An ad hoc tutorial on transformers, starting from the very beginnings for neophytes, and building on it. Comments welcome.
Magic: There’s a common view that transformers are mysterious, with inner workings of electromagnetic field interactions, exotic materials, and strange parasitic effects. To a certain extent and for certain applications this is true. In things like audio output transformers, high frequency switching power transformers and RF transformers, the exact types of materials and the position of every single turn of wire matters.
However, for mains frequency power transformers, much of the magic stuff can simply be ignored, with good results. Someone with a working knowledge of Ohm’s Law and how resistors, capacitors, inductors and diodes work can easily understand power transformers.
Ideal Transformer Model: An ideal transformer is nothing but a ratio. This is the ratio of primary voltage to secondary voltage.
For reasons having to do with several freshman and possible sophomore university lectures on physics and EE plus some calculus, the following is true: every turn of wire around the core of the transformer has exactly the same voltage induced from the start of the turn to the end of the turn. You can ignore the math and physics lectures for nearly all practical transformer work, but you really need to know that (ignoring magnetic field leakage; that's later) every single turn has the same voltage from end to end.
That gets us to the ideal transformer model, expressed as number of turns. If Npri is the number of wire turns on the primary and Nsec is the number of turns on the secondary, then
Vpri/Vsec is equal to Npri/Nsec.
The reason to mess with an ideal tranformer model separate from the real transformer is that you can consider that for every real transformer, there is an ideal transformer inside, but it's hidden from you by the real-world imperfections: the wire resistances, leaked magnetic flux, core inductances, core losses, and so on. To calculate real world results, we need ways to measure the imperfections, then use these to calculate the effect of the real world and ideal stuff. This has proven to work very accurately and effectively in well over a century of transformer technology.
First Approximation Real World: You can get GREAT results by knowing only a very few of the real parameters.
The really screaming big ones to know first are, of course, the turns ratio and wire resistances. Next post will have diagrams and some how-to-measure methods.
Transformers
Moderators: pompeiisneaks, Colossal
Transformers
"It's not what we don't know that gets us in trouble. It's what we know for sure that just ain't so"
Mark Twain
Mark Twain
Re: Transformers
I like this topic! It would be nice to get a sense of how to determine the actual number of turns. Ten turns on the primary and 20 turns on the secondary or 1000 turns on the primary and 2000 turns on the secondary are both 1:2. I was surprised, for example, to learn when unwinding a power transformer that it had a filament winding with only about 16 or 17 turns. It seemed so unrefined to me, yet it worked. The HT winding on that PT had too many turns to count by hand.
Re: Transformers
I'm doing the drawings, but I'll slip in a quickie tech partial answer.
This topic is not as complicated as the physics/EE/calculus stuff, but it is more complicated than performance evaluation. Here's a super quick overview.
You have to know:
>the lowest frequency and highest voltage AC that will be fed to the primary
>the core material; this tells you the max flux density the material can support
>the core form factor, as in E-I versus toroid, and how big the laminations stack will be
The resulting core cross sectional area, wire window area, and average length of a primary wire turn can get mixed into a few approximation equations that tell you the number of primary turns. That gives you the volts per turn, and that gives you everything else. There will then be a lot of fiddling with the exact stack, exact wire sizes and so on.
Volts per turn in power transformers scales about linearly with the cross sectional area of the core. For E-I laminations, you try to keep the stack such that the center of the E is approximately square. Certainly between 1:2 and 2:1. Bigger laminations give you more core cross sectional area, and that gives more volts per turn.
I can do a post on this process; if I forget, remind me.
This topic is not as complicated as the physics/EE/calculus stuff, but it is more complicated than performance evaluation. Here's a super quick overview.
You have to know:
>the lowest frequency and highest voltage AC that will be fed to the primary
>the core material; this tells you the max flux density the material can support
>the core form factor, as in E-I versus toroid, and how big the laminations stack will be
The resulting core cross sectional area, wire window area, and average length of a primary wire turn can get mixed into a few approximation equations that tell you the number of primary turns. That gives you the volts per turn, and that gives you everything else. There will then be a lot of fiddling with the exact stack, exact wire sizes and so on.
