New answers tagged

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A volt-ampere in electrical terms, means the amount of apparent power in an alternating current circuit equal to a current of one ampere at an emf of one volt. It is equivalent to watts for non-reactive circuits. 10 kV·A = 10,000 watts capability (where the SI prefix k equals kilo) 10 MV·A = 10,000,000 watts capability (where M equals mega) While the volt-...


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Your voltmeter is probably showing DC offset because depending on how you built your Chua's circuit (and what it is doing at the moment) it will have a DC offset. From your linked document: It looks to me like a properly operating Chua's circuit can have offsets in various directions. Outside of that, your circuit may be operating incorrectly (or not at ...


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The answer is within the linked PDF file. This will be our starting point too, but with a slight difference. Electronic components are available "off-the-shelf" in standard values; 7.13 mH, 5.56 nF, 50nF, and 1428Ω are not standadrd values. Therefore, we choose 18mH, 10nF, 100nF, and 18000&Omega as a nearby "standard" value starting point. ...


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This case is usually slightly more complicated than series RC, but in the end is the same concept, if you want to see the equations developed look for thevenin RC parallel circuit online. Plenty of sources should come up. your current equation for R2 looks like this $$ I(R2)=\frac{V}{R1+R2}(1-e^{-\frac{(\frac{R1}{R2}+1)t}{R1C}}) $$ if you multiply this by ...


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Your power equation listed above \$ P = \sqrt{3} V_L I_L \cos(\phi) \$ is for a balanced three-phase system. Your work shows that this is an unbalanced load. For an unbalanced load, we have to use other methods to calculate power. $$ S = P + jQ = VI^*$$ $$ P = Re\{V_{AN}I_1 \angle\phi_v-\phi_i + V_{BN}I_2 \angle\phi_v-\phi_i + V_{CN}I_3 \angle\phi_v-\phi_i\...


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Some things to try: To solve the offset problem try putting a large resistor to ground on the input of the VCVS, like 1e12 or larger. That way the simulation can find the DC operating point on startup. To solve the gain issue, use an actual integrator with resistors Source:https://www.electronics-tutorials.ws/opamp/opamp_6.html


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Assuming that the three feed voltages are balanced and are sine waves i.e. line voltages are identical with exactly 120 degrees between them, any star connected load of equal values will produce 0 volts at the star point. This means that three identical resistors will yield 0 volts AND three identical inductors (or capacitors) will also yield 0 volts. With ...


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A linear system in electronics means a system that can be expressed as a linear differential equation. Hence $$\dot{x} = \mathbf{A} x + \mathbf{B}u,$$ where \$x\$ is the state vector, \$u\$ is the input vector, and \$\mathbf{A}\$ and \$\mathbf{B}\$ are matrices. Let \$x_1\$ be the state from \$u_1\$ and let \$x_2\$ be the state from \$u_2\$. Thus $$\frac{\...


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It is generally said that the superposition theorem is a result that the circuit is linear. I think that is not correct and actually a bit circular. Because one describes the linearity of the input/output relationship of the circuit while the other talks about how to solve the circuit by turning off sources. Here I prove that the superposition theorem in ...


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Your second picture is of a different situation. The two negative terminals of the batteries are at the “same” voltage, aka the voltage between them is 0, so no current flows from one to the other. Same in quotes to show we are in the ideal world. This is the exact reason no current flows in the first image. However if you measured the voltage between the ...


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The n= 1 trig form must equal the sum of the +/-1 exponential terms because those two exponentials are the only terms in that entire infinite summation with the same period (frequency). Without even thinking about the math and the fact that sinusoids of different frequencies are orthogonal, at a purely intuitive level, if two frequencies are to be equal, ...


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This equation contains ALL fundamental frequency terms. It's written to find how a's and b's are connected to c's. In the red box the left side is a general formula how to make an arbitary real valued sinusoidal signal with arbitary phase angle by summing sine and cosine which both have ZERO phase angle. In the right side there is a sum of complex numbers ...


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In the circuit as you show it you require no C1. As pointed out in the prior discussion, the capacitor may turn on the Triac on sudden rises in opamp output. The TL431 is not really suitable for what you are trying to do since it requires a minimum Ik to set the reference (0.4mA). The strange conduction you are seeing is in all probability due to the ...


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The opamp does not provide sufficient current to blow the fuse. The fuse is rated for 200mA (for the lowest current rated fuse in the family), the opamp can only supply 80mA (if running at 5V at 2.7V it's only 30mA) , or less then half the current to blow the fuse. Lets suppose that ground was attached to the other end of the fuse, instead of the crowbar ...


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If you correctly calculated the impedance then: \$I=\dfrac{U}{Z}\$ or simply \$|I|=\dfrac{U}{|Z|}\$. \$Z=\sqrt{R^2+X^2}= \sqrt{1+4}\$ then \$P=I^2\cdot R\$ and \$Q=I^2\cdot X\$ Or yet simpler: \$P=\dfrac{U^2\cdot R}{R^2+X^2}\$ and so on Znači D.


