Analog foundations of digital electronics clear as mud

[DerGegner]

I self-taught a fair amount of digital and a bit of analog circuits. The very computational part (e.g., put in two ones, get out a zero) which is what they have in digital logic texts is easy for me but at the more primitive level I am having a much harder time. See here for instance:

http://www.eng.utah.edu/~cs6710/handouts/AppendixB/appendixB.doc3.html

See the first diagram therein

"A high voltage at the base turns on the transistor. The output F is discharged to ground, getting close to 0 V but never quite reaching it (it reaches a voltage drop away from 0 V).

When a low voltage is placed on the base, the transistor is turned off. The output node F is charged up toward the power supply voltage through the pull-up/load resistor R1."

Could someone please clarify the bolded phrases for me? Furthermore, what is F exactly? Is it a wire that leads to ground? Is it just a point where you would attach one lead of a voltmeter or similar component, and connect the other to ground, thereby measuring the output?

 

[Aerosteon]

I self-taught a fair amount of digital and a bit of analog circuits. The very computational part (e.g., put in two ones, get out a zero) which is what they have in digital logic texts is easy for me but at the more primitive level I am having a much harder time. See here for instance:

http://www.eng.utah.edu/~cs6710/handouts/AppendixB/appendixB.doc3.html

See the first diagram therein

"A high voltage at the base turns on the transistor. The output F is discharged to ground, getting close to 0 V but never quite reaching it (it reaches a voltage drop away from 0 V).

When a low voltage is placed on the base, the transistor is turned off. The output node F is charged up toward the power supply voltage through the pull-up/load resistor R1."

Could someone please clarify the bolded phrases for me? Furthermore, what is F exactly? Is it a wire that leads to ground? Is it just a point where you would attach one lead of a voltmeter or similar component, and connect the other to ground, thereby measuring the output?

What you're not realising is that wires have an ability to hold charge, and transistor bases are controlled be charge but do not consume it.

Imagine this: You've got a switch connecting a 5v source to the base of a transistor. You close the switch and the voltage in the base of the transistor becomes 5v. Now what happens to the voltage in the base of the transistor when you open the switch?
............
If you said it goes to 0v your making a mistake.
The charge in the base of the transistor remains at 5v. The electrons are not being pulled out of the wire by anything.
The idea of a pull down resistor is to drain the charge out of the base of the transistor once your switch no longer applying a voltage.
A pull up resistor is the same idea when you want a signal to be some positive voltage by default, and you turn it off by connecting it to ground.

Many modern chips use push/pull style output pins so they can drive the voltage positive or negative, but many can only drive it one way so you need a pulling resistor.

 

[DerGegner]

That's not quite what I asked, but thanks in any case and the question has been resolved elsewhere

(Really, though, thank you)

 

[Master_Ghost_Knight]

It is a bit tricky but simple. Notice that on the other end of the transitor you have the ground (by default zero volts), the character F is an index for that node (or this node is at potetial k) given that the other node of the trasistor th potential is zero, that would mean that the difference of potential in the terminals of the transistor is k. This is a semi-conductor, you add potential at the gate the resitence between the 2 other terminals either increase or decrease depending of the gate potential (so some times it works as a high risistance the others it works almost like a wire).


The bridge between the analogic transitors and digital system is not that circuitry are capable of maintaining exactly a specific differnce of potential, but because the transitors when put in serie, the difference of potential will stabilize on one of 2 points (linear area tending to zero, or saturated area tending to saturation voltage) and that is mainly how we distinguish one state from the other (and that is why computers work in binary and not in another base system, although there are rare computers that work in trenary [tri-state]).