Hot answers tagged

5

These are the kinds of questions that should be brought here more often. I see that ErikR has correctly identified the flaw in your writing and that you've selected the answer. Great! So I won't belabor those details. Instead, I'll recommend that you learn to use sympy, which is freely available. (Yes, there are some 'instructions' you'll need to follow to ...


5

Your numerator in this: $$ v_{o}=v_{oi}\frac{R}{Ls+R_{L}+\frac{R(\frac{1}{Cs})}{R+\frac{1}{Cs}}} $$ Should be \$Z_R || Z_C\$. That is, $$ v_{o}=v_{oi}\frac{R || Z_C}{Z_L+R_{L}+(R||Z_C)} $$


5

This is a very typical result that basically means you got a garbage output, because something overflowed in the filter synthesis computations, or there was catastrophic loss of precision, or some iterative algorithm failed to converge. Typically, when you start to write filter design software, you'll get such "results". The general idea is to ...


5

C1 and C3 are "standard" power supply decoupling caps. Their purpose is to prevent high frequency noise that might be on the power supply rails from coupling onto the output currents. C2's purpose is a little harder to glean. The datasheet says nothing (that I could find) as to why this cap is needed or what it does, and I didn't have time to ...


4

@Neil_uk is right calling it a poor question. The first part of the circuit attenuates the input signal to bring it within the range of the op-amp and whatever its output goes to. The second part of the circuit is the op-amp itself. It's there to spare the designer the effort involved in applying Ohm's Law when there are several circuit elements involved. It'...


4

Was looking at this not long ago, as a series of steps in the bode plot, logarithmically spaced, exactly like the first comment. Expression below will give a slope of +/- 20\${\gamma}\$ dB/decade, between \${\omega}_0\$ and \${\omega}_f\$. N is number of pole-zero pairs. $$\displaystyle\frac{\left(1+\frac{s}{\omega_0}\right)^\gamma}{\left(1+\frac{s}{\omega_f}...


3

Because you have the transfer function with all parts values you can compare this function with the general function - expressed NOT with coefficients (which have no direct relation to the kind of response) but with the pole data wp (pole frequency) and \$Q_p\$ (pole quality factor). It is in particular the Qp value that gives you information on the kind of ...


3

Such filters have 3 tasks: lowpass CM signals (usually noise), lowpass DM signals (the actual antialiasing) and prevent conversion of CM signals to DM. the DM mode filter is formed by parts between In+ and IN-. It is an RC lowpass made by R1, R2, C1. there are two common mode filters, one viewing into each input. E.g. for In+ you have R1 and then the ...


2

There are several ways you can do it with only analog components. Which method is most appropriate depends on what information in the signal you want to preserve:- Amplitude: You want to discard it and fix the output at 10 V. Frequency: ...must be exactly preserved, or not? What input frequency variation must it handle? Phase: What short and long term ...


2

Referring to the 2nd question: Depending on the values of R1 & R2, This arrangement allows for a higher impedance input, and lower impedance output. In my line of work, the circuit is a simple version of the input stage of a DAQ (data acquisition) system. The voltage divider reduces the input voltage to the Analog to Digital converter, and has a first ...


2

The ‘resistance of water’ isn’t exactly a thing, in the sense that current isn’t always proportional to voltage. You’ll need to decide under what conditions you want to make the measurement - AC vs DC, what distance between the electrodes, what kind of electrodes do you want to use, etc. You may well be measuring more about the electrodes than you are ...


1

There are many flaws in your design. A PWM shall not be used, rather a +/- bipolar square pulse voltage. The RC filter as you depicted of course affects the measurement, you need to buffer the input signal with Opamp, first. You can use a dual voltage PSU, then use and ADC with chopper and external chopper driver, for example AD7730. It drives the external ...


1

First the output signal surpass the input signal (overshoot) Initially the output signal follows the input exactly and goes up by the same amount relative to their current baselines. However the output signal quickly 'sags' back to the baseline as each capacitor in the filter charges, with a small undershoot caused by the second capacitor charging less due ...


1

You are looking at the typical step response of a highpass filter. You can start with the prototype Butterworth highpass transfer function and determine its step response: $$\begin{align} H(s)&=\dfrac{s^2}{s^2+\sqrt2 s+1} \tag{1} \\ h(t)&=\mathrm{e}^{-\frac{t}{\sqrt2}}\left(\cos\left(\dfrac{t}{\sqrt2}\right)-\sin\left(\dfrac{t}{\sqrt2}\right)\right) \...


1

To answer your question, you will need to (a) simulate the filter with an EM simulator, to see where the currents in the copper are largest (often at the corners of the traces) or better yet (b) build the filter and measure the temperature rise when it’s used at high power. A multi physics type of simulation would find the current maximums and estimate the ...


1

LC filters are a pain in this regard: you need to account for all the possible loads, in and out, and once calculated they stay that way. Therefore a change in either the input or the output needs recalculating the values. Then you have an even order Chebyshev. As you probably know, these filters have the peculiarity of equiripples in the passband (or ...


1

how do I find the optimal one (approximations within reason are acceptable)? Not only this sounds very vague, but what you want is a bit of a mission impossible: it's not that you can't build an analog version of an FIR, it's just that it would require the equivalent of the \$z^{-1}\$, which is some sort of a delay line. There are delay lines, but not for ...


1

The problem is with the capacitor. You see, when using a controlled rectifier, if a Capacitor is present, it will charge the capacitor with the input voltage source with the maximum value of \$v_{i}(t)\$, after it is cut by the switching. Since your angle is bellow 90 degrees, you will always pass through the maximum value. As such, I would suggest you to ...


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