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  1. #1
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    Mar 2004
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    Active High/Active Low

    Hi all,
    Just to make sure. What does active high and active low mean? For a normally closed limit switch does this mean that the pin is in a active high state because it becomes active with the high +5 V coming in?
    Thank you all for your help.
    I really appreciate it. :rainfro: :rainfro:

  2. #2
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    May 2003
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    Active high and active low are referenced to the destination circuit and usually mean more positive (high) or more negative (low).
    An active high circuit is turned on when the input is +5V (for instance) and off when the input is 0V. An active low circuit is turned on by 0V and off by +5V.

    robotic regards,

    Tom
    = = = = =
    "If you worship at the altar of success then success will alter you."
    - - Glenn White

  3. #3
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    OK I see. Thanks for the help.
    I really appreciate it. :rainfro:

  4. #4
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    So if I have the limit switches with a normally closed switch, when I push the switch this will open the switch and allow 5 Volts to come into the Parallel port and tell the software that the switch has been pushed. And when the switch is in it's "normal" state with normally closed, this does not give much voltage to the parallel port. This means that the software has to acknowledge it as active high right?
    Thanks.

  5. #5
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    You can use a normally closed (N.C.) switch to get either an active high or an active low output. The N.C. switch SW1 gives an active low signal (when the switch activates - opens - the resistor pulls the signal to ground).

    Likewise, the N.C. switch SW2 gives an active high signal, since the pull-up resistor pulls the signal to +5V when the switch is activated (opens).

    If you replaced SW1 with an normally open (N.O.) switch, that would give you an active high signal - the switch would then close on activation and pull the signal to +5V.

    So you can use either a N.C. switch or a N.O. switch to get either an active high signal or an active low signal.

    Arvid
    Attached Thumbnails Attached Thumbnails pull-down-up.gif  

  6. #6
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    So for switch 1, when closed and completing a circuit, the five volts comes in and goes out to the parallel port because of the resistance of the resistor. Then when is opened, no voltage is going to the parallel port. Is this correct? The ground is just there for what?
    Thanks I really appreciate it.

  7. #7
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    You need to keep the signal to the parallel port - like all logic inputs - firmly at either Vcc (+5 V) or GND. You should never just disconnect an input (leave it "floating"). If you do you might get some very strange results.

    I once experimented with a logic gate (a simple inverter), forgot to connect the input of it and it started oscillating so violently in the MHz range that is totally freaked out my radio! I got just buzz instead of the 103.3 MHz FM radio station I was listening to! When I connected the input the music returned at once, and disappeared again when I disconnected it. So I had made myself a crude radio transmitter! Not exactly what I wanted...

    Anyway, when the switch is closed, the 5 V is connected directly to the parallel port input, and a high signal is detected (the resistor is not needed for this). There's a voltage drop of 5V over the resistor, and the current through it is I = U/R = 5/4700 = about 1 mA.

    When the switch is opened, practically no current will pass through the resistor, and the voltage over it will drop to zero. Since it is connected to ground in one end, there will also be ground potential at the other end (since the voltage over it is zero). And so the input sees 0 V = low level. This is how a pull-down resistor works - it "pulls" the signal to ground when no other voltage source is connected.

    Arvid

  8. #8
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    Quote Originally Posted by arvidb
    You need to keep the signal to the parallel port - like all logic inputs - firmly at either Vcc (+5 V) or GND. You should never just disconnect an input (leave it "floating"). If you do you might get some very strange results.
    Arvid
    Would it be correct then to say that logic circuit inputs are subject to conducting non-predictable resonant currents created by the otherwise normal voltage and current changes of the circuit function and that the pull-down resistor is dampening these small resonant currents by dissipating them as 'micro' heat in the resistor which action is simultaneously settling the voltage at the input pin to ground? Could other passive components be used to accomplish this or would it require too much 'tuning'?

    Chris

  9. #9
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    Chris, In reference to the question:

    " Would it be correct then to say that logic circuit inputs are subject to conducting non-predictable resonant currents....."

    No. When the input pin of a logic circuit is left unconnected (called "floating") it is not conducting anything. However, there is often no (practical) way to predict whether the circuit is going to respond to this floating pin as if it were a high or as if it were a low or oscillate. Since the pin is open and is not conducting, the oscillations take place within the circuit. Now, if you are talking about resonent currents that take place within the integrated circuit between circuit components then yes.

