595 shift register output


tried to google answer but i am unable to get one direct answer.
i would like to know if the 595 output are always +V or always -V or is there a way to determine what the output is.

by the two images attached, is it determined by the way it is connected to the load or is one of the pictures incorrect

Hi. Cannot see your first picture. Second picture looks ok, can't see any major problem with that. It may try to draw a little bit too much current from the 595, depending on the led forward voltage.

Not sure what you mean by "the 595 output are always +V or always -V". There is no negative supply in your circuit, and the 595 could not use it anyway. The 595 outputs are always +V or GND, unless /OE is low in which case they are high impedance, I think.

Paul

Ah! Did you fix the first picture? I see it now. That also looks ok. The leds may be a little dim with 1K series resistors if they are an older design.. But i have some modern leds that are so bright and efficient that they are perfectly visible witb 10K series resistors.

The difference between the two circuits is that in the first, the led will be on if you shift a 0 into the corresponding bit in the register. In the second circuit a 1 will switch a led on.

The 595 outputs are always +V or GND, unless /OE is low in which case they are high impedance, I think.

Other way around - high impedance when OE/ is High. Use PWM for brightness control. analogWrite (pin, 0) is full on,
analogWrite (pin, 255) is full off.

If the LEDs are Red, with Vf = 2.2V, then current = (5V-2.2V)/270ohm = 10.3mA, all 8 outputs on would be overstressing the 70mA absolute max current on the VCC & GND pins.
I prefer TPIC6B595 & TPIC6C595, can handle 150ma & 100mA load on each output pin. Only sinks current, so wire up like the top circuit. Shift in a 1 to enable the output.

aes92000:
by the two images attached, is it determined by the way it is connected to the load or is one of the pictures incorrect

No, both correct. The 74HC595 (unlike the TPIC6B595/ TPIC6C595 which are designed only to "sink" current and pull the output down) can pull outputs either high or low. As "standard" CMOS, the "C" meaning "complementary" or symmetrical design, its ability to deliver current in either direction is by design, pretty much identical, unlike earlier NMOS and TTL or LSTTL logic series.

So you either connect the outputs so they pull LED cathodes down and write them LOW to light the LEDs with the anodes connected (with series resistors) to the supply, or the converse, connect the outputs so they pull LED anodes up and write them HIGH to light the LEDs with the cathodes connected (with series resistors) to ground.

It is symmetrical.

i apologize for late response, work gets in the way a lot here lately.

thank you for answers.
the pictures were from the web that i was using to show different wiring schems. i understand now that it is dependent on the way it is designed whether it is sinked or sourced. that answered my question.

thanx again and all have good day