First order RC Network

[SneaKSz]

Hello all,

While simulating some audio amplifiers, I was having trouble with finding out the voltage across a Cap.


It's just a simple RC network, f=15kHz (1Vptp). I determinted Xc = 1/sC = 1/(2*Π*f*10n)=1060 Ω.

Well I thought, while R~ Xc, that the voltage drop across Xc would be half the applied voltage. So Vc~ 0.5Vptp.

Someone can explain me where my thinking go's bad?

Kind regards

 

[Jony130]

Simply you forgot about the phase shift in capacitor.
For Fo = 0.16/(RC) when R = Xc the output voltage is equal
Vo = 0.707*Vin
https://en.wikipedia.org/wiki/RC_circuit#Gain_and_phase_angle

 

[SneaKSz]

Hello,

thanks for your reply.

I've filled in the values but I can't get the 0.707.

Gc= 1/(√(1+(2*Π*15000*1000*10*10^-9)))=0.727
Θc= tan^-1(-2*Π*15000*1000*10*10^-9)=)-0.7557
Vc=Vin*Gc*e^jΘc =1*0.727*e^(-0.756)=0.341

kind regards.

 

[Jony130]

Oki lets try this method

[LATEX]
Fo=\frac{1 }{2*\Pi*R*C}} = \frac{1 }{2*\Pi*1K*10nF}} = 15.9154943KHz[/LATEX]

Xc for Fo is equal Xc = 1KΩ

And the total impedance

[LATEX]Z=\sqrt{R^2+Xc^2}} = 1.4142135K\Omega[/LATEX]

So the total current is equal

Itot = Vin / Z = 1V/1.41421356KΩ = 707.106782µA

And now we can find voltage across the capacitor

Vc = Itot * Xc = 0.707106782V = 707.106782mV

For F = 15KHz Xc is equal

Xc = 1.06103295KΩ

And total impedance is equal

Z = 1.45800923KΩ---> Itot = 685.866714µA

and

Vc = 0.727727183V

 

[MrAl]

Hi,

There's a very general method for use when there are resistors and capacitors and/or inductors being driven by AC sources. The procedure is the same as for resistors after transforming all of the reactive elements to their complex impedances first.

For the capacitor, C1 becomes ZC1=1/(s*C1).
For the inductor, L1 becomes ZL1=s*L1.
For the resistor, nothing changes, so ZR1=R1, or simply R1.
Then we could proceed as if we had all resistors.

For this circuit for example, we only have C and R so C becomes ZC=1/(s*C) and R stays as R.
Next we do the analysis based on the topology, and since now all we have is a voltage divider we know that:
Vout=Vin*ZC/(ZC+R)
or in terms of the value for ZC we have:
Vout=Vin*(1/(s*C))/((1/(s*C))+R)

and after we simplify by multiplying top and bottom by s*C we get:
Vout=Vin/(s*R*C+1)

Now we replace s with j*w and we get:
Vout=Vin/(j*w*R*C+1)

and all we have left to do now is find the amplitude of the right side, which involves finding the real and imaginary parts and taking the square root of the sum of the squares of those real and imaginary parts. We end up with:
Vout(Amplitude)=Vin/sqrt((w*R*C)^2+1)

and of course we replace w with 2*pi*frequency to get the final answer.

With values of Vin=1v peak, R=1000, C=10nf, and frequency=15kHz, we end up with:
Vout=1/sqrt(0.09*pi^2+1)=0.72772718 rounded to 8 decimal places.

Note that if we used the true 3db down point for this R and C we would have had to use a frequency of: 15915.494309Hz
and this would have led to Vout=0.707107 approx.

Now if Vin was given in volts peak (as above) then the output is also in volts peak, but if Vin had been given in volts rms, then Vout would also be in volts rms.

 

[SneaKSz]

Thanks a lot for the replies! I've forgot that I had to take Z into account, few years ago that I saw that stuff, but know I can move on.

What was the impedance for a parallel circuit of R and C?

Cheers!

 

[Jony130]

What was the impedance for a parallel circuit of R and C?

