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Power electronics design

If you are interested in power electronics design at the board or system level, I would recommend LTspice (note the correct spelling) by far above all the others. In addition to being superb for IC design (Linear Tech uses LTspice to design all their own ICs), it also has been specifically designed to run board level, switched mode simulations.

Because of its robust, excellent performance and because it is available at zero cost, LTspice has become the de facto standard SPICE with by far more engineers using it than any other flavor of SPICE. LTspice allows 100 percent transportability and work sharing, i.e., anyone, even those who have not been previous users, can open your files and run your simulations (the free download is well under 10Mb, installs very quickly and is very system friendly – not cookies, messy registry alterations, scattering of installation folders, etc. – removal, if you so choose, is easy and complete).

Like most versions of SPICE today, LTspice has a fine user interface, but that feature should be low on your list. Schematic entry is NOT where you will be spending most your time when doing serious design work. Beyond a point, desktop eye-candy does nothing to help you understand your design and see its flaws and weaknesses (in fact, too many layers of hand holding can just get in the way of that).

Personally, I never breadboard a design anymore until it has proven itself in LTspice (unlike with a breadboard, a simulated circuit’s internals are ALL easily viewable – a great boon for understanding tricky operation). For me, first hardware is always a complete layout (and matches the simulation every time). Of course, the old axiom “garbage-in, garbage-out” very much applies, which means I often spend a lot of upfront time verifying (and modifying and/or making) models to match their components’ data sheets. In fact, I would recommend doing that as a very worthwhile exercise and as something that should impress a potential employer.

When developing a design in SPICE, you will want to spend your time debugging your design, not your simulation or your simulator, therefor it is worthwhile to learn what a simulator needs to run smoothly (with LTspice, all that means is that the input has to be realistic). It was years of working with simulators and a lot of sweat and aggravation before the keys to problem-free simulations gradually crept into my understanding.

1. If possible, make all nonlinear circuit elements be functionally continuous with continuous derivatives (this is not possible for some component behaviors), and

2. *always* craft your simulations so that the nonlinear bits become linear at high frequencies (this is always possible). Non linear devices should never be strict voltage sources. They should be Nortonized and be shunted with small capacitances such that the capacitances (which are linear elements) dominate at small time steps.

3. Always verify that the building blocks of your simulation behave realistically (GIGO).

Follow these guidelines and you will never see the “time step too small” message (I have never met a simulation that couldn’t be made to run well). Note that many (if not most) vendor supplied models fail to meet these guidelines and will give you nothing but headaches if you try to use them “as is.”

Calculate current setting of overcurrent relay

You can calculate current setting of overcurrent relay by using next expression:

Isetting ≥ (ks*Imaxopam)/(a*pi)
Imaxopam=kam*Imaxoptr

where are:

Isetting-current setting of overcurrent relay
ks-safety coefficient
Imaxopam-maximum operational current under which overcurrent relay shouldn’t to act
a-coefficient of layoffs overcurrent relay (0,85-0,95)
pi-ratio of current transformer
kam-coefficient which describes influence of common starting of all asynchronous electrical motors in the appropriate power network after elimination of fault (1-6)
Imaxoptr-maximum operational current of power transformer

Besides I have already explained meanings of all appropriate sizes, I would like to underline differentiate between Imaxopam and Imaxoptr.
Imaxoptr is maximum operational current of power transformer in normal conditions while Imaxopam is maximum operational current of power transformer after interruption of fault and mentioned current includes influence of common starting of all asynchronous electrical motors in the appropriate power network after elimination of fault. It is very important to say that appearing of fault in power network leads to significant decreasing of voltage what has a consequence deceleration of all asynchronous electrical motors in appropriate power network. After interruption of fault, it comes to appearing of process during which all asynchronous electrical motors are starting in some parts of power network which are still turned on. This is situation which is significant different from situation where asynchronous electrical motors are starting one by one while in this case all asynchronous electrical motors are starting at the same time. Because of this fact, after elimination of fault, value of current isn’t same as value of current before appearing of fault. After elimination of fault, value of current, which I called Imaxopam, is higher than value of operational current of power transformer, which I called Imaxoptr, but under those conditions there is no fault, so current setting of overcurrent relay should to be set on that value of current and mentioned relay shouldn’t act under those conditions.

