A question on passive low-pass filter

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In summary, the conversation discusses the use of resistors in a low-pass filter circuit and the consequences of not including a resistor. The cut-off frequency of a filter is determined by the time constant of the circuit, which is the multiplication of resistance and capacitance. Without a resistor, the only variable to determine the cut-off frequency would be capacitance, which could lead to impractical circuit values. Adding a resistor allows for more reasonable values and prevents potential damage to the circuit. The conversation also touches on the topic of studying analog integrated circuits and available resources for learning about filters.
  • #1
NexusN
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Homework Statement



250px-1st_Order_Lowpass_Filter_RC.svg.png

This is a simple passive low-pass filter stated in the Wikipedia.
I am wondering why we need the resistor to allow it operates as a low pass filter.
Can we just ignore the resistor(by shorting it)? What will the consequence be?
Thank you for reading.

Homework Equations





The Attempt at a Solution

 
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  • #2
The cut-off frequency (aka 3dB point, corner frequency or half power point) is the point in the frequency based attenuation curve where a signal strength has been halved by a filter, and this is one of the characteristics of filters that allows us to compare different designs.

This cut-off frequency is determined by the time constant of the circuit, which is the multiplication of the Resistance (R) and the Capacitance (C). You may have seen any of the following formulae:

RC = 1/wc OR RC = 1/(2*pi*fc) OR tau = 1/wc

where:

tau = RC and is the commonly used symbol
wc = cutoff frequency in radians
fc = cutoff frequency in hertz

If we did not put a resister in the circuit, the only resistance would be the static resistance of a wire (very low) and the capacitance would be the only variable we could use to determine the cutoff frequency.

e.g. assume the wire resistance is 1uR and you want a filter with a cut off frequency of 1kHz.

1uR * C = 1 / (2*pi*1000) --> C = 160 F

... yikes! practical circuit values of capacitance exist in the high pico + nano + micro + low milli range, and an average 1 F capacitor is about as large as 2 standard soft drink cans on top of each other (and costs ~$100 and has its own voltmeter).

If we add a 1kR resister into this circuit the capacitance required is now 159 nF, which is much more reasonable. Using both these variables you can mix and match to get the required value of cutoff frequency.

Additionally, in real circuits, if there was only the resistance of the wire, by Ohms law when you have a very low resistance you have a very high current, and this would possibly melt or burn your wire.

And lastly, when looking at the resistance and capacitance, remember to look at the *total* resistance and capacitance, so combine everything in the circuit together to get the equivalent values (including the load resistance!).
 
  • #3
Zryn said:
The cut-off frequency (aka 3dB point, corner frequency or half power point) is the point in the frequency based attenuation curve where a signal strength has been halved by a filter, and this is one of the characteristics of filters that allows us to compare different designs.

This cut-off frequency is determined by the time constant of the circuit, which is the multiplication of the Resistance (R) and the Capacitance (C). You may have seen any of the following formulae:

RC = 1/wc OR RC = 1/(2*pi*fc) OR tau = 1/wc

where:

tau = RC and is the commonly used symbol
wc = cutoff frequency in radians
fc = cutoff frequency in hertz

If we did not put a resister in the circuit, the only resistance would be the static resistance of a wire (very low) and the capacitance would be the only variable we could use to determine the cutoff frequency.

e.g. assume the wire resistance is 1uR and you want a filter with a cut off frequency of 1kHz.

1uR * C = 1 / (2*pi*1000) --> C = 160 F

... yikes! practical circuit values of capacitance exist in the high pico + nano + micro + low milli range, and an average 1 F capacitor is about as large as 2 standard soft drink cans on top of each other (and costs ~$100 and has its own voltmeter).

If we add a 1kR resister into this circuit the capacitance required is now 159 nF, which is much more reasonable. Using both these variables you can mix and match to get the required value of cutoff frequency.

Additionally, in real circuits, if there was only the resistance of the wire, by Ohms law when you have a very low resistance you have a very high current, and this would possibly melt or burn your wire.

And lastly, when looking at the resistance and capacitance, remember to look at the *total* resistance and capacitance, so combine everything in the circuit together to get the equivalent values (including the load resistance!).

Thanks a lot!
You cleared my question!
I am studying Analog Integrated Circuit, have been learning some amplifier designs like current-mirror and compensated amplifiers, and now comes to the topic of filters.
It looks there are much for me to study.
 
  • #4
If you want to you can take a look at these rf filter specs to see if they help http://www.oscilent.com/catalog/Category/rf_saw_filter.htm" [Broken]
 
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  • #5


In order for a passive low-pass filter to effectively filter out high frequency signals, it requires both a capacitor and a resistor. The capacitor acts as a high-pass filter, allowing only high frequency signals to pass through, while the resistor acts as a low-pass filter, allowing only low frequency signals to pass through. By shorting the resistor, you are essentially removing this low-pass filter component, which would result in a different frequency response and potentially affect the overall performance of the filter. In some cases, shorting the resistor may result in a higher cutoff frequency, allowing more high frequency signals to pass through, and potentially reducing the effectiveness of the filter. Therefore, it is important to include both the capacitor and the resistor in the design of a passive low-pass filter for optimal performance.
 

What is a passive low-pass filter?

A passive low-pass filter is a type of electronic circuit that allows low-frequency signals to pass through while blocking high-frequency signals. It is made up of passive components such as resistors, capacitors, and inductors.

How does a passive low-pass filter work?

A passive low-pass filter works by creating a voltage divider circuit that attenuates high-frequency signals. This is achieved by using a combination of resistors and capacitors to create a high-pass filter in series with a low-pass filter.

What are the applications of a passive low-pass filter?

Passive low-pass filters are commonly used in audio equipment to remove high-frequency noise and distortions, resulting in a cleaner and smoother sound. They are also used in communication systems to filter out unwanted signals and in power supplies to reduce ripple voltage.

What are the advantages of using a passive low-pass filter?

The main advantage of a passive low-pass filter is its simplicity and low cost. It does not require an external power source and can be easily integrated into circuits. It also has a wide bandwidth and can handle high power levels without distortion.

Are there any limitations to using a passive low-pass filter?

One limitation of a passive low-pass filter is that it cannot amplify signals, only attenuate them. It also has a limited frequency range and cannot completely eliminate all high-frequency signals. Additionally, the components used in the filter may introduce their own resistance and capacitance, affecting the overall performance.

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