NOR Gate Circuit 4025

Discussion in 'Homework Help' started by howartthou, Jan 26, 2010.

  1. howartthou

    Thread Starter Active Member

    Apr 18, 2009
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    Hi All

    Its been a while. I am now doing logic gates.

    Attached are two pics of a 4025 tripple tree NOR IC gate. The first pic is just a reference so you can see the gates and the pin numbers.

    The second pic is the circuit I am trying to figure out. Its number one pin is on the top right whereas on the first pic the number one pin is on the bottom left. So on the second pic pins 1 to 7 are on the top row from right to left, and pins 8 to 14 are on the bottom row from left to right.

    I am tring to determine what the voltage is on pins 6, 9 and 13.

    As I don't really understand it here is my attempt, please explain where I am right and wrong.

    Voltage on Pin 6.
    Pin 6 is an output pin. Nothing is attached to pin 6. Its inputs are from pin 3, 4 and 5. Pins 3, 4 and 5 are connected to -9V.

    Ummm, does that mean pin 6 has -9V on it?

    How does the 1kΩ resistor drop the voltage? How do you calculate the voltage drop in this circuit? Do I need to in order to determine the voltage on pin 6?

    Voltage on Pin 9.
    Pin 9 is an output pin. Nothing is attached to pin 9. Its inputs are from pins 1,2 and 8. Pins 1 and 8 go to ground. Pin 2 feeds off -9V but there is a 1kΩ resistor and 8kΩ risistor on the line. No idea what voltage is going in here so how do I know what is going out on pin 9?

    Voltage on pin 13.
    Pin 13 is an input pin. Pin 13 is connect to the line with +5V. So I would guess that pin 13 has +5V. But there are 2 resistors on the line and I do not understand the circuit so I have no idea if pin 13 really does have +5V?

    Help :(
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    look at this schema
     
  3. howartthou

    Thread Starter Active Member

    Apr 18, 2009
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    Hi Jony
    Great to see you are still here going strong.

    I did some more reading and now realise I have to work out which output is HI or LO and that is determined by NANDing the inputs, but of course you have to know which inputs are HI and LO before you can NAND them.

    So, with this in mind, for CMOS:

    Vss = 0
    Vdd = 5v
    X = 0.3(Vdd - Vss) = 1.5v

    So:

    Input LO = 0 to 1.5V
    Output LO = 0 to 0.1V

    Input HI = 3.5 to 5V
    Output HI = 4.9 to 5V

    So I can do the NAND logic applying HI and LO, basically if one input is LO the output is HI and when all inputs are HI the output is LO.

    Where I get stuck is working out what voltage is on the input pins based on the circuit.

    So pin 6 for example is an output that is either HI (5v) or LO (0v). But I have to know the voltage on pins 3, 4 and 5. Are they all -8v as shown of your diagram? If so a voltage less than zero would ruin the IC.

    So I can't believe -8v is on pins 3, 4 and 5.

    Any idea whats wrong with my analysis?
     
    Last edited: Jan 27, 2010
  4. Wendy

    Moderator

    Mar 24, 2008
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    Your NAND is a NOR.

    The chip can not tell one voltage from another, your ground is only important to you. It is the point your DMV is using to display voltages, but the IC doesn't use it at all.

    Adding one small detail for my own clarification...

    [​IMG]

    +V is a one, and it is +5V, -V is a zero, and it is -9V. Transistion for CMOS is ½ the voltage, or -2V. Anything above this is a one, anything less than that is a zero. From there it is simple logic.
     
  5. howartthou

    Thread Starter Active Member

    Apr 18, 2009
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    Thanks Bill, yes my NAND is a NOR, and I am a little stressed too, no excuses I guess.

    Could you please explain one example as I am still confused.

    So is ground -9v then?

    I still don't know how to read which input pin is 0 or 1.

    Lets look at pin 3 (third from top right), because that goes to -9V does that mean this pin is a 0?

    Please give me one or two examples.
     
    Last edited: Jan 28, 2010
  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    I see that again it is a problem with GND

    [​IMG]

    And If we move GND to V-

    [​IMG]

    And from the gate point of view this two diagrams are identical.
     
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  7. howartthou

    Thread Starter Active Member

    Apr 18, 2009
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    Hi Jony130

    As usual you are right on the button, yes I have concerns with ground and you have cleared most of the confusion with your excellent enhancements to the circuit, thank you, much appreciated.

    But now I have issues with Voltage Divider circuit analysis and KVL.

    I can see how we have 14V and 9V (the 9V + 5V = 14V).

    But the other voltage drops of 11V and 1V that you show elude me. :confused:

    We know that the Voltage drop accoss a series resistor is:

    En = Etotal x (Rn / Rtotal)

    Where:

    En is the Voltage accorss a particular resistor in the circuit.

    Etotal is the total Voltage in the whole circuit.

    Rn is the resistance of the particular resistor in question.

    Rtotal is the total Resistance in the circuit.

    So I figure:

    Etotal = 14V
    Rtotal = 59kΩ (30 + 20 + 8 + 1)

    And lets look at the voltage drop accross the 8kΩ resistor so:

    Rn = 8kΩ

    Subsitute and we have:

    En = 14 x (8000 / 59000) = 1.89V

    So why does your diagram show only 1V?

    Now let Rn = 30kΩ

    Subsitute and we have:

    En = 14 x (30000 / 59000) = 7.11V

    Now let Rn = 20kΩ

    Substitute and we have:

    En = 14 x (20000 / 59000) = 4.75V

    So we have 11.86V.