Volts per turn in power transformers scales about linearly with the cross sectional area of the core. For E-I laminations, you try to keep the stack such that the center of the E is approximately square. Certainly between 1:2 and 2:1. Bigger laminations give you more core cross sectional area, and that gives more volts per turn.
I can do a post on this process; if I forget, remind me.
"It's not what we don't know that gets us in trouble. It's what we know for sure that just ain't so"
Mark Twain
Mark Twain
Re: Transformers
power transformer is easy - there is transformer formula
U=4.44*Bm*f*Afe*N
Bm- max magnetic flux (how much it should be - lets hear experts - this is key factor)
Afe- area of middle column in E lam in m^2
f- frequency- Hz
N- number of turns
U=4.44*Bm*f*Afe*N
Bm- max magnetic flux (how much it should be - lets hear experts - this is key factor)
Afe- area of middle column in E lam in m^2
f- frequency- Hz
N- number of turns
Re: Transformers
Wait, dont be so fast - Bm is NOT the max value from the datasheet of the laminations - and why is not?
Think about what is going on during rectification, and why for the transformer working for DC rectification you must use lower value
Think about what is going on during rectification, and why for the transformer working for DC rectification you must use lower value
Re: Transformers
Yes, bepone, that's one of the equations. We'll get to that. As you note, there are other considerations, and not just whether it's used for rectification or not. Be patient. You've worked with transformers a long time. The beginners this is intended for need to take it step by step.
"It's not what we don't know that gets us in trouble. It's what we know for sure that just ain't so"
Mark Twain
Mark Twain
Re: Transformers
Ideal Tranformers
An ideal transformer is nothing but a ratio. This is the ratio of primary voltage to secondary voltage.
Thhe ideal transformer model in general ignorse resistive losses, magnetic losses, coupling losses, interwinding capacitance and so on. Then the real-word imperfections are added back in as separate components so the designer can isolate their effects and tinker with them if need be. This more-real model isn’t the actual real transformer either, but it’s a whole lot easier to understand and usually provides results very, very near the actual physical-world transformer.
Here's a diagram of an ideal-transformer that I will use to build a example. The general idea is to provide an isolation transformer that takes in 115.000Vac(rms) and puts out an isolated 115Vac(rms) to a resistor load, providing 175W to this load.
For reasons having to do with several freshman and possible sophomore university lectures on physics and EE plus some calculus, the following is true: every turn of wire around the core of the transformer has exactly the same voltage induced from the start of the turn to the end of the turn. This is an important side note to remember later. Given that, and that the example is to provide 115Vac(rms) to 115Vac(rms), we immediately know that the number of primary turns Np will be equal to the number of secondary turns Ns, and that both Vpri/Vsec and Npri/Nsec ratios are equal to 1. An ideal transformer makes the exact number of turns irrelevant. That's a real world thing.
In the imaginary ideal world, there are no limits on currents. The schematic lines are all superconductors with no resistance, the cores don't saturate or heat, and all the power from the input gets to the output.
Getting Real
This starts with anideal transformer model plus three parts; the resistance of the primary winding wires, the resistance of the secondary winding wires, and the primary winding inductance. With only these added parts, the ideal transformer model makes quite accurate predictions of how power supplies work. In fact, adding a few more “imperfection” parts to the ideal transformer can make good predictions of audio output transformer performance. But baby steps first.
I cheated picking the ideal transformer model. I found a commercial transformer that had good specs for an example, and adapted the example to it. I picked the Hammond 169VS. This is designed for 115Vac(rms) to 115Vac(rms) isolation at a power handling level of 175VA.
I added the representations of the real world imperfections onto the ideal model, then tweaked things to compare to the ideal transformer.
