What is impedance?

hello,
I'm new here and since I can't ask in the Theory sections, I figured I'll ask here.. sorry if that's not ok..

I've heard of this concept "impedance" and the teachers I came across told me that it's similar to resistance, so that wasn't hard to get..
Then when we studied op-amps, I see terms like "input impedance" and "output impedance"..

I read this description which seems to be simple and easy to digest
impedance

but I still don't know what we mean by "input and output impedance" when talking about op-amps (or any other devices).. Anyone cares to share what they think is correct?

thanks,
~dyf

Resistance is at DC, impedance is basically the same thing but for AC instead. For an opmap, often they will be the same thing.

The datasheet for an opamp shows that its output impedance is extremely low at low frequencies because of its very high internal gain and negative feedback. But at high frequencies its output impedance is much higher.

Energy Storing devices present a 'resistance' called Reactance to the voltages they are supplied from.It is because of a variety of factors.The impedance comes in to play when you have got alternating voltage in your ckt.For e.g if you connect a Inductor across a DC supply it will act as a short and a capacitor will act as an open ckt.The combination of resistance and reactance is called impedance.Else go to Wikipedia.

This is a great question.

I spent my first couple of years in college, thinking they were the exact same thing, just the difference was one, was for a.c. and one was for D.C. Both physically and mathematically, they are not, at all; and here in lies the problem you are having, IMHO. It wasn't until I understood the two mathematically, that I would understand the differences in them physically.

a.c. is: Z = R + iX <-- This is the difference, and a huge difference.
-vs-
D.C. is: R

Without getting too complicated, you will need to understand the differences in real, and complex number's systems.

*Impedance has a complex component (reactance) plus a real component ( R).
*Resistance is simply a real component.

To directly answer your question:
When it comes to opamps, the input impedance is directly related to a feedback scenario. When open loop, the impedance is equal to R, a real component, when feedback is employed, you begin to add a complex component, a.k.a., a reactive component to it.

A really general analogy I use is: Resistance is physically constant, does not change, and it is INHERENT to the device. No matter what the input or output is, R is ALWAYS THE SAME.

Impedance: Impedance is GENERATED, it will literally change, depending on the input and output. It is not INHERENT to the physical make up of the device, and is based on frequency.


I would HIGHLY suggest learning simple RC circuits (e.g. tank circuits, and other really simple passive filters) before you get anywhere NEAR opamps and impedance as they will help you to understand impedance.

Hope this helps.

Understanding of 'impedance' is not so easy sometimes. At first you have to know that Impedance is 'resistance' in an AC circuit.
You must know these too:
Resistance of a Resistor = Voltage/Current (It is correct in BOTH DC and AC circuits)
Resistance of a Capacitor(Capacitive Reactance) = 1/(2'PI' F C). where F is the Frequence(please note that Reactance exists just for an ALTERNATING current) .
Resistance of an Inductor (inductive reactance) = 2'pi' F L .
As mentioned before, if you connect a Capacitor across a DC supply it will act as an OPEN circuit and if you connect an Inductor across a DC supply it will act as an SHORT circuit. But what will happen if you connect each of both to the ALTERNATING supply? You know that the frequency is an essential part of an alternating supply (like Mains which their frequency is 50 or 60Hz). So when you connect each of those STORING components (I.e the inductor or the capacitor) to the alternating supply, According to the above formulas, those components act AS an IMAGINARY Resistor,which is called IMPEDANCE.
They actually resist against the voltage or current and you can reduce the voltage or the current of a LOAD by them in an alternating circuit. But the Resistance of those both components which we called it impedance does not follow the OHM's law, but yet they reduce the voltage or current by alternating voltage (frequency).
So for an OP-Amp or a speaker, which both use alternating signals we will have IMPEDANCE phrase too.

I'm sure he's REALLY confused now

Especially the IMAGINARY part.
Impedance is a delusion.

metropolis said:

but I still don't know what we mean by "input and output impedance" when talking about op-amps (or any other devices).. Anyone cares to share what they think is correct?

thanks,
~dyf

Click to expand...

metropolis take for an example the amplifier of your Hi-Fi system (if you got one...)

your amplifier has "high input impedance" and it means that it doesn't need much output power from the previous stage (mp3 player) to do it's job.

low input impedance requires much output current from the device you
have connected to the input of your amplifier. usualy the input impedance of an amplifier is few Kohms.

the output is different than the input because it needs to drive a 4ohm or 8ohm speakers so it must have low impedance to give the required current.

