32

I believe your analysis to be good. I've made sallen-key 4th order filters that cut-off around 3 MHz with absolutely no worry about performance. I don't see that 10 MHz is unachievable. It's all about op-amp choice. For a unity gain stage it's easy to ascertain where the gain starts dropping below (say) 0.99 and regard that as the limiting frequency. On the ...


12

General Form, 2nd Order HPF The general form for a 2nd order high-pass filter is (\$K\$ is the voltage gain): $$\begin{align*}\frac{K\:s^2}{s^2+2\zeta\:\omega_{_0}\:s +\omega_{_0}^2}=\frac{K\:s^2}{s^2+\frac{\omega_{_0}}{Q}\:s +\omega_{_0}^2}\label{eq1}\tag{1}\end{align*}$$ Butterworth The next thing is to look up the constants for Butterworth filter. These ...


11

The noise is caused by minute mechanical vibrations of the pot wipers on the rings (the latter are the resistive material.) Since neither the wiper nor the ring material are atomically smooth where they touch, rubbing the wiper on the ring produces slight vibrations. Some of that vibration is perpendicular to the contact patch between wiper and ring. The ...


11

This could arise as the sum of the input offset voltages ... but for 3 opamps, that would only come to 3.9mV worst case, so that isn't it, especially since it's unlikely that the SPICE model has the worst case offset voltage. But the opamp has a pretty large input bias current ... 0.65 uA. Have you balanced the impedances on both inputs to the opamp? If you ...


11

1st stage If you set \$\gamma=\frac{C_1}{C_2}\$ and \$\rho=\frac{R_3}{R_4}\$, then the left-side 2nd order filter (1st stage of two) has \$\omega_{_0}=\frac1{R_4\, C_2 \,\sqrt{{\vphantom{M}}\gamma\,\rho}}\$ and \$Q=\frac{\sqrt{{\vphantom{M}}\gamma\,\rho}}{1+\rho}\$. In the above circuit \$\rho=1\$, so \$\omega_{_0}=\frac1{R_4\, C_2 \,\sqrt{{\vphantom{M}}\...


10

These are very good answers and let me add mine considering the rainy day in Occitanie : ) I am using the fast analytical circuits techniques or FACTs as described in my book. The principle is quite simple: determine the natural time constants of the system when the excitation is turned to zero (a 0-V \$V_{in}\$ source is a short circuit). These natural time ...


9

This is a "well-known" issue with the Sallen-Key configuration. The problem is that the output impedance of the amplifier rises with frequency and starts to look inductive so it's no longer 'fighting' the input signal more as the frequency increases. The output then increases by +20dB/decade until it levels out at the GBW of the amplifier. You can ...


8

The opamp in a Sallen-Key filter is supposed to be a unity-gain buffer. Yours has a gain of +3, so it isn't surprising that it's oscillating. Wikipedia talks about this. If you need that much gain, you need to do it elsewhere. There's also the Application report from Texas Instruments Analysis of the Sallen Key Filter that explains why the gain of 3 or ...


8

Although it is possible to design a Sallen Key filter with a gain higher than unity, this is rather uncommon for a reason. Any gain in it introduces positive feedback into the structure and leads it towards instability. Particularly when you take the amplifiers own poles into consideration. The OPA177 has a gain bandwidth of ~600kHz, at a gain of 3 you have ...


8

My first question: really? A mere 100kHz is already too high for active filters to be practical? No, 100kHz is nothing, but it all depends on the opamp. At some point the Gain Bandwidth Product is going to cause problems. If you had an op amp with a 1MHz or 10MHz GBWP (which may have been typical at the time of the first edition of AofE, maybe they didn't ...


5

How could I alleviate this noise? Probably not what you want to read now, but: don't use pots in audio circuits where they are subject to DC currents. This is the case for the 2 pots you mentioned, but not for the other 4. Due to the mechanical nature of the device, the resistance variation is not "clean" and continuous. If the potentiometer is subject to ...


5

Back of the envelope calculation... 20k/4250 = about 4 It's a second order filter, so at a frequency 4x above its -3dB point it should attenuate the signal by 4 squared, or about 16. Output signal is about 0.3Vpp. Input signal is 5Vpp. 5/0.3 = about 16. So it delivers the correct attenuation. If you want more attenuation, use a lower -3dB point, or a higher ...


4

The main problem with that Sallen Key topology at high frequency is that the output impedance of op-amps rises, so fails to control feedforward of the input signal through the 2C capacitor, trashing the stopband.


4

One is Salena Kay, and the other I don’t know what is called. Sallen Key with unity low pass gain: - Pictures taken from this useful web calculator. Sallen Key with gain setting resistors: - Pictures taken from same calculator page as above - scroll down MFB or multiple feedback: - Note that the low frequency gain of the above MFB circuit is \$-\frac{...


4

simulate this circuit – Schematic created using CircuitLab Datasheets for audio amplifier ICs strongly recommend using a DC-blocking capacitor between the input potentiometer and the IC's audio input in order to prevent the slider noise caused by a very small DC current (on the order of microamps or less) between the IC input and the ground leg of the ...


