# Tag Info

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It appears to be half-wave rectified AC voltage so lots of flicker with LEDs at 600 RPM or 10 or ? 20 Hz Since LEDs will be more efficient, you can measure lamp voltage then choose to rewire lamps to DC if you consider all the other lamps and current load on the rectifier.

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The convention in SPICE world is that two pin elements have a unique direction of current, from pin 1 to pin 2. For 3 or more pin elements, the current is always considered going into the pin. In the link you posted they are using custom symbols, something stated right in the opening paragraph (top of the page): In addition to LTspice IV, this tutorial ...

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It does look like the headlight runs on AC - this is quite common, since filament lamps work well on AC and it avoids having a rectifier in that part of the circuit. So for a LED replacement you will need a rectifier. This won’t resolve the flicker at low revs though, adding a capacitor may help.

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It looks like you didn't check the link in the comments. Using currents with behavioural sources can be problematic, something officially explained in the help (emphasis mine): Circuit element currents; for example, I(S1), the current through switch S1 or Ib(Q1), the base current of Q1. However, it is assumed that the circuit element current is varying ...

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Blockquote All I know is that all the capacitors would act as short circuits Um ... no. A critical aspect of this circuit is that the impedance of the capacitors is ((not)) constant; it varies as a function of the frequency of the signal going through it. This is why (in round numbers) a treble control does not affect the bass frequencies. Write down ...

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If you adjust the bias circuit to just about or few mV above the voltage over a base-emitter junction. I.e. 0.6V where the transistor is close to off, then voltages above that will turn the transistor on whereas negative voltage will not get amplified because the transistor is already off. Simulating different scenarios reveals that this is far from ideal: ...

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I think I should move the Q point on the DC load line to either Ic=Ic(sat) or Vce=Vcc to only allow half the AC signal to be outputted, This answer involves setting the Q point to Ic=Ic(sat) rather than I(cutoff). Although it amplifies half the AC signal, in common emitter configuration, it does not have the nice quality of low quiescent power found in &...

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Change the values of R1 and/or R4 so that the transistor is not quite turned on, ie. the voltage on the Base should be a bit less than 0.6 V (eg. change R1 to 330k and you should get ~0.5 V). The transistor should then be turned off when 'quiescent', and positive halves of the AC signal should turn it on.