    Once, however, you pull the input either high or low, there are no oscillations to dampen, there are no resonent currents. The circuit is just stable, and the internal and external current flows are as designed.

    As for using other methods to pull-up or pull-down, it depends on what you are trying to accomplish. When you combine logic where the output of one gate drives the input of the next gate, it is usually not necessary to pull up or down. When the input of a gate is driven by a switch it is best to use pull-up or pull-down, and resistors are the easiest way. When a gate is unused it should be pulled-up or pulled-down (unused input pins are usually grounded).

    On the other hand, if your goal is actually to get the circuit to oscillate (possible), then you would not want to use a pull-up or pull-down resistors! But thats another subject altogether.
    Patrick;
    The Sober Pollock

  10. #10
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    Quote Originally Posted by sbrpollock
    Chris, In reference to the question:

    " Would it be correct then to say that logic circuit inputs are subject to conducting non-predictable resonant currents....."
    Now, if you are talking about resonent currents that take place within the integrated circuit between circuit components then yes.

    This is what I meant but I probably shouldn't have used the word conduct as it suggests a path beyond the end of the pin. What I was thinking of was that the internal resonant currents are appearing at the end of the pin much as a radio frequency signal might appear along the length of an antenna.

    Once, however, you pull the input either high or low, there are no oscillations to dampen, there are no resonent currents. The circuit is just stable, and the internal and external current flows are as designed.

    It seems that the dampening has to be a continuous event that has the effect of making the circuit otherwise appear stable. Certainly the internal circuit would become unstable again if you were to suddenly remove the pull down resistor.

    When a gate is unused it should be pulled-up or pulled-down (unused input pins are usually grounded).

    Can they (unused input pins vs. used input pins) be grounded directly or do they need to be pulled down through a resistor?

    Chris

  11. #11
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    Aug 2003
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    First.....My mistake:
    I probably shouldn't have said that most inputs are tied to ground. With most of the devices I've dealt with thus far, I tied to ground without a resistor. But, Iv'e been told now that tying to logic high (supply) with a current limiting resistor is more noise resistant.

    Maybe it would help to examine the purpose of the resistors in Arvid's Schematic:

    When switch 1 is open there is a single circuit path between the pin and ground through the 4.7k resistor. Since there will be very little current through this resistor, the voltage drop across it will be zero, and in turn the voltage level on both sides will be ground. The voltage level at the pin will be ground.

    When switch 1 is closed there are now two parallel circuit paths, one between +5volts and ground through the 4.7k resistor, and one between +5volts and the pin. Now there is a full five volts across the resistor, and hence the voltage level at the pin is now +5volts.

    The purpose of the resistor is to limit the amount of current that flows from +5volts to ground while the switch is closed, not to act as a dampener. Picture the same schematic without the resistor, just a straight line to ground. It would be a dead short between +5volts and ground. That is what the resistor is for.

    Resistors do not have any kind of characteristics that would enable them to "dampen".
    Capacitors and Inductors resist "changes" in voltage and current, but resistors do not.

    Not all logic circuits will even oscillate. Some will just change states seemingly randomly for no apparent reason. These pins are susceptible to noise and to static also.

    The reasons some circuit families may go into oscillations are different too. For example, the input gates CMOS circuits made up many Field Effect Transitors (FETs) may experience slow charging of very small amounts of gate capacitance due to very small amounts of leakage current if the input is alowed to float. Once the gate crosses the threshold voltage the transistor conducts by itself spontaineously. However, if you simply hold the voltage of this same gate above or below the threshold, the gate will be stable and will have no tendency to change its state.

    Yes, the moment the pin is allowed to float again the circuit may be unstable again.

    Lastly, In Arvids schematic, these inputs are going to a parallel port on a computer. I think the parallel ports on computers are allready protected from floating, so the resistors in the schematic truly are only to provide N.O. and N.C. operations with the same type of switch. They might not be necessary to prevent oscilation or unexpected operation. I would still use them however.
    Patrick;
    The Sober Pollock

  12. #12
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    I like to look at it this way (and I guess it's about the same as Patrick wrote above):

    Logic gate inputs are often very high impedance - which means they don't pass current, they just "feel" the voltage present. Now if one leaves an input like this unconnected (floating), electric and magnetic fields in the surroundings will eventually build up an electric charge on the input, triggering (or untriggering) the gate, in a seemingly random fashion. When the gate switches, this might in itself give rise to a small electromagnetic "pulse", and if this pulse is enough to again change the state of the input, then we have an oscillator.

