# Tag Info

## New answers tagged circuit-analysis

2

First of all, you wrote: "defined by the current across" but the current is trough a component, voltage is across. Well, we are trying to analyze the following circuit: simulate this circuit – Schematic created using CircuitLab Using KCL, we can write for every node the node equation: $$\begin{cases} \text{I}_{\text{R}_1}+\text{I}_{\text{R}_2}=0\\ ... 0 With the R ratios given, the Voltage gain to RL=100 Ohms output= + 0.1 Vin. The non-INV gain is greater than the INV gain. If the 100 changes to 1k then the gain increases to +1. 0 Right answer, wrong reasons. Just because someone is too lazy to show the power supply rails doesn't mean they're there. You could easily change the values of some of the resistances and make this thing into a comparator with hysteresis. Break the problem down! You have options, but I would set $V_o$ to zero, and solve for $V_{out}$ as a function of $... -1 Well, we have the following circuit: simulate this circuit – Schematic created using CircuitLab Using KCL, we get: \begin{cases} \text{I}_\text{V}=\text{I}_{\text{R}_1}+\text{I}_{\text{R}_2}\\ \\ \text{I}_{\text{R}_2}=\text{I}_{\text{R}_3}+\text{I} \end{cases}\tag1 Using KVL, we can write: \begin{cases} \text{I}_{\text{R}_1}=\frac{\text{V}... -1 Perhaps this simplifies how to look at things. I've grounded the left side of your schematic, below. (We can do this for exactly one node of our choice.) simulate this circuit – Schematic created using CircuitLab I think, now, that you can see that$R_1$doesn't really have any impact at all on$I_{\text{R}_3}$. So you can ignore that part. I ... 1 You are using superposition to solve a circuit by solving each source separately, then adding the results. When you solve the circuit for a given source, you have to remove each OTHER source from the circuit. For a current source, you do this by turning it into an OPEN circuit (i.e. no wire); for a voltage source you do this by turning it into a SHORT ... 0 Your teacher is wrong. Of course there are some currents passing through that resistor. My simulator is showing 24mA will pass. Check your maths please. -1 The process of charging a capacitor via a resistor, even if the resistor is very small, produces an energy (or IIR) loss each cycle. An inductor based switching regulator doesn’t suffer from this theoretical loss because the energy that can be put into an inductor by connecting it momentarily across a supply voltage (as in the case of a boost regulator), is ... 7 The plots below (taken from here) show why calculating ADC 'bandwidth' from peak sample values may be problematic. The plot on the right shows what happens when the sampling frequency is below the input frequency. The signal has aliased down to a lower frequency inside the Nyquist limit, and there is no way to tell what the actual input frequency was. ... 14 The bandwidth is the frequency for which the output is down 3 dB. It it a function of the analog portion of your circuit and does not depend on the sample rate nor the Nyquist rate which are basically digital concepts. Thus, in your experiment, the bandwith will be the frequency, fin, for which the output has dropped by 3 dB. 4 I think you are confused about what happens when the input frequency exceeds the Nyquist limit. The peak-to-peak voltage doesn't change...assuming you collect enough samples...but the apparent frequency of the signal changes. If there is a "bandwidth" beyond which the signal is attenuated, then it is due to the analog electronics of your ADC, not to the ... 2 The extra two optocouplers seem to be there not for ground isolation purposes - their both sides are ground referenced so they don't block ground loops. The extra optocouplers just seem to be doing logic level translation, buffering and driving the thru and out connectors. In a rather unconventional way, and not within the required resistance tolerance ... 5 Optocouplers are used for galvanic isolation. In most cases, this protects against dangerous voltage differences, but in MIDI inputs, it just prevents ground loops. However, the optocouplers used for the MIDI outputs in the linked schematic do not provide any isolation whatsoever, because the grounds and +5V power supplies are connected together (they are ... 2 If you trust the other guy to put one on his MIDI in, it's over-protection. But - in the real world, it's 10 cents well spent. Ground loops are a big problem in audio systems. MIDI is not particularly high speed, so there is no reason not to do it. EDIT : as others have pointed out the opto in this schema has nothing to do with isolation. In fact it is not ... 0 There is only one current -- I -- but there are multiple voltages and voltage drops around the circuit -- VR, VL and U. We know how the current through the inductor changes: \frac{dI}{dt} = \frac{V_L}{L} And from Ohm's law: V_R = IR And because it's a circuit the source voltage equals the total voltage drop: U = V_L+V_R Solving for the ... 