The primary and secondary windings have resistances as specified by Hammond, and this model includes the No-Load-Secondary-Voltage (NLV) and the no-load primary current. The magnetizing current of 0.49A is the price we pay for having a transformer work. If the core is not filled up with magnetic flux, the transformer simply does not work. About all you need to know about this at a beginner level is that the core must be filled up with a certain number of magnetic zoots to work. The "zoot" is an imaginary unit of magnetic muchness. Cores need enough zoots to support their operation, and simply can't hold more than X zoots per volume. The magnetizing current is modelled as a primary inductance with a value that nets through the correct number of zoots.
And that's how I calculated Lpri = 619mH. With 115Vrms across a 619mH inductance, 0.49A of inductor current flows.
An ideal transformer is nothing but a ratio. This is the ratio of primary voltage to secondary voltage.
Thhe ideal transformer model in general ignorse resistive losses, magnetic losses, coupling losses, interwinding capacitance and so on. Then the real-word imperfections are added back in as separate components so the designer can isolate their effects and tinker with them if need be. This more-real model isn’t the actual real transformer either, but it’s a whole lot easier to understand and usually provides results very, very near the actual physical-world transformer.
Here's a diagram of an ideal-transformer that I will use to build a example. The general idea is to provide an isolation transformer that takes in 115.000Vac(rms) and puts out an isolated 115Vac(rms) to a resistor load, providing 175W to this load.
For reasons having to do with several freshman and possible sophomore university lectures on physics and EE plus some calculus, the following is true: every turn of wire around the core of the transformer has exactly the same voltage induced from the start of the turn to the end of the turn. This is an important side note to remember later. Given that, and that the example is to provide 115Vac(rms) to 115Vac(rms), we immediately know that the number of primary turns Np will be equal to the number of secondary turns Ns, and that both Vpri/Vsec and Npri/Nsec ratios are equal to 1. An ideal transformer makes the exact number of turns irrelevant. That's a real world thing.
In the imaginary ideal world, there are no limits on currents. The schematic lines are all superconductors with no resistance, the cores don't saturate or heat, and all the power from the input gets to the output.
Getting Real
This starts with anideal transformer model plus three parts; the resistance of the primary winding wires, the resistance of the secondary winding wires, and the primary winding inductance. With only these added parts, the ideal transformer model makes quite accurate predictions of how power supplies work. In fact, adding a few more “imperfection” parts to the ideal transformer can make good predictions of audio output transformer performance. But baby steps first.
I cheated picking the ideal transformer model. I found a commercial transformer that had good specs for an example, and adapted the example to it. I picked the Hammond 169VS. This is designed for 115Vac(rms) to 115Vac(rms) isolation at a power handling level of 175VA.
I added the representations of the real world imperfections onto the ideal model, then tweaked things to compare to the ideal transformer.
The primary and secondary windings have resistances as specified by Hammond, and this model includes the No-Load-Secondary-Voltage (NLV) and the no-load primary current. The magnetizing current of 0.49A is the price we pay for having a transformer work. If the core is not filled up with magnetic flux, the transformer simply does not work. About all you need to know about this at a beginner level is that the core must be filled up with a certain number of magnetic zoots to work. The "zoot" is an imaginary unit of magnetic muchness. Cores need enough zoots to support their operation, and simply can't hold more than X zoots per volume. The magnetizing current is modelled as a primary inductance with a value that nets through the correct number of zoots.
And that's how I calculated Lpri = 619mH. With 115Vrms across a 619mH inductance, 0.49A of inductor current flows.
"It's not what we don't know that gets us in trouble. It's what we know for sure that just ain't so"
Mark Twain
Mark Twain
Re: Transformers
The differences between real world and ideal are very apparent. The primary and secondary winding wire resistances will cause voltage loss and heating when any current flows, because that's what resistance does. The core inductance sucks some current away from the input before the hidden internal ideal transformer ever gets to work.
Hammond did something that all transformer makers do. They specified that the output will be >>at least<< 115Vac(rms) for loads from zero amperes up to 175VA. The jargon has started.
