Impedance in a DC circuit - is it the same as resistence

[amateur_25]

Hi
I've always been taught impedance only exists in an AC circuit. In an AC circuit it describes the resistance of the circuit as well as it's inductance and capacitance. Datasheets/ the more experienced refer to impedance in a dc circuit. Is that just the same as resistance?

[jeremy]

yep.

[VK5RC]

To me it does!
Are the data sheets referring to something that may also carry or block some RF e.g. power supply rails that may need to be uncoupled?

[nowlan]

isnt it reactance?

[kizzap]

Impedance is the total load that the voltage source sees. It includes the "real" part of the load, the resistance, and the "imaginary" part, which is influenced by the capacitance and inductance of the circuit in question.

If resistance and impedance were in a Venn diagram (those intersecting circles) you would find the circle for resistance will be small, but contained wholly inside the circle for impedance.

-kizzap

[w2aew]

Resistance + Reactance = Impedance.

Impedance is used as a overall term, even if the particular circuit in question is purely resistive, purely inductively/capacitively reactive, or a combination of both.

[KD0CAC John]

For me this is self-taught [ or finding others to learn from ] and my perspective is hugely affected by the lack of math skills , so as a result the heavy reliance on conceptually understanding .
Like physics , the closer we look the more details there are , and then your perspective is part of what details you look at , most of my perspective at this time , is RF / ham radio .
I think of impedance as frequency dependent resistance , resistance changes with the frequency .
DC is like one plane , only varying in voltage/current , so only having resistance .
AC adding oscillation / wave and the frequency of that brings in reactance / impedance - I can not find it now but a old Bell labs video with a wave machine demonstrates this very well .
At least to me a single chart showing most all aspects of this subject is a Smith Chart - learning to read one of these seems to give the most info in one view .
The above is not factual , just my understanding at this time and open to calibration & alignment        

[ejeffrey]

Impedance is just the answer to the question "if I apply a voltage, how much current flows"

As a matter of convention, one reason you often see impedance in DC circuits is when the value isn't due to a 'real' resistor.  For instance, a BJT opamp opamp looks like near infinite input impedance, but that is due to feedback, not a physical resistor alone.  The impedance "looking into" the base of a transistor is the emitter resistor times the current gain, and in an opamp that 'emitter resistor' itself is normally a current mirror that uses active transistors to generate a high impedance current source.

So there wouldn't be anything wrong with saying the the input resistance of an opamp is X megaohms, but it is more conventional to call it input impedance.

[MacAttak]

1. There is no such thing as a purely DC circuit. Even an unachievable perfectly regulated circuit still gets turned on/off at some point, which introduces a (long) wave with sharp edges, and therefore harmonics and reflections.
2. Reactance is frequency-dependent, increasing with frequency and dropping to zero at infinite wavelength (aka DC).

In practice though, for most purposes of DC circuit analysis and design, #1 is not consequential. And since reactance approaches zero in a pure DC system, it is customary to just ignore reactance component of impedance in DC analysis - which means in that case Impedance = Resistance.

But you can in fact do all DC network analysis using Impedance instead of ignoring it. Instead of V=IR, you'd be using E=IZ, and instead of simple voltages and resistances you would need to deal with phased reactances and voltages - which means doing math using complex numerical form which IMO is a PITA.

[brabus]

I was about to reply, when I saw that MacAttack perfectly explained what I wanted to say: there is no such thing as a "pure DC circuit". Brilliant!


For sure the comprehension of this concept is not easy at all, and it also depends on the OP's knowledge on control theory.

[villager]

I was about to reply, when I saw that MacAttack perfectly explained what I wanted to say: there is no such thing as a "pure DC circuit". Brilliant!


For sure the comprehension of this concept is not easy at all, and it also depends on the OP's knowledge on control theory.

Would you explain how control theory can contribute to this topic?

[brabus]

Would you explain how control theory can contribute to this topic?

Sure!

Let's put it in this way: any dynamic system has a load (combination of R-L-C 's) that can affect its stability.
This is true for basic generators (just imagine a pure L-C load applied to a pure voltage source), but in particular for traditional feedback-regulated systems, such as OpAmps.

Let's make an example that we can all try in real life: imagine an OpAmp used as a buffer for a DC voltage. As long as the OpAmp is unity-gain stable, we have a nice and quiet DC circuit, right?
Now, what happens when we connect a capacitor to the output? Well, if we think about being in DC and dealing only with output resistance, we expect a nice and clean 1st order rise, right?
Wrong! The OpAmp can be unstable with that capacitive load, simply because the loop phase delay increases, moving the poles of the system: bingo, we were expecting to be in DC, and have achieved instability instead!

So, control theory is required at least to evaluate the stability of the system, and only when the system is stable (once we define the concept of "stability") we can begin to talk about DC. In fact, if we stick to the control theory, we can achieve much more information on our circuit, instead of focusing just on DC values.

