Talk:RC circuit

This is the talk page for discussing improvements to the RC circuit article.
This is not a forum for general discussion of the article's subject.

Put new text under old text. Click here to start a new topic.
New to Wikipedia? Welcome! Learn to edit; get help.

Article policies

Find sources: Google (books · news · scholar · free images · WP refs) · FENS · JSTOR · TWL

Archives: 1: 2 years

Electronics Mid‑importance

	Electronics portal This article is part of WikiProject Electronics, an attempt to provide a standard approach to writing articles about electronics on Wikipedia. If you would like to participate, you can choose to edit the article attached to this page, or visit the project page, where you can join the project and see a list of open tasks. Leave messages at the project talk pageElectronicsWikipedia:WikiProject ElectronicsTemplate:WikiProject Electronicselectronic
Mid	This article has been rated as Mid-importance on the project's importance scale.

shouldn't 'calculations' be used in the title

or, at least rc calculation formulas.. 12.146.12.2 (talk) 09:10, 23 November 2024 (UTC)[reply]

RC filter not −3 dB

Hello :)

Contrary to what almost everyone claims, a first-order RC filter does not attenuate amplitude, voltage, or power by −3 dB at the cutoff frequency!

BECAUSE: The electric current is the same at every point in a series circuit, which is the case for the RC filter—this resistive and capacitive series divider.

The definition of the bel is clear: The bel is the logarithmic measure, base ten, of the ratio between two physical quantities of the same nature expressing power.

This means that in the gain calculation, we must use only physical quantities that express power. Examples:

Gain = log (P_output ÷ P_input) Gain = log (|S_output| ÷ |S_input|)

Or physical quantities expressing fields that, when combined, result in physical quantities expressing power—this within the logarithmic function. Example:

Gain = log ((U_rms_output × I_rms_output) ÷ (U_rms_input × I_rms_input))

THEREFORE: Since the effective electric current is the same at every point in a resistive and capacitive series divider, the effective output voltage, i.e., across C at the cutoff frequency, being U_rms_RC ÷ √2, here is an example calculation:

U_rms_input = 1 V U_rms_output = 1 V ÷ √2 ≈ 0.707 V Gain = log ((0.707 V × 1 A) ÷ (1 V × 1 A)) ≈ −0.15 bel

The attenuation of an RC filter is therefore about 0.15 bel at the cutoff frequency, whether in effective voltage or apparent power. Example:

|S_RC| = 1 V × 1 A = 1 VA |S_C| = 0.707 V × 1 A = 0.707 VA Gain = log (0.707 VA ÷ 1 VA) ≈ −0.15 bel

In this example, the apparent power, i.e., the modulus or magnitude of the complex power, allows us to retain all the information regardless of the system—not just a part of the power, meaning active and/or reactive power.

If I express it in watts, meaning active power, the active power of C would be 0 W, so in active power, the attenuation would be −∞ bel, which is not very useful for the example.

As shown by an experiment I conducted with an RC filter using R = 1000 Ω and C = 100 nF at the cutoff frequency:

P_RC ≈ 499 μW P_R ≈ 499 μW P_C = 0 W Q_RC ≈ −499 μvar Q_R = 0 var Q_C ≈ −499 μvar |S_RC| ≈ 706 μVA |S_R| ≈ 499 μVA |S_C| ≈ 499 μVA

SO TO SUMMARIZE: Let's avoid confusion. My advice is to include, along with the effective voltage in the logarithmic function, either the effective current or the apparent electrical impedance. Example: 50 Ω if the input and output of a device have this impedance.

OTHER EXAMPLES: If we use only electrical voltages without additional information in this logarithm, it no longer means anything because another physical quantity is missing to define power. Examples:

U_rms_input = 1 V I_rms_input = 1 A U_rms_output = 2 V I_rms_output = 1 A Gain = log ((U_rms_output × I_rms_output) ÷ (U_rms_input × I_rms_input)) ≈ 0.301 bel

U_rms_input = 1 V I_rms_input = 1 A U_rms_output = 2 V I_rms_output = 2 A Gain = log ((U_rms_output × I_rms_output) ÷ (U_rms_input × I_rms_input)) ≈ 0.602 bel

The examples show that if we use only a single physical quantity expressing a field instead of power, like voltage alone, the information about current is missing—which, by the way, can be identical from input to output in a system. This gives:

1 V × 1 A to 2 V × 1 A ≈ 0.301 bel 1 W to 2 W ≈ 0.301 bel

Whereas:

1 V × 1 A to 2 V × 2 A ≈ 0.602 bel 1 W to 4 W ≈ 0.602 bel

Below, I have gathered equations that help avoid any errors.

Calculation of gain in bel based on the output apparent power relative to the input apparent power:

Calculation of input apparent power based on output apparent power and gain in bel:

Calculation of output apparent power based on input apparent power and gain in bel:

If we step outside the bel scale, meaning the logarithmic measure, base ten, of the ratio between two physical quantities of the same nature expressing power, and we use only the logarithm function, then we can use this function for anything we want, just like any other mathematical function. Sylvainmahe (talk) 13:31, 20 March 2025 (UTC)[reply]

This article uses the voltage ratio of input to output.

dBs = 20 log_10 (voltage ratio) + 20 log_10(0.707) = 3.01. Constant314 (talk) 15:14, 20 March 2025 (UTC)[reply]

This is false and mistake, it makes no sense, reread my explanation. Sylvainmahe (talk) 12:07, 21 March 2025 (UTC)[reply]