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? **broken link removed**
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 number you're looking for is in the data sheet: "DC current drain per VCC, GND". It's 75 mA. That's the total that the power/GND pins can carry without damage. Since all drive current passes through Vcc or GND, it then becomes the absolute maximum total drive current.
As noted, this is a damage threshold and function isn't guaranteed.
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.
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.
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.
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.
Good catch; I forgot all about the 74HC595 losses. And the ULN2003 would be driving from 1 up to 40 segments; its saturation voltage would be 0.7V for one, and more like 1.2V for all.
So you get about 12mA for one segment alone, and 7mA each when they are all on. Lots of flicker.
*My LED needs 20mA current to light up to its maximum brightness.
*I multiplex 7 LEDs (7 rows) then the duty cycle will be 14.28%.
*So I have to supply 7X current (100/14.28) to light my LED to maximum brightness.
*That is 20mA X 7 = 140mA current I have to provide.
The problem is ULN side it can sink this current.But 595 side it cannot supply this 140mA current because its rated only for 35mA max.
So my question it is not worth reducing the resister either.Can you tell with this 35mA current for the 7 rows how will be the duty cycle?My figures show it will be 25% duty cycle.
35mA/7= 5mA. The brightness will not be 1/7th because brightness is logarithmic. 10mA is only a little dimmer than 20mA and 5mA is only a little dimmer than 10mA.