# Tag Info

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I believe you have a handle on it. P = I^2 * R. The thermal rise in a resistive device will be the square of current. That puts the square in RMS. This is why RMS has the relationship it does with peak. This is the AC voltage that, when applied to an incandescent light, will provide the same light, heat and service life as that DC voltage.

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B is correct. Thought experiment: There's a (much bigger) transformer feeding your house. What happens if you switch off everything else except one little light bulb? A: The transformer over-provides, the lightbulb takes everthing it's given and burns itself out. B: Only the required amount of current is taken by the lightbulb and everything operates as ...

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B, power supply is capable of providing more amps than the load requires.

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Let's call your small transformer the "apple" transformer, and your large transformer the "orange" transformer. You would expect that if you're operating at the same voltage but higher currents, that all of the inductances of the "orange" transformer are proportionally lower. So the individual coil inductances, and the leakage inductances, would go down. ...

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I would look at using a low-side gate driver IC, which will give you level shifting and an output power stage capable of operating at 10V or more. Also, they are inexpensive. For example, the Microchip (nee Microsil) MIC44F18/19/20 will drive 100mA with a couple hundred mV drop, and typically draws less than 10mA idling at 1MHz.

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I fear you have misunderstood how op amps work. You have overlooked the fact that zero times any gain remains zero. That is, the portion of the input waveform which has a zero value will produce a zero output. As a result, the op amp (which has a nominal gain of 4) will attempt to produce an output from zero to 13.2 volts. Since Vdd is only 10 volts, the ...

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The cheapest way would probably be using a CMOS hex inverter 40106 which can be operated at up to 15V and biasing the input near below the switching threshold using a voltage divider and either AC-coupling the input signal via capacitor or DC-coupling through another resistor. The signal will get inverted. If this is a probem just feed it into another ...

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I think the main problem is that your simulated transformer has unrealistically low winding resistances. This results in high rms current because the winding resistance is not limiting peak rectifier current sufficiently. I simulated a simplified version of your circuit in LTspice. Using primary and secondary resistances of 12Ω (a guess based on a ...

3

Lets's discuss only about what you said "I have understood" Not sure how you consider those those historical non-measurable quantities named EMF, but the result is ok for DC: There's linearly increasing current in an ideal coil which is connected to a constant DC voltage. There's no need for the EMF. If some mechanism, say an ideal voltage source forces ...

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Consider an ideal inductor with a sinusoidal voltage source, $\small v_S=V\:cos\: \omega t$. The induced EMF must exactly match this, therefore we have: $$\small L\frac{di}{dt}=Vcos\;\omega t$$ solving the differential equation for $\small i$ gives: $$\small i=\frac{V}{\omega L}sin\:\omega t$$ This is the magnetising current, which is also present ...

1

You don't want to know the VA of the transformer, at least not as a single figure. The core details and your working flux level will give you the turns per volt and the core heating, which are more or less independent of current. You want to know the load rms current, which gives you the copper heating. Modify your Spice simulation slightly to compute the ...

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There are a lot of confusing posts out there about this. Part of this comes from people not being clear which component of current they’re talking about: The reactive current, 90 degrees out of phase with the primary voltage. It’s real, it’s measurable, it has important impact when designing the transformer, but it transfers no power. The current in phase ...

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If you carefully read through this transmogrification of your question it MAY help: Note carefully that a transformer with more turns on the secondary than on the pimary is a current "step down" transformer. This is the main point that you have to appreciate in your "visualisation". Changing "voltage" to "current" and "up" to "down" in your original ...

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In general, rated values are the values for which the transformer is designed and can be expected to perform as expected over the expected lifetime when operated continuously. For the primary voltage, the transformer will overheat if operated continuously with a voltage that exceeds the rated value. It can tolerate 5% or 10% above rated, but the lifetime ...

7

$$P_{OUT} = P_{IN} - losses$$ Ignoring the losses: $$P_{OUT} = P_{IN}$$ $$V_{OUT}I_{OUT} = V_{IN}I_{IN}$$ Rearranging: $$I_{OUT} = \frac {V_{IN}}{V_{OUT}}I_{IN}$$ So, for a step up transformer where VOUT > VIN the output current is less than the input current by the voltage step-up ratio. From the comments: So if the voltage increases on ...

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If the voltage at the secondary is more than the voltage on the primary, then the current on the secondary will be LESS than current on the primary. A transformer changes both voltage and current in such a way that the output power equals the input power (less a little for losses in the transformer).

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What do the hollow circles at the top and bottom portion of the circuit represent? Those circles are the connections of the different elements indicated by the curly brackets below the schematic. Each element is represented by an electrical equivalent schematic and these circles show the border/connection of each schematical representation. They also show ...

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A couple of things: The transformer is there to provide the required step-down ratio but also to make the circuit isolated from the input side. If you notice, you've grounded the lower terminal of C2 (an all of the other legs there) on the secondary side of the transformer. So, in some way, that is interacting with what you have going on on the primary ...