After calculation of current setting of overcurrent relay, you need to check coefficient of sensitivity of acting of overcurrent relay by using next expression:

ksens=Ifmin/(Isetting*pi)

where are:

ksens-coefficient of sensitivity of acting of overcurrent relay
Ifmin-minimum fault current (1 phase fault to the earth or 2 phases fault)
Isetting-current setting of overcurrent relay
pi-ratio of current transformer

Value of coefficient of sensitivity of acting of overcurrent relay should to be higher or equal with 1,5 in case when is fault at opposite busbars (busbars where isn’t overcurrent relay) or higher or equal with 1,2 in case when is fault at the end of the longest feeder which begins at those opposite busbars. Time setting should to be selected like that overcurrent relay needs to wait acting of another protection which is on the feeders (for example distance protection).

Insulating resistance measurement

Please remember that Insulating resistance (IR) measurement and associated polarization index tests is just one of the many tools used for insulation system integrity analysis. Its value and repeatability is dependent on the environmental condition at the time it is taken; as mentioned temperature, humidity contaminations all contribute/effect the reading.

The baseline figure should be obtained from either factory or during initial commissioning (as per factory condition). So performing commissioning in the rain, dirty surface, high humidity may result in low values for both dry type and oil filled equipment. Low reading in itself does not indicate bad insulation where the machine cannot be returned to service.

The bottom line is assessment lacking or other data would be:
1. The machine was running at it was running ok before the test.
2. The leakage value at operating voltage will be V/R; therefore the heat loss will be I^2R. Is that OK or warrant some corrective measure.
3. PI may approach 1, is that OK or not? Is this mtruly and indication of wet insulation or of resistive value but will still be OK when energized as per 3. above?

IR, PI measurement along with Cap bridge / dissipation tests, PF test and others are performed to ensure the insulation integrity for maintenance and commissioning.

If cable and equipment have gone through routine maintenance, it is good practice to perform these tests and making sure no ground are left before energizing.

Please read a “a stich in time” by Megger.

It by itself is just a test. The test is meaningful.

Excitation system in generator

The excitation system requires a very small fraction of the total power being generated. If we could simply increase the excitation (a very small amount of power) and increase the generator’s real power output, the world’s energy problems would be solved, because we would have a perpetual motion machine.

In the case of a generator connected to a large grid, the generator will inject any desired amount of power into the grid if its prime-mover is fed the desired power (plus a small additional amount of power to take care of losses). This is true, regardless of the total load on the grid, because the generator’s output is an extremely small fraction of the total grid power, and it alone cannot make drastic changes to the grid’s frequency.

Normally, the load varies by a very small fraction of the total grid power. If the load increases, the frequency of the entire grid (including the generator in question) lowers a very small amount, generally less than one-hundredth of one Hz. The frequency slew (that is, the rate-of-change of frequency) is very low, because there is a massive amount of energy that is stored as the kinetic energy of the rotors of all of the generators. At this point, nothing needs to be done; the system simply runs a little faster or slower.

Over time, as the load changes a greater amount, the frequency moves further from the nominal frequency (50 Hz or 60 Hz). When the difference between the actual frequency and the nominal frequency becomes greater than about 0.01 Hz, action is taken to make changes to the output of the grid’s generators.

The specific action may be determined by the regulating authority (for instance a power pool in the US) and it is usually based on economics, subject to other constraints. If the load has increased (and the frequency is less than the nominal frequency), the generators that have the lowest incremental cost of power will be asked to increase their output, or if all generators are near their limits, new generators (with the lowest incremental cost) are asked come on line. It’s important to note that a generator’s limit is usually 80% or 90% of its rating. The 10% or 20% of unused capacity is the system’s “spinning reserve”, which is used to maintain grid stability for sudden, large power variations.

The same thing happens with a generator connected only to its load or a weak grid with just a few other generators. However, because there is relatively little kinetic energy stored in the rotors of the one or few generators, the change in frequency associated with a load change is much greater, so frequency variations are much greater and corrective actions may not be implemented before the frequency varies by more than a few Hz.

Phase rotation errors

Phase rotation errors are not as rare as they ought to be. I’ve seen more than one building with a systematic phase rotation error. This can be prevented by carefully following the color coding system (Yellow Orange Brown and Red Blue Black for 480 volt and 208 volt systems in the US for example) and tagging feeders at both ends to assure proper connections.

