Experience with 74hc595

[buracek]

They can obviously take some abuse - the evidence is all around us.

There's another limiting factor which is that the voltage output drops with increasing current.

Look at VOH in the datasheet. With 6V Vcc and 7.8mA draw, the voltage drops to 5.2V.

It doesn't specify values for higher currents than 7.8mA but with Vcc=5V and 20mA draw it will probably drop down below 3V.

At 3V a Blue LED won't allow 20mA to pass. If you connect a blue LED to a pin you'll never reach 20mA because there won't be enough volts to push that much current through the LED. It will reach equilibrium at a value less than 20mA.

The same sort of thing might happen to the whole chip, that's why they survive. As an experiment you could set it up to draw 20mA on each pin then turn the pins on one at a time and measure the current going into the chip (and on each pin). Draw a graph of the result and post it.

Just because the chip survives doesn't mean it's good engineering. The LEDs will get dimmer, the chip could heat up...it might even damage it in the long term.

Thank you, this is the sort of answer I was looking for. I'll do the test when I'll have a few minutes. I wonder myself what will be the voltage drop, maximum current and power dissipation on the chip.

[buracek]

So you are proposing sourcing 70mA with 4 high output pins to Gnd outside of the chip,
and sinking 70mA with 4 low output pins from Vcc outside of the chip?
That would probablyt be okay, chip internal control currents are negligible. Got a feeling the chip would get warm. 70mA comes in VCC pins, goes the Gnd else where.  70mA from elsewhere comes in and goes out the Gnd pin.

This is correct. I wonder if anyone tried this. Seems to be OK according to the datasheet, unless it will exceed the max. power dissipation as you mentioned.

[CrossRoads]

Well, absolute max is 70mA, no good engineer is going to run it that hot.
If you want LEDs to be powered with 20mA, than a better choice is cd74AC164, +/-24mA output per pin, or TPIC6B595, can sink 150mA per pin.

[nickgammon]

Yep. There's no confusion over it or "different answers". The specification says 70mA per chip.

Per pin, in what I linked.

The pins you're referring to are "Vcc" and "GND", not I/O pins.

It seems an odd qualification to make, since there is only one Vcc pin and only one Gnd pin.

I'm puzzled now about how my LED sign is working. I presume it was designed by professionals. At a given moment one 595 can be driving (sourcing) 8 LEDs, via a 150 ? resistor (each), and running at 5V. From my calculations above that should be 20 mA of current each, so that would be 80 160 mA  and not 70 mA for the chip.

One slightly mitigating factor might be that as the multiplexing switches to different rows of LEDs some outputs will not necessarily be driven high all the time, so if a "cooling off" period was required the chip would get that.

However previous discussions have indicated that you are not supposed to exceed maximum levels, even momentarily.

One more question, from the datasheet from reply #4, what does it mean ±70 mA? What is the difference between that and 70 mA, since the chip won't be outputting negative voltages?

(edit) Corrected error. 8 x 20 mA is 160, not 80.

[CrossRoads]

+/- 70 refers to VCC or GND pin, so I'd read that as 70mA in on VCC pin (70), and 70mA out on Gnd pin (-70).
What kind of voltages do you see across the LEDs & current limit resistors, with a 'scope?

[fungus]

+/- 70 refers to VCC or GND pin, so I'd read that as 70mA in on VCC pin (70), and 70mA out on Gnd pin (-70).

I never thought of it like that before.

So...theoretically I can connect 4 of the LEDs backwards then run 8 LEDs simultaneously at 17mA and be under the limit?

[CrossRoads]

Backwards? So only 4 on at time, but switched quickly? Sure, that could work.
Would have to use OE/ (G/) to make both the outputs turn fully off then, otherwise one or the other will always be on.

[nickgammon]

+/- 70 refers to VCC or GND pin, so I'd read that as 70mA in on VCC pin (70), and 70mA out on Gnd pin (-70).
What kind of voltages do you see across the LEDs & current limit resistors, with a 'scope?

Hmmm. I measured (as best I could) around 600 mV drop over the 150 ? resistors. I also found out that the circuit is actually powered by 3.3V, not 5V as I assumed.

So theoretically that should be:

Code: [Select]


0.6 / 150 = 0.004  = 4 mA


The voltage drop over the red LEDs seems to be 2V, as expected, so the sizing would seem to be:

Code: [Select]


(3.3 - 2) / 150 = 0.0087 = 8.7 mA


There must be some loss through the transistors or something.

Based on that though, 8 x 4 mA = 32 mA so that seems to be within spec for the 595 chip.

[nickgammon]

This is the exact part number:

[CrossRoads]

ST brand part
http://www.st.com/st-web-ui/static/active/en/resource/technical/document/datasheet/CD00000339.pdf

Same numbers:
Symmetrical output impedance:
|IOH| = IOL = 6 mA (min.) for QA to QH

IO DC output current  35 mA
ICC or IGND DC VCC or ground current  70 mA

[fungus]

Backwards? So only 4 on at time, but switched quickly? Sure, that could work.

No, simultaneously.

4 LEDs go from +5V to 4 of the 74HC595 I/O pins, with a resistor to allow 17.5mA to pass.
4 LEDs go from the 74HC595 I/O pins to GND (same resistor, oriented in the opposite direction).

If I set the first ones I/O to 'LOW', they'll turn on and 70mA will go through the 74HC595 GND pin.
If I set the second ones I/O to 'HIGH', they'll turn on and 70mA will go through the 74HC595 Vcc pin.

I'm not exceeding the limits but I can light up 8 LEDs simultaneously :-)

[Grumpy_Mike]

Yes you can do that but the words in not backwards it is sourcing or sinking current.