2 If you take only the output from the final amplifier, then the transfer function is: At some value of input, the output stage will saturate and not increase any further. If we then add the other outputs, we will get (not to scale): As each amplifier saturates, then the slope reduces until the next amplifier in the chain saturates until all 4 amplifiers ... 0 Well, first of all we can look at the wiki-page of 'Operational amplifier applications' and look under 'Inverting amplifier', to find that your TF is given by: \mathcal{H}\left(\text{s}\right):=\frac{\text{V}_\text{o}\left(\text{s}\right)}{\text{V}_\text{i}\left(\text{s}\right)}=-\frac{1}{\text{R}_1}\cdot\frac{\text{R}_2\cdot\frac{1}{\text{Cs}}}{\text{R}... 1 You are correct that the current in the inductor can't change instantly. But it certainly can in any of the resistors. So, immediately after the switch closes, you have this situation: simulate this circuit – Schematic created using CircuitLab where I1 is the same as the coil current before the switch closed. Now, what's the current through V1? 1 To determine the transfer function of this Wien-bridge oscillator, you can try the fast analytical circuits techniques or FACTs. This is the documented problem number 9 actually. The principle is quite simple: you consider the transfer function denominator as a combination of the circuit time constants determined when the stimulus is turned off. Basically, ... 0 There is no practical limit. The idea that charge pumps are impractical or even inefficient has no basis in reality, at least not today. It might have been superficially true at one point many many years ago before higher capacity ceramic capacitors existed, and most electrolytic caps were basically garbage. Actually, most electrolytic caps today are ... 0 The impedance of that circuit will be proportional to the impedance of your capacitors at the frequency of your AC source. I don't know how exactly it will relate in the circuit you are using. That arrangement isn't one I've fiddled with. Since you are using LTSpice, you can easily find out how the impedance of the doubler relates to the impedance of the ... 1 Both have 0 as the steady state condition. You are confusing transfer functions for signals. A transfer function describes a system's behavior; it is a characteristic of some other thing. A signal is more or less a function that evolves in time.$H_1$and$H_2$both have a DC gain of five, so if the systems they represent are excited by a signal that ... 1 Without any reactive components (ie. capacitors or inductors) in your feedback circuitry, the cutoff frequency will be set almost entirely by the characteristics of the op-amp itself. As Caleb commented, the gain bandwidth product from the datasheet will give you an estimate. If you have a specific op-amp in mind I'd directly measure your cutoff frequency. ... 1 It's essential to link points A and B. Even if the capacitors have the same value, if they have different leakage currents, then under DC conditions, the potential of point A, which was initially halfway between the rails, will drift towards one or other rail, over-volting one of the capacitors. R1 and R2 should conduct at least 10x the expected leakage ... 0 If R1 shorts out and V1 exceeds the rating for C2, then the C2 could be damaged. Same for R2 and C1. You could put a bleed resistor between A and B. 1 I decided to investigate how changes in time domain affect the s-domain to see if I could find something that would help to solve my question. And I found something! Let's start with$i_1(t)$and investigate the values in both domains that produce the same current: \bbox[8px,border:1px solid black] { i_1(t)=20\cdot e^{-t}H(t)\,\,\,\text{[mA,ms]}\,\, ... 3 Nothing to do with Laplace. Just some mathematics. I found that you have dimensional inconsistency while you converted units. For instance in this equation :- I_1(s)=\frac{20}{s+1} The dimension of current is milliAmperes, hence the whole RHS should evaluate to 'milliAmperes' units as well.$s$has units 1/ms as you explained. Okay, so in the term$(... 1 So you have selected two bulbs with identical resistance. When the switch is open, these two bulbs form a voltage divider. In your case, the voltage is divided equally due to same resistance. Also, same current is flowing through both of those bulbs. So the circuit consists of two resistors in series. And the current flows accordingly. Now, you close the ... 1 Only Ohm's law is enough I guess. Assuming both bulbs are ideal and of same resistance. In the first case when switch is at open position, both the bulbs are in series. Equal amount of current will pass through both bulbs, which will be:$$I_1=\frac {V_{battery}}{2R} In the second case when switch is closed, the second bulb is shorted across and hence ...