See definitions below.
Making that specification come true requires that the transformer design take into account the losses in voltage from current flowing in the primary and secondary wire resistances and current "lost" to the magnetizing currents. The transformer designer increases the voltage ratio to make up for the current*resistance voltage losses in the winding resistance. The idea is to make the output voltage sag down to the specified 115Vrms or a little more when the output current is the full 1.522A(rms). The turns ratio is kicked up from 1:1 to something like 114.35/120.6.
I came up with that weird ratio from the numbers on the spec. The spec says that the no-load secondary voltage (NLV) will be 120.6Vac when the load resistance is effectively infinite and the primary is driven with 115Vac(rms). Under these conditions, the secondary current is effectively zero, so the voltage loss in the secondary resistance is zero. The voltage lost in the primary is the magnetizing current times the primary wire resistance of 1.32 ohms, or Vlost = 0.49A(rms) * 1.32R = 0.647V. So the hidden ideal transformer primary does not get 115Vac, it gets 115 - 0.647 = 114.353Vac(rms) as an input. This has to make an output of 120.6Vac(rms) on the secondary, so the voltage/turns ratio on the hidden ideal transformer has to be 120.6/114.35 = 1.054629.
Notice that the manufacturer told you that the output voltage will be higher than specified with no load, and will sag to no less than the specified output voltages under full load. This sag from no load to full load is also sometimes called "regulation" by transformer makers.
An unexpected consequence of how wire gauges and core stacks work is that regulation (actually sag from no load to full load) gets worse the smaller the transformer. A small 5VA/W power transformer may have regulation/sag of 20-30%. A sneaky way of saying this is to simply not mention the no-load voltage, but merely to specify the guaranteed output voltage at full load, and specify the "regulation". A transformer specified at, for instance, 12Vac at 50ma with 25% regulation will have 12*1.25 = 15Vac at no load.
RMS: Root of the Mean Square; this is part of the calculus stuff you can put off knowing for a while. RMS is important because for any waveform, the calculus-figured-out RMS value of the waveform would cause equivalent heating in a resistor to a DC value. That is, if you have a resistor of 10 ohms, and apply 12Vdc to it, you get heating of
12V*12V/10R = 144/10 = 14W
An AC sine wave 12Vrms supplied to a 10 ohm resistor also causes 14W of heat. The peak value of an AC rms voltage can be calculated by multiplying by the square root of two (1.41...). A 120Vac(rms) sine wave will have a peak value of 120*1.414 = 169.7Vpk. Waveforms other than sine have other conversion factors.
VA : Volt-Amperes; transformers often power things like motors that have big phase shifts between the voltage and current, unlike resistive loads. An overall inductive load may pull lots of current, but not cause much heating like a resistor would. The transformer supplying this still has to pass that much current, although the power is less than "real". VA is another way of saying "this many volts times this many amperes, even if the amperes and volts are phase shifted from one another.
For this example, we'll use resistor loads, so VA = watts, and 175VA at 115Vac(rms) is equal to 175/115 = 1.52173913... amperes; call it 1.522A(rms)
Hammond did something that all transformer makers do. They specified that the output will be >>at least<< 115Vac(rms) for loads from zero amperes up to 175VA. The jargon has started.
See definitions below.Making that specification come true requires that the transformer design take into account the losses in voltage from current flowing in the primary and secondary wire resistances and current "lost" to the magnetizing currents. The transformer designer increases the voltage ratio to make up for the current*resistance voltage losses in the winding resistance. The idea is to make the output voltage sag down to the specified 115Vrms or a little more when the output current is the full 1.522A(rms). The turns ratio is kicked up from 1:1 to something like 114.35/120.6.
