I have a circuit, using a 5V supply. which drives LEDs by connecting them direct between IC pins with no series resistor. The ICs are 74HC595N and one chip is providing the anode supply and another is providing the cathode supply.

The LEDs are some 3mm green individual and there are 10 5161BS 7 segment displays which are multiplexed so each is powered only one tenth of the time.

What would be the current in a green LED and in a segment of the multiplexed display?
Is this OK for the chips pins?

 

Is this OK for the chips pins?

No, even if it looks like it works, I would not do it. It could very well be damaging the chip slowly. Use a resistor. If you can’t get enough current that way, use a higher current buffer (and a resistor).

Edited to add: How much current is the output rated for, and how much do you need?

 

The green LED you are using has a specifications curve plotting the current and its forward voltage.
If at 10mA its Vf is 2.0 V; the led can be driven by either 10mA with a limiter resistor or with 2.0V without a limiting resistor from the IC pin.

The supply voltage of the IC being 5.0V needs to drop 3V. Four Si diodes in series to the LED does it. Or two green LED instead of one plus one Si diode does it. (if no Vdrop inside the chip)

+5V------------74HC595-------------LED---------\/\/\/\/\300Ω---------------gnd
or
+5V------------74HC595-----------LED-----|>|----|>|----|>|----|>|----gnd
or
+5V-----------74HC595-------------LED----LED-----|>|--------------gnd

Where gnd can be another 'low' pin of the same IC.

Try it with an adjustable voltage supply : Apply 2.000V to your green LED with no resistor: will drain 10mA.
Or, feed 10.00mA from an adjustable current supply to your green LED : will show 2Vf drop.

So yes, a LED can be driven without current limiting if its Vf is fixed.

 

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So yes, a LED can be driven without current limiting if its Vf is fixed.

And you don’t care what the current is. The 10mA at 2.0V is typical, any specific LED might need 1.9 or 2.1 at 10ma, and might be 5 or 15 mA at 2.0V.

 

There will be a voltage drop on both the anode and cathode supply to the LEDs inside the 'HC595 but I can't get what value that would be from the datasheet. In the 'features section it says "±6mA output drive at 5V"

 

And also if the 74HC595 is driving some other part of the circuit, connecting LEDs is a no-no.
The LED is going to load down the outputs and the next part is not going to see a valid logic voltage.
Use a separate buffer to drive LEDs.

 

there are ICs that include both shift register and suitable driver. TPIC6B595 for example

 

I don't think there's any output current limit protection with 74HC devices. Safer to use current limiting resistor(s).

 

From the datasheet, the Absolute Maximum continuous output current rating is 35mA for any output pin and 70mA for either supply pin, so external current limiting seems necessary to prevent damage to the IC or LEDs.

 

IMHO there's no substitute for data in answering the original question. Start with a resistor, measure the current. If it's low-ish, reduce the resistor value. Repeat until you can remove the resistor entirely, or have proven to yourself that you cannot.

 

IMHO there's no substitute for data in answering the original question. Start with a resistor, measure the current. If it's low-ish, reduce the resistor value. Repeat until you can remove the resistor entirely, or have proven to yourself that you cannot.

And do that for every chip that you bought, as the output current spec is quite wide.
Or use 4000 series CMOS.

 

I have a circuit, using a 5V supply. which drives LEDs by connecting them direct between IC pins with no series resistor. The ICs are 74HC595N and one chip is providing the anode supply and another is providing the cathode supply.

This is one of those reasons why the products nowadays don’t last. I mean, SMT resistors in quantity are below cheap. To save a few tenths of a penny between the resistor’s cost and its placement, the product’s service life is compromised.

Just last week my granddaughter came to me with her favorite talking toy which had ceased to work. I took a cursory look and found exactly the same issue: no LED limiting resistors anywhere.
Now you might say: it is a toy. Those are disposable anyway, purchase a new one. The problem with this mindset, in addition to generating E-waste, is that this beyond-stupid practice often finds its way to products which are not toys.
And copied by scores of newbies who don’t know better.

 

And also if the 74HC595 is driving some other part of the circuit, connecting LEDs is a no-no.
The LED is going to load down the outputs and the next part is not going to see a valid logic voltage.
Use a separate buffer to drive LEDs.

The only thing these outputs feed is the LEDs.

 

This is one of those reasons why the products nowadays don’t last. I mean, SMT resistors in quantity are below cheap. To save a few tenths of a penny between the resistor’s cost and its placement, the product’s service life is compromised.

There are a total of 10x7 + 60 = 130 LEDs and as this is an existing board there is no way to find space to fit in all those resistors. I think I will have to go with it as is. The multiplexing will limit the average current for each LED.

 

It is not the average current thru the LEDs, is the average current thru the ICs driving those LEDs.
And, can’t you find the space to fit 0603 or even perhaps 0402 resistors? It doesn’t have to be one resistor on each individual LED, but on the IC’s outputs which, if multiplexed, will be far less than 130.

But you do you.

 

I see that no one mentioned that the forward voltage is dependent on the temperature, so it means that in a hot day it will consume more current for the same voltage, and with that comes larger dissipation of the chip driving the LED, which gets even closer to failure.
I stand with using some limitation too, because it's not only that someday it will burn the chip/LED, it will remain the cause of failures until you make another PCB with corrections.

 

if space is at premium and there is a lot of LEDs or 7-seg displays to drive, why bother with shift registers? it is more practical to use LED matrix driver IC (led display driver). one such IC can replace bunch of 595s. for example MAX7219, LP5860, TM1637, TM1638 etc. many of them have built in current control so no external current limiting resistors are needed.

 

There are a total of 10x7 + 60 = 130 LEDs and as this is an existing board there is no way to find space to fit in all those resistors. I think I will have to go with it as is. The multiplexing will limit the average current for each LED.

Then use a serial input LED driver IC with constant current outputs, such as TLC5952 (24 outputs) or TLC5962 (48 output)

 

if space is at premium and there is a lot of LEDs or 7-seg displays to drive, why bother with shift registers? it is more practical to use LED matrix driver IC. one such IC can replace bunch of 595s. for example MAX7219, LP5860, TM1637, TM1638 etc.

It drives those LEDs with just 6 uC pins. It is a very cunning design.

 

It drives those LEDs with just 6 uC pins. It is a very cunning design.

TLC5952 can drive them with 3 microcontroller pins.