When Using Integrated Circuits - WHAT RESISTOR

Discussion in 'General Electronics Chat' started by CoderJohn, Mar 18, 2016.

  1. CoderJohn

    Thread Starter New Member

    May 22, 2014
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    So I am dabbling with some integrated circuits and need some help....

    Say the Voltage Input High Min Value is 2V, the Max is 5 V.... If I had a 5v source and wanted to approximate the voltage on the pin to be 2.4 Volts would I calculate it as

    -5 + r(0.005) + 2.4 = 0
    r(.005) = 2.6
    r = 520 ohms

    Or would that be wrong.... I know the Input Leakage current is 10uA so really nothing to worry about... but lets say I wanted to pull it down, what is a reasonable way to do this formula wise... I don't want to be told to just use a fairly large resistor, rather I would like to know how it is calculated.....

    would you want it so that....

    VLmax > Ilkg * Rgnd > VLmin or what?
     
  2. Dodgydave

    Distinguished Member

    Jun 22, 2012
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    Use a resistor divider with two resistors, both at the same value will give you half supply 2.5v, try 100K and 96K.
     
  3. #12

    Expert

    Nov 30, 2010
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    Where 2 r4sistors are in series and they are called R1 (connected to 5 volts) and R2 (connected to R1 and ground).
    2.4V = 5V x R2/(R1+R2)
    Where leakage is 10 ua, the current through the resistors is at least 1000 ua for a 1% error.
    Rtotal = 5V/.001A
    R1+R2 = 5000 ohms or less.
    2.4 = .001A R2
    R2 = 2400 ohms
    Therefore R1 must be 2600 ohms.
    Then you adjust for resistors you can actually buy.
    For instance, (2) 2400 ohm resistors and a 220 ohm resistor.
    That puts you about 2% off the target.
     
  4. crutschow

    Expert

    Mar 14, 2008
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    First of all, why do you want 2.4V for the input?
    What's the circuit and what type of input is it?
     
  5. CoderJohn

    Thread Starter New Member

    May 22, 2014
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    It is a Z80 Processor and 2v is the minimum logic level high. I chose 2.4 as the min just for extra safety.
     
  6. dl324

    Distinguished Member

    Mar 30, 2015
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    Is this for pulling up inputs that aren't being driven?
     
  7. TheButtonThief

    Active Member

    Feb 26, 2011
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    If the chip works with a 5V supply and your available infeed is 5V, why not just supply the ship with 5V?
     
  8. crutschow

    Expert

    Mar 14, 2008
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    You misunderstand the spec.
    2V is indeed the minimum voltage for a logic high but a normal logic high is 5V and that is what you should use.
    No need to use a resistive divider to lower the voltage.
     
  9. CoderJohn

    Thread Starter New Member

    May 22, 2014
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    I am not trying to figure out what I need so much as the math and what not....
    It is for if I wanted to use a pull-up resistor OR a pull-down in either case. I prefer to know how to select one my self. I know KVL, KCL, Voltage Division, and Current Division, but I am only in Circuits 1 and we are doing the boring nitty gritty stuff at the moment and digital circuits isn't until circuits 2... Which I can't take until I transfer and I have always liked the Z80 assembly language, so I figured I would attempt to mess with this and see what I can make. The only problem is, a lot of the IC is a black box....so I have some questions below.


    Here is the components data sheet:
    http://pdf1.alldatasheet.com/datasheet-pdf/view/25733/STMICROELECTRONICS/Z8400BB1.html

    1.) Can anyone explain how you would use that to calculate the resistor range for a pull-up and the resistor range for a pull down?

    2.) Are the internal components of a IC negligible enough to consider the input as ground and not account for any extra resistance on the inside?

    3.) The Ilkg is 10 uA max... Vil Max is .8 volts. I read that the input pins of integrated circuits and other things are generally 10kOhms to 1MOhm... Does that mean that the impedance of the input pin could be represented by .8/(10^-5) = 80,000 ohms

    I am either so lost or this data sheet lacks some information.

    In regards to above, so if I want a pin to stay on at all times, do I need a resistor from the 5v to the pin?

    The only time I need a resistor is if the high signal will ever be shorted to ground to disable the pin correct? As to limit the current being sent to ground?
     
  10. dl324

    Distinguished Member

    Mar 30, 2015
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    Pull up/down resistor values are determined by the logic family used and appropriate value is a combination of the sink/source capability of the driver and the input leakage current.

    For standard TTL, 1K pull-ups were common.
     
  11. crutschow

    Expert

    Mar 14, 2008
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    Actually, applying a steady-state voltage to a CMOS circuit input at near the mid-range between the supply voltage and ground, can cause excess current to flow in the input transistors, since they both can then be biased on, rather than one being on and one being off as occurs for the normal 0V and 5V logic levels.
    So applying a voltage other than 0V or 5V to the input with a voltage divider is not only unnecessary, but could be detrimental to the circuit.
     
    InspectorGadget likes this.
  12. #12

    Expert

    Nov 30, 2010
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    I did that in post #2. You are allowed to ask questions if I wasn't very clear by answering mostly in the language of math.
     
  13. CoderJohn

    Thread Starter New Member

    May 22, 2014
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    Well I know how to calculate resistors if you know what you want/need but you didn't really account for the resistance of the input pin. Is that because you treat the input pin as ground? Like that is really my question I suppose...

    For a PullDown you want a high resistor to prevent wasted power to ground correct? Either way the IC will receive 5 volts because with a pull down it could be seen as a parallel circuit...

    With a pullUp it is seen as a series is this all correct so far? Cause with a pull up you are running power through a resistor and into the IC. The only time it derails from that path is if you short it to ground in which case it ignores the IC path.

    With the pull down the path is from the input pin to ground. However, when the power is activated it goes through both paths(resistor to ground and input pin.) Thus, making it parallel.

    Is this all correct? If so then the pull-up needs to be large enough so that your current isn't exactly high but it doesn't need to be so high as to absorb a significant amount of the voltage. Which really isn't a big deal if typical IC's have resistance of 10 kilo-Ohms to 1 mega-Ohm

    So assuming I was using just a pull-up and just wanted the current to be 1mA at most when the input was shorted to ground then 5/.001 = 5000 ohms like the guy above stated earlier. Assuming the impedance of the input pin has a minimum of 10kilo-ohms than that would mean my pull-up resistor consumes only 1.6666 volts leaving 3.3333 volts to the input pin. Which is more then enough to create a logic high from a pull up.

    However, if I want to leave the input at a constant on state I should just connect directly to the stable 5v source.

    If I have stated anything wrong please correct me... This is just what I have taken from what you guys have said.
     
  14. dl324

    Distinguished Member

    Mar 30, 2015
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    The input to your Z80 is a gate capacitance with a small leakage current. The equivalent resistance is 100's-1000' of Mohms and will have an insignificant affect on pullup/down resistor calculations.
     
  15. #12

    Expert

    Nov 30, 2010
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    Figuring 10 ua leakage at 5 volts, you can calculate an impedance of 500K. The calculation of 5000 ohms is to insure a 1% compliance with the intended voltage. The real purpose of the pull-up or pull-down resistor is to jerk the input pin one way or the other. For any input except an ADC, 1% isn't necessary. You could use 10K to get the input within 2% of the 5V rail (or ground) or 47K to get the input within 10% of the 5V rail (or ground). Most ppl use 10K. It's all good if you're close to the rail voltage. Hanging your input in the middle causes the problems crutschow mentioned.
     
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