4

The effect as mentioned by Spehro Pefhany is even more severe because of the following: For rising frequencies the capacitive impedance in the feedback path gets lower and lower - and an increasing portion of the input signal arrives DIRECTLY via this feedback capacitor at the opamps (increasing) output impedance. This works against the "normal" ...


4

The response of the whole chain got a strange behavior @1MH it come from the sallen & key low-pass filter. I don't where the increase of gain came at 1 MHz and I don't know how to correct this. I would step back and consider if that elaborate cascaded filter system even makes sense. You have an ADC with a 3.6 MHz sampling rate and target bandwidth of ...


4

The transfer function of the first active block has - of course - a 2nd-order denominator \$D(s)\$. The transfer function has the form $$H(s)=\dfrac{N(s)}{D(s)}=\dfrac{s\dfrac{R_3R_4}{R_2}C_1+\dfrac{R_3+R_4}{R_2}}{s^2R_3R_4C_1C_2+s(R_3+R_4)C_2+1}$$ Therefore, it is not a classical 2nd-order lowpass and also not a bandpass (both in multi-feedback topology). ...


3

Is that because of some capacitance effects ocurring in op-amps or something else? Your sim uses the cranky old 741 op-amp and that device fails to become effective even as a unity gain amplifier around 1 MHz. You have about 20 dB pass band gain and so you are asking each 741 to roughly provide a gain of 3 and now the effective point where the 741 gives ...


3

The fact that your circuit is oscillating at such a high frequency suggests very strongly that you have ground/decoupling issues. JRE commented that you need a solid ground plane, and I agree. Admittedly, this means you'll need to get creative about routing -Vcc. Additionally, your schematics do not include the decoupling caps which have clearly used. Please ...


3

First, I'm setting aside the fact that a \$1\mathrm{pF}\$ cap just isn't realizable in practice on a circuit board. The input capacitance to the op-amp is probably greater than that. There's a reason it's hard to find capacitors with values less than \$4.7\mathrm{pF}\$ or so. You are doing your calculations assuming an ideal op-amp. However, the expected ...


3

Get rid of most of the capacitors in that filter, leave only small (10-100pF) ones across the negative feedback resistors. You'll see a similar 4mV output, I bet, so this has nothing to do with the filter. The filter is a red herring. Now, you may not like that 4mV offset, but that's a DC problem, so it will need to be solved like such DC problems are solved ...


3

That is not a simple lowpass filter, it is a Bridged-T notch filter in a negative feedback loop, followed by an 8.7 kHz single-pole lowpass filter. The T network produces a peak in the circuit's frequency response. The narrowness and height of the peak are determined by the ratio of C1 to C2. I've seen this configuration as a low-distortion oscillator. ...


2

There are several one-opamp circuits with Chebyshev II or Cauer behaviour, for example: double-T-feedback circuits with positive (fixed) gain, Boctor-filter (based on Multi-feedback topology), Scultety-structure GIC-based structures (Hint: Google for Boctor and Scultety) For example, see here: http://www.schematica.com/active_filter_resources/...


2

Do you think the AC analysis extracted the correct impedance? Yes. it looks likely that the graph starting at about 30 kohm and falling to 3 kohm is precisely the input impedance of the filter design. The only problem is that your filter design is unrealizable because the bare-bones input capacitance of the AD8029 is 2 pF (typically) and your model is ...


2

TI has a 10MHz design App Note. It is based on their THS4001 low-cost 270 MHz -3dB Op Amp. Op Amps have an open loop output impedance much higher than your 50 Ω signal generator. This makes them stable with their short circuit protection. The higher GBW is used to lower the Zout = Zoc/GBW. The breadboard ESL (0.5nH/mm) and stray capacitance will need ...


2

I have had some Moog synthesizers with noisy pots. After replacing them with new, clean pots the noise remained. Turned out the electrolytic capacitors had gone bad and were leaking DC on to the pots causing it to sound just like dirty pots. I replaced the ecaps and the noise was gone.


2

On a more practical note, it's possible the pot is simply dirty. If the pot is not sealed, you can apply a small amount of Deoxit Green, to both wafers, cycle the pot several times, and see what happens. Amp techs and hobbyists have eliminated scratchy pots on many an amplifier this way.


2

You have your sallen-key topology wrong. Even if it was correct, I don't like the circuit because attaching a load will mess with the transfer function. Ideally your load attaches to the output of the op-amp or a buffered node, as then attached loads will not draw current from the filter but the op-amp itself, preserving the filter's characteristics. ...


2

Analytic Form The first thing you need to be aware of is that there is an analytic form and that's what you are looking at, right now. The general, 2nd-order, low-pass filter form is: $$H\left(s\right)=\frac{h\:\omega_{_\text{C}}^2}{\omega_{_\text{C}}^2+2\zeta\,\omega_{_\text{C}}\,s+s^2}=\frac{h}{1+2\zeta\,\frac{s}{\omega_{_\text{C}}}+\frac{s^{\,2}}{\omega_{...


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