I came up with that weird ratio from the numbers on the spec. The spec says that the no-load secondary voltage (NLV) will be 120.6Vac when the load resistance is effectively infinite and the primary is driven with 115Vac(rms). Under these conditions, the secondary current is effectively zero, so the voltage loss in the secondary resistance is zero. The voltage lost in the primary is the magnetizing current times the primary wire resistance of 1.32 ohms, or Vlost = 0.49A(rms) * 1.32R = 0.647V. So the hidden ideal transformer primary does not get 115Vac, it gets 115 - 0.647 = 114.353Vac(rms) as an input. This has to make an output of 120.6Vac(rms) on the secondary, so the voltage/turns ratio on the hidden ideal transformer has to be 120.6/114.35 = 1.054629.
Notice that the manufacturer told you that the output voltage will be higher than specified with no load, and will sag to no less than the specified output voltages under full load. This sag from no load to full load is also sometimes called "regulation" by transformer makers.
An unexpected consequence of how wire gauges and core stacks work is that regulation (actually sag from no load to full load) gets worse the smaller the transformer. A small 5VA/W power transformer may have regulation/sag of 20-30%. A sneaky way of saying this is to simply not mention the no-load voltage, but merely to specify the guaranteed output voltage at full load, and specify the "regulation". A transformer specified at, for instance, 12Vac at 50ma with 25% regulation will have 12*1.25 = 15Vac at no load.
RMS: Root of the Mean Square; this is part of the calculus stuff you can put off knowing for a while. RMS is important because for any waveform, the calculus-figured-out RMS value of the waveform would cause equivalent heating in a resistor to a DC value. That is, if you have a resistor of 10 ohms, and apply 12Vdc to it, you get heating of
12V*12V/10R = 144/10 = 14W
An AC sine wave 12Vrms supplied to a 10 ohm resistor also causes 14W of heat. The peak value of an AC rms voltage can be calculated by multiplying by the square root of two (1.41...). A 120Vac(rms) sine wave will have a peak value of 120*1.414 = 169.7Vpk. Waveforms other than sine have other conversion factors.
VA : Volt-Amperes; transformers often power things like motors that have big phase shifts between the voltage and current, unlike resistive loads. An overall inductive load may pull lots of current, but not cause much heating like a resistor would. The transformer supplying this still has to pass that much current, although the power is less than "real". VA is another way of saying "this many volts times this many amperes, even if the amperes and volts are phase shifted from one another.
For this example, we'll use resistor loads, so VA = watts, and 175VA at 115Vac(rms) is equal to 175/115 = 1.52173913... amperes; call it 1.522A(rms)
"It's not what we don't know that gets us in trouble. It's what we know for sure that just ain't so"
Mark Twain
Mark Twain
Re: Transformers
This is great. Thanks for your time and effort!
Re: Transformers
Thank R.G. most in formative. Where are we in terms of the full dissertation? 10%?
Ange
Ange
Re: Transformers
Yeah, probably 10% of the power transformers stuff. I think there's one more resistive-load diagram, then it heads off into rectifiers; after that, maybe some basic core selection and turns calculations.
I really hadn't planned for this to go off into output transformer design or switching power supply transformer designs, largely because I think most people would be long ago put to sleep. If you meant transformers in general, that is a very deep well. For that, we're a fraction of a percent at best into it.
"It's not what we don't know that gets us in trouble. It's what we know for sure that just ain't so"
Mark Twain
Mark Twain
Re: Transformers
Resistive loading example:
Here's the Hammond 169VS wannabe as calculated and modeled.
This process is partially theoretical, partially based on experience. It is possible that the exact numbers are off in the last digit from rounding as I went; this is a holdover from learning this stuff back when "calculator" and "computer" meant "slide rule" for engineering students.
Hammond promises that if you buy this transformer model, it will put out about 115Vac(rms) minimum at 175VA. Using a resistor as a load, no funny phase shifts get into the calculations (in the secondary at least).
> 175W at 115V(rms) is a current of 175/115 = 1.521739..., which I rounded to 1.522A(rms).