To check for proper phase rotation sequencing (ABC and not ACB) you can use a phase rotation meter. Without that you can bump a three phase motor that should be correctly connected to see if it turns in the right direction. If it’s wrong, reverse any two phase wires from the source to the distribution equipment. However, if you have a tie breaker and intend to operate the secondaries of two transformers in parallel by closing it that is not good enough. Both transformer distribution networks have to be connected correctly on all three phases. You have to check the voltage across each corresponding pair of terminals on the tie breaker and be certain they are all about zero volts. If you don’t and there is an error, closing the tie breaker if that is possible at all (some electronic breakers may lock you out) will result in a phase to phase bolted fault that can result in severe damage to your distribution equipment. Phase rotation errors are invariably the result of incompetent installation, inadequate specifications for feeder identification, and inadequate inspection.

There are times when the phase rotation error is made on the primary side of the transformer. If this happens it can be compensated for by reversing the phase rotation error from the secondary side. This is less desirable but it will work. If you have multiple phase rotation errors in the same distribution network you have quite a mess to clean up. It will be time consuming and expensive tracking all of them down to be certain you have eliminated them. False economies by cutting corners on the initial installation of substations and distribution equipment will result in necessitating very expensive and inconvenient repairs. If it is not corrected you risk severe damage to three phase load equipment.

Moving data around within memory of an individual PLC

The first question would have to be – why do want to do it? If the data already exists in one location that is accessible by all parts of the program, why are you going to use up more PLC memory with exactly the same data?

Well, there are a couple of candidate reasons. One might be recipe data. You have an area of memory with a set of stored recipes for different products, and at an appropriate moment you want to copy a specific recipe from the storage area to the working area. The first thing to be said about that is that if your recipes are at all complex and you have a requirement to have a significant number of different recipes, then PLC memory is probably not the right place to be storing them. The ultimate, these days, of course, is that recipes are created by techies on PCs away from the production area, in nice quite, comfortable labs or whatever, and are stored on a SQL server. Only the recipe for today’s actual production run gets transferred to the PLC. But there are some applications where there is a limited number of different recipes required and the recipes themselves are quite simple, when it can be reasonable to store the recipes in PLC memory.

A second reason for copying memory areas within the same PLC is for procedures, sub-routines or whatever. But again, these days, all PLC languages have some sort of in-built facility for procedures – what Rockwell uniquely call Add On Instructions, what everyone else calls UDFBs – user defined function blocks. In any case, the point is that these facilities usually make all that memory management stuff transparent to the programmer. You just configure the UDFB and call it as required. The compiler takes care of all the memory data moves for you.

Another reason for copying memory, actually related to the previous, is a technique much used by PLC programmers where they use an area of memory as a ‘scratch-pad’. So they will copy some unprocessed data to the scratchpad area, all of the operations performed on the data take place using the scratchpad, and at the end, they copy the processed data back again. Again, it is questionable how much this technique is actually required these days, I would suggest that it most cases, there probably is a better way using a UDFB. But I have seen some programmers who routinely include a scratchpad area within any UDFBs they define.

Sub-transmission network

Q:
What factors determine current distribution between two 33kV feeders feeding a 33/11kV Substation within a sub-transmission network.

A:
Try the voltage divider rule. Take the R + X of each feeder (resistance and reactance) and find the Z. Remember that square root of R² + X² = Z. Now that you have the Z of each feeder, now find the Z of the two in parallel. To do this we have Z = 1/(1/Z1 +1/Z2). So, if Z1 = 2.16 ohms and Z2 = 1.67 ohms, then our Z of the two in parallel is 0.94 ohms. Now we pass the current of the entire substation between these two feeders. Let’s say that the current is 240 amps. Now we have 240Ax0.94ohms = 226 volts (IxZ=V). And since voltage divided by impedance gives us current (V/Z = I) we now take the voltage drop across the two feeders in parallel and divide each of the feeder impedances to get the separate feeder current. So we get 226V/2.16ohms = 105 amps (feeder 1) and 226V/1.67ohms = 135 amps (feeder 2). I have not tried this with your exact situation. Having different voltages from two different substations will change things, but at least this way you have a good start on the problem.

Since one end is tied together and the two other ends are from different substations, then you will have the classic voltage sending and receiving formula. Since the load is the one substation, then their will only be one power factor of the one load, so I would think this formula would apply: Es = Square Root of ((ErCosƟ + IR)² +(ErSinƟ +IX)²), which is square root of ((Receiving voltage times the cosine of the current phase angle plus current times resistance of the line)² + (Receiving voltage times Sine of the current phase angle plus current times reactance of the line)²). The voltage drop across each line would be VD=I(RcosƟ +XsinƟ) where R is the line resistance and X is the line reactance and the Ɵ is the phase angle of the load.