Hey Nigel,
I didn't say him that Impedance can have a DC component yet like PhillDubya did

wow!! those are awesome replies.. thank you each and everyone..

i've been reading through the thread more than once so i don't miss anything and i can see it better now, especially when whiz115 put it in an example.. i'll probably do some practical circuits to have 100% grasp at it..

cheers

Capacitive reactance is 1/(2'PI' F'C)
wizard left off the C. I'm sure he knows better

whiz115, impedance should not be confused with power.
Generally it will be true that a high impedance circuit node will provide less power than a low impedance node but it is not part of the definition of impedance.

Another definition of impedance is anything that gets in the way or hinders someone in the act of dancing.

flat5 said:

whiz115, impedance should not be confused with power.Generally it will be true that a high impedance circuit node will provide less power than a low impedance node but it is not part of the definition of impedance.

Another definition of impedance is anything that gets in the way or hinders someone in the act of dancing.

Click to expand...

Hi flat5!!

probably the way i explained it wasn't very good... if the input impedance for an amplifier
is low then it will require more current from the source device you have connected to the amplifier
thus the output power of the source device must be enough to withstand the requirments!

but if the input impedance of an amplifier is high then everyone is happy!

flat5 said:

Capacitive reactance is 1/(2'PI' F'C)
wizard left off the C. I'm sure he knows better

Click to expand...

Oh, Sorry I forgot to put 'C' and 'L'.

Thanks for reminding

P.s. You are much more clever than me, I am sure.

Hey! You're the wizard. I just have too much time on my hands, so I can Google if something does not look right.

Wiz115 said:
"if the input impedance for an amplifier
is low then it will require more current from the source device you have connected to the amplifier"

Not if the amp will work with less input current.
A low Z output stage may not be able to source more power than a high Z stage. My main point is it should not be in a definition of impedance.

flat5 said:

Not if the amp will work with less input current.
A low Z output stage may not be able to source more power than a high Z stage. My main point is it should not be in a definition of impedance.

Click to expand...

what do you mean with the above?
sorry but i don't totaly understand it... you think i'm saying wrong things to metropolis?

metropolis asked "I still don't know what we mean by "input and output impedance" when talking about op-amps (or any other devices)"



and i gave him an example so he can understand the basics...

impedance Z is the resistance on AC

The input impedance of an amplifier like in my example or for an op-amp is very high (some Mohm for the op-amps), so the input of the op-amp can draw as little current as possible from the source.

knowing the input Z of an amplifier you can know the load you place to your source! connecting low Z output to high Z input it is called bridging and insures good match between the two of them... meaning less stress to the source, low distortion etc.

hi guys...

so i did this lab experiment in school. we're supposed to construct this circuit with R1 and R2 = 1KΩ:

**broken link removed**

and these are the results that i got.. first the input voltage is constant at 1Vdc, we just change the load resistance, then repeat the measurements when input voltage is 10Vdc.

**broken link removed**

I think we're supposed to show that the amplifier works well only with low load resistance (impedance).. that's what i was able to pull out from the results at least. what do you guys think?

PS: i already submitted the lab report, so this is not 'doing my homework'.. I'll ask the instructor about it, but I doubt she'll give me a satisfactory answer, but I'll do it anyway..
I don't want to just pass the courses, i would love to understand how things work because in the field, the design doesn't care how many marks you got, but how well you understand stuff..

Thanks again for the great support!

Your numbers are backwards.
An ordinary opamp has a max output current of about 25mA.
You have 1.021V across a load of 10 ohms which is 102mA. Impossible!
You also have 10.26V across 10 ohms which is 1.026A which is also impossible.
You have 21.56mA in 1000 ohms which is 21.56V. But the supply voltage is only 15V and the max output voltage of most opamps would be only 13V.

Which opamp? Maybe you have a power amp.

audioguru,
i think the minus sign is because i was using the inverting input, so the output has to be inverted. It's an ordinary 741 amplifier.
but it's likely that i screwed up somewhere.. if i get the chance, i'll do them again.. but thanks for your input.. i got something to think about now, heh

0.268V across 10 ohms is 26.8mA.
0.273V across 10 ohms is 27.3mA.
2.407V across 100 ohms is 24.07mA.
4.62V across 200 ohms is 23.1mA.
6.94V across 300 ohms is 23.1mA.
9.41V across 400 ohms is 23.5mA.
So the opamp limits the current to about 24mA.
See how your numbers are backwards?