big batteries and step up DC converters

[antiscab]

The next step is to get the inductor from a DC controller, mine comes from such a source [90 microhenry @300A]. They have a gapped iron cct so they don't saturate, the energy is actually stored in that air gap

out of curiosity, where did you get that industrial DC controller from?
I've always wondered where you could get big inductors from?

Matt

[Richo]

300A inductor still sounds too small.
144v x 300A is 43kW.
Remember power in is the same as power out ignoring losses.
Don't forget you will have some form of duty cycle too which will reduce the peak 43kW.
Energy is stored in the created EM field.
An air gap creates fringing of an EM field and so will loose some of this field/energy.
So will not be efficient as a toriodal.
But power is the name of the game right!

Another consideration will be it's construction.
If you are using laminated steel this will limit you to lower frequencies.
Which is fine but means you will need bigger caps to remove the ripple.

Are you sure the 90uH just wasn't used to provide some filtering on the DC input of the controller?

[T2]

Sheesh I almost wish I hadn't brought it up !

The inductor was sourced from a 400 Amp Willey DC controller. Of laminated costruction with its volume approximating a 4" cube the output inductor was rated for only 4khz operation. I have previously experimented with it and a single device (MJ11031) at around 50Amps and 40v @ 14khz with mixed results for a controller of my own design. In retrospect maybe I should have started at 4khz and worked my way up in frequency when I had obtained better competency at the 50Amp Icmax level. But that's history now.

The Willey used an ancient USG3524 PWM control chip with both outputs wire or'd. By cutting a trace one output could provide the needed duty cycle limited to about 45%. The feedback input op-amp would need modding to supply a constant voltage output. I might keep the original circuit which used a Vcesat sense line on the power switch to terminate the pulse in case of trouble.

As some here might remember you had to use 16 transistors with the same current gain for the power switch else they were inclined to blow up. The lower gain devices always being the faster to turnoff, would leave the slower devices to be trapped carrying all the current. As they commutated to the flyback diode this would put them outside their RBSOA rating with the eventual and predictable result of their destruction. It didn't help that the capacitor bank was not on the main PCB a no no of course. The leads to the caps had stray inductance that couldn't be snubbed.

Energy is stored in the created EM field.

Well the iron laminations do help contain the field but more ampere turns are expended getting the flux to cross the gap, than traverse the iron cct. So that's where the energy is - makes sense to me. The gap dimensions are fairly critical. For power use one could choose a ferrite core of the requisite magnetic properties and avoid gapping altogether.

I wasn't aware of flux fringing as a problem here, though I was always concerned that if the air gap was too wide you wouldn't get the flux and since the inductance is porportional to the square of the flux linkages then the resulting energy storage capability would be suppressed. OTOH if the gap was made too small the iron will saturate early i.e. at a lower current so the linkages will no longer increase with current - the iron will begin to behave as a huge air gap - with the effect that the inductance drops off drastically along with the energy storage.

Are you sure the 90uH just wasn't used to provide some filtering on the DC input of the controller?

There was no input filtering but actually that wouldn't have been a bad idea for Willey to have included a battery input choke. As an EV controller this was an unsuccessful design in other ways too.

[Richo]

Sheesh I almost wish I hadn't brought it up !

I think it's always good to bring up all options regardless.
Otherwise we all end up narrow minded.

[Squiggles]

Sheesh I almost wish I hadn't brought it up !

I think it's always good to bring up all options regardless.
Otherwise we all end up narrow minded.

Damned Right!

[T2]

I think it's always good to bring up all options regardless.

As options go this one is probably worth pursuing. Sure the Upconverter adds further complexity but it reduces battery pack complexity big time. The reduction in the number of battery interconnects, contactor safety disconnects and BMS modules is worthy of consideration.

Given the choice a standard battery pack optimised for storage rather than arbitrarily selecting the voltage as if it were a 47Kohm resistor - is the way to go. For Prius, Toyota already appears to be taking this route.

Battery capacity in the 2010           3G     Prius is 6,5Ah @ 201,6V (~15 mOhm)   ZVW30 Gen 3
Battery capacity in the 2004-2009 2G     Prius is 6,5Ah @ 201,6V (~15 mOhm)   NHW20 Gen 2
Battery capacity in the 2000-2003 classic Prius is 6,5Ah @ 273,6V (~20 mOhm)   NHW11 Gen 1
Battery capacity in the 1997-2000 Japan Prius is 6,0Ah @ 288V (25-30 mOhm)   NHW10 Not released into US market

When cruising along with the Upconverter shutdown, the relatively low voltage of the battery will oblige the 3-ph inverter to cope by moving towards a High modulation index. It is to be assumed this will be a more efficient output waveform.