> with 1.522A coming out of the ideal transformer secondary, the current in the primary of the ideal transformer has to be 1.522A(rms) * Vsec/Vpri = 1.522 * 1.0546567.... = 1.60518... which I rounded to 1.605A(rms).
>The 115Vac(rms) source feeding the primary has to provide that 1.605A plus the primary inductance current of 0.49A. This means that the input current is 1.605 + 0.49 = 2.095A, right?
Wrong. The inductor current is phase shifted with respect to the referred secondary current. 1.605A at 0 degrees + 0.49A at 90 degrees is 1.678Arms at an angle that's neither 0 degrees or 90 degrees, but between them. RMS currents (and voltages) add by squaring each term, adding the squares, and then taking the square root of the sum. For angles of 0 and 90 degrees, this is all you need to do; at other phase angle differences (related to the term "power factor", later), you have to compute another term before taking the square root, but we only have one resistor and one inductor. So the total primary current is SQRT( 1.605^2 + 0.49^2 ) = 1.678A.
> The 115Vac(rms) input supplies this 1.678A through the primary wire resistance of 1.32 ohms, losing 1.32R * 1.678A = 2.215V in the primary resistance. The ideal secondary only gets 115 - 2.215 = 112.85Vac(rms) at its primary winding. Yeah, the diagram says 112.78Vac. I did some rounding step slightly differently somewhere. But it's really, really close, and that's what we're after. The secondary voltage is then 112.78 * Vp/Vs = 118.949... which I rounded to 118.9. This 118.9V supplies 1.522 A to the load, which is the secondary wire resistance and the external load resistor. This is R = 118.9/1.522 = 78.153 ohms, of which 1.61 ohms is the wire itself, so the load resistor can be 78.153-1.61= 76.5 ohms.
> The 76.5 ohm resistor has 1.522A running through it, so the output voltage is 1.522A * 76.5 = 116.443Vac(rms). So Hammond is justified in saying that the transformer will provide at least 115Vac(rms) at 175VA. It's actually a little better than that, which is always a good thing in power circuits.
A bit of a review: we started with published specs on a transformer, and used those specs to derive a model of the transformer using an ideal transformer with resistor and inductor imperfections that can be used to predict the performance of the real world transformer. We used only addition, subtraction, multiplication and division, and in one case taking the square root, along with Ohm's Law to do this. It is possible to do a similar model for most real-world mains-AC transformers you will encounter.
About phase shifted current addition: I used some trigonometry in calculating the sum of primary magnetizing current and reflected secondary load. The resulting load on the 115Vac source has to be vector/trigonometrically/calculus added to reflect what the real world does. In the picture below, the AC voltage is the dark black line. Any resistive current follows this, scaled by I = V/R, so it's always in phase with the voltage. In inductors, the inductor current lags the voltage across the inductor by 90 degrees; the inductor current reaches a peak just as the AC voltage driving it goes back through 0V. It turns out from the maths describing this that you get the right answer for RMS (heating equivalent) current if you square the resistor current, square the inductor current, add these, then take the square root. This gets more complicated for partial phase shifts, but that's not important to understand at this point.
Here's the Hammond 169VS wannabe as calculated and modeled.
This process is partially theoretical, partially based on experience. It is possible that the exact numbers are off in the last digit from rounding as I went; this is a holdover from learning this stuff back when "calculator" and "computer" meant "slide rule" for engineering students.
Hammond promises that if you buy this transformer model, it will put out about 115Vac(rms) minimum at 175VA. Using a resistor as a load, no funny phase shifts get into the calculations (in the secondary at least).
> 175W at 115V(rms) is a current of 175/115 = 1.521739..., which I rounded to 1.522A(rms).
> with 1.522A coming out of the ideal transformer secondary, the current in the primary of the ideal transformer has to be 1.522A(rms) * Vsec/Vpri = 1.522 * 1.0546567.... = 1.60518... which I rounded to 1.605A(rms).
>The 115Vac(rms) source feeding the primary has to provide that 1.605A plus the primary inductance current of 0.49A. This means that the input current is 1.605 + 0.49 = 2.095A, right?
