As someone who only picked up the hobby a year ago, I just want to say THANK YOU!!!!!!!!!!!! Your explanations are beyond helpful and have greatly contributed to early projects.
To answer your question for why is the second shift register one bit away from the first. I think it's because you would have the output from the first going to the input of the second shift register but only after the next consecutive clock pulse will the output be triggered! Quite neat.
05:10 it's more reliable here to define constants instead of using variables. variables are stored in the RAM of the microcontroller, but constants are always in the code section and won't spend limited space in microcontroller's RAM.
I’m still very new to Arduino coding and tbh, copied this code from someone else’s project. I had to look everything up to see how it worked. Thanks for adding your knowledge. I hope it helps someone out.
@@maker_karen1785 Ah, are you the person in the video? :) It would be nice to also credit the person who has written the original code in the description.
@@MarinusMakesStuff Hi, yes I'm Karen in the video. I do all the research, writing, filming, and editing for The Learning Circuit. (Well most of the editing. Max makes it look shiny.) That's a really good point. I got a lot of my reference material from www.arduino.cc/en/tutorial/ShiftOut. But I'm looking at their example code and it doesn't match mine at all. I tend to piecemeal information together for my videos, so I'm not sure if I directly copied code from anywhere or just borrowed bits. I move through making these videos so quickly that the information can be a blur sometimes. I always keep a References list of URLs but none of the ones for this episode have code like what I used. I will try to keep that in mind in the future and try to make sure the original code authors are referenced in the description.
also the output enable explanation is a little inaccurate. Pulling the pin high will make the outputs go high Z or essentially open circuit, it disconnects the output drivers. However you can use PWM on the pin to control LED brightness etc (I use it on the jumbo clock to control the Aux LED brightness, I use the signal from the "blanking" pin used on the CD4511 which I use to control the digit brightness, inverted using a spare ULN2003 and a pull up on the OE of the register controlling the LEDs)
When the purpose of the exercise it to explain the shift register, it suffices. When the purpose is to add more outputs, try the I2C bidirectional ic MCP23017, it’s much more versatile... You can add 8 of the to the I2C but giving you 128 ports that can individually set as input o output. Couple outputs to a ULN2803 and inputs thrugh an opto-coupler, and you’re in business.
50PCS = € 49,15 / lot MCP23017 50PCS = € 4,03 / lot Hc595ag Shift Register If you need 50PCs and you don't care about 45€ being thrown away, then it doesn't matter what to use. I was always wondering about MCP23017 why it cost as MCU, but has only a tiny part of functionality.
This is the best explainer video for me, thank you for this great work. Now I have a question: what software did you use to create the animation from 1:10
Great tutorial! 👍 I can't help but notice the LEDs are set up backward for counting up in binary LOL. 😝 Could just turn the board around or flip the video. 🤔 😁
8:40 Can you tell me about the swing machine sitting behind you? It looks like digital one. Can you tell me the brand or any details that I could found similar one ? BTW very good explanation of shift register. thanks
Thanks for the video! Two quick questions: 1) How would you make it so that the second shift register doesn't start receiving bits (or lighting up) until the first one is full (all LEDS are lit)? 2) I know you can use bytes to tell the shift register what pattern to display (ex. 0b01001010), but does this carry over to the second shift register? If I put in a 16-bit number, would it show that pattern using both shift registers? Thanks!
For Q1, it depends on what you wish to do. A quick and dirty way is to call serialOut() twice before the digitalWrite(latchclock, high) is called. That way the data in the first serialOut() is in the second shift register, while the second serialOut() is in the first shift register when the latch is 'activated'. For Q2, int is already 16 bit, so it is already 16 bits and no change is needed! However, if you are looking to transfer all 16 bits in one call of serialOut() you will be disappointed. From the arduino reference for shiftOut() under notes and warnings, shiftOut() only supports transferring 8 bits each time it is called.
This is a pulldown resistor. When the pin on the arduino is brought low, its possible for a voltage to remain on the pin. the resistor gives it a path to ground.
@ Now I've started reading up on it. That Arduino code line "Serial.begin(9600)" makes a lot more sense now. You may have just sent me down a rabbit hole. ;-)
I've used shift registers to drive some BCD - 7 segment decoders (CD4511), each shift register can drive 2 digits, I have cascaded 4 shift registers to drive six CD4511 7 segment display drivers which drive ULN2003 open collector drivers and and 8 extra LEDs (by omitting the 4511's and driving the ULN2003 off the shift reg outputs). Did I mention the LED displays in this are made from LED strips, one whole digit is around 130mmx220mm. I made them into a clock, which I call the "jumbo clock". I have some smaller displays that use a similar thing, 2 shift registers for 4 digits (or 3 and a few directly controlled segments). I made a few 18b20 based thermometers using some 44.5mm LED displays
@@maker_karen1785 It was either that or "Hodor!". I may not have immediate use for the topic at hand, but I want you to know that I appreciate your work.
Really it doesn't have ANYTHING to do with decimal. They are simply switches. Using decimal representation of binary for that is simply more convenient. Those are basically dip switches.
I broke my head thinking about that HC595, until i realised that the outputs SOURCE current when high, but also SINK current, when low. Please tell me that is correct, otherwise I will again not sleep over it.😂
@@alfredkaserekasivanzire9488 This video uses a 595. There is another video coming up that uses a 166 to show adding inputs. But I made that video at the beginning of July so I forget exactly which chip I used. That video should be released in 3 weeks.
05:57 isn't a great explanation of how a for loop works. A for loop typically has three parts that work with a variable: 1. what should the variable be initialized to? 2) when do you want me to break out of the for loop? and 3) how quickly should I increment or decrement the variable? These three values are separated by a semicolon. Technically the rules of how you set up a for loop don't _require_ these three things, but by convention that's typically how they're coded. In the example that was given, the loop will count from 0 - 255, then terminate. Since the code is in the function loop(), it then starts over at 0 because loop() gets called by the Arduino base code over and over again.
Video focused on ShiftRegister and C++ syntax is extra. On the screen you have working example of loop and for more details check book "C++ in 21 Days" Karen is more familiar with ICs and in Arduino/C++ she is relatively new. www.amazon.com/Sams-Teach-Yourself-Days-5th/dp/0672327112
The /OE pin has nothing to do with shifting or latching, and taking it HIGH will most certainly change the parallel outputs - they will be tri-stated! Who edits this stuff?
@@maker_karen1785 Surely. www.ti.com/lit/gpn/SN74HC595 Notice that on the first-page Logic Diagram /OE is the top input, and is only connected to the output-enables of the parallel outputs - it is asynchronous to the clocks and does not interact with the shift logic. If I am incorrect my apologies; please point out my mistake.
The way she said it was confusing. You are correct. The /OE pin has nothing to do with the shifting or the latching. It has everything to do with the output. She was referring to the output not changing until /OE is high. Generally if you are not waiting for other things to complete you just tie it high and your output will change when the latch changes. However you might not want to show your register change and /OE can be controlled by some other Logic input to it. Chances are that your output might be going to a bus so the high /OE is a high impedance state to make sure that you do not put your data on the bus until it is time to.
