As far as I learned, the rule of thumb for the ratio of output impdance to load(input impedance)is 1 to 10. Thevenin resistance(output resistance) Rth should be very small compred to the load resistance Rload. Rth/Rload must be small in order to avoid “loading the circuit”. A maximum ratio of 1/10 is often used as a design rule of thumb.

I was trying to understand how this would result in RC circuits. I came across to the following link which is part of a text book where it explains rule of thumb in RC case (see page 19):


Basically the text says:

Since impedance of an RC circuit is:

Z = Xc // R = [R * (-j/(w * C))] / [R + (-j/(w * C))]

we have a frequency dependent impedance here and we should apply the worst case scenario.

To make my question clear I add the following circuit:

enter image description here

In above circuit filter A is driving filter B. Considering the circuit, here is what I understand from his arguments on input and output impedance:

1-) We want ideally Zout (in filter A) to be as low as possible. That means in worst case Zout will be high which happens when frequency is very low. And he approximates filter A to R here.

2-) We want ideally Zin to be as high as possible. That means in worst case Zin will be very low which happens when frequency is very high. This approximates Filter B to R. I can see that from the Z = Xc // R relation. I can picture this in my mind when I think with "phasor representation". If the frequency goes too high, Xc goes to zero and Z becomes R. This looks more straightforward.

My question is about the first argument above. I will repeat again:

Lets consider Filter A again in above circuit. We want ideally Zout to be as low as possible because we don't want loading. That means in worst case Zout will be too high which happens when frequency is very low.

But when I think with phasor representation. If the frequency is set to too low, Xc goes to infinity and Zout doesn't become R.

Where am I wrong here?


Here is how I derive mathematically so far(let me know if agree):

lim[f-->0]: Zout = ?

lim[f-->0]: Xc --> infinity

lim[f-->0]: (-j * R * Xc) / (R - j*Xc) = (-j * R * infinity) / (R - j * infinity)

|Zout| = (R*infinity) / sqrt(R^2 + infinity^2)

|Zout| = (R*infinity) / infinity

|Zout| = R

  • \$\begingroup\$ I don't think you're missing anything. You're just discovering why active filters are popular. \$\endgroup\$
    – The Photon
    Jun 17, 2016 at 2:34
  • \$\begingroup\$ I'm sorry but don't understand how he approximates to R for the output impedance Zout. He takes worst case scenario when freq. is too low(to make Zout high as possible). If I think the impedance equivalent of this RC circuit it is: sqrt (Xc^2 + R^2) and since Xc goes to infinity with low frequencies , Zout doesnt become R. How do you explain this part? \$\endgroup\$
    – user16307
    Jun 17, 2016 at 9:02
  • \$\begingroup\$ i might have understand this. i tried to derive it. see my edit let me know if you agree. \$\endgroup\$
    – user16307
    Jun 17, 2016 at 9:51

2 Answers 2


We can't achieve the 'ideal'. But if Zin is 10x Zout of the previous stage then the loading, attenuation, and interaction between stages will be minimal.

If the frequency is set to too low, Xc goes to infinity and Zout doesn't become R

Zout becomes R plus the source impedance. Provided we stick to the 1:10 rule (Rb = 10 x Ra) we are OK. The only problem is that each stage multiplies the impedance by 10, so 3 stages is about the limit for a practical circuit.

However, this only applies if you need low attenuation and minimal interaction between stages. If you don't mind a bit of loss and roll-off in the passband then a lower ratio may be acceptable.

  • \$\begingroup\$ Sorry but I dont understand how it approximates to R. He tajes worst case scenario when freq. is too low. If I think the Thevenin equivalent of this RC since Xc goes to infinity Zout doesnt become R. How do you explain this part? \$\endgroup\$
    – user16307
    Jun 17, 2016 at 7:43
  • \$\begingroup\$ I attempted to derive it mathematically myself. See my edit. If right it means I got what he means. \$\endgroup\$
    – user16307
    Jun 17, 2016 at 9:49
  • \$\begingroup\$ Suppose the generator's output impedance (resistance at DC) is 100 Ohms, Rout is 1k and Rin is 10k. The total resistance looking into the output of filter B is 10k + 1k + 100 Ohms = 11.1k. So it's not R(in), but close to it. \$\endgroup\$ Jun 17, 2016 at 11:21
  • \$\begingroup\$ I'm not talking about generator's output impedance effect. Please take it as zero. My question is about how Xc at low frequencies yields a Zout as R. Can we show how at low frequencies itapproximates to R? Basically Im asking how can we write an equation for a low pass filter's output impedance so we can see how it changes wrt frequency. \$\endgroup\$
    – user16307
    Jun 17, 2016 at 11:38
  • 1
    \$\begingroup\$ @user16307 As you stated, Xc tends to infinity as frequency tends to zero. When you have a large impedance in parallel with a small impedance, the result is approximately equal to the small impedance. \$\endgroup\$
    – user57709
    Jun 17, 2016 at 13:53

Z = Xc // R = [R * (-j/(w * C))] / [R + (-j/(w * C))]

divide numerator and denominator by 'w'

Z = [R * (-j/(w * C))/w] / [R + (-j/(w * C))/w]

Z = [R * (-j/(C))] / [R/w + (-j/(C))]

at very high freq, 'w' -> infinity, R/w term vanishes

Z = [R * (-j/(C))] / [(-j/(C))]

Z = R (worst case at high frequencies)

At very low freq, the source impedance is R//Xc and Xc is very large (tends to infinity) the result

Zout = R*Xc/(R+Xc)

Divide numerator and denominator by Xc,

Zout = R/(R/Xc+1)

R/Xc term can be dropped as Xc is very large,

Zout = R (low freq worst case)


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