Wrong. The inductor current is phase shifted with respect to the referred secondary current. 1.605A at 0 degrees + 0.49A at 90 degrees is 1.678Arms at an angle that's neither 0 degrees or 90 degrees, but between them. RMS currents (and voltages) add by squaring each term, adding the squares, and then taking the square root of the sum. For angles of 0 and 90 degrees, this is all you need to do; at other phase angle differences (related to the term "power factor", later), you have to compute another term before taking the square root, but we only have one resistor and one inductor. So the total primary current is SQRT( 1.605^2 + 0.49^2 ) = 1.678A.
> The 115Vac(rms) input supplies this 1.678A through the primary wire resistance of 1.32 ohms, losing 1.32R * 1.678A = 2.215V in the primary resistance. The ideal secondary only gets 115 - 2.215 = 112.85Vac(rms) at its primary winding. Yeah, the diagram says 112.78Vac. I did some rounding step slightly differently somewhere. But it's really, really close, and that's what we're after. The secondary voltage is then 112.78 * Vp/Vs = 118.949... which I rounded to 118.9. This 118.9V supplies 1.522 A to the load, which is the secondary wire resistance and the external load resistor. This is R = 118.9/1.522 = 78.153 ohms, of which 1.61 ohms is the wire itself, so the load resistor can be 78.153-1.61= 76.5 ohms.
> The 76.5 ohm resistor has 1.522A running through it, so the output voltage is 1.522A * 76.5 = 116.443Vac(rms). So Hammond is justified in saying that the transformer will provide at least 115Vac(rms) at 175VA. It's actually a little better than that, which is always a good thing in power circuits.
A bit of a review: we started with published specs on a transformer, and used those specs to derive a model of the transformer using an ideal transformer with resistor and inductor imperfections that can be used to predict the performance of the real world transformer. We used only addition, subtraction, multiplication and division, and in one case taking the square root, along with Ohm's Law to do this. It is possible to do a similar model for most real-world mains-AC transformers you will encounter.
About phase shifted current addition: I used some trigonometry in calculating the sum of primary magnetizing current and reflected secondary load. The resulting load on the 115Vac source has to be vector/trigonometrically/calculus added to reflect what the real world does. In the picture below, the AC voltage is the dark black line. Any resistive current follows this, scaled by I = V/R, so it's always in phase with the voltage. In inductors, the inductor current lags the voltage across the inductor by 90 degrees; the inductor current reaches a peak just as the AC voltage driving it goes back through 0V. It turns out from the maths describing this that you get the right answer for RMS (heating equivalent) current if you square the resistor current, square the inductor current, add these, then take the square root. This gets more complicated for partial phase shifts, but that's not important to understand at this point.
"It's not what we don't know that gets us in trouble. It's what we know for sure that just ain't so"
Mark Twain
Mark Twain
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martin manning
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Re: Transformers
I don't see the no-load primary current on the data sheet, so you must have measured that, no?R.G. wrote: ↑Tue Jul 09, 2024 10:18 pm The primary and secondary windings have resistances as specified by Hammond, and this model [169VS] includes the No-Load-Secondary-Voltage (NLV) and the no-load primary current. The magnetizing current of 0.49A is the price we pay for having a transformer work.
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Re: Transformers
Iex, top row of the 'electrical data' table.martin manning wrote: ↑Sat Jul 13, 2024 6:41 pmI don't see the no-load primary current on the data sheet, so you must have measured that, no?R.G. wrote: ↑Tue Jul 09, 2024 10:18 pm The primary and secondary windings have resistances as specified by Hammond, and this model [169VS] includes the No-Load-Secondary-Voltage (NLV) and the no-load primary current. The magnetizing current of 0.49A is the price we pay for having a transformer work.
I guess it means 'excitation current'.
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martin manning
- Posts: 14057
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Re: Transformers
Ah, yes, of course.