30

Look at the extremes. DC and very high frequencies. For DC you can remove the capacitors and short the inductors. For high frequencies you can short the capacitors and remove the inductors. By looking at the resulting circuit it should be easy to tell whether low or high frequencies can pass.


26

What you call "normal" is a simple two-stage RC filter with very bad selectivity (two real poles only). In contrast. the Sallen-Key topology is capable of producing a second-order lowpass response with much better selectivity (higher pole Qp) and various possible approximations (Butterworth, Chebyshev, Thomson-Bessel,...). However, there is one big ...


22

The capacitor does not "short out", it has charged up to a constant voltage by storing energy as electrical charge, and if something external tries to change the voltage over the capacitor, it means that more or less charge is needed to change the capacitor voltage up or down, and moving charges means is current flowing. So in short, a capacitor wants to ...


21

I wouldn't think of a decoupling capacitor as a filter in the way you describe. Like an RC filter like this, where the source of the noise is the power supply and your "decoupling" capacitors are helping to filter that out before it reaches your chip. simulate this circuit – Schematic created using CircuitLab It's not keeping noise from getting to ...


19

It's a compromise. With R at 1000 ohm and C at 100nF (cut-off frequency = 1.59kHz), the driving voltage at the input may be required to drive signals with frequencies well above 1.59kHz into what is getting close to a 1000 ohm load. Consider what the impedances are at 1.59kHz - R of course is 1000 ohm and C's impedance also has a magnitude of 1000 ohms ...


13

H(w) =1/ (square root of 2) That sounds a bit confusing, we usually refer to the cutoff point as the -3 dB point. That is the same though, -3 dB is half the power. Let me explain: take your \$H(\omega) = \frac{1}{\sqrt2}\$ That means that at that \$\omega\$ the voltage is divided by \$\sqrt2\$, if this voltage is applied across a (load) resistor at the ...


12

The cap near the power pin is not to protect the part from noise, but to keep the part from generating noise as the logic switching causes rapid changes in supply current. Ideally the cap would supply instantaneous demands for more current without increasing current all the way back to the power source. The sum of the impedances on the PSU side of the ...


12

can anyone explain why this circuit behaves like a low pass filter despite having no capacitor in the feedback loop? All op-amp amplifiers will behave like low pass filters due to parasitic capacitance between the output and, in your circuit, the inverting input. If there is 2 pF present then it forms a low pass filter with the 400 kohm feedback resistor ...


11

The inductor plus capacitor form a frequency-dependent voltage divider. \$ \dfrac{V_{OUT}}{V_{IN}} = \dfrac{Z_C}{Z_C + Z_L} \$ For DC and low frequencies the impedance of L1 (\$Z_L\$) is low, and that of C1 (\$Z_C\$) high, so the input voltage won't be attenuated much. At high frequencies it's the other way around: \$Z_L\$ is high, and \$Z_C\$ is low. The ...


11

It's simply an AF (audio frequency) coupling capacitor, removing any DC that is on the signal.


10

The first RC does indeed act as a low pass filter. The second capacitor (the one on the right), combined with the load impedance presented by whatever is connected to the audio out, acts as a high pass filter. Together, these two filters act as a band pass filter.


9

Some really good answers. My take is that the goal of the L-C filter isn't filtering out power supply ripple. That is best done with stiff (low ESR) caps on the power lines/planes and picking the right regulator part to start with. Plus, if you power your Arduino from a USB port the low frequency ripple sort of noise would be negligible.A cheapo wall wart is ...


9

If you have an ideal OpAmp the 10k resistor would carry no current and could be replace by a resistor with an arbitrary value, including a piece of wire. Therefore the answer is no, the 10k resistor plays no role wrt. the low pass filter.


9

The correct place to ground it is to 0 V. It is the reference for all voltages in your circuit. simulate this circuit – Schematic created using CircuitLab Figure 1. For perfectly regulated power rails C2 and C3 would perform in this application exactly the same as C1. Theoretically you could also connect the bottom of the capacitor to -5 V or even +...


8

For frequencies this low placement doesn't really matter. Switching R6 and C7 will make the layout somewhat cleaner though.


8

I am afraid, changing the opamp type will not help. The observed effect (less damping for rising frequencies) is the typical disadvantage of the lowpass Sallen-Key topology. The reason is as follows: For rising frequencies the "classical" output signal from the opamp decreases (as desired) - however, at the same time there is a signal arriving at the ...