When acceleration is called for the motor will need to see its full V/Hz exercised. At this time the Upconverter will turn on to provide the necessary voltage headroom for the inverter.

The idea hinges on the economics of fewer and larger batteries.
Its enabling technology is the interface to an inverter that is tolerant to a varying bus voltage

[coulomb]

When cruising along with the Upconverter shutdown, the relatively low voltage of the battery will oblige the 3-ph inverter to cope by moving towards a high modulation index. It is to be assumed this will be a more efficient output waveform.

Well, I think the goal is surely that the combination of up-converter and inverter combined is more efficient, so the overall battery consumption is reduced. When the up-converter is turned off (pass through), it should be quite efficient, effectively being just a turned on switch (or diode, with just the diode, IGBT or MOSFET voltage drop contributing a loss; no switching losses in the switches or inductors etc). When the load demands a higher voltage than can be obtained with the up-converter turned off, it has to be turned on.

I doubt that they play games with how much to turn the up-converter on; I imagine it would be fully on or fully off. But I suppose that's possible too. Suppose you need about 110% of the battery voltage for a long time, say a long gentle hill with no dips. It might be more efficient to up-convert to say 120% of the battery voltage, rather than up-convert to the maximum voltage (say 250% of bus voltage) and inverting that down to what the motor needs. Higher up-converter ratios are supposed to be less efficient, so an up-converter designed for 120% would be more efficient than one designed for 250%, everything else being equal. However, I don't know if an up-converter capable of 250% but running at 120% is any more efficient than the same up-converter running at 250%.

Maybe it's simpler (and possibly more efficient) to tell the inverter to run at 100% modulation index (or whatever its highest index is when producing the highest possible output voltage amplitude) whenever the required motor voltage is higher than can be obtained with the up-converter off. Then the up-converter would be used to set the motor voltage. That way, the voltage ratios are always at their lowest (in both the up-converter and the inverter, except when the up-converter would be required to convert down).

Edit: changed the last parenthesised text to hopefully make more sense.

[coulomb]

I just thought I'd update Weber's boost converter design with some figures we discussed over email. For reference, here is the circuit again:



[ Edit: I once said: "The capacitors here are small snubber caps for protection of the IGBTs." This is wrong; the capacitors are needed to supply current to the load (or absorb it from the source when regenerating) when the upper transistors are off. ]

Relative to the bottom rail, the output voltage can never go below that of the DC bus. You get this with 0% PWM (you can even leave the top transistors on indefinitely; not so the bottom transistors ). For some concrete figures, let's consider 1200 V IGBTs. The usual working limit for these is 900 V, allowing up to 300 V of ringing or spikes before the IGBT's ratings are exceeded. So let's use a Vbus of 300 V. So the output voltage can range from 300 V (at 0% lower transistor PWM) to centre point of 600 V (50% PWM) to 900 V (67% PWM), all relative to the negative end of the DC BUS. Relative to the positive end of the DC bus, it's 0, +300, and +600 V.

So we get 600 V swings from a 300 V input, a 2:1 boost. (The output can be +- 600 V because it's differential drive: the output is essentially +- 600 V from output to output).

But in a conventional design, we could have had 900 V swings (from a 900 VDC bus, ignoring IGBT drops). As far as I can see, the current through the IGBTs is the same (boost compared to conventional), so the penalty of this design is 2/3 silicon usage (crudely, you pay for 900 V output, and get only 600 V output). [ Edit: it's worse; see next post. ]

So here are the three negatives of this design:
1) 2/3 silicon usage, assuming a 2:1 voltage boost. With higher boost ratios, the usage would get worse. [ Edit: it's actually 1/3; see next post. ]
2) You need the three large inductors. You can't use the inductance of the motors as "free" inductors.
3) The PWM ratios are not linear to the output voltage, and you have to ensure that the PWM ratio never goes below a certain limit (considering the PWM ratio from the lower transistor's point of view).

The main competitor would be a conventional topology with a boost stage. The negatives of this are:
1) You need a large inductor, but this time only one
2) You need two extra IGBTs (one if you don't want regen)
3) You get (I think) poor utilisation of the booster's two IGBTs; they see output voltage and also input current. That's possibly one reason that the Prius booster is limited to ~20 kW, while the outputs are 50 or 60 kW.

I guess which one is better would depend on the relative cost of the inductors verses the IGBTs, and how much space and weight is allowed / desirable.

