mbrooke
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What is this phenomenon and what is the theory behind it? How is it different from induction? ////////////////////
Reflecting Wave Effect
- Thread starter mbrooke
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It has to do with the fact that waves partly reflect and partly pass through at any point when the impedance of the medium changes. In the case of VFDs the motor impedance is significantly different from the cable. At the motor the reflected wave (pulse) can be nearly the same voltage as the forward wave leading to twice the DC bus voltage at the motor terminals at around 100-250 feet. With even longer cables the first pulse might not be fully attenuated before another pulse hits, leading to 3-4 times DC bus voltages.
This is basic transmission line theory (distributed RLC). Induction is part of it but self inductance would cause the issue even without it.
The fix is one of two options. The first is to add an RC circuit at the motor terminals to match the impedance to the line so that a reflection does not occur. The other approach is to add an LC filter (dv/dt) filter to remove the higher frequencies on the drive output, or a multipole filter (sinus filter) for longer distances.
This is basic transmission line theory (distributed RLC). Induction is part of it but self inductance would cause the issue even without it.
The fix is one of two options. The first is to add an RC circuit at the motor terminals to match the impedance to the line so that a reflection does not occur. The other approach is to add an LC filter (dv/dt) filter to remove the higher frequencies on the drive output, or a multipole filter (sinus filter) for longer distances.
mbrooke
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Is this in any way related to quantum mechanics or general relativity or am I over thinking it?
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You are overthinking it. It is simple classic E/M theory, easily modeled by classical mechanical wave systems (like the horizontal Slinkies used in AP physics classes.)
Anyone who has worked with RF transmission lines or high speed data lines will be familiar with the phenomenon.
Anyone who has worked with RF transmission lines or high speed data lines will be familiar with the phenomenon.
mbrooke
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winnie
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Step 1 in understanding this is to have an idea of what a transmission line is. For most of the work that an electrician encounters, you don't have to think about the speed of light. Voltage applied at one end of the wire is _instantly_ (in this approximation) seen at the other end of the wire. But once frequency is high enough (or the wires long enough) you can't ignore the speed of light; you apply voltage at one end, and it takes some time before you see that voltage at the other end.
Imagine a very long circuit. Say a pair of wires 1" apart and 180,000 miles long. (Make the wires superconducting, we don't want to worry about voltage drop!!) You have a lamp at 1 end of the wires and a power supply at the other end. You apply power at one end, and current starts flowing into the wires. Power is flowing into the wires, but the lamp is off. A second or so later the light turns on. When you turn off the power, it takes a second or so before the lamp turns off. If you were to pulse the power on for 0.1s, the lamp will pulse on for 0.1s, but about 1 second later.
Now remove the lamp and pulse the power at one end. The other end of the wire is 1 second away, so the source end simply can't see the fact that the lamp is gone. Power will flow into the wires for that 0.1s pulse. When the energy of the pulse reaches the other end, there is an open circuit. But the energy has to go somewhere, so it reflects back. 2s after the pulse gets launched, you will see a voltage pulse coming back on the wires.
The issue of 'reflections' is that energy exiting at one end of the line is delayed in time versus that energy applied at the other end.
The rest is understanding the mathematics of the above, figuring out the speed that the pulse will propagate, the 'resistance' seen at the input end of the wires which depends on wire size and spacing, and what it means for the load to be 'matched' at the far end.
In a nutshell: when you calculate out the distributed inductance and capacitance of the line you find that the input of the transmission line looks like a resistor. The load at the opposite end of the line has to show the same resistance or you will have some amount of reflection. But that is in the details, not the general concept.
-Jon
Imagine a very long circuit. Say a pair of wires 1" apart and 180,000 miles long. (Make the wires superconducting, we don't want to worry about voltage drop!!) You have a lamp at 1 end of the wires and a power supply at the other end. You apply power at one end, and current starts flowing into the wires. Power is flowing into the wires, but the lamp is off. A second or so later the light turns on. When you turn off the power, it takes a second or so before the lamp turns off. If you were to pulse the power on for 0.1s, the lamp will pulse on for 0.1s, but about 1 second later.