I am sure that this brief example is not enough, there is a lot more to say; I just hope my previous statement about control theory is now clarified.

[LvW]

Let's make an example that we can all try in real life: imagine an OpAmp used as a buffer for a DC voltage. As long as the OpAmp is unity-gain stable, we have a nice and quiet DC circuit, right?

I think, also in this case, we have no "nice and quiet" DC circuit. Even if the opamp is unity-gain stable we will observe a kind of overshoot or even ringing - dependent on the actual phase margin.

Now, what happens when we connect a capacitor to the output? Well, if we think about being in DC and dealing only with output resistance, we expect a nice and clean 1st order rise, right?
Wrong! The OpAmp can be unstable with that capacitive load, simply because the loop phase delay increases, moving the poles of the system: bingo, we were expecting to be in DC, and have achieved instability instead!

Isn´t it better to say (instead of "moving the poles of the system") that the capacitive load - in conjunction with the finite opamp output impedance - creates a NEW pole and, thus, can heavily degrade systems stability?

[brabus]

Let's make an example that we can all try in real life: imagine an OpAmp used as a buffer for a DC voltage. As long as the OpAmp is unity-gain stable, we have a nice and quiet DC circuit, right?

I think, also in this case, we have no "nice and quiet" DC circuit. Even if the opamp is unity-gain stable we will observe a kind of overshoot or even ringing - dependent on the actual phase margin.

Yes, we have all the turn-on transitories, which depend on the dynamics of the opamp (in extreme simplification). To keep my previous statement valid, we can suppose that every transitory is finished, and the OpAmp is giving just DC plus noise.

Now, what happens when we connect a capacitor to the output? Well, if we think about being in DC and dealing only with output resistance, we expect a nice and clean 1st order rise, right?
Wrong! The OpAmp can be unstable with that capacitive load, simply because the loop phase delay increases, moving the poles of the system: bingo, we were expecting to be in DC, and have achieved instability instead!

Isn´t it better to say (instead of "moving the poles of the system") that the capacitive load - in conjunction with the finite opamp output impedance - creates a NEW pole and, thus, can heavily degrade systems stability?

Your observation is right, if we ignore the inverting pin input capacitance (plus stray capacitance).
Anyway, this capacitance is tipically very small, and the pole is far from influencing stability... tipically; nevertheless, I remember having issues with this; what a nightmare .

In any case, once again this proves that: there it comes, Control Theory again...


p.s.: bist du von Bremen? 

[LvW]

p.s.: bist du von Bremen? 

Yes - I am. How did you know?

[Thilo78]

Yes - I am. How did you know?

I might think that your avatar gave it away 

[LvW]

I might think that your avatar gave it away 

You think, it is known all over the world?

[Thilo78]

At least across Europe. I'm quite sure of that. 

[EEVblog]

I've always been taught impedance only exists in an AC circuit. In an AC circuit it describes the resistance of the circuit as well as it's inductance and capacitance. Datasheets/ the more experienced refer to impedance in a dc circuit. Is that just the same as resistance?

In a DC circuit, yes, impedance is just the normal resistance, and it's not uncommon to call it that. The reason is that in an AC circuit, you talk in terms of impedance, not resistance, so even a pure resistor in an AC circuit is usually called impedance as well. So it's fairly common to continue to use the term impedance in a DC only circuit as well.

[Thilo78]

Why not put in formulas?

Z = R + jX, where Z is the (complex) Impedance, R the (ohmic) Resistance, and X the Reactance.

X depends on the frequency in the circuit, and consist of capacitive and inductive reactance.

In case you are looking at DC, you can practically neglect X, unless you want to look at transient operation (e.g. switching).

Assuming that, you are left with Z = R or Impedance (@DC) equals (ohmic) resistance.*

Though, a word of advice: If you're looking at switching operations or operation or timers and opamps, you should not easily neglect reactance, as they are vital for the operation and forgetting to take, say, large capacitors in your thought, you might left wondering why nothing happens on your breadboard. 

As Dave says: "A trap for young players".
I fell for that several times 

PS: Please correct me, if I'm wrong somewhere. 
*PPS: This is only half true. With capacitors in line with your DC voltage, Z tends to become quite large. That might effectively block DC voltage and current.

[LvW]

Z = R + jX, where Z is the (complex) Impedance, R the (ohmic) Resistance, and X the Reactance.

Shouldn´t we mention the restriction that this applies to LINEAR conditions only?
For the real part of non-linear I-V relationships we have to distinguish between the static and the dynamic (differential) resistances, don´t we? 

[Thilo78]

Shouldn´t we mention the restriction that this applies to LINEAR conditions only?
For the real part of non-linear I-V relationships we have to distinguish between the static and the dynamic (differential) resistances, don´t we?

Yes, sounds right.
From my power distribution point of view, I usually neglect that part. 

Of course, you could also mention negative ohmic resistance, but I think that would lead to far here