New answers tagged

0

You should not get the same Volts reading between Live-Neutral and Live-Earth. Both Earth and Neutral cables are connected to earth. But as you can see in the following drawing Image from electricalbaba.com between "Ground" and "Neutral tied to Ground", is (usually) soil, which induces more resistance than the "direct" ...


0

It depends on your power sequence. If 12V can come on before the 8V and 5V, and they don't interact, then this can work. If however there is some control interaction between the 5V and 8V and the stuff powered by 12V (like speakers, motors, solenoids, etc.) then you need to make sure that stuff is turned off until the other rails are stable. One technique is ...


0

Sounds like a USB power bank might fit your needs, though if you attempt to drive all the motors at once under load it might be too much current. At least the voltage will be regulated to 5V and the battery charge and protection will be taken care of. The average and peak currents will determine how long a charge will last and whether the maximum output ...


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

Your best bet is some battery source and a regulator to put it down to 5V (it seems that all your gear is more or less 5V powered). It was already commented that some onboard regulator could need 7V, for example. I would, however, worry more about your current budged, powering something like 20 servos will need probably more than 1A (check your datasheet)


1

Think about the phasor diagrams as differences. If E1 = E2, then Er = 0. A synchronous machine can work as a generator or motor. Used as an alternator (generator), the synchronous machine is driven by a prime mover. Once synchronized, prime mover controls real power (kW) and excitation controls reactive power (kVAR). If you take two alternators (...


0

If your intent is to preserve the motorcycle as much as possible, probably the least-invasive way to do that is to use a 12V-6V DC-DC converter inline with the headlamp hi/lo switch. Then this will boost up the current enough for the 6V headlamp. This is a common item as it turns out. Choose one with adequate power for your headlamp (60W or so.) Otherwise I'...


0

I don't get why only the active component of the currents are affected The other way round - active components are not affected. Why? Because active components of the currents determine active power output. Active power output cannot change because power received from prime movers is unchanged. You have to keep power balance.


0

You just fell (slightly) at the final hurdle: - You correctly arrived at 33.33 volts but that is peak voltage. The RMS is 23.568 volts i.e. \$\sqrt2\$ lower.


2

The circuit you show is generally fine. I only have little concerns/recommendations: Add at least some resistance at GPIO 20 to limit the current required to charge the gate capacity. I would increase the pulldown resistor at the base. 470 Ohm will result in 7mA, which usually is not needed in such a scenario. I believe, some kOhm would be sufficient and ...


0

The main thing to keep in mind is that inserting the inductor reduces the motor voltage and the torque capability of the motor as shown in my answer to: Speed control for PSC induction motor. The diagram provided there shows that the torque required to drive a fan reduces drastically with speed. The required torque is proportional to speed squared. As a ...


1

You can use the formula: $$V_{\text{rms}}=V_{\text{peak}} / \sqrt{2}$$ for: pure sine waves (with zero mean) fully rectified 1., with ideal diodes Image adapted from: https://en.wikipedia.org/wiki/Rectifier The other formula is just a consequence of: $$V_{\text{peak}}= V_{\text{peak-to-peak}} / 2$$ for any waveform with symmetrical min. and max. values.


0

You are correct in that the peak voltage and peak-to-peak voltages are the same for half-wave and full-wave rectified signals. However the relationships you cite between RMS and peak voltage are only valid for sine waves.


1

To me, a plug has male contacts (pins). In the specification table for that connector, the "Backend type" line indicates that "40M" is the manufacturer's code for that style of PCB-mounting connector. Somewhere in the catalog there may be a description of how the connector part number is created - what all the bits of the number mean. I ...


0

Q1 Would it be best to use a buck xfmr instead? If the blower is running the way you like with a series inductor, there is no need to change that. It is a wash whether the transformer would be more or less efficient. One disadvantage of the inductor method is that it affects power factor, as you are aware. Did didn't specify the power of your blower, but my ...


5

This is a bit like asking if you can fill a 42 foot high tank through a spout 12 feet off the ground. No. Your source has to be higher than the destination or you need a pump (a voltage booster in your case).


1

Here are two threads that discuss this type of power adapter, AC-DC Adapter - Output lower than rating and what's this 'cont' output for? https://www.eevblog.com/forum/beginners/canon-acdc-transformer-with-24v-cont-gnd-cont/ They both come to the same conclusion, that the Cont pin should be pulled up to the 24V pin and the voltage will then ...


