Reading old 7400 datasheet - why negative input current? - Page 1
Forum section: electronics / beginners

author: bitman
author_rank: Unknown
timestamp: August 30, 2023, 02:26:08 pm
content: I'm trying to do what I thought was simple, but it's just confusing me. I need to place pulldown resistors on an 74LS245 on the input side. And I want to figure out what the MINIMUM current I need to provide for the pulldown to work (ie to figure out the resistance). When I look at the datasheethttps://www.mouser.com/datasheet/2/308/74LS245-1190460.pdfI see the I(il) LOW Level Input Current marked as -0.2 mA for max current. I'm not sure how to read this AT ALL. I understand it's measured at 0.4V but given that's positive, why is the current marked as negative?And it still doesn't tell me what the minimum is. I'm poking around on "beginner sites"https://www.electronics-tutorials.ws/logic/pull-up-resistor.htmlto figure out how to calculate the resistor value - and what I've found has statements about 400uA current source for Low - this is NEGATIVE 200uA? Am I just taking that sign too literal?

author: retiredfeline
author_rank: Frequent Contributor
timestamp: August 30, 2023, 02:31:28 pm
content: Negative current means out of the pin.

author: Terry Bites
author_rank: Super Contributor
timestamp: August 30, 2023, 03:20:31 pm
content: If you look at the input structure, you can see that current must me drawn away from an emitter to turn on Q1.That's the TTL way.

author: Kim Christensen
author_rank: Super Contributor
timestamp: August 30, 2023, 03:35:14 pm
content: Quote from: bitman on August 30, 2023, 02:26:08 pmPullupresistors are the norm for TTL. Pull down resistors have to be fairly low value to get an acceptable logic low level. What's the reason you're using pulldown resistors?

author: wasedadoc
author_rank: Super Contributor
timestamp: August 30, 2023, 03:45:41 pm
content: Quote from: Terry Bites on August 30, 2023, 03:20:31 pm@Terry BitesWhat you have written is theinverseof the TTL way. Current drawn away from either of the two emitters in your diagram turnsOFFQ1.

author: bitman
author_rank: Unknown
timestamp: August 30, 2023, 04:38:15 pm
content: Quote from: Kim Christensen on August 30, 2023, 03:35:14 pmI know that pullups are standard, and it's my default. In this case, the small circuit/PCB I have controls status lights, and the default needs to be off when the input is floating. The 245 seems to make high-Z into a pullup internally and it provides the wrong status lights for the small application of this PCB. And I cannot use an inverted buffer as the input signals when provided will not be inverted. I really didn't want to use pulldowns - the current draw almost doubled when I did that. I found an ACT (instead of LS) chip for 245 and presto I am back to acceptable current draw for the little status led circuit. 1uA vs. 200uA is quite the difference that works for me.Now that I understand what's meant by negative current - reading the details I knew it was from the IC to something else, but I didn't make the connection. Thankfully I have this forum to remind me of the simple things I missThanks all!

author: bitman
author_rank: Unknown
timestamp: August 30, 2023, 04:46:17 pm
content: Quote from: edavid on August 30, 2023, 04:37:09 pmOther 7400 LS/HC chips. The exact model/part differs. My current design doesn't have the buffer and it just draws too much on some chips - particular when it comes to fanout - barely being able to drive a single IC - let alone a handful, when LEDs are put directly on the line (yeah, dumb design). It's why I like doing my "lessons learned" using simple TTL/CMOS chips - it helps me understand more complex designs a lot better.Which is why the 2K resistor pulldown the site I linked calculated (with the 200mA instead of 400mA) worried me. Not too different from an LED with a resistor around 680 ohms and hence not really solving the problem.

author: bitman
author_rank: Unknown
timestamp: August 30, 2023, 05:06:30 pm
content: Quote from: edavid on August 30, 2023, 04:49:42 pmOutside of it "just" being bi-directional (ie. to me the same), it's what I had in the drawer. And for breadboard stuff (prototyping) it's pin layout SUCKS.

author: Gyro
author_rank: Super Contributor
timestamp: August 30, 2023, 07:22:50 pm
content: Quote from: edavid on August 30, 2023, 04:49:42 pmActually you want to be using a 74HCT541. The 244 pinout is horrible for flow-through applications (it was originally designed for implementing 4 bit transceivers).

author: BillyO
author_rank: Super Contributor
timestamp: August 30, 2023, 10:13:18 pm
content: Quote from: wasedadoc on August 30, 2023, 03:45:41 pmNo. He was right. Pulling either emitter to ground forward biases the B-E junction turning the transistor on. If Q1 is on then the output transistor (call it Q2) will be off and the gate output will be high. Pulling both the emitters high will turn off Q1 turning on Q2 pulling the gate output low.