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Powering my IC

Discussion in 'General Electronics Discussion' started by Davidzb, Mar 3, 2017.

  1. Davidzb

    Davidzb

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    Mar 3, 2017
    Hi all,

    My first post so here we go. My SN74HCT00 IC specs are VCC = 4.5 V to 5.5 V and input current of 1 µA Max

    What would be the best way to regulate my 9v supply to meet these conditions? Voltage divider? Zener diodes?

    Thanks in advance for any help

    David
     
  2. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    Hello David. Use a 5 volt regulator. Let me know if you need help finding one.
    Thanks
    Adam
     
    HellasTechn likes this.
  3. BobK

    BobK

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    Jan 5, 2010
    Or use one of the many 5V USB chargers you probably have lying around like most of us.

    Bob
     
  4. AnalogKid

    AnalogKid

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    Jun 10, 2015
    If your only power source is the 9 V supply, then a resistor and 4.7 V or 5.1 V zener diode with work fine. Be sure to place a decoupling capacitor across the IC power and GND pins. The IC current draw might be much greater than 1 uA, depending on what you are doing with it.

    ak
     
  5. Davidzb

    Davidzb

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    Mar 3, 2017
    Thanks everyone for the help, it's much appreciated.

    David
     
  6. KMoffett

    KMoffett

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    Jan 21, 2009
  7. Davidzb

    Davidzb

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    Mar 3, 2017
    Hi Ken

    I was just going to play around with a couple of the NAND gates on the IC to familiarise myself with the operating principles of ICs as I've not used them before. Thank you for the help
     
  8. HellasTechn

    HellasTechn

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    Apr 14, 2013
    I agree . The simplest and safest is an 7805 Vreg.
     
  9. Audioguru

    Audioguru

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    Sep 24, 2016
    The 74HCT00 is designed to be very fast (the H in its numbers means High speed) and to be compatible with 40 years old TTL logic (the T in its numbers). If it is connected wrongly then its high output current can destroy it.

    Since you have a 9V battery for the power supply then you should use a CD4011 quad NAND gate IC that is a little slower and with low supply voltages its output current is low enough to avoid damage but it works with a power supply that is from 3V to 18V.
     
  10. Davidzb

    Davidzb

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    Mar 3, 2017
    Hi all

    Thanks for all the help so far, hugely appreciated. As a learning exercise I've decided to continue with the SN74HCT00 and use the resistor and Zener diode combo to regulate power to the IC

    I am double checking my IC specs on the datasheet at Texas Instruments' website

    From there is get that voltage should be 4.5V to 5.5V, OK i understand that

    But current-wise, it mentions max input clamp current ±20 mA on one page and then a max ICC of 20µA elsewhere. And in another place it mentions a Low Input Current of 1 µA Max

    Confusing :(

    Would anybody be able to advise on this and what the max current allowed for the SN74HCT00 would be?

    I have decided to run an LED (with a max 30mA forward current) from the output of the NAND gate to have something nice to look at so as KMoffett said I will need to add this to the current required to run the IC.

    Then I can get to calculating the resistor value required

    (I would post the datasheet link but not sure if that is allowed)

    Thanks a lot again and sorry for the long post

    David
     
  11. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    May 8, 2012
    David, yes it's allowed.

    Chris
     
  12. BobK

    BobK

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    Jan 5, 2010
    The current used by the CMOS chip will be negligible compared to that used by the LED.

    Bob
     
  13. Audioguru

    Audioguru

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    Sep 24, 2016
    The datasheet for the SN74HCT00 says that its absolute maximum allowed output current is 25mA then 30mA in an LED will probably destroy the IC and the LED. Most ordinary LEDs are spec'd and used at 20mA.

    When the supply voltage is 5V then the maximum allowed input voltage is a little more than 5V so the inputs have clamp diodes that clamp a high input voltage to near the +5V or clamp it to near 0V. The clamp diodes are tiny then their maximum allowed current is 20mA. Did you see the Note 1 in the datasheet that says the input voltage can be higher than the supply voltage if you limit the current (to no more than 20mA through the clamp diodes with a resistor)?

    The input current is very close to zero and the operating current is also very close to zero but can be high if the frequency is high then it is charging and discharging stray capacitance.

    See Note 3 where it says that all unused inputs must be at Vcc (+5V) or 0V by your circuit.

    Did you see that the IC can make oscillators using two gates then your LED can blink? Here is an oscillator circuit:
     

    Attached Files:

    Davidzb likes this.
  14. Davidzb

    Davidzb

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    Mar 3, 2017
    Thank you, this is very helpful
     
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