As I know, high speed transceivers always use a 0.1uF or 0.01uF capacitor for AC-coupling.

                end resistor

Capacitor and end resistor are a high-pass filter.

If end resistor is 50 ohm.

I calculate bandwidth with $$ f = \frac{1}{2 \pi R C} $$ if \$C = 0.1\mathrm{\mu F}\$, then \$f = 31.847\mathrm{k Hz}\$

if \$C = 0.01\mathrm{\mu F}\$, then \$f = 318.47\mathrm{k Hz}\$

But I always see 0.1uF and 0.01uF both used in the 1GHz, even 10GHz circuit.

As I calculated above, bandwidth should not enough.

I'm confused.

  • \$\begingroup\$ What kind of transceiver is this? Is the signal a modulated RF carrier, or is it gigabit per second binary? \$\endgroup\$ – The Photon Feb 20 at 4:29
  • \$\begingroup\$ "But I always see 0.1uF and 0.01uF" - because people like nice round numbers like 1 and 10, rather than the optimum value for the job. And manufacturers and distributors like it that way too, because then they don't have to stock a huge range of different values. \$\endgroup\$ – Bruce Abbott Feb 20 at 4:53
  • \$\begingroup\$ @ThePhoton: Gigabit Ethernet or PCI-E \$\endgroup\$ – curlywei Feb 20 at 6:00
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    \$\begingroup\$ Part of your mistake is that you mix "bandwidth" and "frequency". A frequency band runs from one frequency to another, f1 to f2. The bandwidth is f2-f1. For bands that start at DC, the bandwitdh would be f2-0Hz=f2. But you need to make certain what your band boundaries are to calculate a bandwidth. If there's no upper bound (like here), the filter has infinite bandwitdh. \$\endgroup\$ – MSalters Feb 20 at 16:21

It's a HIGH-pass filter, not a low-pass filter. That cutoff frequency you calculated is the LOWEST that can get through, not the highest. Frequencies lower than that are blocked, and frequencies higher than that are passed.

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  • \$\begingroup\$ Hi @DKNguyen : Ok, I'm kown. But if I'd like to used in above gigabit, I think I should select lesser then 0.01uf, because I can get good gain when my circuit work on 1GHz.Why I always see 0.1uF and 0.01uF both used in the 1GHz? I'm confused. \$\endgroup\$ – curlywei Feb 20 at 4:17
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    \$\begingroup\$ @curlywei All things being equal, for higher frequencies you could select a smaller capacitance though you don't have to. But things are never equal in practice because smaller capacitors tend to have lower parasitic inductance and it might be necessary to use a a smaller capacitance to reduce the capacitor's parasitic inductance which would impede your high frequencies. \$\endgroup\$ – DKNguyen Feb 20 at 4:19
  • \$\begingroup\$ Thanks @DKNguyen, now I know. \$\endgroup\$ – curlywei Feb 20 at 4:22
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    \$\begingroup\$ While there are protocols that extend almost down to DC and have GHz edge rates, they are a pain to deal with (SDI springs to mind, you need typically 4.7uF of coupling cap that is good to GHz rates, expensive). Most modern high speed line protocols have DC balance (typically 8B10 or similar coding) so do not have any low frequency content, and as such the low frequency corner is not at all critical. Most designers have many, many 10 or 100nF caps used for decoupling scattered around, and something you already have on the BOM trumps something ideal that is an extra BOM line most of the time. \$\endgroup\$ – Dan Mills Feb 20 at 18:30

A binary data signal isn't a single frequency. If it's truly random, it will have frequency content from near DC all the way up to about half the data rate (i.e. a 1 Gbps signal has content from DC to ~500 MHz).

Using a smaller capacitor value would block some of the low-frequency content of the signal, resulting in excessive wander of the binary signal. For example, consider if the data stream happened to have 16 ones in a row at some point. With a too-low blocking capacitor value, the signal would tend to drift back towards zero during this long run of ones.

To alleviate this, it's possible to use encoding to limit the low-frequency content of the signal. For example, Gigabit Ethernet uses 8b/10b code to limit the maximum run length and ensure that the signal is balanced between ones and zeros. Given a maximum run length of zeros or ones, it's now possible to choose a minimum capacitor value that allows you pass the high frequency content in the signal without allowing excessive wander during the (limited length) runs of similar values.

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RC forms a high pass filter whose cut off frequency should be in the range of minimum frequency you want pass. For example if the minimum frequency for your signal is 200MHz. For this the 3dB cut off frequency of high pass filter should be little below 200MHz to pass it without any attenuation.

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Mohit has the right idea. The series capacitor is used to block DC and pass the higher frequency AC signal. The term "transfer function" is used to describe the ratio of Output to Input in a circuit. In terms of voltage, this would be Vo/Vi and is dependent on the configuration and values of components within a circuit. Here the circuit may be thought of as a simple frequency-dependent voltage divider. The voltage at the transmitter is Vi and Vo is the voltage at the receiver. The transfer function here would be: $$\frac{V_0}{V_i}=\frac{R}{\sqrt[2]{R^2+(1/2\pi f C)^2}}$$

due to a 90 deg phase shift of V & I in a cap.

At f=0, the impedance of the capacitor is infinite, and Vo goes to zero, blocking DC. At very high frequencies the impedance of the capacitor approaches zero and Vo approaches Vi (R/R --> 1). When $$f=\frac{1}{2 \pi RC}$$, the impedance of the cap is equal in magnitude to that of the resistor but because of the 90-degree phase shift, the composite impedance =R*SQRT(2). This frequency is considered to be the 'breakpoint' of the circuit. At this point, the transfer is -3dB as Mohit correctly states.
The dB value = 20*log|Vo/Vi|. That is Log base10. If you were to swap the positions to R & C in this circuit, you would have a low-pass filter with the same breakpoint. If you are interested in this subject I would suggest looking into Laplace transforms and Bode Plots for circuit analysis.

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  • \$\begingroup\$ I can't read you equation. You can use MathJax to format your equations. The line breaks have mangled yours to the point I'm not sure what you meant to write - I was going to just go ahead and fix it, but realized I couldn't figure out what you meant. \$\endgroup\$ – JRE Feb 21 at 7:01

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