LRC circuit impedance formula.

[arvinfx]

What is the formula to measurement this :

 

[Mike odom]

ZL = (2∏fL)
ZC = 1/(2∏fC)
ZR = R

Zeq = 1 / ((1/ZC) + (1/ZR))

Z = ZL + Zeq

I = V / Z

 

[arvinfx]

Thanks, but what about a triangle curve current ?

 

[MrAl]

ZL = (2∏fL)
ZC = 1/(2∏fC)
ZR = R

Zeq = 1 / ((1/ZC) + (1/ZR))

Z = ZL + Zeq

I = V / Z

Hi there Mike and arvinfx,



Sorry to say Mike but you can not combine reactances like that. The impedances are complex. So this means we need this:

ZL=j*w*L
ZC=1/(j*w*C)
ZR=R
( j is the complex imaginary operator, w=2*pi*f, f the frequency in Hertz )

and then we can put ZC in parallel with R like you did and then add ZL to the result. So we get:

Zpar=1/(1/ZC+1/ZR)=1/(j*w*C+1/R)=R/(j*w*C*R+1)

Then:

Z=ZL+Zpar=j*w*L+R/(j*w*C*R+1)=(j^2*w^2*C*L*R+R+j*w*L)/(j*w*C*R+1)=(R-w^2*C*L*R+j*w*L)/(j*w*C*R+1)

which simplified a little comes out to:
Z=(R+j*w^3*C^2*L*R^2-j*w*C*R^2+j*w*L)/(w^2*C^2*R^2+1)

That's the impedance of the circuit and we see it is complex, and then as you said the current is:
I=E/Z


arvinfx:
The impedance is shown above, but if you want the response to a triangle wave that is a time domain problem. Is that what you want?

 

[Mike odom]

you did the same thing I did, except replacing 2∏f with jw...

 

[Ratchit]

arvinfx,

What is the formula to measurement this :

Test instruments are used to measure, formulas are used to calculate. What are you trying to do? Calculate impedance, current, voltage, what?

Ratch

 

[arvinfx]

Hi there Mike and arvinfx,


Z=(R+j*w^3*C^2*L*R^2-j*w*C*R^2+j*w*L)/(w^2*C^2*R^2+1)

So, could you put value and find Z , please?

arvinfx:
The impedance is shown above, but if you want the response to a triangle wave that is a time domain problem. Is that what you want?

I want to calculate a sine wave inverter output impedance, which is about circuit , But the current wave is triangle not sine wave...

 

[MrAl]

Hi,

Well for example with R=2, C=10uf,L=1mH, and f=50Hz, we get:
Z=1.999921+0.3015934*j
which as i mentioned earlier this is a complex impedance.

But if you want to figure out the output with a current that is triangular, then we have to do a time domain analysis. But you also have to specify where your output is taken from...probably across the capacitor right?

Also, it might be better if you specify what the voltage is rather than the current. The voltage is probably pulsed right and so that makes it a little easier too. The current isnt really triangular in that case. It's close to a triangle, but not exactly.

So you next have to specify if you would rather analyze the voltage across the cap (the output presumably) with a pulsed input voltage, or you have to specify some other type of analysis but try to be specific so that we dont have to keep asking you more questions about it. The more specific you are the faster we can get you an answer

Also it might help if you specify some typical values for R, L, and C, and the voltage pulse amplitude.

 

[MrAl]

you did the same thing I did, except replacing 2∏f with jw...

Hi Mike,


Yes that's pretty much it. But the impedance has to be represented with the complex form (including the 'j') otherwise we cant get the right result.

Im waiting for the reply from the OP so we know what kind of excitation he wants to use.

 

[arvinfx]

This is full schematic:



And this is output voltages from this filter (before is blue, after is red):



L=0.05h
C=.000001F
R=12Ω
V = 310V
And R is the load which could be inductive or capacitive but i put only a resistor ...

 

[MrAl]

Hi,

So what kind of solution are you looking for here?

 

[arvinfx]

Look mral, I have to calculate the pick of current on switch and select a suitable one for this h-bridge, and all the questions were about Ipk.

 

[MrAl]

Hi,

Ok then maybe you were on the right track in the first place. Using the impedance you can calculate the peak AC current and then use that to estimate the max current.

I'll be back in a few minutes with more details...

LATER:
Since your output load is 12 ohms and the filter capacitance is so low in value (1uf) at 50 or 60Hz the resistive load dominates quite a bit, so the current through the resistor is 120/12=10 amps and the current through the cap is very little assuming 50 or 60Hz operation. That means the current through the inductor is also 10 amps, so the peak is about 14.1 amps. You should add a little more for safety anyway.
Be advised that with an inductor that large 310vdc may not be enough or might just barely supply enough current to drive the output voltage to 120v even at 50Hz. 315vdc would probably be the minimum requirement. Lowing the inductance relaxes this requirement and then you dont need so much overhead voltage.