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An ideal transformer has a defined relationship between the primary and secondary voltages that is equal to the ratio of the number of turns in the windings. An ideal transformer doesn't care about time or frequency or flux.


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It depends what you deem to be an ideal transformer. If the ideal transformer has infinite primary inductance, no saturation, and no primary resistance, then it can carry an arbitrary primary current, and indeed a primary voltage will create a secondary voltage. A secondary load will then cause a secondary and primary current to flow. If the ideal ...


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Yes, that's the basic effect of the ESR (equivalent series resistance) and the capacitance of the cell. If you use any free circuit simulator and play around with reasonable values of R and C, you will get a better understanding of what's going on inside the cell. If you measure the given voltage drop with a known current and the time taken to recover, you ...


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You are correct, their are two ways 3-Phase Meters can determine Import/Export Wh totals. I am a Electrical Engineer and have programmed EDMI smart power meters for customers in NZ who provide reconciled Import/Export kWh data to retailers for billing purposes. Option A is sometimes referred to as Net Metering and Option B, Gross Metering. Net metering is ...


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It's kind of a half-a** approach compared to proper supply sequencing - consider what may happen if there are brief power interruptions of various durations- can you make the circuit fail to properly sequence-- probably. And what discharges the capacitors- probably some parasitic path within the regulators that is not well controlled. Maybe you don't ...


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5 ICL's can mitigate the surge current. Did you want to do power down sequencing too? Another solution may be a ring Johnson counter with an enable input. But LDO's are usually low current so a low ESR intermediate bulk cap is all usually needed with DC OK logic signal if critical. The cons to RC controlled logic std logic are variations in slope RC ...


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The question is erroneous. given Source= 14V; Load = 115A battery load = 10 A is impossible at 14V since the current direction will always be charging above 13.5 +/- 0.x and discharging below a voltage 13V toward 12.5 to 12.8 where SoC=100% therefore the assumptions and conditions supplied are invalid. Conclusion: Invalid question. More valid is low RPM ...


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Why limit the alternator to producing 105A when the load requires more than that? So, allow the alternator to function with the "normal" regulation and it will cover the load as well as charging the battery. If the engine is under-sized and cannot produce sufficient power then the alternator won't be able to produce full output. Also, if the alternator is ...


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The characteristics of a passive network like this are very much dependent on the load impedance it sees. Obviously, the step-up scenario has half the load impedance of the step-down scenario, and this provides greater damping (lower "Q"). If your converter had feedback and dynamic control of the switching duty cycle, you could pretty much eliminate these ...


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Since you have 6 Li-Ion batteries in series, you could charge them at 25.2V (Since the max charge for a Li-Ion is 4.2V * 6 = 25.2V). The current used the charge the cells depends on the battery itself. Batteries usually have a part number printed on them. So if you find the datasheet, then you can determine the maximum current you can charge the batteries at....


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I'm appreciative of the solution by @Transistor which does work. I put this solution into practice and found that under no load the relay gets over 50 degrees C and emits a lot of EMI. I tried several relays to make sure it isn't a manufacturing defect. Darn. I replaced the relay with a DPDT switch instead. If this helps anyone else, it turns out that it ...


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N is the number of turns in a single coil of the motor windings. The maximum value of flux is the flux that the iron core for the coil can carry without being saturated. N and Phi are selected by the motor designer. Once the motor has been designed, everything that is of interest to the user of the motor is marked on a rating plate that is permanently ...


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A way to get around this would be to use a sense resistor, maybe one with 0.01Ω. The sense resistor is placed between the load and the DC to DC converter. Then connect the DMM to the leads and measure the voltage. 6A would give you 0.06V and allow you to still measure with the DMM in mV mode.


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You can use an efficient LED that operates from (say) 3mA and then your series resistor would dissipate 1.5W so you could use a suitable rated resistor or resistors (rated voltage as well as power dissipation). You could consider using two (or more) resistors in series. To get a safe circuit you can mount the resistors and LED in close proximity to the HV ...


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You can certainly use a transistor and trigger a triac (or SCR). Typically that takes tens of mA or more at a couple of volts. So a -5V supply and a resistor switched by a BJT works nicely for triacs, and +5 for SCRs and maybe acceptably for some triacs that are rated for Q4 operation (MT2 negative gate positive). But think of a low-cost phase-control ...


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Insert a rise time and fall time for the pulsed voltage source. There are default values used when you input 0 that are realistic but >0, if you want to see more ideal behavior use a value like 1n. PULSE(0 5 10u 1n 1n 1.5434u 10u)


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Triacs (and SCRs) only need a about 1.2V to trigger, so, yes, a 5V circuit could trigger one. Diacs have a kind of "snap action" when the threshold voltage is reached their voltage drop reduces to about 2V (allowing more curren to flow) until the current stops flowing. After breakover of the diac the current surge that flows into the triac gate is ...


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No, current flow does not drop in saturation. It increases very slightly as VDS increases. I think you already know this and are just wondering where the carriers go since the channel is pinched off. Basically, they spread out in the two dimensions at right angles to the channel so that conduction is no longer just taking place through a narrow channel.


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As the channel current increases, so does the drop across it (the channel is like a resistance), and as a result the drain voltage rises above source voltage. This does start to constrict the channel but as the channel gets pinched the drop gets limited as well and so a form of equilibrium is reached which limits the maximum current the channel can carry.


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I think this table is pretty accurate (even if technologies have improved) when considering the parameters (first column). Here the PowerMOSFET depicted is 500V/200A rated, which make them completely different to an RF "power" mosfet. When in doubt you can always go look online to electronic component resellers. Here is a common "High Power" MOSFET 100V/...