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how would C1 discharge since only I1 will be passing through the mosfet? Because it had been charged when the mosfet was not conducting as shown in your edited question, picture 6. In picture 2, when the mosfet is open, and C is being charged through D1. This is simplified in picture 6: the current \$i_1\$ is charging C1 and \$i_1 = i_{C1}\$. When ...


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The low side sink current switch causes a high release voltage after clamp to 0V.


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Although the solution using the keyword 'startup' seems to work, I would recommend not to use the "startup" condition (unless it is explicitely desired to have the DC source also starting up from 0V). It's better to use initial condition directive .ic i(L1)=0, so voltage source V1 is 5V at time = 0 seconds. The "startup" option lets DC sources ramp from 0V ...


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This is an inductor current graph, not an output capacitor voltage graph, so it's difficult to tell from it whether the output capacitor discharges through the load (it does). The current between \$DT_s\$ and \$T_s\$ is not constant, but the derivative of the current is proportional to the voltage difference between the inductor terminals, which during "off"...


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Suppose you have a transistor having a \$h_{oe} \$ equivalent to a 10k resistor. You wish to make a measurement of \$h_{oe} \$ at a DC collector current of 100uA. To prevent loading \$h_{oe}\$ in your measurement test jig, you must supply a DC collector bias voltage through a much larger resistor...say 100 times larger....like 1Meg ohm. In this manner, the ...


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If a transistor is a current sink you expect infinite impedance output, but due to the Early Effect shunt resistance and output capacitance and layout stray impedances including probing method, this can be >1M and <1pF depending on size , ratings and frequency applied. Using a short circuit current with Rc=0 is the easiest way uA to Amps but then Vce= ...


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Suppose you want to characterize the transistor at 1GHz, yet that node has 10pF or 20pF parasitic capacitance what with 1) scope probe 2) biasing circuit 3) mechanical structure to hold the device At 1GHz, 1pF is -j159 ohms. At 1Ghz, 20pF is -j 8 ohms. Thus the realworld prevents "open circuit" tests at high frequency.


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You can consider each opamp in isolation. Each opamp can be seen as a negative impedance converter within its linear region. Consider the first one, lets apply 1V and see what current we get... With 1V on pin 3 the opamp will (within its linear region) act to drive pin 2 to 1V, which will mean there is 1V across the 3.3k resistor R6, causing 1/3.3 mA to ...


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You are missing the circuit in your question. Anyhow... Anything which safely isolates your AC supply from your output supply - such as using a transformer with full bridge rectifier and a regulator, should work to keep you safe, should you touch the 9V DC (i.e. none of the AC will get to you from there). A transformer is commonly used as its a good way to ...


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A fairly common example of when RLC circuits can produce an excessive voltage is a 2nd order low pass filter used in conjunction with a voltage regulator. I've fallen foul of this myself so I know it can be a problem. I designed a 2nd order filter to remove noise from the input feed to a voltage regulator. The voltage regulator was nominally fed about 20 ...


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Consider that the circuit board itself and/or any insulated wiring has a breakdown voltage. After breakdown, any nearby components might be adversely affected (exceed specified maximums on random pins, etc.)


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The high Q of Series RLC circuits amplify current according to f and Q= 2pifL/(DCR+ESR+Rs) ratio. Parallel RLC circuits amplify voltage at fo, according to shunt R for high Q=R/(2pi*fL) at fo. Damage depends on component ratings for ripple current induced power dissipation or BDV. Any questions?


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A common emitter amplifier is a single transistor amplifier where current entering the base of the transistor is multiplied by a factor (typically called beta or hfe) to give the current entering at the collector. The important features distinguishing a common emitter circuit from other single transistor amplifiers are that: The base terminal is an input ...


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Depending on the FET fabrication method, you may find Rgate and Rsubstrate are adequate to dampen the resonances.


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YES,YES 1) See best datasheets for test methods. 2) All traces have ESL starting at ~0.5nH/mm and scope probes are also notorious for ground ESL resonance with 10:1 coax xx pF, for resonance <50 MHz, thus tip/ring method is best between to nearby pins <1cm apart with gnd clip and probe tip both removed.


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For circuit 1: it will discharge until \$ V_C = V_{BAT} \$ $$ V_C = (V_{C0} - V_{BAT}) \times exp \left( \frac{-t}{R1*C} \right) + V_{BAT} $$ For circuit 2: it will discharge to zero but \$ V_{C1} \$ depends on the previous equation $$ V_C = V_{C1} \times exp \left( \frac{-t}{(R1+R2)*C} \right) $$


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THIS CIRCUIT WILL KILL YOU Or your family Or your friends The capacitor MUST be X or Y rated for AC mains use. If it is getting hot you are using the wrong cap. Regardless of what formulae and calculations say, if the resistor is getting too hot you must use a larger wattage one or cool it better. The same applies to the zener diode. This circuit ...


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With 12Vdc drawn from 230Vac, you have a current-limit by 100Ω and 2.2uF @ 100Hz = -j800 Ohms for an apparent current of ~170 mA . with 230mA rms thru 100 Ohms the power dissipation is I^2R-Pd=5.3W exceeding its rating and operating a finger burning max temps of 150'C+ Conclusion You must use an e-Cap rated for > .4A ripple current with low ESR for ...


