What does 50va mean on a 24 volt transformer.
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What exactly goes into a transformer's VA rating? At a fundamental level it seems that there has to be a limit to what the VA rating can represent.  For example if I draw 2 Amps-RMS from the secondary of a 30VA rated transformer at 15 Vrms this makes sense as the maximum Complex Power I can draw. Still, at a power factor of 1 my maximum current in the transformer would be sqrt(2) * 2 Amps.   But what if my power factor is abysmally low (ie in a DC rectifier) due to harmonic distortion and while I still draw 2A-RMS I have brief current spikes of 15 Amps? Surely the transformer's magnetics couldn't handle these spikes as I'd expect the core would saturate.  My question is what is the practical limit for VA ratings of transformers? Would brief current spikes that reduce to only 2A-RMS in this case be fine? Obviously the secondary/primary copper losses play a part in the rating but how do the magnetics factor in?  I ask this because I am designing a simple bench power supply using a 125VA transformer with a 24Vrms secondary and am trying to figure out how much power I can draw before the current spikes into the inductor are too much for the transformer. I also don't want to have to resort to active PFC as I don't want to risk messing up any HVAC PCB design.  Thank you
28697
I have the following circuit, where I put R52 = R53 = 2700 Ohms :    Triggering levels are of 1V and 2V approximately according to my computations, that is okay according to the following figure showing input in green and output in blue :    But what I find really strange is that the maximum voltage output is around 3.3V and not at 5V. How would you explain that ? And last detail : how would you explain that the output is not exactly at 0V when the input is high?  Thank you very much for your help.
96647
Why is HV kept open in OC test of transformer? Also, why is LV shorted in SC test of the transformer? (Generally this is preferred - why?)
47215
I have the following ammeter:   The seller recommends 75mV 100A shunt be placed with it. I want to know what will happen if I change the voltage going through the shunt. Should I be changing the shunt as well to read amps correctly?   I want to build a 1-40V (or maybe 50V) adjustable DC power supply and use this ammeter to measure current. What the right value of shunt that I should use to get the right measurement? Changing shunts in a lab bench power supply is impractical, of course.  EDIT: I am planning to measure only up to 5A or so. Sorry I forgot to mention this previously.
107643
I asked a similar question  about this, but I still don't  quite understand.  Say I grab onto two wires from a standard 120V 20A wall socket. I would fry. I would have 20A at 120 forced on my body.  But, when speaking in terms of circuits, as in my linked question, if I have a power supply outputting 100mA maximum at 5v, for some reason my device only gets exactly what it needs? Why is this?  Most things flow from an area of high concentration to low concentration, taking the path of least resistence. If you connect two wires to a power supply, why don't the maximum number of electrons want to go through your circuit?  If current behaves this way...why do you have to regulate voltage?
68514
This is a question which I seem to have tackled multiple times, solved each time after reading a dodgy internet explanation, then partially forgotten about and retackled half a year later. It is time to put an end to this.  So first, at the power plant, a set power is produced. Let's call this $P_g$. Then there is a transformer that can say hypothetically transform this power into any voltage, and therefore following the equation $V = IR$, any current desired. Cool so far.    Consider the case where you simply condense all the power lines into a single resistance, R. Now we can relate the power lost from the resistance $P_l$ as $P_l = I^2R$. By inspection, as the resistance is a constant, lowering the current drastically lowers the power lost. This is great, as lowering the current means raising the voltage, which seems all handy dandy.  But wait: here's my problem. Can't the Equation for power loss be translated as thus:  $P_l = I^2R$  $P_l = IV$  $P_l = \dfrac{V^2}{R}$  BAM. Now the power seems to depend upon the voltage alone, as the resistance does not change, so it seems pretty logical that REDUCING the voltage reduces the power lost. Nuts. Evidently I have done something wrong. Lets take a simple case, $P_g = 10$. Lets say I set the voltage to 10, and the current to 1, and the resistance to 1 as well. That seems ok, as $P = VI$. Now lets use the two equations we have to determine the loss of power.  $P_l = I^2R = 1$  Seems logical, the current is 1 definitely, and the resistance is 1 definitely, so the power loss is one. Makes sense too.  $P_l = \dfrac{V^2}{R} = 100$  WHAT IS LIFE!! But this seems pretty logical to me too, the voltage across the resistance is 10 (is it not? I might be forgetting something fundamental here? not sure), and the resistance is 1.   Any help is helpful, I'm probably just being silly at one step or another.  EDIT  Well I was right, I WAS being silly i think. I pretty much completely ignored the fact that $V = IR$, so in the case where $P_g$ is 10, and the resistance was 1. I seem to have run into a complication now. If I set the voltage to 10, according to $P = VI$ the current should be one, but according to $V = IR$ the current should be 10. I run into a conundrum? I am now thoroughly confused. Any help is helpful.
12596
Just what the title states. The house, located in India, uses old-fangled . The typical domestic supply here in this part of the world is 250V AC at 50cps. I am in the middle of replacing the fuses with Miniature Circuit Breakers.   MCB appear to receive the line wire from the top connector, the bottom connector being the protected device/s. Is this universal/standard? Does the protected device connect to the top, or bottom of the MCB?