0

The circuit is what it is, and the calculation correctly shows that the power dissipated in a matched load is kTB. (Power is not “developed across” a load, it is dissipated in a load.) Without more context, there is no particular reason to choose a matched load. It leads to the largest noise power dissipation in the load, but there is no reason, in general, ...


4

There is a linear regulator such as LM1117 on the Arduino (and clone) boards you mention. The MCU chip itself does not draw much current, but if you draw current from the GPIO pins or connect something that draws a lot of current to the 5V pin the regulator will get very hot if the input voltage is too high. So, while you can use 12V (and the LM1117 can ...


-5

You can use VIN and GND pin, also I think 12 V will decrease Arduino lifetime, I suggest to use between 6-9 V, and I use 7 V for my Arduino, also Arduino voltage range supported is 6-20 V and recommended voltage is 6-12 V


0

Increasing both width and length of a transistor by the same factor will not increase the overall W/L ratio. However, the length of the device is proportional to the early voltage (VA). In other words, if the length is doubled, the early voltage will also be doubled. This will cause the drain current to decrease by a factor of 2 and the transistor's output ...


4

I don't know about the video, but the Euler's relation \$e^{jx} = \cos(x) + j\sin(x)\$ where j = \$\sqrt{-1}\$, x = \$ \omega t\$ ...means that the real portion Re(j\$e^{j\omega t}\$) = -\$\sin(\omega t) \$ or Re(j\$e^{-j\omega t}\$) = \$\sin(\omega t) \$ So I think there are two things: Take the real part There is a missing minus sign in the book.


0

This OpAmp configuration produces good results. simulate this circuit – Schematic created using CircuitLab


0

Ok gents, I got the voltage divider done with 200 and 66.7ohms, The output changed the original voltage from .6v - 3.3v To .4v - 1.98. And I wanted to achieve 1.89v MAX voltage so Great! Although after it was all done, I took the Subaru for a spin and instantly the ECU knew that it was being fooled and left me stuck at the gas pumps for about 5min until ...


2

This works better simulate this circuit – Schematic created using CircuitLab


9

Assuming you actually grounded the negative side of the +5V supply, you provided the op-amp with adequate supply voltages, and closed the loop around it to make it a voltage follower, it would still likely give you considerably less than 2.4V with only about 2mA flowing, for most US part number zener diodes. The "knee" on low voltage (less than ...


1

You can and must connect the grounds together. It's the common reference. The only other thing wrong with your Fritzing picture is that you are feeding the ADC 12V voltage, divide it down with a simple voltage divider with two resistors. You don't need to measure the GND with the ADC.


3

One interpretation is that rating is not there to give you any indication of how the transformer should be used in a flyback application, but only to tell you what peak voltages it was designed to work with. The likely reason for this is there are too may conditions, caveats, and gotchas with different flyback applications to give you a simple number rating. ...


3

I think you are confused about transformers in general and flyback transformers in particular. Transformers are not fed with DC typically. While you could use half-waved rectified "DC" (and it's not really DC) but full-wave rectified DC would do nothing but generate heat in the primary winding of the transformer. Without AC you get no changing ...


0

Based on the question you said the motor requires 5v thus you need a source that can provide 5v with a lot of current, I'd recommend you get a 6v SLA battery, it may be 6v but it's still safe to be used on 5v applications yet at the same time SLA batteries have a very low internal resistance thus there barely will be a voltage drop when you connect the ...


1

Assuming you're operating from something like a power bank, and consumption is less than a few hundred mA, you can use one of the inexpensive modules based on the Xian Aerosemi MT3608. They generally have this schematic (from this question: Adjust the pot to set the output voltage to +9V before connecting it to the other circuitry, of course. You can either ...


0

Two 5V batteries can be recharged as OP requires when wired in parallel. Add a switch to place them in series once recharged, for 10V to operate your pedal. If 10V is too high, use a diode or two to drop the voltage.


3

You need a thing called a "boost converter". If you were a product designer you'd probably design one onto your board, but that's a job for the experienced -- if you need to ask, it's probably beyond you. You can get cheap unreliable ones from eBay or hobby electronics sources, or you can get reliable ones from distributors. Basically, if it ...


-2

This is essentially an algebra problem involving square roots. It may have more than one answer. It is also a series RLC circuit. The current is the same flowing through all three components. The phasor voltages add vectorially to give the line voltage which is 200 volts rms. In other words : sqrt( (resistive voltage)^2 + (total reactive voltage)^2 ) = ...


