The enclosed attachment is the circuit I have some questions about:



I see this as an opportunity to learn some things about electronics that heretofore I have somehow missed out on. For one thing, the negative rail is omitted (probably understood but I don't understand it). So, where is the negative rail to (a) and the positive rail to (b)? That is my first question. My second question has to do with reading schematics. My reading of texts and my college work on the subject did not include this data (this part of the education). Therefore, I must from this site learn how such data can be supplied, filled in, understood, what have you? Can you make a reference or simply explain how such omitted data can be gleaned? My last question is the assumed switching of Input A and B here. Could someone draw that in, as well? I assume (for sake of example) that it is like so many other lesson types: a double pole, single throw. Still, how is that filled in to the schematic to make a complete circuit? Thanks ahead of time!

 

The circuit is referenced to the ∇ "rail".

V(a) and V(b) are driven from perfect voltage sources that produce either 5V or 0V. I will leave it as an exercise to the student as to why V(o) is not driven all the way to 0V.

 

The enclosed attachment is the circuit I have some questions about:

View attachment 96487

I see this as an opportunity to learn some things about electronics that heretofore I have somehow missed out on. For one thing, the negative rail is omitted (probably understood but I don't understand it). So, where is the negative rail to (a) and the positive rail to (b)? That is my first question. My second question has to do with reading schematics. My reading of texts and my college work on the subject did not include this data (this part of the education). Therefore, I must from this site learn how such data can be supplied, filled in, understood, what have you? Can you make a reference or simply explain how such omitted data can be gleaned? My last question is the assumed switching of Input A and B here. Could someone draw that in, as well? I assume (for sake of example) that it is like so many other lesson types: a double pole, single throw. Still, how is that filled in to the schematic to make a complete circuit? Thanks ahead of time!

Only the positive rail is shown for (a) and only ground is shown in the circuit on (b) because the connection from the missing rail is actually coming from the input. The input is either at ground (0V) or at supply voltage (Vcc).

V(a) and V(b) are driven from perfect voltage sources that produce either 5V or 0V. I will leave it as an exercise to the student as to why V(o) is not driven all the way to 0V.

View attachment 96491

@MikeML

The discrete version is used to avoid limitations on max voltage. The voltage can be anything up to the limits of the transistor, diodes or current limit of the resistors at supply voltage levels.

 

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I see this as an opportunity to learn some things about electronics that heretofore I have somehow missed out on. For one thing, the negative rail is omitted (probably understood but I don't understand it). So, where is the negative rail to (a) and the positive rail to (b)?
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The negative rail (ground) is implied in (a) as the reference for the source voltage for inputs A and B and also is the reference for the (+) voltage and the Y output voltage.

There is no positive rail for (b).
Inputs A and B are referenced to ground and provide the power to the circuit.

 

The circuit is referenced to the ∇ "rail".

V(a) and V(b) are driven from perfect voltage sources that produce either 5V or 0V. I will leave it as an exercise to the student as to why V(o) is not driven all the way to 0V.

View attachment 96492

First, my hearty thanks to all for helping. Okay, as to your statement about Vo never quite reaching 0V, I would see that as a feature of the conductance characteristic of the BJT. Some argue that it won't "turn on" (or off) until about 0.60 volts whilst others tell me it is rather around 0.72V to 0.75v (the average I have heard is 0.70V for silicon units). Therefore, instead of my babbling, it is I believe the transistor's "on" threshold being exceeded. Hmmm, I also notice from the graph that the output is inverted. So, the BJT is in the common emitter mode. Much to digest. Again, my thanks to all for their wonderful assistance. By the way, what is this "thumbs up" and "thumbs down" stuff? Thanks. Have great evenings, everyone.