[ Edit: added third disadvantage of the separate DC boost topology. ]
[ Edit2: added notes to see next post re utilisation correction. ]
[ Edit3: Corrected PWM ratios, and made them all refer to the lower IGBT. ]

[coulomb]

But in a conventional design, we could have had 900 V swings (from a 900 VDC bus, ignoring IGBT drops). As far as I can see, the current through the IGBTs is the same (boost compared to conventional), so the penalty of this design is 2/3 silicon usage (crudely, you pay for 900 V output, and get only 600 V output).

Oops!

While it superficially looks like the IGBTs carry only output current (which of course is 1/2 the input current for a 2:1 boost ratio), they actually have to carry the input current, and spit the inductor's energy that results from the input current out over a smaller period of time.

What this means is that the silicon utilisation is actually 1/3, not 2/3. So I guess that's the killer for this topology.

[coulomb]

For those who aren't sick of this, I'll post how I think Weber's design would work, and why the capacitors aren't optional as I first thought.



First consider only the U phase, during the first part of the cycle when the lower IGBT is on, and the upper one is off. I'll consider a PWM ratio of 67% (2/3), with 90 A @ 300 V being converted to 900 V at 30 A (voltages relative to DC-). The input inductors will be sized such that with 300 V across the inductor, the current ramp will be 30 A per PWM period.

Current (red arrow) flows from the pack through the inductor and lower transistor. The transistor is on for 2/3 of the PWM period, so the current will ramp up 20 A.

Now consider the second half of the PWM cycle, with the lower transistor off and the upper transistor on. The current through the inductor now flows through the upper transistor to the capacitor and the load. I've drawn the V phase in this state; the current is in blue. In the steady state, the output will already be at +900 V (relative to DC-). That means that the voltage across the inductor is now 600 V, and is of the opposite polarity to what it was in the first part of the PWM cycle. So now the current will ramp down twice as fast as it ramped up (so it will ramp down at 60 A per PWM period), but it will do so for only 1/3 of a PWM period, so that's 20 A. So the current will end up where it started, and the cycle repeats.

Of course, the very first cycle, the output voltage will be zero, but that means that the inductor still has 300 V across it of the same polarity as when the current was ramping up, so it continues to ramp up at about 30 A per PWM cycle. This will charge the output capacitor, which will reduce the voltage across the inductor in the second part of the next PWM cycle, until eventually the capacitor reaches over 300 V, at which point the current will start ramping down in the second half of the PWM cycle. It will eventually reach equilibrium with the output voltage at 900 V.

I'll temporarily consider the capacitor to be part of the load now. With equilibrium reached, and when the output current is 30 A (achieved by suitable adjustment of the load and output voltage, let's say), this is delivered as a "pulse" of 90 A for the last third of the PWM cycle, and no current (upper transistor is off) for the first two thirds of the cycle. So the average current during the last third of the cycle is 90 A, so it is actually ramping down from 100 A down to 80 A. During this time, the pack is supplying this current; it is in series with the inductor. During the first two thirds of the PWM cycle, the inductor current must therefore be ramping up from 80 A to 100 A.

In both parts of the PWM cycle, the average current from the 300 V pack is 90 A. Power and therefore energy are conserved, as of course they must. So we can see that both IGBTs are handling 90 A average (the output current). When the lower IGBT is off, it sees the output voltage (900 V). When the upper IGBT is off, it sees the output voltage (900 V) as well. So both IGBTs see full output voltage, and full input current.

So that's why the capacitors are needed; we can't have pulses of 3x current, then no current, into the load. So now we consider the capacitors to be part of the converter, not the load, and they supply the missing current when the upper IGBT is off, having absorbed the excess current in the second part of the PWM cycle. This is shown in green in the diagram (for the first part of the PWM cycle).

As mentioned earlier, in fact the output swing is only from 300 V to 900 V, further reducing the IGBT utilisation, compared to a traditional buck-only (but with regen) topology.

[ Edit: added paragraph about why the capacitors are part of the converter, not the load. ]

[thingstodo]

Another alternative:

- 12V DC to AC modified sine wave inverters are available off-the-shelf in multi-KW ratings
- they contain a DC/DC converter, with a decent sized heat sink and fan
- if they are powered by separate batteries, you can add the voltages in series

It may be faster (less expensive?) than a custom design.

I have tested two 1375W continuous, 2750W peak, DC to AC inverters, with the DC sections chained together. About 150VDC each feeds a 208 VAC test VFD. The test batteries could not supply enough current for a rated current test. But the test proved (at least to me) that the DC outputs can be added in series.

[weber]

- 12V DC to AC modified sine wave inverters are available off-the-shelf in multi-KW ratings
- they contain a DC/DC converter, with a decent sized heat sink and fan
- if they are powered by separate batteries, you can add the voltages in series.

That's clever. But I'm pretty sure these DC/DC converters would not be bidirectional and so they would not allow regenerative braking.

