circuit with such an N-channel transistor that has its sou rce at 0 volts and its gate at 3.5 volts that results in its drain being at 0 volts? And to design a circuit with such a P-channel transistor that has its source at 5 volts and its gate at 1.5 volts that results in its drain being at 5 volts?

One of us is confused. What you describe here is normal operation of the F ETs. When the N channel device has a high on the gate, the channel is clos ed and the source and drain are connected. So why would this not be possib le??? Are you asking about the gate voltage not being at the rails? 3.5 v olts is enough to turn on the FET if it is designed that way. The switchin g threshold is a design aspect of a FET.

and the output {rs} is defined as the value of {lw} if {pv} is low, and al ternately {hg} if {pv} is high. I have N-channel transistors {pivotHigh}, { amplifyHigh}, and {negateLow}; and P-channel transistors {pivotLow}, {ampli fyLow}, and {negateHigh}; and I have wires {weakResult}, {lowNegated}, and {highNegated}.

I'm sure these names mean something to you, but they are arbitrary to me an d make it hard to follow what you are talking about. Multiplexers have a c ontrol input and two signal inputs and a signal output. You didn't say if you are controlling digital or analog signals.

w} to the source of {pivotLow}, {hg} to the source of {pivotHigh}, the drai ns of each of {pivotHigh} and {pivotLow} to {weakResult}, {weakResult} to e ach of the gates of {amplifyHigh} and {amplifyLow}, the source of {amplifyH igh} to ground, the source of {amplifyLow} to high voltage, the drain of {a mplifyHigh} to {highNegated}, the drain of {amplifyLow} to {lowNegated}, th e other end of {highNegated} to the gate of {negateLow}, the other end of { lowNegated} to the gate of {negateHigh}, the source of {negateLow} to groun d, the source of {negateHigh} to high voltage, and the drains of each of {n egateLow} and {negateHigh} to {rs}.

Yeah, I'm definitely not wading through that description. There are tons o f diagrams showing how a mux is made from transistors. Pick up a TTL data sheet and see what they did. Some are very clever and designs have changed over the years... or maybe I'm thinking of digital FFs. I was digging int o the behavior of FFs when both reset and set are asserted at the same time . Turns out there is no expected output and each design works differently.

nes whether the transistors {amplifyHigh} and {amplifyLow} turn on and off. If that threshold is somewhere between 1.5 volts and 3.5 volts, then it se ems like this circuit should work as a multiplexer.

The threshold is not abrupt and is defined by your transistor process. Tha t's why they usually use a buffer in the path to give the switch FETs a ful l range of gate drive.

plementations of multiplexers, they usually have involved at least fourteen transistors, not six, three NAND gates with four transistors each and one NOT gate with two transistors.

Again, is this a digital or analog mux?

volt value to turn on an N-channel transistor? Or that noise can cause a 2.

5 volt value to turn off such a transistor? And vice versa for a P-channel transistor? If so, is there some way to bundle up the four transistors {piv otLow}, {pivotHigh}, {amplifyLow}, and {amplifyHigh} to reduce the risk of noise keeping the multiplexer from functioning correctly?

Why don't you make a drawing and post it somewhere so we can see what you a re talking about? Muxes are typically made using FETs as switches for the data/signal paths, but as gates for the control element. One big differenc e from using all logic gates is that logic gates have delay while the signa l path through the pass switches is very fast. So the control input will n ot be as fast, but the data path will.

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Rick C. 

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