1

This is a series resonant circuit. The reactive impedance of the capacitor is subtracted from that of the inductor. You have the voltages for the overall circuit, the resistor and inductor. Since the voltage is proportional to the resistance/reactances you effectively know the information you need to know to do the calculation. Draw a diagram of the overall ...


-2

Basic LCR rule Vs=SQRT((Vl-Vc)^2+Vr^2) plus some algebra. This should be in the first few paragraphs of the LCR in series chapter of any AC theory textbook. Have a look here... Page 3 has the information you need to solve this problem with only the voltages. learnabout-electronics.org/Downloads/ac_theory_module09.pdf Also here... https://byjus.com/physics/...


2

I have been told this can be done with a Diode. Well, yes, but. A small-signal silicon diode will drop around 0.7 volts always -- which means that when your battery is almost fully discharged, your radio will still be getting 0.7V less than the battery. It would be much better to get a voltage regulator. 3.3V is a common voltage, and it should work well ...


5

If I understand you right, you mean something like this: Where Q1 is a transistor, which opens or closes 5V supply to the load. R1 is 1k-100k (standard is 10k, but it doesn't matter which one, it's a pullup for the gate). Q1 can be a PNP, whose base is similarly pulled up by R1 so that it doesn't conduct in normal condition. If you use NPN or PNP, don't ...


0

I've seen that often in linear supplies and they usually did it with relays. I guess that doing a tap switch with TRIACs or SCRs (essentially the same thing) would be a good solution. It could be a little inconvenient to drive the gates, however, and for 3A probably sticking to relays is better (unless you need to continuosly switch range). An SSR (which is ...


1

Other than simply bigger diodes (just heat sink thems!), some manufacturers make 'ideal diode controllers' that drives MOSFETs to do the same thing. TI and Analog for sure, with Analog making even a full bridge rectifer thingie driving mosfets (disclaimer: probably even Infineon, On Semi and all the others make these. I just know better these product lines). ...


1

There are avaible big bridge rectifiers. They are very cheap - I would suggest using one of these with two diodes in parallel connected internally. They also come in different physical sizes as well as versions with possibility to attach to a heatsink. There are versions for 50A or even bigger currents.


3

The voltage on the drain is independent of the voltage on the gate. To control the device, you control the electric field between the gate and the source. When you apply an electric field to the source-gate region of the device, you change the conduction of the source-drain region by rearranging the charge carriers in that region.


1

Many electronic appliances with motors require several times more current when starting up (for the first few seconds) than they do to run continuously. To accurately calculate your power requirements, use an amp-hour or watt-hour meter to determine how many watt-hours the machine uses per hour. Multiply that number by the desired runtime in hours to get the ...


-1

Are you asking the question because you only have some kind of a meter that just measures AC? Why would you want to use it? You can buy a DMM with a 1000 VDC range for $6.79, and it will also measure AC voltage and resistance in several ranges. You must know this: you have an internet connection, so it must be a question to settle a bet or idle curiosity. ...


1

Theoretically it should show the correct voltage of the source, because average, root mean square and maximum of the constant voltage is just that voltage itself. What else the AC voltmeter could be supposed to indicate? However this depends. For instance, if the single diode is used to rectify the AC, the device still can be calibrated to indicate the ...


1

Super thanks to @Seth for explaining why this phenomenon happens (TIL!) To summarize for future reference, 2 techniques to avoid the issue: (1) can check continuity of nets by Ctrl-click on any trace. Alternatively, from the menus, Inspect -> List Nets, and then click on the node in question from the list, and it will highlight the path in the same way. (...


1

Although there are right answers I would like to explain why these two circuits are not equivalent so you can avoid these mistakes later on. Two one-port circuits are equivalent, when for a given V we get the same I and vice versa (i.e. same V-I characteristics) like between A and B in the following: You are applying this to a two-port circuit which is not ...


3

In order for traces to connect to the zone, they need to have an endpoint inside the zone. Here, you have a trace (the vertical Vdd) that begins and ends outside of the zone fill. This is logged as https://gitlab.com/kicad/code/kicad/-/issues/1800 in the KiCad issue tracker. Unfortunately, so far the fix for this will impose performance penalties too great ...


1

My question is, do these formulas always hold? The answer is really "yes" and "no". Other answers have explained the "yes" answer, but they all depend upon a capacitor or inductor being "ideal". Real capacitors and inductors have "stray reactances" and resistances. But even if we ignore these, real ...


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