    But if the input is connected to some well defined voltage, either directly or through a resistor, then any electric charge will be conducted away, and the level of the input will stay where it's supposed to.

    Arvid

  13. #13
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    Thanks Patrick and Arvid. I get it now. The resistor drains the electric charge (rather than dampens it) and any oscillation that occurs is way more random than I had originally imagined being mush less a product of the circuitry and more a factor of unknowns.

    Chris

  14. #14
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    Sep 2004
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    To All
    NOTICE: Do not use the circuit with SW1 in it as it connects the ciruitry to positive voltage without any current limiting. I have done this in the far distant past and found it allows the smoke to get out of the little critters that live on the circuit boards.

    All kidding aside the circuit was for explanation and is not good practice. Simply use the circuit with SW2 in it and choose a normally open switch or normally closed switch to get active low or high inputs.

    A normally open switch when closed will represent an active low.
    A normally closed switch when opened will represent an active high.
    The resistor is a 10,000 ohm or so. It is not critical 4,700 ohm to 10,000 ohm will work.

    Just remember if the action being sensed produces a low it in a active low. The reverse holds for active high.

    c_nut

    In the swamps 1 mile north of mile post 223 I 10.

  15. #15
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    c_nut,

    that is just not correct; both circuits work just fine. Inputs by definition have high impedance (otherwise they are outputs), so the current is limited by the impedance of the input itself.

    If the smoke escaped you did something else wrong, like connecting the output ("TO PARALLEL PORT") to another output, or so.

    Arvid

  16. #16
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    Sep 2004
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    Arvid

    Let me first say I think you are great for sharing your knowledge with fellow enthusiasts. It takes time to do this an I am sure all of the guys you help appreciate it.

    Been doing this a long time (electronics) and all of the books and all of the real circuits I have seen use a pull up resistor to protect the chip input. I am not saying if you use the circuit it will fail instantly but you are pushing the parameters and Murphy likes that, it gives him a chance to exercise his law.

    Why skate on the edge. With all of the things to go wrong during a prototype stay on the safe side. Resistors are 2 cents apiece and are hard to break, chips cost more and are fragile. Using the chip to do the current limiting is not the better choice here.

    Just matters of outlook both are correct.

    C_nut

    In the swamps 1 mile north of mile post 223 on I 10.

  17. #17
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    Sep 2004
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    Quote Originally Posted by OCNC
    Thanks Patrick and Arvid. I get it now. The resistor drains the electric charge (rather than dampens it) and any oscillation that occurs is way more random than I had originally imagined being mush less a product of the circuitry and more a factor of unknowns.

    Chris
    The logic circuitry in the chip is a amplifier. It has a unity gain ie. 5 volts in = 5 volts out. As with all amplifiers if the output signal gets into the input, a feedback occurs. Yes that whistle in a PA system is feedback, uncontrolled feedback. First a little of the output gets in the input and a small whistle is heard. If the feedback is not stopped the whistle will increase to a painful level.

    If the input pin of a logic chip is left floating it’s high impedance makes it sensitive and it will pick up the output of itself. After all the output and input are only separated by a very small distance on the chip itself. This sets up the feedback path and the chip begins to oscillate. Connecting the input to some voltage or ground does not allow the feedback to build to the uncontrolled state.

    This is what the other guys said I just rephrased it from an analog circuitry point of view.

    Keep asking questions, it is the best way to learn …..for both of us.

    C_ nut

    Mile post 233 on I 10 in the swamps 1 mile north.

  18. #18
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    Quote Originally Posted by c_nut
    Arvid

    Let me first say I think you are great for sharing your knowledge with fellow enthusiasts.
    I think we all are! It's a great forum and it's from discussions like these one learn.

    Quote Originally Posted by c_nut
    Been doing this a long time (electronics) and all of the books and all of the real circuits I have seen use a pull up resistor to protect the chip input. I am not saying if you use the circuit it will fail instantly but you are pushing the parameters and Murphy likes that, it gives him a chance to exercise his law.
    On a circuit board, when one have an input (that comes from the outside, "uncontrolled" environment), it's usually a good idea to have a series resistor: Physical connector - decoupling capacitor to ground - series resistor (100 ohm or so) - possibly pull-up or pull-down resistor - logic gate.

    The series resistor in combination with the capacitor gives the logic gate some protection against ESD and general abuse of the input. Perhaps this is the resistor you are thinking about?