8

The fundamental nature of a capacitor is that stored charge = capacitance x voltage: - $$Q = CV$$ We also know that current is the rate of change of charge with time hence, if the formula is differentiated with respect to time we get this: - $$I = C\cdot\dfrac{dV}{dt}$$ The impact of this formula is that if a current is injected, the capacitor's terminal ...


8

The capacitor does not become an actual short circuit. When we say that a capacitor is a short circuit at high frequencies, we are talking about the impedance of a capacitor when a sine wave voltage is applied to it. The impedance is the voltage divided by the current (similar to resistance). $$ Z_C = \frac{V}I = \frac{1}{2\pi jfC} $$ It's an imaginary ...


7

Two degrees of freedom and one specification means that you can only fix the product \$RC\$. As has been pointed out in answers and comments to similar questions, the answer is: it depends - it depends on some other constraint. Seriously, a question like this is almost unanswerable without additional context. Here are some considerations that might go ...


7

The short answer: A capacitor alone is good for delivering power when the MCU power draw changes fast. The RC filter is used to block unwanted high frequency signals. The looong answer: The two different circuits are used for different purposes. As you have stated, the voltage across a capacitor cannot change instantly. I'm sure you know that An MCU ...


7

First, that's a high-pass filter. Secondly, if you do not load Vout (on the second circuit), then I think you can see that the series resistance R1 + R2 is equivalent to R1 (on the first circuit), so you can find the -3dB frequency. The output voltage at high frequencies (second circuit) will be \$ V_{OUT} = V_{IN} \cdot \$\$R_2 \over R_1 + R2\$, and the ...


7

The LM324, while a brilliant achievement with 1970s transistors, has one well known bug - actually documented in its datasheet. This answer is based on a guess that you are running into this bug. Some people sneer at it because of limitations like this - but it is still a fine opamp if you design to its limitations. Its Class B output stage is specifically ...


7

The trace from the power source to the capacitor has inherent parasitic inductance. This provides the series element to make a low-pass filter. A more complete model of the circuit looks like this: simulate this circuit – Schematic created using CircuitLab I've drawn the load as a time-varying current source, because the main goal of the bypass ...


7

LM358 has very low gain-bandwidth product (1MHz) therefore the Sallen-Key filter topology you used might not work well as a lowpass above a few tens of kHz. What happens is the HF jumps over the opamp through C5, and at HF it doesn't have enough feedback to keep its output pin low impedance, so the HF just goes on to the output. Also as Scott says in the ...


7

It's just a conversion from radians to degrees. The expression for a capacitor's impedance uses units of radians/sec. for frequency. The simulator is reporting frequencies in Hertz. 10 rad/sec is 1.6Hz


7

basically, every physical system in this world has a finite bandwidth and hence exposes some low-pass behaviour, so in the larger scheme of things, this is to be expected You use an ancient (LM)741 series opamp. These things have very modest bandwidths – in fact, the current Texas Instruments LM741 datasheet only claims 0.5V/ms as slew rate, and doesn't ...


6

Yes, it actually looks like this (doing a numerical integration of the differential equation): At equilibrium, the peaks of the sawtooth are at \${1\over 1+e^{T\over 2RC}}V_S\$ and \${e^{T\over 2RC}\over 1+e^{T\over 2RC}}V_S\$, where \$T\$ is the period of the square wave (different from the \$T\$ in the question plot). For this example, that's about \$0....


6

Spehro gave you the answer, now let me tell you why you could (should?) have known that by looking at the circuit. Two resistors in series are indistinguishable from one resistor with the sum of the two resistances. (I hope you knew this?) Hence when we take Vout in the second circuit from the R1/C1 junction, we have exactly the same circuit as the first (...


6

What is the purpose of this simple 1 capacitor 2 resistor [high] pass filter? The extra resistor gives you an extra degree of freedom. For the standard RC high-pass filter in the first schematic, the transfer function is $$H(j\omega) = \frac{j\omega R_1C}{1 + j\omega R_1C} $$ So, the asymptotic high frequency gain is 1 and the corner frequency is \$f_c=...


6

Your reasoning not to use an op-amp is basically flawed. here I quote you: - Of course a voltage follower will be applied at the output of the passive filter. This voltage follower can in fact be made into a 2nd order low pass filter using the sallen key topology: - As you can see the op-amp is configured as a voltage follower (unity gain) but overall,...


Only top voted, non community-wiki answers of a minimum length are eligible