Complicated impedance and pure resistance of RC poles

Thread Starter

Mozee

Joined Jul 23, 2016
85
Hello there,
This is not a homework, but while reading Floyd's book I can across something similar and decided to solve it! I just wanted to calculate the total impedance that the op-amp will see when connecting these 3 poles through a resistor of 6.2K to the non-inverting input to form an RC phase shift oscillator, also, I wanted to figure out the value of Rin to find the Rf required for the op-amp so that I can get an Av of 29 or more!
In this RC phase shift oscillator, I'm trying to calculate the total impedance of the 3 RC pairs!
Now the C3 and R3 have a Z3 of sqrt(Xc3^2 +R3^2) then we will have a parallel R2 with this Z3, can I directly add R2//Z3? as in R2//Z3 = Z2
Z2= 1/(1/R2)+(1/Z3) ????
if that was correct, we will face an Xc of C2 in series with the resulting Z2, can we add Xc2 with Z2 in series
Finally, should the Rin2 be included in the calculation ar all?

Thank you.


IMG_1175.JPG
 

MrAl

Joined Jun 17, 2014
6,469
Hello,

If your question is can we compute two impedances in parallel with:
1/Ztotal=1/Z1+1/Z2

and thus Ztotal is the inverse of that, then the answer is YES.
Of course when dealing with complex numbers you have to be able to find the inverse of a complex number which is not really that hard to do.

There are also other tricks you can use when you have networks with repeated sub circuit connections like that one. Thevenin and Norton theorems make it simpler because you can actually work from left to right, incrementally including the very next component in the circuit, and end up with a final value on the right.
You might also note that in this circuit your resistor Rin2 is in parallel with the last resistor in the RC network.

You might also want to compute the optimum gain if you intend to get a sine wave output, and provide some clipping to keep it stable.
 
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Thread Starter

Mozee

Joined Jul 23, 2016
85
Thank you, my friend, the thing is I don't know how to calculate the impedance trying to find Z2,
Z2 contains the previous branche's impedance plus the resistor R2, so I have an impedance of Xc3+R3 parallel with R2,
How can I find that out?
do they add normally or what is the exact process?
 

Marc Sugrue

Joined Jan 19, 2018
60
If you simplify the schematic showing the impedances it may help. Basically the network Z3 is in parallel with R2 so as Mr Al says you can calculate the parallel networks working backwards
 

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The Electrician

Joined Oct 9, 2007
2,737
What is the Floyd book you refer to?

If you want to derive the impedance the opamp output sees as a symbolic expression in the various Rs and Cs, it will be very complicated. It will be a good example of what Professor Middlebrook called a "High Entropy expression": https://pdfs.semanticscholar.org/7696/20ac16aba2720f4709e6a9ad8970ccd6aca7.pdf

If you make all the Rs and Cs the same rather than allowing separate values as you show (R1, R2, R3, C1, C2, C3), the expression becomes much simpler.

Here's the expression for the circuit you show:

PhaseNetworkZin.png

Of course, if you substitute numerical values for the components it becomes a single complex number.
 

MrAl

Joined Jun 17, 2014
6,469
Thank you, my friend, the thing is I don't know how to calculate the impedance trying to find Z2,
Z2 contains the previous branche's impedance plus the resistor R2, so I have an impedance of Xc3+R3 parallel with R2,
How can I find that out?
do they add normally or what is the exact process?

Hello again,

Well if you start with Rin (shorter notation for Rin2) then you put that in parallel with R3:
Rin||R3

and now you have one resistor not two, which we can call Rp.
Now Rp is in series with C3 which has impedance Z3, so we add that:
Rp+Z3

and we can call this RpZ3.
Now this in turn is in parallel with C2 which has impedance Z2, so we have:
RpZ3||Z2

which we can call RpZ3Z2.

Now that is in series with C1, which has impedance Z1, so we add again:
RpZ3Z2+Z1

So you see how this works?

However, these networks for this purpose are rarely found to have different values for C1, C2, C3 and R1, R2, R3, so we almost always see:
R1=R, R2=R, R3=R
and:
C1=C, C2=C, C3=C

and that makes it a lot simpler in the final expression.

See attachment for the total Z.