    A pull-up or -down resistor on the other hand, does not protect the input, it only gives it a known state when unconnected. And I really cannot see the need to add a series resistor to the switch circuit above - logic gate inputs simply does not need one since their high impedance limit the current. To add one in case one happened to abuse the input - well, what stops one from applying the abuse on the wrong side of the resistor? It should reside on the mother board after the DSUB to really be effective, and hopefully the mother board manufacturer has already done their job there.

    Arvid

  19. #19
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    Aug 2003
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    Back To The Original Question

    I want to take a moment to address the original Question of this thread, before going on about pull-ups, pull-downs, floating pins, and current limiting resistors.

    Sanghera asked:
    "Just to make sure. What does active high and active low mean?"

    First a few definitions:

    "High" and "Low" are generic terms that refer to the signal levels for the logic family in question. For Example:

    For TTL logic Devices:
    Any voltage over 2.0 volts will be recognized as a "high"
    Any voltage under 0.8 volts will be recognized as a "low"

    For CMOS logic Devices:
    Any voltage over 3.7 volts will be recognized as a "high"
    Any voltage under 1.3 volts will be recognized as a "low"

    There are many families and each has their own set of "threshold levels".

    A graphic explanation can be found here:
    http://www.interfacebus.com/voltage_threshold.html

    You could even apply these terms to some of the older industrial equipment I've worked on where, before we updated them with PLCs, everything was controlled by 110 volt AC relays. On these machines anything above approximately 90 Volts AC was a "high", and anything below approximately 15 volts AC was a "low". (This is only an EXAMPE)

    "Active" high and "Active" low are both in reference to the OPERATION of the circuit or device who's input they are accociated with. Look at it from the point of view of: "Under which condition will the circuit perform its ACTion?".

    For example:

    Example 1:
    Let's say we design a circuit to run a pump. This circuits "ACTion" is to run a pump. It has one input called "runPump". We could do this one of two ways.

    Active High:
    Whenever the input called "runPump" is high, the pump runs.
    Whenever the input called "runPump" is low the pump does not run.

    Active Low:
    Whenever the input called "runPump" is low , the pump runs.
    Whenever the input called "runPump" is high the pump does not run.

    From the example you see that with active high the circuit runs the pump (performs its ACTion) when the input is high, and with active low the circuit runs the pump (performs its ACTion) when the input is low. Hence the names "ACTive High" and "ACTive Low".

    Note: I've often seen these refered to as "Positive Logic" and "Negative Logic" also.

    Example 2:
    Lets say we have a device (or it could be a whole machine for that matter) that has an "enable". When the device or machine is "enabled" it will run, and when it is not "enabled" it will not run. It has one input called "enable". So, we could make our enable active high or active low. Lets look at the two possibilities.

    Active High:
    Whenever the input called "enable" is high, the device or machine is enabled and it will run.
    Whenever the input called "enable" is low, the device or macine is NOT enabled and it will NOT run.

    Active Low:
    Whenever the input called "enable" is low, the device or machine is enabled and it will run.
    Whenever the input called "enable" is high, the device or machine is NOT enabled and it will NOT run.

    From this example you can see that with the enable as active high, a device or machine will only run when this input is high. And, with the enable as active low, a device or machine will only run when this input is low.

    You should be able to see in this second example that from a SAFETY standpoint one of these is safer than the other.

    If you set up a machine with an enable that is active low, you could actually cut the wires to your enable switch while the machine was running and it would stay enabled and would keep running!

    On the other hand, if you set up a machine with an enable that is active high, any type of malfunction in the enable circuit such as a loose or broken wire, a dirty switch not making contact, etc., will keep the machine from running.

    You could look at it this way: With an active high enable, if the machine is running and a malfunction causes it to begin to destroy itself, when it rips the wiring for the enable switch out of the control cabinet, it will stop!
    Patrick;
    The Sober Pollock

  20. #20
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    Aug 2003
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    Active High/Low On DataSheets

    When reading data sheets and other info from the semiconductor manufacturors, active low is indicated by a "-" or "bar" above the name of the input or output pin.

    In the example below, three inputs "sleep", "reset", and "Enable" are active low.
    The fact that these are active low is indicated by the straight line or "bar" above their names in the drawing. The other inputs such as "Step", and "Dir" are active high since they have no line or "bar" above their names in the drawing.
    Attached Files Attached Files
    Patrick;
    The Sober Pollock

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