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Logic function performed by 7432 integrated circuit

  1. Feb 10, 2013 #1

    s3a

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    1. The problem statement, all variables and given/known data
    The problem and its solution are attached.

    2. Relevant equations
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    3. The attempt at a solution
    All I think I know is that pins 7 and 14 are used for powering the integrated circuit because that's what I believe I read but, I don't fully understand what's going on. I believe the pins that get power are those connected to little ground symbols For example, is it possible for pins 4, 5 and 6 to also take in power?

    Most importantly, I don't understand why “the 7432 [integrated circuit] performs as a 2-input OR gate when wired as shown in [the figure attached to this forum post]”.

    Could someone please elaborate about all this for me?

    Any help in getting me to fully understand this problem would be greatly appreciated!
     

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    Last edited: Feb 10, 2013
  2. jcsd
  3. Feb 10, 2013 #2
    The 7432 is a quad OR gates: 4 OR gates into a single package. Each gate has 2 inputs and an output wich is the OR of the inputs.
     
  4. Feb 10, 2013 #3

    s3a

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    What parts of the 7432 integrated circuit consists of OR gates? Is it within that rectangle with the 14 pins or is it where points A and B are or what?

    Also, can you explain the wiring/pin/ground/battery stuff to me please?
     
  5. Feb 10, 2013 #4

    CWatters

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    Did you forget to attach the drawings to your original post?
     
  6. Feb 10, 2013 #5

    s3a

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    Silly me, I just added it now. :)
     
  7. Feb 10, 2013 #6

    CWatters

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    Perhaps this will help. Google images is your friend.
     

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  8. Feb 10, 2013 #7

    CWatters

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    So you can see from the above drawing that pins 4 and 5 are inputs to one of the four OR gates in the 7432 and pin 6 is the corresponding output.

    Pin 14 is connected to VCC = battery +ve
    Pin 7 is connected to GND = battery -ve
     
  9. Feb 10, 2013 #8

    s3a

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    Thanks for your answers, CWatters.

    It helped but, I'm still a little confused as to why there are ground signs all over the place and why they are where they are and why the battery is where it is and how exactly the switches A and B work. If I'm correct, I get that when a switch is directed upward (in the orientation that people would normally look at the drawing), it means that the input is 1 and when the switch is directed downward (in the orientation that people would normally look at the drawing), it means that the input is 0 but, how exactly (directly and indirectly) needs to happen for the switch to be directed upward or downward (in the orientation that people would normally look at the drawing)?

    P.S.
    Is the integrated circuit the entire Fig. 3-26 or just the part that is the rectangle with the pins?
     
  10. Feb 10, 2013 #9

    CWatters

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    The ground symbols are just a form of short hand. They avoid having to clutter up the diagram with lots of lines joining all those points together. I have redrawn the circuit another way. See enclosed.

    There is no particular reason why the battery is shown on the left hand side.

    Ok when you move switch A up it connects pin 4 on the 7432 to the battery +ve. That makes that input pin a logic 1. When you move switch A down it connects pin 4 to the battery -ve. That makes that input pin a logic 0.

    The same applies to swich B and pin 5.

    The integrated circuit (7432) is just the rectangle with 14 pins. The rest of the diagram shows how that integrated circuit might be connected.
     

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  11. Feb 20, 2013 #10

    s3a

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    Thanks again. I'm almost there! :D

    I just wanted to ask about the resistor's "purpose". Why was that 150 Ω resistor placed where it is? What would happen to the 7432 integrated circuit if the resistor were removed ror had it's value reduced or increased?
     
  12. Feb 21, 2013 #11

    CWatters

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    The resistor is used to limit the current flowing from the 7432 through the LED. In short it sets the "operating point" of the LED. The operating point is a combination of voltage and current values that make the LED work as required while protecting the 7432 which can only deliver a limited amount of current (after all it's a logic gate made from tiny transistors not a massive power station).

    An LED is not like a light bulb, it has a different Voltage vs Current curve....

    http://www.radio-electronics.com/in...-light-emitting-diodes/led-voltage-curves.gif

    Note how the current rockets up once the voltage exceeds a certain value (called Vf). This value tends to be different for each colour of LED. To turn on an LED you need to choose an operating point on that steep part of the curve BUT you should also limit the current so that a small change in Voltage doesn't cause a massive change in the current or damage the 7432.

