The 74HC595 is a popular choice for driving LED signs, it can source 35ma per pin according to the Philips datasheet. But what is the maximum for the package?

hi Bill,
The datasheet says a maximum dissipation of 500mWatt

Well looking over a few data sheets from different manufactures of this chip it seems like you might reconsider the use of this chip unless added drive help is given.

The +/- 35 ma is a maxium DC spec where going above will damage the chip. Recommended DC drive is +/- 6ma.

Also the max package total dissipation spec limit is 750mw. So 35ma x 5v = 175mw X 8 = 1.4 watts, well about the 750mw and that does not include power for the chip logic that it consumers. It would have a short if not dramatic life if driven will all outputs on

Lefty

Edit: The 750mw rating is for the HC595A version I found in a motorola data sheet, Eric's 500mw is for the non A version.

The output transistor will have 5V across it only when it is shorted. Driving a resistor the output voltage with 35mA will probably be 1V for its saturation voltage loss. Then it dissipates only 35mW.

I saw the 75ma and that's what I thought it meant. So the design is a poor one. Not sure what you mean audioguru.
The 2003 darlington drivers have a fair voltage drop if I recall...
So at 5V VDD, 1.2V per LED and 1.2V (assumed) for the 2003 and a 150ohm resistor I calculate 17.33ma per LED or 130ma which is way over the spec. Seems like a poor design.

Originally Posted by mneary

I agree it's poor, and I wouldn't manufacture it, but as an alphanumeric display on the hobby bench, it's not really very far over the edge. You don't want to ever try to light an entire block.

Same problem on the 2003, whose saturation voltage rises past 1.6V at 500 mA (14 * 40 = 560 mA). The higher sat voltage would starve the segments down to 14 mA or so.

And, of course, the AT90S2313 is obsolete.

Never thought of that, the old Elektor design used 74LS164 with only 8ma max segment drive (sink 330ohm & PNP drive transistors)

I think the 74HC595 will have an output saturation voltage loss of 0.7V. Red LEDs are 1.8V and the ULN3003 also has an output saturation voltage loss of 0.7V at this low current.
So the remaining 1.8V across the 150 ohm current-limiting resistors produces an LED current of only 12mA.

Aren't the LEDs multiplexed?

The LEDs are multiplexed, and at 12ma per LED the 74HC595 is sourcing 96ma and the ULN2003 will need to sink 40x12ma 480ma.

Originally Posted by blueroomelectronics

The LEDs are multiplexed, and at 12ma per LED the 74HC595 is sourcing 96ma and the ULN2003 will need to sink 40x12ma 480ma.

hi Bill,
Have a look at the TPIC6595 S/R.

Originally Posted by ericgibbs

hi Bill,
Have a look at the TPIC6595 S/R.

Nice part for the column drivers.

Originally Posted by Suraj143

Hi all in the above diagram I'm going to use a 47R instead of 150R.Is my value ok for a good brightness?

The resistors are supposed to be 150 ohms so that the 74HC595 ICs do not get too hot inside.

For more brightness then the resistor value can be reduced if a betterdriver IC is used that can pass the additional current without getting too hot.