Now remove the lamp and pulse the power at one end. The other end of the wire is 1 second away, so the source end simply can't see the fact that the lamp is gone. Power will flow into the wires for that 0.1s pulse. When the energy of the pulse reaches the other end, there is an open circuit. But the energy has to go somewhere, so it reflects back. 2s after the pulse gets launched, you will see a voltage pulse coming back on the wires.
The issue of 'reflections' is that energy exiting at one end of the line is delayed in time versus that energy applied at the other end.
The rest is understanding the mathematics of the above, figuring out the speed that the pulse will propagate, the 'resistance' seen at the input end of the wires which depends on wire size and spacing, and what it means for the load to be 'matched' at the far end.
In a nutshell: when you calculate out the distributed inductance and capacitance of the line you find that the input of the transmission line looks like a resistor. The load at the opposite end of the line has to show the same resistance or you will have some amount of reflection. But that is in the details, not the general concept.
-Jon
mbrooke
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Great explanation.
Basically current is flowing in an open circuit for a brief moment? Is this some type of electron "elaticense"? Magnetism plays no role, right?
Basically current is flowing in an open circuit for a brief moment? Is this some type of electron "elaticense"? Magnetism plays no role, right?
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Current does not flow into an open circuit in this case any more than you can say that current charging a capacitor is flowing into an open circuit.
A transmission line, whether power distribution or data/signalling, or the power feed from a VFD to a motor, has two characteristics which are dependent on the physical layout of the wires, the insulation, and any surrounding shielding. Those are the bulk capacitance between the conductors, so many fraction Farads per foot, and the bulk inductance of the wires themselves in fractional Henrys per foot. Those two characteristics combine to cause current proportional to applied voltage to flow into the transmission line, which forms an effective resistance. Which is called the characteristic impedance of the line. But that flow only takes place undisturbed until the leading edge of the applied voltage waveform reaches the other end.
Since the bulk/distributed inductance plays a role you have to say that magnetism does indeed play a role.
There is energy present in the magnetic field along the transmission line, just as there is energy stored in the capacitance along the way.
mbrooke
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I'm confused though... the above make it sound like current is briefly flowing in an open circuit. And it would make sense if electrons entering a wire result in a delay in electrons leaving a wire.
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That sort of works in envisioning the reason for the time delay, but it does not give you a handle on the relationship between applied voltage and the resulting current.
And the key point (I think) in my description above is that the current is flowing into the distributed capacitance, not piling up inside an isolated wire.
mbrooke
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Your terms don’t really follow. Nor does your attempt at explaining it. The problem is you are trying to make sense out of a transmission line which simplifies a very complex issue.
In electrically short cables we just ignore the cable properties and assume wire is a short circuit. If we approach current limits we maybe treat it as a resistor but that’s all we do. But when the wavelength of a sine wave or the speed that a pulse turns on and off approaches the physical length of the cable, we cannot use a simple short circuit model. Why this happens is pretty simple.
If we had no connection at the end nothing flows. So your explanation is way off.
In a transmission line we have at least two conductors. Even in an antenna the Earth itself counts. The two (or more) conductors are separated by an insulator...that’s the definition of a capacitor. Call is stray capacitance or whatever you want. Imagine each foot of wire is a tiny capacitor in parallel. Normally these little capacitors don’t matter but as I said we have long cables so hundreds of tiny stray capacitors in parallel add up.
