# Tag Info

## New answers tagged ac

0

Use a relay. simulate this circuit – Schematic created using CircuitLab Figure 1. A possible circuit. The relay separates your 3-phase from your 3.3 V Raspberry Pi circuit making it a safe system.

1

A DC controller "inside" a bridge rectifier may work. Feed AC via load bridge ACin and bridge AC out to AC2. If you short Bridge + & - outputs the load runs on full AC. Now place a DC speed controller between bridge DC+ and DC-. The controller (pot or PWM short or ...) sees rectified full cycle DC but load sees AC. PWMing the DC load control (...

0

"Just a wire", and for that matter the "textbook electrical circuit", only exists in "electrically short" setups where the longest wire is shorter than the wavelength of the highest frequency component in use by an order of magnitude or more. Beyond that, you have to always think of pairs of wires as transmission lines - loose ...

1

The formula features $C$ in the $1/C$ reciprocal configuration because in fact the impedance of a capacitor to the flow of AC current in fact decreases with increased capacitance. For instance, a 1 $nF$ capacitor will not pass 60Hz AC very well at all; it appears nearly an open circuit to that frequency. A 100 $\mu F$ capacitor, much larger, passes ...

1

There's a trick for making a low capacitance, high reactance, capacitor: just twist two wires together. These "gimmick" capacitors were perhaps more common in the past, but may still be found in the wild. So, yes, wires have capacitance to other conductors.

0

Examine the capacitance of Twisted Pairs (2 wires) versus capacitance of a Coaxial Cable (a wire inside a tube). A wire does have very high capacitance reactance as it interacts with stored charges in its surroundings. As you try to send a signal from A to B, a wire in isolation requires the lowest amount of charge. On the other hand, a wire over a sheet of ...

0

Breadboard is likely to have higher parasitic capacitance between tracks than either of your filter capacitors (10 pF and 5 pF) - this means trouble The 100 uH inductor is quite possibly self-resonating below the frequency you are aiming for and this means you won't get sustained oscillations. For both of the above try lowering the inductor value (L1) by ...

3

A straight wire will, as you say, have a very low capacitance and therefore a very high capacitive reactance (dependent on frequency). What you may be overlooking is that the capacitive reactance is only one part of the wire's impedance (R + jX), the other parts being inductive reactance and resistance. As the resistance in a wire will be significantly lower ...

6

The same formula applies. Wires have low capacitance to their surroundings so they have high Xc. Capacitors have high capacitance in comparison so they have low Xc. Thus the wires are a smaller AC load (less current flows) and capacitors are a larger AC load (more current flows). It is no different from resistance, high resistance means small load and small ...

4

The actual connector part will fit without problems - a 4mm banana connector always fits the same across all vendors. Unfortunately the plastic protection is not standardized - I experience problems every now and then. Either the plastic is too long or the thicker part of the connector starts too early or the tiny plastic cap in the middle of the banana plug ...

0

They're all the same 4mm diameter banana hole same diameter shroud and a compatible springy pin. For a phase sequence meter choosing plugs with a shroud that's deep enough to fully protect the pin is probably important.

0

no there's nothing special about this type of adapter. As long as all Earths, all Neutrals and all Lives wires are connected only to each other, and not mixed up in any way, then it should be fine. You may have crossed over one to the other and this may have caused your problem in the first place, but then you'd be the only one knowing if this was the case ...

1

Yes they exist, however i dont recommend you do them yourself as mistakes are deadly. Just purchase a proper Y wire online. Your mistake was probably shorting the live and neutral wires together. This should be avoided at all costs. The commercial wires dont have anything special in them, just doing the wiring the correct way arround and with proper ...

2

The motor current rating indicates that the motor is not a 5 Hp motor by international standards. You should be looking for a 3 Hp VFD. There VFDs on the market at that power level that can provide the required output voltage and current with single-phase input power. There are models rated for 120 V, single-phase input and 230 V, three-phase output. You don'...

1

If you locate a sensitive audio system behind the wall of the panel, the twisting may matter. I'm not sure the heavy gauge 1cm diameter incoming wires can be twisted at 1 twist per inch. However, even at 1 twist per 6", the flux cancellation at moderate distances away (6" or further) will become a bunch of dB. The steel of the breaker box may be ...