585
I realize connecting two different voltage sources in parallel is a contradiction (in an ideal circuit). But if I were to connect this in practice and measure the voltage across points A and B, what value of voltage would it show? And how much current would be sunk by the 5V battery? (Non ideal conditions)
99772
I came across this switching power supply circuit and I noticed the transformer is connected in a way I had never seen before. What is the benefit of connecting it like that? And why does the circuit use a capacitor voltage divider?
18129
I am reading the  From the table on ELECTRICAL CHARACTERISTICS (page 2) (TA = 25°C unless otherwise noted) (Each Diode)     Forward Voltage mVdc     IF = 1 mAdc 715mV   IF = 10 mAdc 855mV   IF = 50 mAdc 1.0V   IF = 150 mAdc 1.25V     On page 3 there is the following VI curve graph:  Clearly at 25°C - 100mA is Vf = 0.9V. There is huge difference between the table and the IV curve.  Which source of information I should trust? Is this (typical) and (maximum) where the table guarantees the diode won't exceed Vf 1.0v at 50mA while the graph is just typical values?
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So can anyone explain me how the collector voltage(V (output)) will vary with variation in base current with respect to the following figure?         &ndash; Schematic created using
103351
I want to know what size transformer for LM317 at full 30 VDC. I know its a 24 VAC center tapped transformer but is 1 A or 5 A transformer best suited for a linear power supply of 1.5-30 VDC?  I will be using a standard LM317 variable power supply circuit with the 2200 uF 50 V smoothing and the 240 ohm drain resistor and potentiometer. I have a homemade oversized aluminum heat-sink that is quite large.
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I recently got a desktop LED lamp, and I feel a "strange" slight vibration when touching the metal case. I found that other devices also have high voltage on their case.  What is the reason/trade-off to design the PSU that way? Are "real" transformers inefficient? Even if it's harmless, why is the high voltage connected to the metal case, and not just through insulated wires?    When I turn the lamp on, the setup in the picture measures the power line (220V) 214Vpp (59Vrms). When I also touch the metal case, the voltage drops down to around 30V.  AC adaptor:
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Transformers have hundreds of turns on both the secondary and primary winding, and as a result use very thin copper wires for each.  But, why do they not just use fewer turns on each winding and get the same voltage ratio?   More importantly, why not use fewer turns of a thicker wire for an increased VA? (instead of 1000:100 turns of 22 awg wire, why not 100:10 turns of 16 awg wire if this would increase VA)
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Below is an NPN transistor symbol and the voltages at its terminals are Vb, Vc and Ve with respect to the ground:    I read that: during the saturation the Vce = (Vc-Ve) settles to around 0.2V and the further increase in base current will not make Vce zero.  But why doesn't Vce become zero?  As far as I know: when the transistor is saturated, the base-collector junction turns on, like a diode, so the collector voltage will follow the base voltage increase, only it will be a diode-drop below. But the same thing happens between the base voltage and the emitter voltage. So at saturation and beyond one can write the following(?):  Let's call the diode drop as Vd between p and n junctions, so the collector and the  emitter voltages can be re-written in terms of the base voltage as:  Vc = Vb- Vd  Ve = Vb -Vd  Vce = Vc -Ve = 0  Where am I wrong here?
69660
I have some "NDV rev1.1 4248 back lights" from a television.   There is a print that sais "94V" on the strip, and on the wire supplying the strips it said "96V".   Every strip has 9x buttons which have 3 LED's in them. So i devided 94 by 9 and figured that every button should have 10.4V roughly, which makes sense, because a white LED is normally 3.5 volts.  I have a 12v Lithium battery which measured 11.1V so i tested one of the LED's with it and it lasted less than one second.   Are the LED's are rated for 10.4V roughly and not more?
84214
The circuit is an altered Joule Thief of my design that is driven by an AA or AAA battery. I’m observing the current at the battery anode with my DMM in series. (Extech EX330) This circuit is extremely efficient (In relation to other Joule Thief I've build.) and does some weird stuff - like, it needs a pushbutton sometimes.    Measurements: It is drawing 9.21 mA DC and 6.48 mA AC in series with the battery. It shows 1.364 VDC and 0.010 VAC in parallel with the battery. Output on the LED is 2.572 VDC and 0.011 VAC with no measurable frequency. (with my DMM at least) All measurements are taken with the same configuration and voltage.    But how is it drawing AC and DC and how do I calculate power consumption? --> Could it be because the emitter pin of the NPN is directly connected to ground? --> Or is the circuit just "switching on and off"?     Download schematic:  (I tried to simulate the circuit in LTspice. Thanks for making me aware of this amazing program, Sredni Vashtar!)     What the pushbutton is used for: The pushbutton is used to "kickstart" the oscillation when it wont start by itself. For example: I have to manually "oscillate" using this button to start the circuit if the capacitor is too big and battery voltage is low. Pushing the button releases the energy in the transformer into the capacitor. If it reaches LED forward voltage the circuit keeps oscillating on it's own. It does work without manual intervention when then capacitor in parallel with the LED is not too big, like &lt;5000 µF, and it's a fresh battery.    . Thank you for your answers! I'm learning so much! :D