Pull up resistor

digital-logicembedded

I'm software, not hardware, and would ask our electronics guy, but he's away on a site visit. Someone on StackOverflow suggested I repost this here.

I get the basic idea of a pull up resistor as explained on the first few links Google throws up, but as they're used in the circuit diagram I'm looking at I'm not 100% sure. The examples I've seen all show them used along side a switch that closes to ground, so they hold the voltage up until the switch is closed. Fair enough. But here I have not a switch but the output from a flip-flop latch.

Now, when the output from the latch is low/zero, is that the same as ground? I hadn't thought so, but I don't see how else the input at the other end would otherwise ever be taken low…

                       VCC
                        |
                        |
                 PULL UP RESISTOR
                        |
                        |
                       \|/
LATCH OUTPUT ---->---------> ON/OFF INPUT FOR OTHER COMPONENT, ACTIVE HIGH

Apologies for the ASCII circuit diagram…

So assuming VCC is always on, how is the component ever switched off? The only way I can think is that the latch output being low is equal to it being a ground, but I'm not sure that's the case.

The latch component is 74AHC574 (it comes straight up on Google).

This started out as a software investigation, but we're not sure the component on the right is being power cycled as the code suggests.

Best Answer

A schematic would be helpful as there is no way to know what "the component on the right" is; however here is what I can tell you:

From the datasheet for the 74AHC574, you can see that the 74AHC574 is "OCTAL EDGE-TRIGGERED D-TYPE FLIP-FLOPS WITH 3-STATE OUTPUTS". The tri-state (3-state) outputs are probably the reason for the pull-up resistor. The pull-up resistor holds the output high when the flip-flop is tri-stated (/OE high). When the /OE signal is low, the output is driven either high or low depending on the output of the flip-flop. In the case where the output of the flip-flow is low, it acts just like your switch that "closes to gnd".

It should be noted that since a tri-state flip-flop is used, there could be another device that is supposed to drive the output when /OE is high. Once again, a schematic would help.