The cable also is a one turn coil. It has self-inductance in series in each foot. Not just mutual inductance which in transmission lines we typically ignore. And some resistance. So again on short cables we can ignore this but not on long ones. Mutual
Inductance between the two conductors also exists too so we have two inductance terms in the model. Again very small. But we have tens if not hundreds of feet of transmission line so we have to treat it as a long distributed system. So the schematic would look like this:
It turns out that we can simplify the math a lot. Within the transmission line the wave travels essentially unchanged...in equals out. All we have to worry about is propagation velocity...waves travel slower than the speed of light in anything other than vacuum. And we calculate the impedance of the cable, source, and loads. At the end point we can calculate what happens to the wave based on the ratio of the impedances. Some passes through to the load, some is reflected. Then we do the same calculation again at the other end at the source. Then again at the load. Usually after 2-4 cycles the transmitted and reflected waves are too small to matter. The real explanation has to do with those hundreds of little RLC circuits each of which charges and discharges distorting the input pulse but actually modeling a cable with hundreds of RLC components is completely impractical and the cable is largely made up of an infinite series of small sections. So we don’t try to model it directly. Transmission line calculations give us the simple answer directly in something we can do with “napkin math” (by hand).
With an oscilloscope you can actually see it. At the motor terminals we don’t just get a nice square pulse. What we actually see is a ringing waveform, like this:
The ringing is caused by each of the reflections bouncing back and forth, arriving at a different point in time. As they add and subtract from each other they gradually settle out until all we see is the DC bus voltage. But in the process of getting there the first reflection can vastly exceed the surge voltage rating of the motor or the cable, causing an arc and breaking down the insulation. As cable length increases the voltage increases though there is a limit of twice the DC bus voltage. On one pulse. As the cable length increases further we can reach a point where the reflections of one pulse have not completely died down before another pulse starts, leading to up to 4 times the DC bus voltage.
DC bus voltage is 1.45 times the RMS input so on a 480 V drive that’s 696 V. Double that is 1392 V. Standard duty motors have a minimum surge voltage of 1000-1200 V and this level of voltage appears at around 100 feet of cable. So we are OK at 230 V but not 460 or 575 V. An inverter duty motor is rated 1450+ V. Most test at around 1650-1750. #14 THHN with the minimum 15 mils insulation tests at around 2800 V. Starting around 200-250 feet is when we start to see voltage tripling.
So my advice is at 50-100 feet you need to start thinking about reflections. Some drives are worse than others. Yaskawa and Allen Bradley have little filtering and are particularly bad. You can just use inverter duty motors if you are careful to always do that and go up to about 200 feet. After that a dv/dt filter is required up to around 500-600 feet (check manufacturer) at which point only a sinus filter will do. You can use a load reactor but it adds heat to the cabinet, lowers efficiency, lowers maximum power output, and doesn’t do as well as the dv/dt filter for about the same cost.
AB does sell a line matching filter aka 1204 motor terminator that you can install on the motor end. This works by adding impedance to the motor to match the line, eliminating reflections. It’s best used if there is a space issue on the drive side and it’s fairly inexpensive but it does have to be matched to the motor size and line length. These things are very simple but GE has patented the simplest version so as the patent runs out perhaps we will see this option appear instead of big heavy reactor solutions.
Capacitor at first is a dead short then charges eventually acting like an open circuit. This allows the next one to charge and so forth. Similarly the magnetic field of the mutual and self inductors at first blocks current then gradually lets it through. If we stand back and look at the distributed circuit elements it looks like a delay slowing down the pulse. It is not blocked...an inductor is a short to DC and a capacitor is an open, but it slows down the initial wavefront. The effect is frequency dependent too...higher frequencies (faster changes) are affected more.
mbrooke
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This time around I'll admit that I have no clue that I have no clue
Alright.
Probably is, but I was just going by post #6.
Understood.
Understood.
Ok. So what happens to the electrons when the wave reflects back? Do they change direction briefly? Or just slow down? This is where my imagination kind of derails so to speak. I mean I can picture energy oscillation between reactors and capacitors in series... but this is a bit more complex for me. At least from my mental process.
Practical information noted!
I had no idea electricians needed to be worried about this, but I guess they do! Never stop learning.
mbrooke
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Alright, this I can envision. Even through fluid representation I'm iterating in my head.