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You may(1) find the information in the datasheet. Some give the information directly, for example the TCAN330 from TI Other datasheet give it indirectly, for example the SN65HVD23x from TI Here, the current required for the IC itself is given by \$I_{CC}\$ Supply current. Note the "No load" condition. Typically, the bus load is 60 Ω. In dominant ...


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the receiver interface circuits will be monitoring the bus, either to detect collisions during transmitting, or for external data packets; this requires an analog comparator with about 20 nanosecond delay time, with stable hysteresis, and with input voltage-dividers on both of Vin+ and Vin- pins of the comparator, to level shift DOWN into 0/5volt rails from ...


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According to this article from Texas Instruments, a typical CAN bus driver output stage will look something like this: Thus, we can expect the bus to draw significantly more current in the dominant state. Unless we know the termination (load) resistance, differential voltage (CANH-CANL), and maximum output current of the drive IC, we can't determine the ...


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Triacs are nothing like two transistors and a diode. Read the Wikipedia article or (better, I think) the GE SCR manual ca. year 197x which will explain things better than anyone here is likely to spend the time doing. There are four possible quadrants of operation. The commonly shown two-SCR "equivalent" (which would be like four transistors) is also very ...


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The view of an SCR, from what I read (and built, and tested) as a kid, is this simulate this circuit – Schematic created using CircuitLab Each side of the model provides voltage gain, and to "turn on" the SCR requires the GainSide1 * GainSide2 be > +1. The gain of either side is gm*Rload, where gm is the transconductance and Rload is 1Kohm for this ...


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I think the bigger issue is total current draw of the circuit. As a general rule you shouldn't plug a power strip into another power strip because someone could overload the capacity of the circuit / power strip if they didn't know what they were doing. In a commercial setting you would probably even fail an inspection by the local fire department if they ...


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1) What is Z0? There are no transmission lines here. The figure shows the power divider connected to three transmission lines. The parallel lines drawn extending away from the points marked \$V_1\$, \$V_2\$, and \$V_3\$ are representations of transmission lines. It's unfortunate for introductory material that they put the "Port" labelings at the far ends of ...


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Well, according to @Neil_UK answer, I think I have a little bit freedom in the way to present information in an academic level. So I think that could considered as a free space for trainers/teachers and university professionals to present the information required in the course the way they want but taking into consideration the main rules that the topic is ...


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With the voltage source connect as shown in the schematic, the zener diodes are reverse biased and will (try to) clamp the input voltage to about the sum of their individual zener voltage. If for example D1 and D2 are a 4.7V zener and D3 a 6.2V zener, the clamp voltage will be about (4.7+4.7+6.2)V = 15.6V. If the input voltage is standard 15V, the zeners ...


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Circuit breakers are mainly intended to prevent fires. They don't react quickly enough to block ESD pulses, and they don't react quickly enough to prevent many failure modes due to over-current in the protected equipment. What they do is prevent excess currents from flowing for minutes, hours, or days, which could overheat the protected wiring or equipment ...


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There is a mechanism that kicks the circuit breaker into action, and it relies on current (not voltage) in the traditional circuit breaker. When the current reaches a certain level a coil triggers the spring loaded switch to turn off. Some breakers also have a bimetallic that triggers when it gets to hot in response to current. Source: https://electronics....


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On the input side of \$C_c\$, you have a voltage source (I assume of value E0) connected through \$R_c\$ to three capacitive loads \$C_s\$, \$C_t\$ and your signal \$\Delta\$C(t). In the steady state, since there's no DC path to ground and \$\Delta\$C(0)=0, the junction of \$R_c\$ and the capacitors will be E0. However, when \$\Delta\$C(t) changes, the ...


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It depends why you want to break them into groups. If you simply want to partition the course into more easily managed teaching sections, then an alphabetic grouping is just as rational. Put all the power devices into alphabetic order, and split them into two. If you want to group them by the way they're used, then the groups you've chosen aren't too bad. ...


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It depends on what you want the failure mode to be. X caps fail short, while Y caps fail open. Since your snubber resistor will most likely not be rated for continuous operation you most likely would choose a Y cap. Thus, if it fails the HF noise may be present but the snubber resistor will not burn up. See this article for more information.


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Metallized paper & film or poly capacitor is suitable for high frequency filtering


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Multiplying \$I_{RMS}\$ by \$V_{RMS}\$ gives apparent power. This is the hypotenuse of the power triangle, where the other two sides are the real (average) power (in watts) and the complex power (in VARs). If, and only if, the load is purely resistive then the complex power is zero and the apparent power will equal the real power. In general, multiplying \$...


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Multiplication of Vrms and Irms gives average power into a resistive load only. Into other loads which result in out of phase waveforms, capacitive, inductive, rectifier for instance, the result is not average power. If you multiply Iavg by Vavg, the caveats are similar. For well behaved waveforms like DC, the result is average power. It's easy however to ...


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Using P = V^2/R, at Power = 0.001 watt, and R = 50 ohm, you can compute Vrms = 0.223 volts. Then scale up by 2 * 1.414, to compute Vpeakpeak = 0.632 volts. The +12dBm input is 4 factors of 2X power (3dB) stronger than 0dBm. The +12dBm input is 2 factors of 2X voltage (6.01dB) stronger than 0dBm. Thus we know the +12dBm is 2 * 2 * 0.223 voltRMS.


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build a current mirror with 5 outputs connect one to each supply control pin add resistors to pull the control pin in the opposite diorection.


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Becasue there isn't a max voltage, only a max power. You can put in whatever voltage you want, as long as the power referenced to 50Ω is kept below 12dBm. The requirement forces you to look at the frequency content of your signal and ensure that the total power of that signal does not exceed the rating. So measure or calculate the total power of whatever is ...


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