9

Your question is certainly a matter of opinion. What sings in one's mind varies widely from individual to individual. That said, there is a singular book on the topic. This is Modeling The Bipolar Transistor from Dr. Ian Getreu. This book was originally written and published internally by Tektronix for customers of their STS systems, as well as for anyone ...


7

How spice managed to solve such a huge system of nonlinear equations quickly and without running out of memory ? Google sparse matrix solver. A good SPICE very likely defaults (or knows when to switch to) a sparse matrix solver, since large circuits typically produce sparse matrices (each node connects to only a small fraction of the branches) it would ...


7

I tried your circuit(s) and I realized you're modeling the PWL source normally, but then you're inverting it with the first opamp. So I added a minus sign to the behavioural source, and tweaked the parameters a bit. Here's what came up: I simplified a bit the circuit, just for the purpose of making and simulating it, but it's equivalent with your values. ...


6

Only elements that have both inputs and outputs can be considered to be linear. "Linear" is a term that applies to a system with inputs and outputs, since it describes a particular kind of relationship between them. This means that simple voltage and current sources (when seen as ideal circuit elements) are neither linear nor non-linear. In essence, a ...


6

Yes, it is possible to use the Thevenin equivalent of a voltage divider driving an emitter follower. The subcircuit you are replacing by a Thevenin Equivalent is a linear circuit (just the part consisting of 15V constant voltage source and the voltage divider; you are not replacing any of the non-linear part). Therfore using the Thevenin Equivalent for the ...


6

If the non-linear circuit has a steady-state voltage, any resistor to the same voltage can not inject any current and thus no matter what resistance it is, it will be perceived as an open circuit. But if some external load changes which might affect that voltage, then current will flow and affect the circuit. Ridiculous Example: A tiny coin cell at 3.05V ...


5

I found a great article at this site to model transformers. It explains "The work of Hsu, Middlebrook and Cuk 2 develops the relationship of leakage inductance, showing that relatively simple measurements of input inductance with shorted outputs yield the necessary model information." They developed this for IsSpice, but it could be applied ...


5

If you look at a push pull amplifier with some amount of cross-over distortion it looks like this: - In other words it creates a symmetrical shaped signal in amplitude but generates odd harmonics that can be significantly reduced by correct biasing and sometimes negative feedback. Now take a look at an example of even harmonic distortion: - You should be ...


5

You are mistaking "linear" with "flat over frequency". The "attenuation vs frequency" curve has nothing to do with linearity. Mathematically, "linearity" means that when if input signal A produces output signal B, then n x A produces n x B. That is, the output is proportional to the input. Linearity DOES mean that if the input signal is sinusoidal, the ...


5

Load lines are useful in understanding and calculating the operating point of a circuit which would otherwise require complex calculations. Diodes have a non-linear relationship between I and V and we often use an approximation such as "2 V drop across the LED" when making calculations. We can do better if we have a load line chart. Figure 1. A simple LED ...


4

The Wiki article is using superposition correctly. It starts by assuming the output is in one state, then assuming the output remains in that state, calculates the comparator input with respect to the circuit input. It uses this relationship to determine the input level (producing zero volts at the comparator) to determine the trigger point at the specified ...


4

Sure, scale it down in current by a factor 100 and then use an opamp buffer to buffer the output voltage: simulate this circuit – Schematic created using CircuitLab This will give you the same voltage but at less power consumption. Oh, you want greater precision ? Then why not do that this way, see circuit 2 or 3 Of course adding the opamp is ...


4

The professor in that video is trying to demonstrate the effect of feed back on noise and other non-linearities. Lets start by assuming the amplifier is ideal: infinite gain, infinite bandwidth and capable of driving any load. I know these do not exist but it makes it easier for us to do a thought experiment. Taking the circuit of Figure 1 then the ...


4

There are 2 obvious possibilities for falling "gain" with increasing input level : you're clipping (output is hitting either supply rail, or both. Unlikely with gain=10, and 5mv in). You're making DC measurements, and confusing DC offsets with input signal. As DC offsets are likely to be around 5mv with low grade opamps, that's my bet. What you expect: ...


4

Do the slew rates combine in some way? YES Is unity gain instability the reason for the second decrease in slew rate? NO Is there a variable that affects slew rate I am not considering? YES Follow the recommended Figure 1 test conditions in the datasheet then understand why. Slew rate is current limited into either Miller capacitance or load ...


