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Very long RC time contant!


Kevin Weddle
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The RC time constant must be long enough with repect to the signal time to cleanly pass the signal through the capacitor.

Is this statement true or false?

Answer : True.

I wanted to revisit this topic because I was sharply denounced last time. Now, I know the truth. Any takers?

You are probably talking about an audio coupling capacitor.
The time constant of the capacitor with the impedance of the input of the next stage is usually calculated to form a highpass filter, so at the cutoff frequency the signal begins to be impeded and is not cleanly passed.
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It's not seen using a sinewave.

Of course a sinewave or any other waveshape is affected by the time constant of a coupling capacitor.
At the frequency where the reactance of the capacitor equals the impedance of its load, the circuit's output level is down 3dB which is a loss of -0.707.
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I don't think you meant equal impedance, that would be half the voltage. Your thinking resonant frequency.

That is not true. A capacitor has phase-shift that causes the response to be 0.707 instead of half the voltage. Half the voltage is created by two equal resistors that don't have phase-shift.
Look at RC filters in Google.
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What is funny, if you start a sinewave on an RC just up to 90 degree and stop, the capacitor still gets voltage, rather than remaining at zero. It's not going to wait until the input starts to go down, because it may not go down.

I plotted RC phase shift on paper one time using a triangle. I set the resistor equal to the capacitive reactance. Let's say they are both 100 ohms. If the input is 20Vpp, then capacitors about 14Vpp and the resistors about 6Vpp. When the input reaches +10V, the resistor must be at +3V and the capacitor at +7V. The resistor can't be at +4V because it doesn't go that high. It could be +2V, that would put the capacitor at +8V, but the capacitor doesn't go that high. The resistor could be +2V and the capacitor +7V, But that is only +9V total.

Since the inputs at +10V, the resistors at +3V, and the capacitor is at +7V, don't they all appear to be in phase?


But everywhere else, besides the peak, like in this example, the different phases and voltages seem plausable.
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  • 2 weeks later...

At what frequency did you calculate the capacitive reactance? You are aware that triangular waves are a bunch of sine waves at different harmonic frequencies and amplitudes summed together, right (a.k.a. Fourier Analysis)? Phase shift only pertains to voltage in relation to current or vise versa. Voltage or current by themselves don't have any phase shift.

Recall "ELI the ICE man" voltage (E) leads current (I)in a Inductor (L) and Current (I) leads Voltage (E) in a capacitor © by 90 degrees. Or if you preffer, current (I) lags voltage (E) in an inductor (L)... etc.

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  • 4 weeks later...

Of course a voltage divider made with a resistor in series with a capacitor to ground have a phase shift.  At the frequency that the reactance of the capacitor is the same as the resistor's value, the level is 0.707 times, not half. It is called a lowpass filter. 0.707 times is 3dB down.

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Wow, not once have I ever read anything about how you arrived at that. If I had to include phase into the voltage divider, I would be stuck. And so would my books author. Just remove 2pi from the reactance equation. It works perfect and with no surprise. And you know, discovering this a bit late in life doesn't bother me one bit. That equation is about as useless as alpha, IIR, and cycles per second.

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A diagram of response vs phase-shift can be shown with phasors, but I couldn't find one.
All tutorials show that the response of an RC lowpass filter is down 3dB (times 0.707) at the cutoff frequency where the reactance of the capacitor equals the resistance of the resistor:

post-1706-14279142861773_thumb.png

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