Current draw of Darlington Array

[ivaylopg]

Hi everyone,

I tried to find the answer to this on the forum but I don't think anyone had asked this exact question yet, and I think I got the answer from the datasheet but I'm a bit of a n00b when it comes to reading those things.

ANYWAY:
I've got an Arduino Uno hooked up to 16 Shift Registers (74HC595) which in turn are hooked up to 16 Darlington arrays (ULN2803). So, basically I've got each output of the shift registers feeding directly to an input of the darlington arrays. The Shift Registers are powered from the Arduino's 5V out, and the Darlington Arrays are switching small incandescent bulbs (12V/40mA) and so have COM hooked up to a 12V source.

1) What is the total current draw of EACH Darlington input? I want to make sure I do not exceed the 200mA max of the Arduino. If I'm reading the datasheet correctly (which I'm probably not), then each input is drawing at max 1.35mA, so 1.35 x 16 registers x 8 = 194.4mA.

2) To avoid this being an issue at all, I'm assuming I can just power the Shift Registers from a separate 5V supply instead of the Arduino and link the ground lines together. Am I right?

3) Am I missing something obvious?

Just to clarify, I haven't built this yet: I have a prototype built out with 16 shift registers driving 144 LEDs to help me figure out my code. Now I'm making a second prototype and bringing in the darlington arrays instead of the LEDs. I made a mockup with 4 registers and 4 arrays. I just want to make sure I'm not going to blow everything up when I expand it to 16 registers and 16 arrays.

Thanks!
Ivaylo

[Zener]

I came up with 213mA which is very close to your answer. You may be right, or we may both be wrong! How did you calculate your answer?

[ivaylopg]

The datasheet refers to the "input current" being between 0.93mA and 1.35mA, so I just assumed the highest current and multiplied by the number of outputs. How did you come up with your number?

Datasheet here: http://www.ti.com/lit/ds/symlink/uln2803a.pdf

[westfw]

I want to make sure I do not exceed the 200mA max of the Arduino.

Isn't that oft-quoted number the max output of the Arduino CHIP? Since you're not driving the darlington arrays directly from the chip, it doesn't matter. (OTOH, there's a board maximum current as well, which is less obviously defined (it depends on the input voltage.)

The ULN2803 has a very simple internal circuit; each driver has a 2.7k resistor leading to the base of a darlingtonton transistor, so the current is just (5-2*Vbe)/2700)

Yes, having a separate power supply for the shift registers should solve any problem.

1.35mA per output is (significantly) less than your LED mockup is drawing, isn't it?

[Zener]

westfw is correct. As for the difference in the input current numbers, the data sheet is specifying those at a Vin of 3.85V where you are actually using 5V. And westfw and I were targeting typical values vs the max spec you were using. So, using westfw's correct formula, and using .7V for Vbe (a reasonable guess) we get (5-1.4)/2700 = .0013333..., *128 = 171mA typical. What would cause this to rise toward a "maximum" spec would be if the 2.7K resistor value was lower, and also if the temperature of the device goes up, the Vbe's would drop a bit, which would increase the current also.

[ivaylopg]

Thank you both for your help and your clear answers.

1.35mA per output is (significantly) less than your LED mockup is drawing, isn't it?

In hindsight, yes, i should have really realized this before asking this question.

Thanks again!