Are the capacitor values important for a capacitor block DC?

I need a block DC to transform an initial signal of 1.98 Vpp (1 kHz to 100 kHz) with the voltage ranging from 0 to +1.98 V:

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For a signal of 1.98 Vpp (1 kHz to 100 kHz) with the voltage ranging from -0.99 V to +0.99 V:

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What would be the most appropriate value for my case? And why?

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    \$\begingroup\$ Yes, the value is important. The capacitor (along with other components) creates a high pass filter in order to block DC, and you need the corner frequency of the filter to be below the lowest frequency of your signal. You need to show your circuit, though, in order for anyone to help you determine the correct value. \$\endgroup\$ – Null Feb 14 '17 at 20:31

If you want to use a capacitor as a DC-blocking element (i.e., in series with the signal source) you should choose its capacitance value according to:

  • AC signal frequency f;
  • Equivalent Resistance Req seen from "NODE A" (see figure below) to GND.


simulate this circuit – Schematic created using CircuitLab

Why that? As someone else put it already, the role of the capacitor is to implement an high-pass filter, meaning that high frequency components are passed and low-frequency ones (like the DC) are blocked.

Unfortunately, this filtering is not restricted to the DC (0Hz) component, since in most cases you have a load of finite value to drive (resistance Req in the schematic), so also low-frequency AC components will be attenuated by some extent. Remember that the -3dB frequency of an RC filter (high-pass or low-pass) is: $$f_{-3dB}=\frac{1}{2 \pi C R_{eq} }$$. This means that at that frequency your signal gets attenuated by 3dB (and at lower frequencies the attenuation will be even stronger). Therefore, the capacitor value should be chosen large enough in order not to cut off the lowest frequency component of interest (1 kHz in your example). In other words, you must end up with a -3dB frequency sufficiently lower than 1 kHz , so that this frequency is not attenuated too much.

A good way to make sense of this is to consider how a capacitor is built: basically, there are two conductive plates separated by a dielectric. It is straightforward to figure out that a DC current won't never flow through it... Rather, current through a capacitor is possible if the voltage across it changes over time (and that's the case for higher frequency signals). Roughly, the higher the frequency and the capacitance value, the larger the capacitor current, so the smaller the signal blocking capability.

  • \$\begingroup\$ Why implement an High-Pass Filter? A only capacitor as a DC-blocking element is not good? If I understand correctly, the HPF attenuates frequencies which are below the cutoff frequency, and the DC component is 0Hz, ie it will be removed in any high pass filter, regardless of cutoff frequency, right? I want to "Decrease" the value of the sinusoidal voltage, from 0 to 1.98V to -0.99V to + 0.99V. Is this -0.99V the DC component? \$\endgroup\$ – VF35468 Feb 14 '17 at 22:08
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    \$\begingroup\$ Ok got it :) you are right in saying that any capacitor blocks the DC (and that's what I pointed out in my answer). However, you are not going to have problems only if you measure the output WITH NO LOAD (i.e., with infinite resistance----> zero cutoff frequency). However, if you are driving a real load you have a finite resistance which may cause the cutoff frequency to rise..... and you'd not want to kill your 1kHz sinusoid.. \$\endgroup\$ – NotANumber Feb 14 '17 at 22:40
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    \$\begingroup\$ So.... let's say that if you have an infinite load impedance, well... no need to worry about the capacitor value: even a 1pF will work fine without deteriorating your AC signal. However, the presence of a load (was it desired or not) makes things a little bit trickier \$\endgroup\$ – NotANumber Feb 14 '17 at 22:53
  • \$\begingroup\$ Okay, I understand the cutoff frequency, but what exactly is the DC component in my initial sine wave (0 to 1.98V)? The capacitor removes the DC component, ie, "Decreases" 0.99V from the initial amplitude. How exactly does the capacitor select the 0.99V, which is exactly half the amplitude, and transform the signal into 2 parts, positive (+ 0.99V) and negative (-0.99V)? \$\endgroup\$ – VF35468 Feb 14 '17 at 23:26
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    \$\begingroup\$ The DC component is the time average value of a signal ,so 0.99V. Any finite-energy signal can be represented as a sum of several sinusoidal and cosinusoidal oscillations (Fourier series/integral). You start from 0Hz + 1kHz... as I pointed out the capacitor is an OPEN CIRCUIT for DC, so no constant current will flow through the load causing a constant voltage to show up. Therefore, only time varying signals can be observed at the A node. This is due to basic physics, i.e, charge mirroring at the capacitor plates \$\endgroup\$ – NotANumber Feb 15 '17 at 0:01

The DC-blocking capacitor acts as a high-pass filter when combined with a (series) source impedance and (parallel) load impedance. Use these resistances and your frequencies of interest to choose a capacitor value.


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