    Lets assume it's a RED LED. Looking at the curve you can see that the recommended operating point (dotted line) is about 2V at 20mA. Lets choose that as the operating point we are aiming for when the LED is ON.

    Then back to your circuit... How do we calculate the resistor value?.....

    If the output of the 7432 is at 5V (logic 1) and the LED voltage drop is 2V that leaves 3V across the resistor. If we want the current to be 20mA then using Ohms Law...

    V = IR
    so
    R = V/I

    = 3/0.02
    = 150 Ohms.

    In other words with a 150 Ohm resistor the LED voltage will be about 2V and the LED current about 20mA.

    If you wanted to change the colour of the LED to Orange you would go back to the LED curves.... Note how the recommened operating point for an orange LED is about 3V at 20mA. The recommended resistor would then be..

    5-3=2V
    R = 2/0.02
    = 100 Ohms
     
  13. Feb 21, 2013 #12

    CWatters

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    Take a look at the LED curve. What would the current be if the 5V output was connected to the LED without a resistor? 5V isn't even on the chart :-). The current would certainly be very high (off the top) and probably damaging to the 7432. Even if the 7432 survived the higher power consumption would reduce the battery run time.

    Regarding the last part of the question... Have a think about what would happen if you used a 150 Ohm resistor with a Blue LED.
     
  14. Feb 21, 2013 #13

    s3a

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    Thanks. That gave me a lot of useful information.

    For the blue LED, is this correct?:
    5 V – 4 V = 1 V

    V = RI
    (1 V) = (R_blue)(0.02 A)
    R_blue = (1 V)/(0.02 A) = 50 Ω

    I googled and, it seems that the average of (or most) LEDs is 2 V so, that's most likely why the book chose 2 V, right? Would that mean that most LEDs emit red light?

    Also, another part of my book says that a high logic level for the fourth pin is NEAR +5 V. Why is it near and not exact at 5 V? Also, is it the case that all high logic levels, regardless of the pin, are near and not exactly at the 5 V value?

    Edit:
    Also, in real life, is the resistor part of the LED or do the engineers/designers/etc put some other resistor in series?

    Edit #2:
    Actually, I believe they are separate because the resistor needs to "steal" current from the LED since that's what the engineers/designers/etc want. Am I correct in thinking this?
     
    Last edited: Feb 21, 2013
  15. Feb 21, 2013 #14

    CWatters

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    Correct.

    Correct but for the wrong reasons. Red LEDs are the cheapest to make. It's to do with the type of semiconductor material needed to make them and the volume sold. I've forgotten the exact dates but when I was growing up Red LEDs were much cheaper than other colours and there were no Blue LEDs available at all. Blue LEDs may have existed as prototypes but were too expensive to be mass produced.

    This is because the transistors/FETs used to make logic gates are not perfect "ideal" switches. FETs have an "ON resistance" thats not zero Ohms. Transistors have a minimum saturation voltage of perhaps 0.3V so they can't "pull" the output pin all the way upto 5V. If you look at the data sheet for the logic family you will find a section on the characteristics of output pins. This should tell you how much current it's designed to deliver while still managing to maintain the minimium definition of a logic 1.

    Some logic families are better at sinking current then sourcing current (eg they are able to pull the output pin down to 0V better than they are able to pull it to 5V) so you will find some circuits where the LED is connected between the output pin and 5V. In that case a logic 0 on the output is needed to turn on the LED.

    Normally the resistor is a seperate component. Here is a small circuit board and you can see 6 LEDs on the right (these particular LEDs have clear lens) and just to the left of them you can see 6 resistors. There are actually 8 resistors on the board but the 6 in line with the LEDs will be the current limiting resistors...

    http://www.raspberrypi.org/phpBB3/viewtopic.php?t=24048&p=288450

    You can buy LEDs with a built in resistor but it's generally cheaper to buy seperate LEDs and resistors. Here is one with an internal resistor...

    http://www.maplin.co.uk/5mm-5v-leds-35775

    Quote "5V integral current limiting resistor - no external current limiter required with a 5V supply".

    Sorry no. The manufacturing processes required to make semiconductors and resistors are different so in general it's just cheaper overall to have the resistor seperate.
     
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