# DIGITAL LOGIC BASICS?

I'm trying to learn the basics of digital logic, and I'm confused as to what adding a bubble at the end of an and/or gate does to the output I uploaded a picture to clarify what I'm asking Does a bubble at the end of a gate change the voltage from high to low/low to high? Or does the bubble have no effect on voltage level of the output? Does it only change the output logically? Or voltage wise as well? If you could tell me if 1, 2, or 3 is correct in the picture, and maybe explain the purpose of the bubble, I would greatly appreciate it thank you

Samantha

Adding a bubble as you call it, is just a shorthand way of saying that a NOT gate has been added to the output of the gate: and in so doing, it inverts the output of the gate. So, an AND gate on its own may give an output of 0 but, with the bubble, it becomes NOT 0, meaning 1. If the output from the AND gate is a 1 instead, then with the bubble added it becomes NOT 1 meaning 0. With the bubble added to an AND gate we have an output that is NOT AND or simple shourtened to NAND. A AND gate can be represented by an AND gate with an NOT gate at its output and so the two gates merged into one with a bubble representing the NOT part. The bubble simply means that a NOT gate has been added to the end of the gate. AB ….. AND ... NAND 00...….. 0 ... ….1 01...…. 0 …….. 1 10...…. 0 ……..1 11...…. 1 ……. 0 Your top gate is a AND gate Gates 1, 2 3 are all NAND Gates The outputs of 1,2 and 3 all give a '1' at the output. It looks like you are making this way too complicated!

Philomel

The bubble at the end is the same as adding an inverter to the circuit but without the gate delay. It simply changes the output state; H>L or L>H.

qrk

The "bubble" inverts, thus a zero becomes a one, or a one becomes a zero. An AND gate with a low & high at the inputs gives a low output. A NAND gate (AND gate with a bubble on the output) with a low & high at the inputs gives a high output. When the bubble is at the output of the symbol it merely inverts the output.