6

There is no electrical reason for the twist; it merely serves to keep the two wires physically associated with each other once the Romex jacket is removed. It both "looks neater" and is easier to do subsequent work on. There is no increased capacitance (the distance between the wires doesn't change), except for the fact that the twisted wires need ...

-2

Some electricians in Canada and the US di not use a light or a meter to test if a circuit has 120VAC. Instead they use their fingers.

1

Instead of using a slightly larger number of inexpensive multiple-sourced parts, you could consider using a part designed for this kind of application, the TI ISO1211. See Figure 3. I think values of Rsense = 560 ohms, Rthr = Rshunt = 1K and C = 100uF/50V would work, but check that yourself. It will run off the 3.3V Pi power supply. You might want to use a ...

2

Assume capacitance, as you sat, created by 30cm by 30cm area, thru 1cm of carpet, into underlying conductive underflooring/concrete. Using C = E0 * Er * Area/Distance, with E0 = 8.9e-12 farad/meter and Er = 5, we now have C = 45e-12farad/meter * Area/Distance C = 45-12 * (30cm * 30cm / 1cm ) * 1meter/100 cm C = 45e-12 * 900/1 * 1/100 == 45 * 9 pF = 405 pF. ...

2

You could have felt the shock only in your finger because that small point was where it all entered and therefore had the highest current density, and once it was inside you, the charge spread out while flowing through you thereby reducing the current density. I would think if you were standing on just a tippy toe you would also have felt something there too....

0

Some good analysis about gain is already in the answers given. One additional thing to watch out for your quiescent current and power dissipation in the MOSFET. From the datasheet: These values vary a bit from one manufacturer to another, but this one, from ON Semiconductor, shows max continuous drain current of 115 mA for the 2N7002. Your quiescent drain ...

3

Because the circuit first_stage has resistor in the source, we can immediately ESTIMATE the first_stage gain as R2/R1, or 30 ohm / 10 ohm == 3X. That is the maximum possible. This initial estimate is also useful for bipolar stage (common emitter) design. I view these circuits (FET, bipolar, vacuum tube) as two parts: (1) converting input voltage to output ...

2

Your input frequency is 250 Hz - I see roughly a time-base of 4 ms. At this frequency, the 100 nF (C3) has an impedance of 6366 ohms and is a pointless addition in parallel with R1 (10 ohms). That's my first observation. There's no point fitting it either. My second observation is that at 250 Hz you will not achieve a gain higher than three for this circuit. ...

3

Under the European Low Voltage Directive (which applies across the whole EU, the EEA and countries that follow CENELEC standards) all appliances brought to market have to be safe to use in either polarity. Plugs are reversible. Figure of 8 appliance connectors are reversible. It's also possible that you may have a neutral fault on a TN-C/TN-C-S system as ...

1

Marcus has given a very general answer, which is completely correct. But I think we an actually give some useful answers for the specific model you've presented. How is capacitance of the diode dependent on the frequency of source? Ideally, it shouldn't. Practically, the model might fail at very high frequencies due to the inductance that Andy mentioned in ...

3

all answers to your 3 questions are the same: That's up to the model you're studying here, not universally fixed. We can't tell you how the capacitance depends on the frequency in your model, because we don't know much about that model. To be exact, we only know that this is some model you've found, it applies to a forward-biased diode with a small AC signal ...

0

It happens because of the cheapest circuit possible, too little bypassing, sometimes too much stress for the components (some parts may be operating around limits), pretty much if you just try to sketch a charging circuit, you get a cheap charger. All those thermal compensation stuff and noise filtering is extra cost and extra brainwork. Yes, it's bad for ...

1

Some additional general comments meant as a complement to what's already posted. High-frequency 'squealing' noises coming from power supplies generally implies incorrect / abnormal behaviour. Magnetorestriction caused by high pulsed currents (i.e. the power supply hiccupping energy into a shorted part) can cause acoustic noise. When you hear this, ...

1

The CR2550-R is a cheap-ish current transformer specifically designed to directly drive a LED indicator. It can drive an optoisolator just as easily. Use a bidirectional optoisolator, and it becomes bridge rectifier, amplifier, and pulse remover all in one part. http://burningsmell.org/current/current-monitor.png Above 2000PF at the base, the rise time can ...