Impedance_20180921_065449.gif
 

Thread Starter

Mozee

Joined Jul 23, 2016
85
Definitely, That all have the same values.
Great, But still I am missing something, I need to find the Phase angle as well.
all C=10nF and all R=6.2KΩ, With that being said,
See, C3 & R3 produces Z3 which is calculated as Z3=√(XC3^2 + R3^2) ... the phase angle will be ϑ=tan-1 Xc/R = 83 degrees

Now how do I add this Z3 which is in parallel with R2? Do I consider R2 to be an Impedance by itself and Z(R2//Z3)= the product over sum of that argument, and keeping the phase angle as it is 83 degrees?

I do know how to analyze a parallel RC and series RC circuits when BOTH reactive and resistive components are available. But here, I have a pure resistance element and therefore I have no sufficient knowledge of how to deal with this situation.

Similarly, After finding the Z2 which is the total of (XC3 + R3) // R2, This resulting Z2 will be in SERIES with the XC2, Can I add XC2 directly to Z2 and say Z1=Z2+XC2 without considering phasors?

What I am missing is the rules of adding Z to R to Xc etc .. Just Explain this to STEP BY STEP me through this example and I will be the happiest person alive :)

BTW I am teaching myself electronics through Floyd and Buchla's book of the Fundamentals of Electronics Engineering 8th ed.
 

Thread Starter

Mozee

Joined Jul 23, 2016
85
Similarly, After finding the Z2 which is the total of (XC3 + R3) // R2, This resulting Z2 will be in SERIES with the XC2, Can I add XC2 directly to Z2 and say Z1=Z2+XC2 without considering phasors?

Forgot to say, Again here the problem is that I have a branch which will have a pure reactive component XC2 in series with an Impedance of Z2
how do I find the total resultant impedance, Do I say Ztot= Z2+ XC2 ? or Ztot=√(Z2^2 + XC2^2) or what? also, how will I find the Phase angle?
Do I say θ=tan-1 (XC2/Z2). ?????
 

MrAl

Joined Jun 17, 2014
6,469
Hello,

Do you understand complex numbers?

A pure resistance is represented with a real part and zero imaginary part such as:
10+0*j

and that would be a 10 Ohm resistor.

So to add two elements, no matter what they are, you just add the real parts and imaginary parts separately
For an example when adding, say we have two complex impedances:
0-1*j
10+0*j

To add these two, first add the two real parts 0 and 10:
0+10=10

and that is the resulting total real part.

Next add the two imaginary parts -1 and 0, and that gives us:
-1+0=-1

and so the resulting imaginary part is -1, so the total result is:
10-1*j

and that is the sum of the two.


Now since capacitors have impedance -j/(w*C), the impedance of a pure cap is:
0-j/(w*C)

To keep this simple, let's say we have a cap value of 2 farads and w=5. The impedance is then:
0-j/10

When added to our pure resistance from above, we get:
10-j/10

as the sum of the two impedances.
Again all we did was first add the two real parts, then next added the two imaginary parts, and keep the real part separate from the imaginary part, and then formed the result as:
realpart+j*imagpart.

If the imagpart came out negative it would be something like above where we ended up with 10-j/10.

Now in parallel, for two general impedances:
a+b*j, c+d*j,

we get total impedance:
((b*j+a)*(d*j+c))/(d*j+b*j+c+a)

and to deal with that a little better you can multiply the top and bottom by the complex conjugate of the denominator.
The result would be:
((b*j+a)*(-d*j-b*j+c+a)*(d*j+c))/(d^2+2*b*d+c^2+2*a*c+b^2+a^2)

Example of a cap in parallel with a resistor:
cap: 0-j/10
res: 10+0*j

two in parallel is computed with:
Ztotal=Z1*Z2/(Z1+Z2)

so we get:
(-j/10)*10/(10+(-j/10))

which is simplified a little as:
-j/(10-j/10)

and simplified more:
(10*j)/(j-100)

the complex conjugate is -100-j so multiply top and bottom by that.
For the denominator we get:
(j-100)*(-j-100)

and for the numerator we get:
10*j*(-j-100)

Simplify the numerator (note j^2=-1):
10-1000*j

Simplify the denominator (note j^2=-1):
10001

so the final result is:
(10-1000*j)/10001

and now expand:
10/10001-j*1000/10001

and that is the final result, real part 10/10001 and imag part -1000/10001 and that is usually as far as we go until we use that in another calculation, which would be a series combination which requires another addition.
 
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