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In signal theory, the Laplace Transformation is linear with respect to units like voltage and current. If you have a function/signal like f(kI, t) and I is the input current level, the LT is of the form kF(I,s), k being a constant, I is the current, t is time, s is complex frequency. But the Laplace Transformation of a function of the form f(t) is not ...

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I did not get context of having limited current of $2.7\, \mathrm{mA}$. however, regarding the capacitor: The capacitor recommended value is $4.7\, \mathrm{uF}$. Other than that, voltage rating of the capacitor should be more than the expected input voltage. Since, the IC input voltage is rated upto $42\, \mathrm{V}$, a $50\, \mathrm{V}$ ...

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RMS should be used to represent signal having zero average (due to symmetry of signal to the positive side and the negative side ). Periodic waves like sine waves usually have zero average. Average value gives the DC content of the signal.

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Assume you are powering on a bulb with a AC RMS voltage of 10 V. It will create same brightness as that of 10 V DC. RMS value is used to represent the strength of the AC signal or equivalent of DC signal. A periodic Sinewave for example, with peak to peak of 28 V will also create same brightness but the average value of the signal is zero (no ...

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A resistive heating element gets hot when connected across 220V AC mains, even though the mean (average) voltage of the peak voltages (+311V and -311V) is 0 volts. Many loads don't care that the voltage reverses polarity with each half cycle. The 220VRMS value gives a more practical way to estimate the real power delivered to the load. For non-ohmic (...

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Yes. Resistors are made in many different ways but consider a wire wound resistor. As you move from your reference terminal 1 to terminal 2 you increase resistance and potential difference.

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An ideal resistor is indivisible, so you can't really talk about voltages inside the resistor. A real resistor comprises some kind of resistive material: a carbon slug, a long wire, a thin metal or carbon film, for example. If you could trace a path from one terminal of the resistor, across this resistive material, to the other terminal of the resistor then ...

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Start at point b and walk to point a, adding up the voltages across components as you go. You have two paths, so if you're being thorough walk both paths and make sure you get the same answer -- if you don't, you made an error.

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You can't solve these kinds of problems with a simple equation because the diodes are non-linear. You need to consider (analyze) four separate cases: both diodes conducting, neither diode conducting, only D1 conducting, and only D2 conducting. Find the currents through the diodes and the voltages across them in all four cases. If you assumed that a diode ...

1

Your equation for figure of merit M, AKA the Q of the circuit, is correct as long as you recognize that $f$ is the resonant frequency, and $C$ is the capacitance that sets the resonant frequency. In this case the resonant frequency is set by the motational capacitance and inductance, $C_1$ and $L_1$. So $\omega = \frac{1}{\sqrt{C_1 L_1}}$, and $... 2 The given RLC circuit is overdamped as:  \frac{R}{2L} > \frac{1}{\sqrt{LC}} Hence the general solution is the sum of two decaying exponentials: i(t)= K_1e^{xt} + K_2e^{yt} where x,y are roots of the characteristic equation: s^2+\frac{R}{L}s+\frac{1}{LC}=0 ie., s^2+1250s+250000 = 0 \implies (s+1000)(s+250)=0 Hence, i(t)=K_1e^{-250t}+... 1 Although I think that @Big6 answer is a good one, I'd like to answer the OP from the Laplace transform point of view, for the sake of completeness. So, suppose you have a linear SISO system that has a generic transfer function$H(s)$which satisfies: Is of order$N$, so it has$N\\$ poles. Is stable, so all its poles have negative real part (they lie ...

2

Imagine the Vdd is brought up from 0V to nominal operating voltage very slowly relative to the RC time constant of Rs*Cs so that node B is essentially at Vdd potential at all times. MS1 could then never turn on. So this is possibly an unreliable start-up circuit because it depends on rapid application of Vdd. Keep in mind that on-chip resistors are ...

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I know this is a late answer, but I've just come across this question and wanted to add my point of view. Suppose we have a generic dynamic (and linear) circuit with one or more external input sources and with some initial conditions at the dynamic elements. That implies that we have chosen an arbitrary time instant as our "initial" time. We don't have to ...

0

What are AC and DC transients? How is transient analysis different from AC and DC analysis? Let's start from what are transients rather! First of all recall that capacitor and inductor are devices which are capable of storing energy into electrostatic or electromagnetic fields respectively (they are not meant to permanently store energy like batteries, in ...

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Remove all components except for D1 and C1. Analyse the circuit behaviour. Now add in D2 and C2. What's different? What's the same? Now add D3 and C3. What's the same and what's different? Do you notice a pattern? Generalise to adding an arbitrary number of Ds and Cs in the same pattern.

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The Vf drop of the diode does have a significant, cumulative effect. Each stage reduces the output voltage by one Vf drop. In the example below, there are two stages. With a 10Vp-p input, the output voltage is 18.66V instead of 20. The difference is two Vf diode drops of 0.66V each (1.33V total.) simulate this circuit – Schematic created using ...

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As Bimpelrekkie explained, whether or not the voltage drop of the diodes is significant or not depends on what voltage you are working with. Since I've been playing with these things for the last few days, I'll post some of the diagrams and measurements I've made. I think the voltage traces explain fairly what happens to the output when the AC is not at ...

1

I do not understand if the voltage drop across the diodes in conduction is negligible or not It is not a case of "negligible or not", as with many things in electronics: it depends. The diodes have a certain forward voltage, typically between 0.5 V to 1.0 V depending on how much current flows though the diode (more current => more voltage drop). If I ...

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A Figure of Merit (FOM), for anything, is a benefit/cost ratio that's found useful to express the 'value' of a particular component. You choose a wanted thing, a benefit, that comes with an unwanted thing, a cost. For a FOM to be useful, those costs and benefits should be easy to scale, so that you can compare the FOMs for different versions of the component....

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In addition to Justin's answer here's a mathematical way of looking at it: Using Vout = A(Vp - Vn) The incorrect assumption that I was making was that Vout == Vp == Vn which is wrong as: Vp - Vn = 0, which is a contradiction as this would make Vout = 0 Looking as Justin's answer we can see that Vout (and therefore Vn) start at 0v and rises with time until ...

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