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I found this picture of a tri-state buffer:

enter image description here

When the enable signal is 0, the output is disconnected from the voltage sources. But why is this called high-impedance? From what I understand impedance is the same as resistance when we look at alternating current? So why is it called high impedance?

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    \$\begingroup\$ Because there is only a high-impedance path from your output to anything else. That's all, that's literally the meaning :) Sometimes things are easy! \$\endgroup\$ Commented Nov 11, 2023 at 16:29
  • \$\begingroup\$ From what I understand impedance is the same as resistance when we look at alternating current no, not really, but for the problem at hand that definition is good enough! \$\endgroup\$ Commented Nov 11, 2023 at 16:30
  • \$\begingroup\$ user394334 - Hi, In order to comply with the Stack Exchange rule for referencing, please edit the question & add a link to the webpage / PDF / video etc. which was the original source for that image (and please remember it's your responsibility to do that in future). Thanks. \$\endgroup\$
    – SamGibson
    Commented Nov 11, 2023 at 16:43
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    \$\begingroup\$ @Circuitfantasist Isn't it so that when we have high impedance it is neither connected to low or high? \$\endgroup\$
    – user394334
    Commented Nov 12, 2023 at 13:51
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    \$\begingroup\$ @Circuitfantasist I am not sure, can you please tell me? \$\endgroup\$
    – user394334
    Commented Nov 12, 2023 at 13:56

4 Answers 4

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Resistance is the real part of impedance.

When the output is disabled, it draws very little current if something else tries to change the voltage. It doesn't significantly load any circuits connected to it.

It will have a small capacitance to ground, which will demand a charging current from a driver trying to change the output voltage quickly.

Perhaps we say it's high impedance rather than high resistance to acknowledge that there is this capacitance, and to assert that it's small enough not to be significant.

From what I understand impedance is the same as resistance when we look at alternating current?

No, not 'the same'. Equations that are true for resistance with DC circuits can be generalised to AC circuits by using impedance instead. Resistance is a real number, always imposes a zero angle between voltage across and current through a component with pure resistance. Impedance is complex number, with typically a phase shift between voltage across and current through a component.

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  • \$\begingroup\$ First, +1. Next, this might be off-topic but, just wanted to ask. Impedance is complex number, with typically a phase shift between voltage across and current through a component. yes, but when it comes to calculating the RMS current through a combination of resistive and reactive components (e.g. series-connected R-C but can be anything else), we take the magnitude of the impedance i.e. \$|Z|=\sqrt{R² + X²}\$. Now what is this called? I'm sure it's neither resistance nor reactance despite its unit being Ohms. If it has a name I'd like to know. \$\endgroup\$ Commented Nov 11, 2023 at 18:03
  • \$\begingroup\$ @RohatKılıç If we are being fussy, then it's the 'magnitude of the impedance'. However, in context, when used informally, just 'impedance' will often do, especially if it's being used with scalar voltages and currents, rather than complex, vector or phasor voltages and currents. \$\endgroup\$
    – Neil_UK
    Commented Nov 11, 2023 at 19:30
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When one of the output drivers is on, we can say there is pretty much short circuit from output to either supply or ground, so in real life, a very low impedance anyway.

When both of the output drivers are off, the output is basically connected nowhere, so impedance is infinite, or pretty high anyway.

And impedance is used in place of resistance, because it covers resistive, capacitive and inductive cases.

So high impedance might mean 10 megaohms at DC, but as there is at least a couple of picofarads of stray capacitance anyway, it is called impedance.

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From what I understand impedance is the same as resistance when we look at alternating current?

That is not correct.

The impedance of an ideal resistor equals the resistance value but, an impedance at AC may also include the reactance of ideal capacitors and ideal inductors but, both these components have no dissipative resistive elements.

So why is it called high impedance?

It's more accurate to state high-impedance because it covers all types of reactance and resistive components. Thus it implies it is high-impedance at both DC and, at high operating frequencies.

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The problem

Isn't it so that when we have high impedance it is neither connected to low or high?

Yes, but how do we know that it is in this state and not in LOW? An interesting problem is how to detect the third (high-impedance) state. Let's see what the problem is by the help of a conceptual circuit of CMOS stage implemented by two mechanical switches ("PMOS" and "NMOS") in series. The output voltage Vout is controlled by a voltmeter.

Experiments

No pull-up resistor connected

LOW: When NMOS is closed, the output is connected to ground, and the output voltage is zero.

schematic

simulate this circuit – Schematic created using CircuitLab

HI: But when both switches are open, the output is "floating", and the output voltage is also zero.

schematic

simulate this circuit

The problem is how to distinguish between the two "zeros".

HIGH: There is no such a problem with this state. There is only one case when the output is connected to Vdd, and the output voltage is 10 V.

schematic

simulate this circuit

Pull-up resistor connected

Since in the HI state the output is "floating", we decide to connect a "pull-up" resistor R between the output and Vdd. Let's see if this will work.

LOW: When NMOS is closed, the output is connected to ground, and the output voltage is zero although there is current flowing. Figuratively speaking, there is an unequal "tug if war" between the NMOS switch with zero resistance and the resistor R with 1 M resistance. Of course, the resistor wins.

schematic

simulate this circuit

HI: When both switches are open, the output is "floating", and the resistor "pulls" it up to 10 V.

schematic

simulate this circuit

HIGH: The problem is that when PMOS is closed, the output voltage is also 10 V, and we cannot distinguish between this and previous state.

schematic

simulate this circuit

"Pull-to-middle" resistor connected

So, we conclude, the resistor has to "pull" the output to some different voltage, best to the middle. Following this "wisdom", we produce this voltage of Vdd/2 by a voltage divider R1-R2, and connect the resistor to its middle point. If the resistances are relatively high (as in the schematic below), there is no need of a resistor; the R1||R2 Thevenin's resistance will serve as the resistor R.

LOW: NMOS shunts R2; Vout = 0.

schematic

simulate this circuit

HI: The "CMOS" output is "floating", and it does not affect the voltage divider; so Vout = Vdd/2 = 5 V. So we understand that the output is in a high-impedance state.

schematic

simulate this circuit

HIGH: PMOS shunts R1; Vout = 10 V.

schematic

simulate this circuit

Analogical example

Here is an analogical fancy situation from life. Imagine you enter an unfamiliar house with a voltmeter in hand, and you see a wire coming out of the wall. You want to know if it is dangerous and you measure its voltage referenced to the neutral wire. The voltmeter shows 0 V. What does this mean - that the wire is connected to the neutral wire or that it is not connected ('floating')?

In order to understand, we can, figuratively speaking, "pull" the wire with some resistive element (resistor, lamp, whatever...), i.e. connect the wire to a line wire through it.

Another solution is to connect an imperfect voltmeter between the wire in question and the live wire. Thus it will act both as a "pull" resistor and voltmeter.

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    \$\begingroup\$ "the output is floating, and the output voltage is also zero" is misleading. The output voltage is unknown \$\endgroup\$
    – Ben Voigt
    Commented Nov 14, 2023 at 17:12
  • \$\begingroup\$ @Ben Voigt, I understand what you are trying to say but let's not make things so complicated. This is a conceptual circuit and is assumed to work under ideal conditions (no leaks). It is normal to imagine that if we connect a voltmeter to a piece of wire not connected to anything, it will show 0 V. But just in case, I have reduced the internal resistance of the voltmeter in this schematic you are talking about to 10 kΩ :-) \$\endgroup\$ Commented Nov 14, 2023 at 19:34

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