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I'm trying to study for my intro circuits class final and always found the idea of impedance and filters confusing.

Take these two circuits:

enter image description here

enter image description here

The first confusion is where are we even measuring the voltage across from? Voltage across the capacitor? Across the resistor? Is there some kind of convention we normally measure on?

Finding the impedances we get:

\$ Z_C = \frac{ \frac{1}{jwC} }{R + \frac{1}{jwC}} = \frac{1}{RjwC + 1}\$

\$ Z_R = \frac{ R }{R + \frac{1}{jwC}} = \frac{RjwC}{RjwC + 1}\$

(where \$ w \$ is frequency and \$ j = \sqrt{-1} \$)

The impedance \$ Z_C \$ increases as we decrease frequency, so the voltage drop across the capacitor decreases when frequency is low. Doesn't that mean that the capacitor is letting through all the low frequency signals... Why is the lower picture showing us that all the high frequency signals are going through the capacitor?

So shouldn't the capacitor be a high-frequency blocking element? Why is it known to be a DC blocking element then?

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4 Answers 4

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The first confusion is where are we even measuring the voltage across from? Voltage across the capacitor? Across the resistor? Is there some kind of convention we normally measure on?

We measure the voltage with respect to the ground symbol, unless the explicit phrase 'voltage across the component' is used. The diagram falls short by not showing a ground line running continuously from the input terminals to the output terminals.

There is another confusion though, those arrows are intended to show the current passing through the components, not the voltage across them.

The impedance ZC increases as we decrease frequency, so the voltage drop across the capacitor decreases when frequency is low.

No, the voltage drop across a capacitor increases when the frequency of the current through it decreases.

Doesn't that mean that the capacitor is letting through all the low frequency signals... Why is the lower picture showing us that all the high frequency signals are going through the capacitor?

All the high frequency current is passing through the capacitor, none is exiting the output terminal and passing through the load.

So shouldn't the capacitor be a high-frequency blocking element? Why is it known to be a DC blocking element then?

A capacitor shunted across two terminals blocks a high frequency voltage from appearing across them, the capacitor creates a low voltage across its terminals.

A capacitor in series with a signal line blocks the flow of low frequency and DC signals, by allowing a large voltage to appear across its terminals.

There is yet another shortcut that the teacher is taking when talking about 'signals'. A signal comprises both a voltage and a current, on a pair of wires, which forms a complete circuit when joined at the source end through the source, and the load end through the load. It's usually obvious in context whether one is operating on the current or the voltage of the signal with shunt or series components, once you have a clear mental picture of the situation. You've not got that clear mental picture yet. You'll get there, with questions like this.

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  • \$\begingroup\$ Thanks for the encouragement! I really do appreciate it and the clear explanation. It dawns on me that there's one question I forgot to ask. In my lower diagram above(the low pass filter) we see that the high frequency signals move through the capacitor into ground while the low frequency signals are the assumed output. $$$$ But right at the point above the capacitor the signals have yet to be filtered, right? How can we be sure that the high frequency signals move into the capacitor? Is this just an oversimplification in the diagram? \$\endgroup\$
    – FafaDog
    Commented Dec 16, 2020 at 15:43
  • \$\begingroup\$ @BigBear The presence of that capacitor causes the high frequency current to flow to ground. That current causes a large voltage drop in the resistor feeding it, the voltage of the high frequency signal on that capacitor node is therefore very low. With low frequency signals, little current flows in the capacitor, little voltage drop across the resistor, so most of the low frequency signal voltage appears on the capacitor. As you can see, filtering has already happened at that capacitor node, large low signal voltage with respect to ground, small high frqeuency voltage. \$\endgroup\$
    – Neil_UK
    Commented Dec 16, 2020 at 16:20
  • \$\begingroup\$ Oh! I think I'm starting to connect the dots. $$$$ Since \$ Z_R = \frac{ R }{R + \frac{1}{jwC}} = \frac{RjwC}{RjwC + 1}\$ we have very little voltage drop across the resistor for low signals and the higher frequency signals experience a bigger voltage drop. So now we know that the higher frequency signals are already attenuated (I feel so proud using that word) Then the high frequency signals are sent to ground because the capacitor let's the higher frequencies through without much impedance (see above equation) whereas the lower frequencies are blocked. \$\endgroup\$
    – FafaDog
    Commented Dec 16, 2020 at 19:46
  • \$\begingroup\$ Thank you so much! \$\endgroup\$
    – FafaDog
    Commented Dec 16, 2020 at 19:47
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For those two circuits, you are measuring the output voltage at the stub on the right and comparing it to the input voltage coming from the left. The impedance of the capacitor drops as the frequency of the applied voltage rises, as you state, which means that it lets through higher frequency signals easier than lower frequency ones. In the first circuit, the capacitor is between the input and output, so high frequency signals will transfer between the input and output better. In the second circuit, the capacitor is between the output and ground, so high frequency signals will be attenuated rather than reach the output. You can also think of it as a sort of voltage divider with the capacitor acting as a frequency-dependent resistor.

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  • \$\begingroup\$ If I were measuring the voltage across the capacitor \$ V_C\$ then I would only see the low frequencies right? \\ \$ V_C = Z_C * V_{In}\$ since \$ Z_C \$ is non-zero at lower voltages \$\endgroup\$
    – FafaDog
    Commented Dec 11, 2020 at 18:26
  • \$\begingroup\$ In the drawings you have, the voltage across the capacitor would be Zc/(R+Zc). Plug in your input signal or the different components and analyze individually. You'll see for low frequencies, Zc/(R+Zc) approaches 1 and for high frequencies, it approaches 0. \$\endgroup\$
    – vir
    Commented Dec 11, 2020 at 18:52
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Capacitors can be low pass high pass filters because their impedance changes with the frequency of the input signal. If we create a voltage divider of 1 stable impedance element (resistor) and 1 variable impedance element(capacitor) we can filter out low frequency or high frequency input signals.

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I'll try to explain it without relying on any AC concepts.

high-pass

For high-pass, imagine changing the voltage on the left side of the cap very slowly. The resistor will keep draining the right side to ground and its voltage won't change. But if you suddenly change the left voltage then the right side will take some time to drain out.

(And to paraphrase what @Neil_UK said, whatever circuit you connect to the output wire can't possibly know what the voltage across the capacitor is. The voltage across the cap is pretty much irrelevant. You care what the voltage on the right-hand-side is.)

low-pass

In the low-pass circuit, if you suddenly change the voltage on the left, then the cap won't fill up to that level for a long time. If you slowly change the left voltage, then the cap can keep up with it.

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