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For a general RLC circuit - 1.Indentify the output asked(given) in question ! 2.Find transfer function i.e $$T(s)=Vout(s)/Vin(s)$$ using KVL, KCL , Mesh , nodal etc. Methods 3.And once you get the Transfer function(which will be 2nd order most likely ) then convert that Transfer function into standard form i.e $$\mathrm{ H(s)=K \frac{\omega_n^2}{s^2 + (2\... 0 This is exactly what a function generator is meant to do. The obvious solution is to figure out what takes time when sending a command to the function generator – it really should take milliseconds. If it's some stupid driver issue in labview: most function generators that can be controlled from a PC listen to SCPI commands, and there's libraries to generate ... 0 The GND shown in the circuit is not PE (protective earth) but is the ground for the driver circuit. The driving circuit will float at mains potential so needs to be isolated from anything that a user might contact. -1 Can't you add a couple of diodes in series on the middle pin of the voltage regulator (assuming the OP was including a regulator) to increase the voltage to what is required? 0 i belive by storing the kinetic energy become potential energy is the best option rather than stroing with batteries that turns it into DC, consider all those power loses 2 In an induction motor, the stator and rotor magnetic fields rotate in the same direction at exactly the same speeds. The rotor (and thus also its conductors) rotate at a speed and direction that is determined by the operating torque. The rotor turns at a speed that is less than the speed of the magnetic fields Unless the rotor driven by an external machine ... 2 Please see this page for information about SPICE: SPICE Information In there is a discussion about inductor loops: You have just this construct in your netlist (L1 and L2) and for no reason that I can discern. You can combine your two parallel inductors into a single inductor with 1/2 the L value and this problem will be resolved. 0 Since you are asking about an AC generator, you need to consider the generator's inductive reactance. That is responsible for most of the voltage drop that remains when the speed is maintained to provide the rated frequency. With a wound-field synchronous generator, the voltage can be more closely maintained by adjusting the field excitation. That is not an ... 0 The flyback diodes were too expensive to buy, and we had no other varistors here for these voltages required. So we ended up changing the AC valve with a DC one of 24V. But the alternatives of using a higher value varistor or back to back zener diodes, would also work. 1 An isolation transformer with a relatively small core will limit the maximum energy available to the circuit. For example, a 100 W transformer would limit the short circuit current to approx 1 A (2 x the core rating). A fast-acting magnetic circuit breaker also will limit the damage caused by a circuit error. If you cannot find an isolation transformer ... 1 Note these are some risk of my circuit that I think of. High voltage jump across some gap. With clean dry surfaces expect arc to sharp edges at 1kV/mm and use a 3:1 safety factor Shoot through via half bridge (Short circuit) This requires a design deadtime for a known reactive delay and load Q. However deadtime also causes flyback. Design appropriate ... 3 My circuit is 230 V BLDC driver using diode bridge to rectify 230 V AC to 230 V DC ... No. The DC link voltage will be $230 \sqrt 2 \ \text V$. When something go wrong it might really dangerous what should I do? Use an isolation transformer. Connect an incandescent lamp in series between the transformer supply and your test circuit. The lamp will ... 1 At least using an isolation transformer 230V/230V. By the way, the rectified voltage is something approx. 325VDC and not 230VDC. If you would like to use a scope, then a HV differential probe is also a good option. Yet another good option would be using a HV DC power supply with current limit setpoint. -1 when something go wrong it might really dangerous what should I do? If you are unsure what you should be doing with 230V AC, then you should not be doing experiments with 230V AC. Period. In the US, accidental electrocution is the 6th leading cause of work-related deaths. 1 It really depends on the architecture of you op-amp. Amps designed for bipolar supplies usually clips in the same way on both positive and negative limits. Single supply amps often can reach really near the ground (which is really useful for DC signals) Output rail-to-rail can be build in different ways, they are either using an output stage which can go at ... 0 The signal you collect with "ant" comes from AC wires through the capacitance between AC wires and "ant". To get some input current the GND should be at least capacitively connected to the earth of the mains AC system. That's seemingly already said by others one minute earlier. Making the circuit slower by increasing capacitance C7 can ... 0 Nice concept but ultimately an opto-isolator might have been simpler. The antenna generates a voltage at its output but in the absence of an earth the whole circuit moves up (or down) by the same voltage meaning that the circuit does not operate. You could try to use the fact that you get an AC voltage across the ends of the antenna. Couple the unconnected ... 0 The AC voltage rating is the maximum Voltage the capacitor can handle divided by the square root of 2. So this capacitor can handle 630V / 1,414 = 445,5V I wouldn't go higher than 400VAC though. 0 Getting the correct RMS value from mains can be very challenging project. First you do need a quite high sampling rate, then use a lowpass FIR filter, but preferably a bandpass so it eliminates also the DC drift. Then you square, integrate, calculate square root. Preferably you do integrate only full periods, so zero cross detection is also needed. If you ... 0 If you really wish to not use a transformer, one possible solution is to use a isolation power supply for your amplifier circuit. However, your design is not going to work because you didn't isolate the ground. The whole point of using an isolated power supply is to isolate both the ground and power. A true isolation power supply will have two input, power ... 0 Could be that the thrust bearings, that keep the motor from moving forward and backward along it's axis, have failed. So as soon as you energize it, the magnetic field makes the motor move forward and jams against the housing. There might not be an economical way to fix that, meaning the fix may cost more than a replacement fan. 1 for a sine wave $V_{ac~rms}=\dfrac{V_p}{\sqrt{2}}$ Vac signal with DC offset = $\sqrt{V_{dc~rms}^2+V_{ac~rms}^2}=\sqrt{V_{dc}^2+\dfrac{V_p^2}{2} }$ 0 Since the rate of change of voltage is steady ... if V = -ΔΦ/Δt , ΔΦ changes linearly over time so the induced voltage is constant). As mentioned in the comments, the rate of change in a sawtooth is constant on the sloping portion. The rate of change is infinite on the falling portion. 0 I am a heating contractor and have performed several jobs over the years. Once I installed an outlet and had 120v but nothing operated. I recently installed a furnace and the old one ran not the new one didnt e cept for a very short time. The pro lem is improper grounding. Sure your meter says it's there but depending on situation and/or equipment being ... 0 Try this configuration from the schematic 'A' and it should do the trick. Good Luck! KB 0 / WATTS 150 W 120V. Standard IEC Plug Is this your model? I suppose the isolation transformer is necessary for preventing 0Vdc from being offset from Neutral or miswired Hot to neutral swaps. Yes I would use a 65W min 16V universal power supply. It has a non isolated DC to Dc converter, so other than possible EMI crosstalk, and charging up 3.3mF cap on ... 3 Earth terminal. It seems to be missing a screw. 1 In the case of a (pure) DC signal, you have, at a frequency f :$$ f \ne 0 => P(f) = 0  And since your signal is pure DC, it is not a finite energy signal. So it's Energy Spectral Density is not defined.