1

When a transformer blows, it interrupts electrical service, regardless of type. Interrupted service does not control your breaker. A transformer blows because it's internal oil insulation ignites before tripping upstream or internal over-current breakers due to an insulation failure , not necessarily an external event such as an unprotected lightning ...

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There is always a parasitic capacitance across the switch and also between wires in conduits. This capacitance is small but if the wires run together in a conduit, it can be enough to let a few µA AC current through. The internal resistance of the multimeter is very high in voltage mode, but it is not infinite. What it is measuring is the output of a ...

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The multimeter is such a weak load that it can measure capacitively coupled voltage. So yes, it sounds normal.

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That isn't how transformers work. Putting a fixed-voltage variable duty cycle signal on the primary will simply give you a fixed-voltage variable duty cycle signal on the secondary. If you want to translate that secondary signal to a variable voltage, you'll need a completely different kind of filter between the transformer and the regulator.

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If you don't want to receive power from the switch -- leave this pins disconnected; If you want to -- you need PoE-unit, which receives power from these pins, and converts PoE voltage level to 12V or so. You heard right -- transformer used for isolation. So your PoE unit should have isolation too. It means that it will also have it's own transformer inside ...

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I believe the baluns used in UHF TV applications were actually transmission line transformers. Twisted pair transmission lines, specifically, but you can use coax as well. Bandwidth is high, but available ratios are limited. For example, using Ferroxcube 4S2 material, it was possible to make a 1MHz to 5GHz 1:4 transformer (Image from Horn, J., & Boeck, ...

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This was such a good answer, posted by Deeds in July 2016 and then deleted, that I decided to CCbySA resurrect it. Deesd said: Slayerz... I love em. 4k on the base, 7 Turns on the primary. 1000 turns on secondary. These are all good transistors, 2n6547 was the best. You MUST have a heat sink or you'll cook the poor things. 2n6547 2SC5387 BUF420AW ...

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The transformer primary and secondary are isolated from each other. What have you defined as the reference potential on each side? That will determine the polarity of math. You can assign both sides in any combination you want, as long as you keep the math consistent with your definitions. Remember the transformer is just DRAWN with dots on the opposite ...

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The manufacturer always specifies voltage in a certain frequency range. Why is this parameter important? Because these are the parameters the transformer has been specifically designed for. As depicted, this transformer's primary input of 10-40v @100kHz will produce X to 5v/3A at the output. (Since the input ranges from 10-40vAC, then 10 is 1/4 of 40, so ...

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The 500V is a voltage used for insulation test between different coils, not voltage over a single coil. Otherwise the transformer works best in the frequency and voltage range it is designed for. If the volt × second is higher then the core starts to saturate.

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My understanding is that baluns such as those you wound are used on the RF and IR ports of a double-balanced passive mixer (i.e., diode ring), and sometimes also on the LO port for a double-double balanced diode ring, but not on gain-type mixers such as the SA602/612 devices. If your purpose at this point is to learn about RF and mixers, I wonder if you ...

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Have here a crude SPICE model of that mixer. With inputs single-ended (as in your first circuit), and looking at single-ended pin 4 as output, unloaded. 24 MHz oscillator is not modelled properly, it is a simple single-ended square wave large enough to switch the upper quad transistors fully on-off. The model uses 2N3904 transistors everywhere, with output @ ...

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The transformer is a typical automotive coil: LP = 8mH RP = 15Ohms RATIO = 1:100 LS = LP * POW(RATIO,2) RS = 100kOhms I did not have time to put proper models in for suitable transistors and diodes, so pick a 27V zener diode in a large package. A push-pull driver rather than a single PNP will produce larger voltages as it's the rate of change of current ...

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The 2009 Cooper Crouse-Hinds catalog for airport runway lights indicates some of the lamps require a constant current power supply, i.e. a special transformer designed to prevent the inrush current spike that shortens lamp life. Look to the fixture manufacturer for a replacement. On the other hand, if you're just experimenting with a lamp, likely surplused ...

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Tungsten Bulbs have a well known positive temperature coefficient or PTC where the filament resistance rises x10 from cold to hot. Halogen gas bulbs have higher MTBF at elevated temperatures with a metal vapor recycling mechanism than std. Tungsten filaments which just use less expensive inert Argon gas. Thus when random startup occurs at peak voltage, ...

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Is winding a balun considered rocket science? Or the problem lies somewhere else? Or I have some foundational misunderstandings about a mixer? Fundamental to the Gilbert Mixer is that you must have a low impedance band stop and high impedance bandpass filter on the output to achieve the SNR expected with the mixer gain and to prevent saturation effects (...

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Aha, I noticed now looking closer that on my perf board I left exactly that lead of VR1 that should short to ground without any connection to ground! This would explain why the potentiometer has no effect, and why the value I calculated from the potential differences is so much higher. It is because it's all pushed through the ZD1 and R4! OK and after I ...

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