Active High/Active Low

[Sanghera]

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.

[ToyMaker]

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

[Sanghera]

OK I see. Thanks for the help.
I really appreciate it.

[Sanghera]

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.

[arvidb]

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

[Sanghera]

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.

[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.

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

[OCNC]

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

[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....."

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.

[OCNC]

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

[sbrpollock]

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.

[arvidb]

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