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Topic: Even if I remove the power, why the shift registers is on? (Read 585 times) previous topic - next topic

118jjh

Hello, from South Korea.

Before starting the question, I am sorry for my bad, English sentences.

(I attached my circuit so that you can understand what I say.)

I would like control 49 solenoid(5V, 1.1A) using  TIP120(transistor) / 7Shift-register 74hc595 / external Power Supply(5v, 60A) and arduino uno R3.

I made the circuit and code. I think the code has not any problem.

but, when I power on the power supply the solenoids work as I did not intend.

some solenoid works even if i did not make any control signals.

and more strange thing is here.

when I just remove 5V line from arduino to Shift register.. the shift register was steel working and (not perfectly working but) better working than before ..

Why It works? I think It should be not working at all, because the shift registers have got no power.

but it didn't

I really don't know why this problem happens.

Could somebody help me, please?

Thanks


rogertee

The cause of your problem you are using either a semi high or low output to control the solenoid and when you remove or first turn on power the outputs are in limbo high,low or in between until the sketch gets running. The shift register uses a clock pulse to make data shift and garbage in = same out. My solution was to make an ON signal as active high and use a 10k pull down resistor. I then fed my output into an opto coupler (4n25 type) diode side so that only a High could turn it on thus no power no solenoid is ON. I ran into your problem with an Android Arduino Bluetooth controller driving AC motors and used a solid state relay.

MarkT

Connect all the #OE signals on the shift registers (pin 13) to a pull-up to 5V rail.  Connect it also to an Arduino pin
that you set to be an output at the end of setup().  Make setup() first clock all zeroes into the shift registers.

That way the shift register outputs are floating until your code has started up and set them to a known state.

The darlington drivers need current to function so a floating input will be the same as a low input to them

This trick doesn't work for MOSFET drivers, note, and extra pull-down resistors would then be needed.

I think you've failed to appreciate that logic chips power up in a random state unless you explicitly
reset them or disable them somehow.  Every logic circuit design needs to have this considered, and
the solution often involves a few carefully chosen physical pull-up or pull-down resistors on key
inputs.
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