In an RC circuit, where the resistor comes before the capacitor, the circuit acts like a low pass filter. Whereas in a CR (I don't know if that is how you say it) circuit, where the capacitor comes before the resistor, the circuit acts like a high pass filter. Why does using the same capacitor just in different order relative to the resistor change the filter type? Thank you!
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\$\begingroup\$ Welcome to the site. Have you "done the math" regarding what Rs and Cs do with frequency? \$\endgroup\$– mike65535Sep 26, 2018 at 23:19
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\$\begingroup\$ We did the math for duty cycle and corner frequency of the RC circuit but I was really confused because it was the same capacitor and resistor so I thought the filter should act the same but Barry and DSWG helped clear it up!! \$\endgroup\$– sjfklsdafjksSep 26, 2018 at 23:23
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1\$\begingroup\$ I recommend reading ElectronicsTutorials - Passive Low Pass Filter. And when you get to the bottom, click Next for the High Pass. \$\endgroup\$– Blair FonvilleSep 27, 2018 at 6:01
2 Answers
For years I tutored the very first electrical engineering course at my uni, so this comes up a lot - don't worry :)
The best way to understand this is to think about all 4 cases:
- An AC (or, you could think about it as a "high frequency") signal through the low-pass filter
- An AC (or, you could think about it as a "high frequency") signal through the high-pass filter
- An DC (or, you could think about it as a "low frequency") signal through the low-pass filter
- An DC (or, you could think about it as a "low frequency") signal through the low-high filter
Key thing is to substitute in what the impedance of the capacitor would be (the resistor has the same impedance as it does resistance, so that doesn't matter). For a capacitor:
- High-frequency would mean the capacitor has a very low impedance (i.e. high-frequency can pass through it)
- Low-frequency would mean the capacitor has a very high impedance (i.e. just like the symbol for it - two open plates - the low-f or DC can't pass through easily)
With these things in mind, let's think about the path the current takes in all 4 cases:
Note: (and for those who are going to call me out) the "output" will actually be some resistor it is going towards, not an open-circut like we always draw when designing filters.
You can see here for the 4 cases:
- For AC, the low-pass filter blocks it from reaching the output as it goes to ground first, because the capacitor offers a low-impedance path to ground (remember: all signals, DC and AC, want to go to ground. That's their end goal)
- For AC in the high-pass, it passes through as the capacitor is an easy path to get through, and the resistor is just there doing it's thing - not offering any easy path to ground
- For DC, the low-pass is just like having a resistor in the way, as it can't pass through the capacitor easily and so the path to ground is not an easy option
- For DC, the high-pass blocks it almost before it even starts flowing, because the capacitor is present and as I mentioned earlier: for DC/low-frequency signals, the capacitor is just like it's symbol - it blocks the signal, and presents a high-impedance path.
I hope this helps. Happy to clarify if my explanation isn't clear enough :)
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\$\begingroup\$ It’s perfect, do you have something similar for LR circuits since they store the energy differently? This really helps a lot and thank you so much again! \$\endgroup\$ Sep 26, 2018 at 23:19
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1\$\begingroup\$ No worries. So an inductor (L) is just the exact opposite: it lets DC (low frequency) through, but blocks AC. This is because at DC the impedance is low (j * 0 * L = 0 ohms) and at AC the impedance is high (j * infinity * L = infinity ohms). So because this behaviour is swapped, the filter behaviour is also swapped (i.e. an LR is a low-pass, and an RL is a high-pass). \$\endgroup\$– DSWGSep 26, 2018 at 23:31
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1\$\begingroup\$ The simplest and easy to understand explanation to date. Enough of that Wo = 2*pi*F stuff. +1 \$\endgroup\$– user105652Sep 27, 2018 at 1:26
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\$\begingroup\$ Cheers @Sparky256, feedback is always appreciated \$\endgroup\$– DSWGSep 27, 2018 at 1:28
It's hard to answer this question without knowing your level of knowledge with respect to voltage dividers and the impedance characteristics of capacitors. I assume you do know that the impedance of a capacitor decreases as the frequency increases. That fact alone explains why the same capacitor can be used as part of a low pass filter and also as part of a high pass filter. In the low pass filter, what you call an RC circuit, the output voltage is taken across the capacitor. At low frequencies, the impedance of the capacitor is relatively high and the output voltage is also relatively high. As the frequency increases the impedance of the capacitor decreases. Since the resistor and the capacitor make up a voltage divider, the output voltage will decrease with frequency, hence the RC works as a lowpass filter. With a CR circuit, the output voltage is taken across the resistor. At high frequencies, the impedance of the capacitor is relatively low and the output voltage across the resistor is relatively high. As the frequency decreases, the impedance of the capacitor increases so that the voltage divider of the resistor and capacitor causes the output voltage to decrease, hence the CR circuit works as a highpass filter. Again the same capacitor can be used, the difference is where the output voltage is taken: across the capacitor for a lowpass filter, across the resistor for a high pass filter.