System with difference neutral

Q:
I have one system with two source. One from genset and the other from PLN (national power supply company) that each system has neutral.
The question is
1. Is there any problem if I connect both neutral directly?
2. Is there any spark when I connect both neutral?
3. How is the best solution to connect both neutral?

A:
1. I understand that the Genset is dedicated for essential load as an Emergency power supply which will be operated by hand (only in Manual Mode).

2. The Control Philosophy for a Generator that intended to be connected to PLN as emergency power source depends on the local service provider regulations.

3. Usually in your case there should be Electrical as well as Mechanical interlocks between the mains incomer & genset main breaker. ie both Sources will never be in Synchronism ( will not feeding the same load simultaneously).This measures will ensure that there will be only neutral point to the system.

How to suppress chaotic operation in a DCM flyback at low load

I would like to share these tips with everybody.
A current mode controlled flyback converter always becomes unstable at low load due to the unavoidable leading edge current spike. It is not normally dangerous but as a design engineer I don’t like to look at it and listen to it.

Here are three useful and not patented tips.

First tip:
• Insert a low pass filter, say 1kohm + 100pF between current sense resistor and CS input in your control IC.
• Split the 1kohm in two resistors R1 to the fet and R2 to the control IC. R1 << R2.
• Insert 0,5 – 1pF between drain and the junction R1/R2. This can be made as a layer-to-layer capacitor in the PCB. It does not have to be a specific value.
• Adjust R1 until the spike in the junction in R1/R2 is cancelled.
You will see that the current spike is always proportional to the negative drain voltage step at turn-on. Once adjusted, the cancellation always follows the voltage step, and you some times achieve miracles with it. Cost = one resistor.

Second tip:
Having the low pass filter from first tip, add a small fraction of the gate driver output voltage to the current sense input, say 0,1V by inserting a large resistor from ‘Drive Out’ to ‘CS input’. The added low pass filtered step voltage will more or less conceal the current spike. You should reduce your current sense resistor accordingly. Cost = one resistor.

Third tip:
In a low power flyback, you some times just need an RC network or just an extra capacitor from drain to a DC point, either to reduce overshoot or to reduce noise. Connect the RC network or the capacitor to source, not to ground or Vcc. If you connect it to ground or Vcc, you will measure the added discharge current peak in the current sense resistor. Cost = nothing – just knowledge.

All tips can be used individually or combined => Less need for pre-load resistors on your output.

Right Half Plane Pole

Very few know about the Right Half Plane Pole (not a RHP-Zero) at high duty cycle in a DCM buck with current mode control. Maybe because it is not really a problem.
It is said that this instability starts above 2/3 duty cycle – I think that must be with a resistive load. If loaded with a pure current source, it starts above 50% duty cycle.

Here is a little down-to-earth explanation:
If you run a buck converter at high duty cycle but DCM, it probably works fine and is completely stable. Then imagine you suddenly open the feedback loop, leaving the peak current constant and unchanged. The duty cycle will then rush either back to 50% or to 100% if possible. You now have a system with a negative output resistance – if Voltage goes up, the output current will increase.

You can see it by drawing some triangles on a piece of paper: A steady state DCM current triangle with an up-slope longer than the down-slope and a fixed peak value. Now, if you imagine that the output voltage rises, you can draw a new triangle with the same peak current. The up-slope will be longer, the down-slope will be shorter but the sum of times will be longer than in the steady state case. The new triangle therefore has a larger area than the steady state triangle, which means a higher average output current. So higher output voltage generates higher output current if peak current is constant. Loaded with a current source, it is clear that this is an unstable system, like a flipflop, and it starts becoming unstable above 50% duty cycle.

However, when you close the feedback loop, the system is (conditionally) stable and the loop gain is normally so high at the RHP Pole frequency that it requires a huge gain reduction to make it unstable.

It’s like when you drive on your bike. A bike has two wheels and therefore can tilt to either side – it is a system with a low frequency RHPP like a flipflop. If you stand still, it will certainly tilt to the left or to the right because you have no way to adjust your balance back. But if you drive, you have a system with feedback where you can immediately correct imbalance by turning the handlebars. As we know, this system is stable unless you have drunk a lot of beers.