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Is this intended to be mass-produced and sold to consumers???? EDIT: Instead of adding fuses... since this is a one off build and won't be subject to professional testing.. please drop the AC high voltage off this board completely. You can make a truly safe design using a common wall adaptor. As a bonus, since power supplies are a common failure point, you ...

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In AC network power flow is controlled mainly by phase angle of voltage (not magnitude as in DC systems). To send power to the grid an inverter must generate EMF shifted relative to the mains voltage. To achieve this you may have an inverter with internal frequency generator. The control circuitry will slowly increase the generator's frequency until power ...

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Powering a flashing circuit directly off the AC line is difficult compared to other approaches. Consider a low voltage flasher circuit such as something using a 555 (there are tons of examples online) powered by a wall wart, that drives a solid state relay to handle the light voltage and current. More parts and larger assembly, but way easier - and way WAY ...

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It’s unlikely that you would get a meaningful power factor - the 1.6A is presumably the maximum at the lowest supply voltage (100V) and probably has a safety margin built in. The difference between the ongoing and outgoing power gives you the efficiency of the device (about 85% is typical). Measuring the power factor is a more complex task; there are a few ...

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If you need to keep Neutral separate from the local ground, then I think you'd be better of using a differential amplifier configuration to drive IN1P and tying IN1N to ground. Something like this: simulate this circuit – Schematic created using CircuitLab

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Yes it does matter - if the common-mode voltage swing goes outside the isolated supply rails then bad things will happen - at best you won't get a meaningful conversion, at worst it could damage the ADC and op amp. Somehow or another you should either bias or just connect the neutral to the 0V if you want meaningful results. As has been pointed out, any ...

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For AC power / Hum in audio systems it goes in stages. Bad: Romex Good: Conduit Better: isolated ground and conduit Best: Isolated ground and conduit. The hot and neutral twisted and the safety ground in the same conduit but not twisted. https://www.jensen-transformers.com/wp-content/uploads/2015/02/AES-Ground-Loops-Rest-of-Story-Whitlock-Fox-Generic-...

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In a 3-phase electromechanical relay, there are three contacts that are electrically isolated from each other but are operated simultaneously by a single electromechanical actuating mechanism. There is no means provided to open or close one contact without the other two assuming the same state simultaneously. There is also very little possibility for a ...

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Which wires should connect to live and neutral on the 220VAC supply and which are live and neutral on the 24V output? Black and red are 230 volts Brown and brown are 24 volts My best guess is that the black wire is used on both the input and output as a common neutral No, the two transformer windings are isolated from each other. Since I have 220V AC, I ...

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