2

A Vienna rectifier is pretty much exclusively used for making power-factor corrected (PFC) three-phase DC supplies. The working principle (as with any single-phase PFC) is that boost control is used to produce a DC voltage that is (usually) significantly higher than the peak AC voltage. Only in this way can proper PFC be achieved. This rectifier circuit ...

0

First, realize these observations: Applying Ohm's law to the resistor in phase a: $v_\text{an} = R \, i_\text{a}$. So, if $i_\text{a} = 0$ at some instant, then $v_\text{an} = 0 \, R = 0$ too. And if $i_\text{a} \ne 0$, then $v_\text{an} \ne 0$. Only one SCR of each parallel group (1 and 4; 3 and 6; 5 and 2) can conduct at a given ...

1

With an input signal amplitude of 200 mV p-p, the diode is ineffective at clamping because it just won't conduct any significant amount at these voltages. Here's a picture from another post that shows how much current a 1N4148 can be expected to take at various low voltages: - The Y axis is the diode forward voltage and the X axis is the current it takes. ...

5

First of all, consider that fiddling with mains voltage can kill you should the discharge go through the heart. SMPS have large caps inside which will keep deadly voltage for hours after you pull the plug. Are you sure you can work safely with it? If yes, you should check if the hot (high voltage) side of the PSU still works. My guess would be that by ...

0

There are a couple false assumptions: Sig Gen is not centred around 0V but rather 100mVpk + 100mVdc so 0 to 200mV. The output is identical starting off as you should expect. The 100mV peak steady state on the diode conducts only about 1uA momentarily then almost nothing the rest of the time so the 250k slowly biases the output to 0V. If that were 10M, you ...

1

During the negative cycles, the capacitor needs to charge so that a positive potential develops on its right side plate. This voltage will add to the waveform during rest of the operation of the circuit. To charge the capacitor during negative half cycles only, the diode needs to be conducting during negative half cycles. -100mV amplitude is not enough to ...

6

You can loop a thin wire many times through the hole in the core to increase the sensitivity. For example, you could use this current transmitter from the same supplier with a 10x loop and get 100mA full scale, which would just accommodate a 10W bulb on 230/115VAC mains (about 40 or 80mA). An alternative would be to purchase a handheld DMM that has an ...

2

10 W at 230 V will draw a current of 10/230 = 43 mA. A 10 Ω resistor in series with the bulb will give a 430 mV voltage drop across it which may be adequate if your multimeter has a 2 V AC range with true RMS measurement. simulate this circuit – Schematic created using CircuitLab Figure 1. Current measurement circuit. Divide the voltmeter reading by ...

16

The answer is clear: Buy a new supply and find something else much simpler to "learn" on. Switching power supplies are outrageously complex and not something you're gonna be able to wrap your head around without a good understanding of electronic theory.

7

Powering up with a random component removed is a bad idea unless you know its function and understand what will happen without it in the circuit. It could be part of a clamp that prevented overvoltage damage to your switching FET for example. So by removing it you could cause damage to other components. (Which it sounds like you did.) Diodes don't create ...

0

~1600 LRAs does sound like a lot of fun. However, I believe how you control them depends a lot on what exactly you want to acchieve. I'm by no means an expert, but what I've read is that the resonant frequency of LRAs is very narrow (high Q). Meaning, that if you want them to be efficient, you have to excite them exactly at resonance frequency. This ...

4

Let go of the 5V -> 0.6V DC/DC idea and design an inverter based on a H-Bridge (or many H-Bridges) feeding a transformer (many transformers). You could then let the inversion step also involve the voltage conversion step (which is a good idea, because DC/DC voltage conversion is harder than AC/AC, which can be done trivially with a transformer). Also, 5V ...

1

Summary: Although I can't analyse the other sensors you mentioned using the photos, switch SW1 is not carrying mains voltage. Doesn't look like there's any voltage conversion circuitry I bet there is. At the very least, the start/stop button won't usually have mains voltage on it, since the power for motors etc. doesn't pass through that, as the power to ...

0

The Vout value can never exceed the rail voltage. So if you have ±5V rails, that will be a max value for Vout. Because of Common Mode Voltage Range (CMVR) in many opamps the output will be lower than that by a few 0.1's of a volt. (rail to rail output op amps do not have this limitation). So if an opamp has ±5V rails with a 10Vpp signal, the most it will ...

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