R||Zc = (R * 1/sC)/ ( R + 1/sC) = R/(1 + sRC)

Or more classic way

the admittance Y is equal to

[LATEX]Y =\sqrt{(\frac{1}{R})^2+(\frac{1}{Xc})^2}} = \sqrt{G^2 + B^2} [/LATEX]

[LATEX]G = \frac{1}{R}[/LATEX]

[LATEX]B = \frac{1}{Xc}[/LATEX]

And finally

[LATEX]Z = \frac{1}{Y} = \frac{1}{\sqrt{(\frac{1}{R})^2+(\frac{1}{Xc})^2}}} [/LATEX]

 

[SneaKSz]

Thanks for the reply and for all the effort! It's clear now

 

[SneaKSz]

Hi again,

Im having a difficulty now wit the series and parallel circuitry :


the impedance of R10 in parallel with C5 =

Z=1/Y=1/(√((1/R10²)+(1/Xc5²)))=2201ohm
then I just add the series resistance and don't reach the simulated value ( Ztot=2.61K).

Do i need to use another approach to get the impedance of this circuit?

like Ztot =( R10*1/(jXc5))/(R10+(1/jXc5)) + R1 = R10/(jXc5*R10+1)+R1=
(R10+R1*(jXc5*R10+1))/(jXc5*R10+1) and get a form like Z\=a+j.b and then use the Z=√a²+b² ?

That a lot of work!

Could someone tell me which approach I should use?

Kind regards

 

[MrAl]

Hi,

The impedance of the parallel cap and resistor is not a single quantity like 2200 ohms, but is a complex quantity like a+b*j where a is the real part and b is the imaginary part of this complex number. You have to work out this value for your circuit first and get the values right. Try again and see if you can do it.

Other examples of complex quantities:
2.34+5.738*j
1+3*j
1298-0.00234*j
-34.5+7*j
-2.765-0.194*j
Note there is always a 'real' part and an 'imaginary' part.

The parallel combination of a cap and resistor will be a complex quantity like this.

The reactance usually denonted with an upper case X is not the same as the impedance. The reactance is a single number like 2200 but we really should use the impedance instead of the reactance.

Once you have the impedance you can add the other resistor to it because it is in series with that impedance, but you can not add the other resistor value to the reactance.

These kinds of circuit require the handling of complex numbers. If we had several resistors in parallel with several capacitors we would have to compute a complex value for each and every one of these sets before we can start to add them together. Sometimes it gets even more interesting when those complex impedances are also in parallel with each other, where we can not simply add them but must compute them as they are in parallel:
Series:
Zser=Z1+Z2
Parallel:
Zpar=Z1*Z2/(Z1+Z2)

I can assure you this gets easier once you do a few of them and get the hang of it. The only difference is that instead of working with single numbers (as for resistor arrays) we have to work with complex numbers.

 

[SneaKSz]

Hello,

I've got it know I think :

Yt\=1/R10+j/2257.55 = 0.0001+j4.4296*10^-4= 4.5411*10^-4<-77.28°
Zt\=1/Yt\= 1<0° / 4.5411*10^-4<-77.28° = 2202<+77.27°

take R1 into account : Zt = 2202<+77.27° + 10k<0° = 485.24+ j 2148.87 + 10K + 0j = 1485.24+j2148.87 => Z circuit = 2611<-55°

Finally got it !

 

[Jony130]

Here you can read about some trick you can use
https://www.electro-tech-online.com/custompdfs/2011/08/ztheorems.pdf
https://www.electro-tech-online.com/custompdfs/2011/08/bode.pdf

 

[MrAl]

Hello,

I've got it know I think :

Yt\=1/R10+j/2257.55 = 0.0001+j4.4296*10^-4= 4.5411*10^-4<-77.28°
Zt\=1/Yt\= 1<0° / 4.5411*10^-4<-77.28° = 2202<+77.27°

take R1 into account : Zt = 2202<+77.27° + 10k<0° = 485.24+ j 2148.87 + 10K + 0j = 1485.24+j2148.87 => Z circuit = 2611<-55°

Finally got it !

Hi,

Looks like you got it. More exactly 2611.34 at -55.34 degrees.