# Calculating impedance using L, C, R measurements?

Discussion in 'Off Topic [BG]' started by bongomania, Oct 21, 2010.

1. ### bongomaniaGold Supporting MemberCommercial User

Oct 17, 2005
PDX, OR
owner, OVNIFX and OVNILabs
Google has only turned up answers that are a bit above my education level. For example: http://www.learnabout-electronics.org/ac_theory/impedance71.php I just don't get it.

Can one of you explain in "educated layman" terms how I may calculate input or output impedance using the sort of data that can be gotten with an LCR meter?

2. ### line6man

(R[SUP]2[/SUP]+X[SUP]2[/SUP])[SUP]0.5[/SUP]

3. ### bongomaniaGold Supporting MemberCommercial User

Oct 17, 2005
PDX, OR
owner, OVNIFX and OVNILabs
And where do I get X from?

4. ### line6man

Inductive reactance is 2*pi*f*L, capacitive reactance is 1/(2*pi*f*C)

5. ### bongomaniaGold Supporting MemberCommercial User

Oct 17, 2005
PDX, OR
owner, OVNIFX and OVNILabs
And f?

6. ### line6man

The frequency.

Weren't you working on an EE degree?

7. ### bongomaniaGold Supporting MemberCommercial User

Oct 17, 2005
PDX, OR
owner, OVNIFX and OVNILabs
Was, yes, but I've had to take a couple of years off, and did not get to this specific subject yet. I plainly said this was above the level I've learned.

F does not always stand for frequency.

8. ### UncleFluffy

Mar 8, 2009
California
Impedance is a complex number - the real part is resistance, and the imaginary part is reactance.

Resistance is the bit that doesn't change with frequency. Reactance is the bit that does change with frequency.

Stealing a picture from Wikipedia:

R is resistance
X is reactance
Z~ is impedance
|Z~| is magnitude (that's the bit that gets measured in Ohms)
theta is phase (which turns up as the phase shift between voltage and current)

Given that reactance is frequency dependent, any circuit that contains inductance or capacitance will have a frequency dependent impedance.

The Wikipedia article is pretty good.

To get back to your original question, you're going to have to measure impedance at DC and at least two different AC frequencies before you can figure out what R, L, and C are. Once you have that, it's just solving 3 equations in 3 unknowns. More data points are useful as a sanity check.

Yep.

10. ### bongomaniaGold Supporting MemberCommercial User

Oct 17, 2005
PDX, OR
owner, OVNIFX and OVNILabs
Bear with me here, I'm not following you too well. How exactly do you mean "measure impedance"? It's really not obvious. Also, I have two LRC meters, but neither of them offers DC as a frequency option. One has a range of 100 Hz to 2 KHz, and the other is fixed at (IIRC) 20 KHz.

11. ### line6man

There is no "DC impedance" or "DC frequency."
Direct current flows in only one direction, so it's 0Hz.
At 0Hz, there can be no reactance (X=0 ohms), therefore, DC impedance is only a resistance.

I think he was working backwards, trying to calculate LCR based on impedance, but you are doing the opposite, calculating impedance based on LCR.

Do you know your LCR yet?

12. ### bongomaniaGold Supporting MemberCommercial User

Oct 17, 2005
PDX, OR
owner, OVNIFX and OVNILabs
I only just got them, and the instruction manuals are no more helpful in this particular regard than the pages turned up by Google. IOW they have a ton of equations, but no explanation of how to proceed through the various tests toward an end result. And I don't yet have the ingrained knowledge that would allow me to take those leaps with no guidance.

I know that there's no "DC frequency", but you said I would have to measure impedance at DC plus various AC frequencies, so I wondered if you meant there would be a DC setting. I get it now that what you meant was "measure the resistance", but that was not obvious before.

13. ### UncleFluffy

Mar 8, 2009
California
Sorry about that, and thanks line6man for clarification.

Please also remember that if you hook up a mystery black box to most LCR meters the amount of L will affect the C reading and vice-versa. Generally they're built assuming that only one of the two is significant.

If you want accuracy (and a better understanding of "what's going on") you're best off with a signal generator, a dual-channel scope, and a precision low-value resistor.

What kind of device are you measuring the impedance of, anyway? That may imply some possible short-cuts or additional complications.

14. ### bongomaniaGold Supporting MemberCommercial User

Oct 17, 2005
PDX, OR
owner, OVNIFX and OVNILabs
Well, I have the signal generator and the dual-channel scope. I'll pick up the precision resistor on my next Mouser order.

What I'm trying to measure is the output impedance of my audio converter, a Presonus Firestudio Mobile. The stock/nominal measurement is given by Presonus, but I specifically intend to use different methods to raise the output impedance, so I need to be able to measure the effectiveness (and actual results) of those attempts.

BTW, thanks to both of you for trying to help me out here!

15. ### karrot-xBanned

Feb 21, 2004
Omicron Persei 8

Bear's with you but he's losing patience.

16. ### bongomaniaGold Supporting MemberCommercial User

Oct 17, 2005
PDX, OR
owner, OVNIFX and OVNILabs
Funny, I usually don't make that kind of homonym mistake!

17. ### line6man

If you have a bear with you there, you've got bigger things to be worried about than calculating impedances.

18. ### UncleFluffy

Mar 8, 2009
California
1% is more than enough precision - no point having a resistor more accurate than the rest of the measuring system.

In that case, I wouldn't worry too much about breaking down the impedance into reactive and resistive components. Assume that they designed it to have a fairly flat impedance across the audio range and only worry about the messy stuff if that assumption proves wrong.

You've effectively got a perfect voltage source in series with the output impedance.

i.e.

Vo - "real" output voltage of your audio converter (into an open circuit)
Zo - output impedance (at a particular frequency)

so if you put these in series with resistor R1 you get:

V1 = Vo * R1 / (R1+Zo)

and then use a different resistor, R2

V2 = Vo * R2 / (R2+Zo)

two equations, two unknowns, so you can solve for Zo. (Best to do R3 as well as a sanity check).

If you do this for a bunch of different frequencies you can check how flat the output impedance is. For something like that I'd expect a fairly flat curve, but you never know...

Pick R1 roughly equal to the claimed output impedance and R2 maybe twice as much (and R3 maybe half as much).

You're very welcome.

19. ### PassinwindI know nothing.Commercial User

Dec 3, 2003
Columbia River Gorge, WA.
Owner/Designer &Toaster Tech Passinwind Electronics
20. ### PhalexSemper GumbySupporting Member

Oct 3, 2006
G.R. MI
I remember something about ELI the ICE man, but things get a little sketchy....