Shift Register

Thread Starter

Zane Finner

Joined Jan 29, 2018
30
Hello, I keep running into problems with the shift register: 74HC595. Knowing myself, I may have just broke it :p. Well, I need help creating a simulation that I can repeat on my real breadboard so I can make sure I m getting everything right! If you want to help me, create an account of TinkerCad and create this circuit(
) or one that is similar, then share it with me. I'm just so frustrated and I really want to see what I'm doing wrong. Thanks xd

Sorry if I'm incoherent
 
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Alec_t

Joined Sep 17, 2013
14,280
Describe the problem you're having.
I doubt members here will want to create a TinkerCAD account. Most of the regulars use LTspice for simulations.
 

MrChips

Joined Oct 2, 2009
30,707
Is the video your video?
We need to see a clear, well focused photograph of your breadboard, not someone else's creation, and not a video.
 

Thread Starter

Zane Finner

Joined Jan 29, 2018
30
Okay, so it's likely a number of things. I may just have a bad breadboard. I connected everything exactly how the schematics showed me, and then I replicated the breadboard from multiple videos. I tested the individual led paths and turned on my 5v power source. Nothing happened each time.
 

Thread Starter

Zane Finner

Joined Jan 29, 2018
30
I found a new shift register breadboard I will try when I get home. I'l leave a message if this doesn't work. The simulation works so if it doesn't go right after I do everything, if must just be my breadboard. I got a budget one so I'm sure it was contributing to at least a few of my problems.
 

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BobaMosfet

Joined Jul 1, 2009
2,110
Schematic is attached here and breadboard image is in video. I'll post a real picture of it when I get home
Well, the first problem is that the schematic you've provided is not showing an actual 74HC595. It's pinout isn't like that. A 74HC595 pinout is thus:

1 QB
2 QC
3 QD
4 QE
5 QF
6 QG
7 QH
8 GND
9 QH'
10 SRCLR
11 SRCLK
12 RCLK
13 OE
14 SER
15 QA
16 Vcc

The apostrophe (') on pin ( is important, that is your output in case you put these in serial. Realistically, you should be looking at a datasheet- it will show you this, and it will tell you how to use it- these are easy chips to use and incredibly useful.
 

Thread Starter

Zane Finner

Joined Jan 29, 2018
30
Well, the first problem is that the schematic you've provided is not showing an actual 74HC595. It's pinout isn't like that. A 74HC595 pinout is thus:

1 QB
2 QC
3 QD
4 QE
5 QF
6 QG
7 QH
8 GND
9 QH'
10 SRCLR
11 SRCLK
12 RCLK
13 OE
14 SER
15 QA
16 Vcc

The apostrophe (') on pin ( is important, that is your output in case you put these in serial. Realistically, you should be looking at a datasheet- it will show you this, and it will tell you how to use it- these are easy chips to use and incredibly useful.
haha thanks I think I got it now. The vid is not mine and its using a different IC than mine but mine is the 74HC595 and I got a simulation of that one that works on tinkercad. It works perfectly :)
 

BobaMosfet

Joined Jul 1, 2009
2,110
You should understand that the 74HC595 is a 2-stage shift register. Single stage shift registers show the outputs blinking as inputs are shifted in. A 2-stage shift register gives you the ability to keep this hidden, and then simply display the full output at once in parallel and it will hold that value until you change it/clear it.

The 74HC595 requires you to use 5-pins to control it (because it's more powerful). Pins 10-14. You should also be aware that you should always consider the current being output on pins by your micro-controller as well. If your Arduino, or any MCU you use doesn't limit current flow on a pin, you should use something like a 4K7 resistor to limit a 5V signal to 1mA of current. That is plenty strong as a signal current, and will conserve the amount of current available to your MCU for use on other pins.
 

hexreader

Joined Apr 16, 2011
581
That breadboard looks like the type that has power rails that only run half the length of the board.

You need to link left and right halves of the power rails together.

.... or maybe move the whole circuit to the far left so that it all fits on one half of the breadboard.

This all assumes that your breadboard is identical to the diagram shown.
 

MrChips

Joined Oct 2, 2009
30,707
TS still doesn't get.

You can stare at a circuit diagram and someone else's video, photo, or simulated graphics all day long and still not see the mistakes that you are making.

You must post your own photo of your own breadboard so that we can see your mistakes. Saying that you did exactly like someone else's demonstration just doesn't cut it.
 

Thread Starter

Zane Finner

Joined Jan 29, 2018
30
You should understand that the 74HC595 is a 2-stage shift register. Single stage shift registers show the outputs blinking as inputs are shifted in. A 2-stage shift register gives you the ability to keep this hidden, and then simply display the full output at once in parallel and it will hold that value until you change it/clear it.

The 74HC595 requires you to use 5-pins to control it (because it's more powerful). Pins 10-14. You should also be aware that you should always consider the current being output on pins by your micro-controller as well. If your Arduino, or any MCU you use doesn't limit current flow on a pin, you should use something like a 4K7 resistor to limit a 5V signal to 1mA of current. That is plenty strong as a signal current, and will conserve the amount of current available to your MCU for use on other pins.
I only have 5k1 and 2k. Is 5k1 too high?
 

Thread Starter

Zane Finner

Joined Jan 29, 2018
30
Breadboard is in attachments. Sorry, I did not make schematics.

MOD: Please only one Thread for same Topic, Merged.

^thanks xd I should be more mindful

I have to take a break. This just keeps frustrating me. I need to work on easier projects. Thanks guys :)
 

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BobaMosfet

Joined Jul 1, 2009
2,110
I only have 5k1 and 2k. Is 5k1 too high?
What do you think? If your voltage is 5V, and your resistor is 5100 Ohms....

I = E/R
I = 5/5100
I = 0.9mA (or 980uA)

This is close to 1mA and will work. You should get yourself some 4K7 resistors-- in fact, I'd recommend a resistor assortment (like from Amazon). Whenever you have a need for a resistor value you don't have, go buy 100. You can find places that will sell them for $0.035/ea, which is a pretty good deal for most folks, and will sell them in packs of 100.
 

Ian Rogers

Joined Dec 12, 2012
1,136
Breadboard is in attachments. Sorry, I did not make schematics.

MOD: Please only one Thread for same Topic, Merged.

^thanks xd I should be more mindful

I have to take a break. This just keeps frustrating me. I need to work on easier projects. Thanks guys :)
Please read Hexreader's post ( number #11) as you are crossing the middle of the breadboard.. The power rails are not linked underneath
upload_2018-3-30_8-45-31.png
 

Thread Starter

Zane Finner

Joined Jan 29, 2018
30
Hey everyone! I finished it haha! I was up last night and my problem was that my breadboards is on the cheaper side, resulting in weak connections within my resisters! I literally wiggled em' around and it lit up :) Here, I took a pic but the lights were to bright to show with it on. I took it apart and I'm going to do this all over again but with a 7 segment. I might try a four digit but I should try normal first. I need to get some soldering equipment and perf boards :p
 

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