Trying to get my mind around gates and their comparative functions -- if any one of you folks can lend a hand, I'd appreciate it
I've read a bit about gates, and often hear "2 switches wired in series and both being closed" will allow 2 inputs to flow to the lightbulb on Q. But that's not saying anything about the 2 inputs themselves playing a part in whether or not the switches are closed.
If I put a low signal through input A flowing to Switch 1, and a high signal to input B flowing to Switch 2, and both switches are closed (completing circuit), and the lightbulb on Q lights, then both signals have not been compared.
As I understand it, this is NOT the full story of the AND gates used in electronic logic.
The logic levels is not just in the status of the switches (the traditional hand closing faucet idea -- as the hand is not identified in that example). Otherwise a low and a high signal could be sent through inputs A and B, and since the switches are closed, this makes Q true (both switch 1 AND switch 2 are closed -- and this is comparing the switches themselves, but not the inputs), but an actual AND gate will only allow 2 HIGH signals through in order to make Q true (on).
Question: What's really happening with 2 switches in series in order to compare 2 high signals to make Q high? Something about the inputs themselves must be closing the switches. Is it correct then, that 2 transistors in series to make an AND gate are applied outside voltage to be made "active" (capable of turning on and off?), but what is *really* closing the switches is *not* this voltage (i.e., NOT some outside force), but the sensitivity of both transistors in *requiring* a high voltage from both inputs in order to close (as opposed to low voltages), *thus* allowing both inputs through and lighting Q?
The logic then is a *combination* of the inputs and their *impact* on the transistors/switches themselves, and not merely in the status of them prior to this. Is it true then the purpose of the outside voltage/current is simply to enable the transistor to be flicked on (or off) by either 5V or 0V coming from the input itself (why, I also don't understand -- since 0V is not really what it is, as 0V can do nothing -- it's more like something around 0 so that there's some force there...?) ... certain transistors are by default "off" when "activated" by an outside voltage, and then constructed so that they're able to be switched in the opposite on/off position by either a 5V or ~0V input?
I've read a bit about gates, and often hear "2 switches wired in series and both being closed" will allow 2 inputs to flow to the lightbulb on Q. But that's not saying anything about the 2 inputs themselves playing a part in whether or not the switches are closed.
If I put a low signal through input A flowing to Switch 1, and a high signal to input B flowing to Switch 2, and both switches are closed (completing circuit), and the lightbulb on Q lights, then both signals have not been compared.
As I understand it, this is NOT the full story of the AND gates used in electronic logic.
The logic levels is not just in the status of the switches (the traditional hand closing faucet idea -- as the hand is not identified in that example). Otherwise a low and a high signal could be sent through inputs A and B, and since the switches are closed, this makes Q true (both switch 1 AND switch 2 are closed -- and this is comparing the switches themselves, but not the inputs), but an actual AND gate will only allow 2 HIGH signals through in order to make Q true (on).
Question: What's really happening with 2 switches in series in order to compare 2 high signals to make Q high? Something about the inputs themselves must be closing the switches. Is it correct then, that 2 transistors in series to make an AND gate are applied outside voltage to be made "active" (capable of turning on and off?), but what is *really* closing the switches is *not* this voltage (i.e., NOT some outside force), but the sensitivity of both transistors in *requiring* a high voltage from both inputs in order to close (as opposed to low voltages), *thus* allowing both inputs through and lighting Q?
The logic then is a *combination* of the inputs and their *impact* on the transistors/switches themselves, and not merely in the status of them prior to this. Is it true then the purpose of the outside voltage/current is simply to enable the transistor to be flicked on (or off) by either 5V or 0V coming from the input itself (why, I also don't understand -- since 0V is not really what it is, as 0V can do nothing -- it's more like something around 0 so that there's some force there...?) ... certain transistors are by default "off" when "activated" by an outside voltage, and then constructed so that they're able to be switched in the opposite on/off position by either a 5V or ~0V input?