    So why does your diagram show only 11V?

    Did you just round down or is my analysis incorrect?
     
  8. Jony130

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    Feb 17, 2009
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    Notice that resistors 8K and 1K are connect parallel to 9V voltage source and 30K, 20K are connect parallel to 5V battery.
    And only 1K is content to the GND, The "GND" for 20K is positive terminal of a 9V battery
     
  9. Jony130

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    Feb 17, 2009
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    With this kind of calculations you calculate the voltage on the resistor, And this voltage is not reference to GND, for this kind of circuit
    [​IMG]

    So 1.89V it is a voltage on 8K resistor
    VR3=2.136V-0.237V=1.89V
    But this circuit is not equivalent of "your" circuit.

    And remember that if we have GND symbol in the diagram then all voltage are measured in relation to GND.
    This means that black lead of a voltmeter is always connect to gnd.
     
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  10. howartthou

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    Apr 18, 2009
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    Hi Jony130

    This is a good tip but the formula "En = Etotal x (Rn / Rtotal)" does not require you to reference ground as it applies the total voltage and total resistance in the calcuation. So are you saying this formula is wrong for any circuit that has resistors in parallel?

    So ground for the 1K is the negative terminal of the 9V battery. Is that correct?

    Regardless, how do you know which ground symbol is your reference point? There are so many of them. For example, would ground for the 30K resistor be 5V + or 5V - ? If you follow the current through the 30K R then through the 20K R you will hit 5V -, if you go the other way you will hit 5 V +, so which one is ground for the 30K R?

    Also note in the orginal diagram the 20K R is grounded to 5V- yet in your version you say that the 20K R is grounded to the 9V+. Why is ground 9V+ and not 5V- for the 20K R? :confused:

    Okay, I will test this for parallel resistors then.

    Lets get the new Rtotal with parallel resistors:

    Rp1 = 8K and 1K:

    Rp1 = R1R2/R1 + R2 = 8 x 1 / 8 + 1 = 1K


    Rp2 = 30K and 20K

    Rp2 = R1R2/R1 + R2 = 30 x 20 / 30 + 20 = 12k

    So, Rt = 1K + 12K = 13K

    Now lets try the voltage accross the 1K R (with no ground reference):

    En = Etotal x (Rn / Rtotal)

    En = 14 x (1000 / 13000) = 1.07 V (This agrees with your diagram that shows 1V for this resistor).

    Now lets try the voltage accross the 8K R (with no ground reference):

    En = Etotal x (Rn / Rtotal)

    En = 14 x (8000 / 13000) = 8.61 V

    Is this correct?
     
    Last edited: Jan 30, 2010
  11. Jony130

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    Feb 17, 2009
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    No , the formula is OK

    [​IMG]

    Voltage between paint A and GND is equal V1=14V.
    Voltage between point B and GND is equal
    V_B = 14V x (R2 + R3 + R4)/ (Rtotal)= 14V x 29K/59K = 6.88V
    And voltage at point C (voltage between pont C and GND)
    V_C = 14V x (R3 + R4)/(Rtotal) = 14V x 9K/59K = 2.13V

    For all voltages in this circuit the gnd is negative terminal of the 9V battery.

    In schematics all GND symbols are virtual connected to the same point.
    See for example:
    [​IMG]
    So in reality this circuit look like this
    [​IMG]

    These resistors are connected parallel to the battery.
    [​IMG]

    So, after simplifying we get this simple schematics:
    [​IMG]

    And of-course voltage at all point are reference to GND.
     
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  12. howartthou

    Thread Starter Active Member

    Apr 18, 2009
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    Thanks Jony, I am learning a lot here.

    What is going on here, these all add to 9V yet KVL says they should add to 14V. Did you assume the power supply is 9V but put 14V on the diagram by mistake?

    If you take ground as point "E" your diagram says that is zero, so where is the missing 5V?

    Also in this diagram are these resistors in series or parallel. They look like they are in series ?
     
  13. Jony130

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    This schematics is jut a example of how you should calculated the voltage with reference to GND. Because it seems to that you assume that all resistor are connect in series to 14V battery which is not true.
    And this diagram has nothing to do with your gate circuit, it's only an example.
    And you have to sum all voltage drop to get 14V, there is non mistake in the diagram.


    So back to your original circuit.
    [​IMG]

    And after simplification we get this simple schematics:
    [​IMG]

    And of-course all resistor are connect in series, but for example 30K+20K thanks to KVL VR1+VR2=5V and VR3+VR4=9V
     
    Last edited: Jan 31, 2010
  14. howartthou

    Thread Starter Active Member

    Apr 18, 2009
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    Hi Jony

    In the original circuit you are showing 1V on the 1K R.

    In your simplified schematic you show point D as 237.29mV

    This has me confused.

    Shouldn't point D be 1V to match the original circuit?

    I still cannot see how you calculated 1 V for the 1K R.

    Can you please explain.

    I can get 1V from your simplified circuit if I use 9V as the total, but you have been using 14V.

    With 9V as Etotal and 9K as Rtotal:

    V_D = 9V x (R4)/(Rtotal) = 9V x 1K/9K = 1V

    This 1V agrees with the original circuit which shows 1V on the 1k R.

    But your sample with the simplified circuit uses Etotal = 14V and Rtotal = 59K.

    So, can you please explain how you calculated 1V accross the 1K R??
     
  15. Jony130

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    Feb 17, 2009
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    To stop confusion.
    This diagram is the correct "simplified circuit" of your original gate circuit
    [​IMG]
     
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