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simulate this circuit – Schematic created using CircuitLab Simulating that shows you get a 15V peak drop across the 100 Ohm resistor. 15V*15V/100Ohms = 2.25W. Whether the resistor can survive that and is rated for it, it's still burning a lot of energy and it's going to get hot. Also, what everyone else said. Be safe.


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Make sure that your cap is a safety cap ,like say X rated 275VAC .Not 4oo VDC.Your Zener will get very hot when the supply is not loaded .If your hot zener goes open circuit the cap will get too much volts and get hot and fail.These non isolated supplies can present a shock hazard .Seek local advice and do not get electrocuted.


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Your circuit does not show LEDs or any other load. If you have LEDs connected to the DC output then maybe they are overloading the resistor and capacitor. You should never touch any part that is in a transformerless circuit to avoid a lethal shock.


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On my iPad here with a BT keyboard. ™ alt-T. Block quote. Alt-Q ∑ Alt W FWIW Mathjax Basically use HTML for best or Mathjax which is “bound by \$ before and after \therefore I am an EE, \$\therefore\$ I think outside the Ω box.


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Did you google it? Because the symbol is not usually available on keyboards and within fonts, and to prevent compatibility issues, users have substituted a colon followed by a period (":.") to represent it. Source: https://en.wikipedia.org/wiki/Therefore_sign


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HPSENSE - High-Pass Sense Input for Instrumentation Amplifier. Connect HPSENSE to the junction of R and C that sets the corner frequency of the dc blocking circuit. HPSENSE is a negative feedback point for the instrumentation amplifier that processes the input and, if you connect the output of said IA to HPSENSE via a low pass filter (yes, that is what ...


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which is the phase relationship between the primary and secondary voltages The dot convention tells you which terminals will have secondary and primary voltages in phase: - Above picture from here. Consider also this picture of two identical windings wound the same way: - For any secondary load current, the primary load current will be 180 degrees out ...


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You are thinking too deeply about this. Consider this: - It has the same formulas as your question does (the parallel one re-arranged slightly, but the same) and, because the capacitance (around the loop) is now fractionally less than for the straight series capacitance scenario (because the two capacitors are in series), it has a resonance impedance peak ...


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I don't understand why I can apply a voltage divider in the 30K and 20K resistors since the base is connected between them. The emitter resistor of 1 kohm projects an impedance onto the base that is beta times higher. So, if beta is 100 and the emitter resistor is 1 kohm, the base will look like a loading impedance of 100 kohm. This loading resistance is ...


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Even if all currents and voltages being 0 is an equilibrium of your system of differential equations, if this system is chaotic, it must be an unstable equilibrium. As soon as even the tiniest current or voltage enters the system (as a result of picking up radio waves, or from static discharge from you touching the circuit while building it, or from moving ...


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Without knowing what a "Chua diode" is, one would get the impression from this question that this circuit is supposed to be a perpetual-motion or free-energy device. The first thing I would look at in that context is the inductor. Unless you place the device in an RF-free zone such as a Faraday cage, the inductor will couple with the magnetic element of ...


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The power comes from Chua's Diode. It is not a passive device. It actively provides power as you can see quickly if you look at its i-v-curve: i-v curve of Chua's Diode (image from Wikipedia) The i-v-curve of Chua's diode occupies the 2nd and 4th quadrant. A passive device would occupy only 1st and 3rd Quadrant: See section "Types of I–V curves" in ...


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IT MUST SUPPLY POWER. Follow this reasoning: A transistor is a negative resistance (NR) device only when it is biased with some power source, so it is not a negative resistance thru the origin of V vs I. The same is true for gas tubes and other passive NR devices. A Power supply with a transistor with gain is a negative resistance combination. If we say ...


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Reality is this - if you apply a voltage source to the primary of a transformer then the phase difference between primary voltage (dot end) and secondary voltage (dot end) is zero. In other words the dots tell you about the phase relationship between primary wires and secondary wires. Because of this, if you have a load resistor connected to the secondary, ...


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Most likely your column (sink) drivers are leaking some current in OFF state. Since I don't have more information on the type of drivers you are using, here are a few possibilities: The column driver has push-pull outputs as opposed to open drain or open collector outputs and the high voltage is lower than the voltage provided by the source driver minus ...


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The voltage will drop at a rate of dV/dt = Ic/C and if current is limited by 1k then Ic (max) =V/R where a power interruption or switch to battery becomes worst case dV=V so the time constant for 64% voltage sag = RC=Tau = 10ms so a 1ms interruption is only 6% and less for a FET bridge. Also any ripple > 1/RCf is attenuated by 20 dB/decade in f, which may ...


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Section 14.1 of the datasheet tells you why - the supply is monitored to allow it to switch over to the backup battery if and when it fails, and needs to not fall by more than a specified rate, such that the monitor circuit has time to make the switch over before power is lost.


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