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I built and simulated a circuit out of 74HC00 series ICs and want to connect it to a microcontroller (ATmega 16).

The development board I use (Atmel STK500) supplies 4.93 V (GND to VTG and GND to output pin). Can I safely connect the 74HC00 series circuit to power supply of the development board without any current limiting resistor or something similar? Can I safely connect the output of the 74HC00 series circuit to pins of the microcontroller as input and outputs of the circuit without any current limiting resistor? It seems that directly connecting an output to an input pin of microcontroller will create a short circuit. Do 74HC00 series ICs limit current?

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    \$\begingroup\$ Those output voltages seem odd; I'd expect it to be a lot closer to the power rails. \$\endgroup\$
    – pjc50
    Commented Sep 18, 2013 at 15:19
  • \$\begingroup\$ I measured 1.72 V between pin 10 (VCC) and pin 11 (GND) of the ATmega16. That seems also odd to me. \$\endgroup\$
    – user29239
    Commented Sep 18, 2013 at 15:40
  • \$\begingroup\$ The ATmega16 was in the wrong socket (previous owner's fault). Thus, I changed the question. \$\endgroup\$
    – user29239
    Commented Sep 18, 2013 at 16:30
  • \$\begingroup\$ Your title mentions 7400, but the text of your question mentions 74HC00. Those are very different beasts! 7400 without any letters refers to the ago-old pre-schottky TTL series. 74HC00 is a relatively modern CMOS series. Next: you built and simulated. In that order?? That figure of 4.93V, is that a power supply, or a logic output pin? Last sentence: output of 74HC to uC pins - OK, that's fine, but then 'as input and outputs', yet the sentence started with 'output'. So make up your mind! This might seem to be nitpicking, but a good answer requires a good question! \$\endgroup\$
    – Wouter van Ooijen
    Commented Sep 18, 2013 at 17:22
  • \$\begingroup\$ I simulated the circuit in Verilog and built it on a breadboard. 4.93 V is the output of the power supply and the a pin of the microcontroller if set to 1. I want to connect pins of the microcontroller to the 74HC00 series circuit as inputs and outputs to the circuit. So if the circuit would a single 74HC00, I want to set the inputs A and B from two pins of the microcontroller and read Y on another pin of the microcontroller. \$\endgroup\$
    – user29239
    Commented Sep 18, 2013 at 17:31

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In general, there is no need to have any current limiting for integrated digital circuits unless it's explicitly stated that it's needed in the datasheet. The IC will only take as much current as it needs from the voltage source, if the voltage is within its operational parameters. There is no need to worry too much about that! Whole situation is nicely explained in this question.

Next, when you're trying to connect the 74HC00 (or whatever you have, it's mostly the same for digital input), the input pin of the microcontroller will be in high impedance state. It will not act as a short circuit to ground and very little current will flow. Most microcontrollers start with their port pins in high impedance state and need to be configured as outputs before being able to source of sink current.

So in normal operation there is no need to have any current limiting at all.

This leaves us with abnormal operation. It can happen due to malfunction of an IC or due to your own carelessness. For example if you have a pin of 74HC00 connected to a pin of the microcontroller, configure that pin as output and set it to low, it will start draining current from the 74HC00 and that can be a dangerous situation. For that reason, I agree with MiljanNC and believe that you should place resistors in series with the lines.

Next, we have the actual value of the resistor. Truth is that 74HC00 can't source much current and that on the other hand ATmega can't really sink that much current. If you want in-depth analyses, you can look up pin schematics in their datasheets and see why. If you absolutely want to be 100% safe, you can imagine the source pin to be an ideal voltage generator and the sink pin to be ideal round and then calculate resistor size using Ohm's law, but in my opinion that would be too conservative value.

Also do pay attention to what datasheets say! The value of 40 mA per pin and 200 mA in total is absolute maximum rating, that is to say the value at which you can expect the chip to malfunction. For normal operation, I would make sure that current isn't above the tested value of 20 mA.

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You can check the datasheet but generally the components you are working with are all TTL compatible.

TTL compatible means that they use standard voltage levels. See also this WikiPedia article on TTL.

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  • \$\begingroup\$ I changed the question, so your answer doesn't apply anymore. \$\endgroup\$
    – user29239
    Commented Sep 18, 2013 at 16:42
  • \$\begingroup\$ That's not really true - this answer is still as applicable as it was before. \$\endgroup\$
    – Chris Stratton
    Commented Sep 19, 2013 at 20:48
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Using a series resistor to perform interconnection between IC-s is a good practice. This will provide you some benefits: - save your uC ports from burning out if you accidentally make a wrong connection to power rails - save your uC ports from accidental error in port config (out instead of in) - provide you some level of signal integrity with serial termination

I'm used to put 33R where ever I find a signal source on schematic :) This way, at least patching of the circuit is easier :)

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  • \$\begingroup\$ If 33R means 33 Ω, at least the ATmega16 is limited to 40 mA per pin and 200 mA total. If I'm not mistaken, the current would be 151 mA if you short a pin. \$\endgroup\$
    – user29239
    Commented Sep 19, 2013 at 8:54

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