Strange behavior on 74HC595 Shift Register IC when its GND pin is used

Question

I was following a tutorial similar to this: https://www.arduino.cc/en/Tutorial/ShiftOut

That is, an array of 8 LEDs that lit up accordingly to the binary representation of a number, which was given in a loop from 0 to 255.

The problem is that when invoked shiftOut() over certain numbers (such as 5, 9 or 13), the LEDs didn't lit as expected.

I even tried replacing the IC, but what finally solved this issue was to disconnect IC's pin 8 (GND) and let only pin 13 (OE) connected to Arduino's ground pin.

Does anyone know the reason this strange behavior happens on 595 IC?

Here's a video showing the circuit finally working (without IC's GND pin connected): https://www.youtube.com/watch?v=NGfpk-kjemo

Answer 1

Disconnecting the ground pin was not a solution.

It is hard to be certain from video but the resistor values are likely too small because I can't see any color on the bands. They are 100?

Look closely at the driver data sheet. It may say an output pin maximum current is +/-35 mA which would suggest that the 100 ohm resistor is okay.

For example, a red LED has a of around 1.6 V. You are using Vcc=3.3, so a 100 ohm resistor limits the current to <17 mA ((3.3-1.6)/100). The actual current will be a somewhat smaller as the Vf increases with current..

But you also stay within the maximum Icc and Ignd on the data sheet. These limits are not necessarily 8 x 35 mA. For example an NXP max Icc is 70 mA. That means you cannot drive all the outputs at maximum current, just 2. This is a logic chip, not a driver.

There is other hint that you should not using all the outputs at the max current spec in the section where the Voh and Vol are described. The current at which the voltage is specified is 6 mA, which is the maximum you should use if all output are going to source or sink at the same time.

A common LED current is 5 mA, so this chip could drive 8 simultaneously. If you want high brightness, you need to add a driver such as ULN2003.

Note that the schematic shows red LED's. The color is important. Different LED's have different forward voltage. The blue LED have the highest in the range of 3.3 V and higher. So with a Vcc=3.3 there is barely enough voltage to light the blue LED. The red and yellow have much lower forward voltage. When the chip tries to light the red and blue at same time, the output voltage is dropping enough that the blue is not visible, but the red is easily lit.

From the video, it looks that you have the same resistor on all the LED's. On different color LED's this will not achieve the same brightness on all LED's. You need different resistors for each color if you want them the same brightness.

The solution is

• Up the Vcc to 5V so that you can reliably light the blue LED's.

• Increase the resistor values start with 510 for red and 220 for blue. This should be about 6 mA for both.