4

You can simply resort to ohm's law: in a resistor, I = V/R. If V is zero, which it will be if the two ends of the resistor are at the same potential, then I must also be zero.


4

Before throwing too much effort in solving the complicated equations, there are a few basic considerations which are easy to do and as seen, it is worthwhile the effort. The common base has a current gain of close to one because the base current is very small, so what goes in to the emitter comes out of the collector - gain close to 1. The input current into ...


3

There are essentially two mechanisms: Internal nonlinearities, arising from non-ideal behaviour of mixers and amplifiers. When passed through a nonlinear device, a modulating signal containing F_1 & F_2 can produce an output signal with side bands located at: $$\{F_1, F_2,\\ (F_1+F_2), (F_1-F_2),\\ 2*F_1, 2*F_2,\\ 2*F_1+F_2, 2*F_2+F_1,\\ 2*F_1-F_2, ...


3

LEDs and light bulbs have an essentially different dependance on temperature. In light bulbs the non-linearity of the V-I char is essentially due to the operating temperature: the hotter the filament, the harder it is for electrons to travel through madly-hot atoms, the higher the resistivity. The key point here is that the temperature rises because ...


3

Most devices have a single solution for the Q or operating point. e.g. NPN transistor (graph 1). However, there are a number of devices that exhibit a region of dynamic negative resistance (dV/dI = -R). In this region you could have three Q points. (Graph 2) Q1 and Q3 would be deemed as 'stable' because in this region they behave as a normal resistance (...


3

This was due to a too high input impedance and/or to short sample time. It did not have time to fill the sample capacitor fully.


3

Inspired by the many correct considerations of Tony Stewart, I'll give my contribution and answer in the same order to your questions. Is this decrease in slew rate due to the fact that the signal was sent through two op amps, each with its own slewing effect? Yes: cascading many stages of amplification lowers the overall speed of their response. ...


3

Use a precision comparator with a really tiny bit of hysteresis to decide whether the input is above or below 1000.5 mV. Use the comparator output to select one of two signals using a transmission gate or multiplexer. Signal 1 is the input potted down by 10:1 Signal 2 is the input potted down by 100:1 Use very good precision resistors and don't expect it ...


3

I have a hard time understanding what an operating point and load line represent and why do we need them at all. Do not worry because it is not just you who have this problem. For example, I first started thinking of this concept as a student in the 70's... and I continue thinking of it even now when explaining it to my students... and even when writing ...


3

As per the manual, x is a special variable and for C it's the voltage across the device (so \$i=C\frac{\mathrm{d}v}{\mathrm{d}t}\$, thus \$v\$ is derived), and for L it's the current through it (so \$v=L\frac{\mathrm{d}i}{\mathrm{d}t}\$, thus \$i\$ is derived). But x is not a mandatory variable, that is, just because it's there it doesn't mean you have to ...


3

Let’s say you have an interharmonic of 1.5. If you scale your assumed fundamental frequency by 0.5 then you will have two integer harmonic sinusoids contributing to the signal - a 2nd harmonic and a 3rd harmonic. Equivalent, of course, to the original signal - just an integer way to look at it. First case: Fundamental = f1 $$ x(t) = \cos{(1*2 \pi f_1 t + \...


3

Is there something I'm not paying attention to, that would greatly simplify this problem? Just start simply. Ask yourself questions about (say) diode D2 which I've marked nodes on called A and B: - The question you should ask is this: could voltage B be greater than voltage A. Think about it - is it possible that B is greater than A? If you conclude it can'...


3

Note: it says "of the form", not "strictly", which means it's an antisymmetric waveform (odd function) due to the \$x^3\$ part. And if you look at the circuit closely, the two Zener diodes with the 2.2 kΩ form a so-called dead-zone, because the input voltage will need to be greater, or lesser than \$\approx\pm V_Z\pm 0.7\$. This makes the ...


3

This led me to the question, what are some examples of 'simple' nonlinear circuits in our daily life? Anything with a diode, or a transistor is a non-linear device. The output does not always scale to the input (some transistors can be put in a linear mode, but saturation prevents them from being fully linear). And, what are the new parameters introduced ...


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