(I still think of electricity as marbles and water pipes)
synchro
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Others have explained much of what there is to say, but here's my contribution:
Waves propagate along a transmission line like coax, twisted pair, etc. because of the successive transfer of energy from each incremental series inductance to the next shunt capacitance, then from that capacitance to the next series inductance, and so on. As mentioned there's a particular series inductance and shunt capacitance per unit length in a given transmission line, but because these are distributed elements you need to break up the line into a lot of individual small L's and C's (aka lumped elements) to accurately model a transmission line with them. A rule of thumb is least a pair of L's and C 's for every 1/10th wavelength of the signal. However, the ratio of each of these little L's and C's (i.e. L / C) stays the same even though both L and C get smaller when the line is broken into smaller segments because L / C = Zo² , where Zo is the characteristic impedance of the t-line (for example, 50 ohm, 75 ohm, etc.).
The speed at which a wave propagates is Vp = 1/ √(LC), which is the speed of light if there is a vacuum (or air for all practical purposes) between the conductors of the t-line. If there's insulation like polyethelene, teflon, etc. then C increases and the propagation velocity goes down.
An analogy to electromagnetic waves on a transmission line is mechanical waves along a string of masses connected together with springs. The springs have stored energy when they are stretched, and the masses have stored energy when they have velocity. So the springs have the most stored energy (potential energy) when they are stretched the most but have essentially no energy left in the middle of their range. In contrast the masses have the most energy when they are moving the fastest (kinetic energy) in the middle range of their movement, but no energy left when they've stopped at the ends of their movement. Because of this complementary relationship of masses and springs, as the wave propagates along each mass hands off its energy to the next spring, this spring to the next mass, and so on down the line.
Inductors and capacitors store energy in a complementary manner just like masses and springs. When a sine wave is applied to them, a capacitor stores the most energy when the voltage is highest and an inductor stores the most energy when current is highest (but the voltage is lowest). So this allows the succesive transfer of energy between inductors and capacitors as the wave moves along a transmission line.
I could go on but that's probably more than enough for now.
Waves propagate along a transmission line like coax, twisted pair, etc. because of the successive transfer of energy from each incremental series inductance to the next shunt capacitance, then from that capacitance to the next series inductance, and so on. As mentioned there's a particular series inductance and shunt capacitance per unit length in a given transmission line, but because these are distributed elements you need to break up the line into a lot of individual small L's and C's (aka lumped elements) to accurately model a transmission line with them. A rule of thumb is least a pair of L's and C 's for every 1/10th wavelength of the signal. However, the ratio of each of these little L's and C's (i.e. L / C) stays the same even though both L and C get smaller when the line is broken into smaller segments because L / C = Zo² , where Zo is the characteristic impedance of the t-line (for example, 50 ohm, 75 ohm, etc.).
The speed at which a wave propagates is Vp = 1/ √(LC), which is the speed of light if there is a vacuum (or air for all practical purposes) between the conductors of the t-line. If there's insulation like polyethelene, teflon, etc. then C increases and the propagation velocity goes down.
An analogy to electromagnetic waves on a transmission line is mechanical waves along a string of masses connected together with springs. The springs have stored energy when they are stretched, and the masses have stored energy when they have velocity. So the springs have the most stored energy (potential energy) when they are stretched the most but have essentially no energy left in the middle of their range. In contrast the masses have the most energy when they are moving the fastest (kinetic energy) in the middle range of their movement, but no energy left when they've stopped at the ends of their movement. Because of this complementary relationship of masses and springs, as the wave propagates along each mass hands off its energy to the next spring, this spring to the next mass, and so on down the line.
Inductors and capacitors store energy in a complementary manner just like masses and springs. When a sine wave is applied to them, a capacitor stores the most energy when the voltage is highest and an inductor stores the most energy when current is highest (but the voltage is lowest). So this allows the succesive transfer of energy between inductors and capacitors as the wave moves along a transmission line.
I could go on but that's probably more than enough for now.
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