[thingstodo]

That's clever. But I'm pretty sure these DC/DC converters would not be bidirectional and so they would not allow regenerative braking.

Correct - no provision for regenerative braking. Using commercial-off-the-shelf (admittedly after taking it apart) is about as Keep It Simple as I can come up with.

Would it make sense to separate the Regen from the DC/DC step-up?

The DC/DC step-down need not have the same current rating. Would a 20 to 40 amp converter per isolated DC bus would be enough braking?

I forgot to mention in the last post - the Boost converters (including conversion to AC modified sine wave) came out to about 90% efficiency. I think it will get up to 92% or so if I had a DC load to test with.

[weber]

Would it make sense to separate the Regen from the DC/DC step-up?

Not sure what you mean here. There's no way to separate them at the VFD. There's just a single bidirectional DC bus. The step-up converter for drive and the step down converter for regen would have to have input paralleled with output and vice versa, so at least one of them had better be a transformer isolated type. And their ratios would have to be slightly different in the right direction so you don't get power going around in circles and getting wasted.

The duplication of circuitry compared to a single bidirectional converter is almost too awful to contemplate. But then so is developing any high-powered switching converter from scratch.

The DC/DC step-down need not have the same current rating. Would a 20 to 40 amp converter per isolated DC bus would be enough braking?

20 or 40 amps at what voltage? I think I'd want regen power to be at least 40% of forward, but 25% could still be useful.

You seem to be considering a rather extreme low battery voltage (2 x 12 V). The battery wiring currents would be impractically high when powering a car, if that is the case.

[coulomb]

You seem to be considering a rather extreme low battery voltage (2 x 12 V).

Err, I think his idea was use only the "back end" of an inexpensive high-power converter designed for outputting 240 VAC. So this would operate at around 350 VDC, to supply 240 VAC output. Even though these usually have "modified sine wave" output, if they could be switched fast enough, they should be able to synthesise a sine wave at variable frequency, for driving a 3-phase motor. Or maybe the idea was to use just two to drive a single phase motor. But single phase motors would not have sufficient power. Or maybe these inexpensive converters contain a full bridge, not a half bridge, so two converters would give you the 6 switches with two spares. Am I on the right track, thingstodo?

[ Edit: now that I reread your original post, I think you are using just the *front end* of these... Perhaps you could give a little more detail on your idea, please? ]

[thingstodo]

Would it make sense to separate the Regen from the DC/DC step-up?

Not sure what you mean here. There's no way to separate them at the VFD. There's just a single bidirectional DC bus. The step-up converter for drive and the step down converter for regen would have to have input paralleled with output and vice versa, so at least one of them had better be a transformer isolated type. And their ratios would have to be slightly different in the right direction so you don't get power going around in circles and getting wasted.

Sorry for the confusion. A separate set of Dc/Dc step-up boards to get the voltage usable and the current to match your motor.

The DC/DC step-down need not have the same current rating. Would a 20 to 40 amp converter per isolated DC bus would be enough braking?

20 or 40 amps at what voltage? I think I'd want regen power to be at least 40% of forward, but 25% could still be useful.

A smaller and lower current Dc/Dc step-down board to drain the regen energy from the DC bus and pump it back into the batteries. I was thinking 20 amps at the DC bus voltage. At ... 48V or 96V? I don't remember what the voltage was ... that would be less than 100A into the batteries. 25% regen could peak much higher for a short time - regen usually does not last that long.

Do you expect to enable the DC/DC step-down based on a minimum voltage rise above the nominal bus voltage?

[weber]

For those who aren't sick of this, I'll post how I think Weber's design would work, and why the capacitors aren't optional as I first thought.

Nice diagram, and correct description as far as it goes. However you haven't shown how it works in regen.

I note that one doesn't need to analyse all these cases at such a low level every time, to know that it will work. (Whether it will be practical is another question). You only need to consider the high level model of a PWM half-bridge with inductor and capacitor as a DC-variac -- a DC auto-transformer as in the following diagram.



By choosing the PWM duty cycle you are simply tapping off (at the outer inductor terminal), a certain percentage of the voltage across the half-bridge. Power will flow in any direction between any pairs of terminals to try to maintain that relationship, and by varying the duty cycle you can produce (or consume) low frequency AC on top of the DC, provided the resulting voltage never actually reverses polarity.

By using two of these half-bridges and putting the load (or source) between them you can cancel out the DC component and produce (or consume) pure AC (single-phase). By using 3 of them you can do 3-phase AC.

For this reversible topology I always give the duty cycle (percentage of ON time) for the _high_ side of the half-bridge no matter what direction power is flowing. Whereas coulomb has apparently adopted the convention that is used for the non-reversible topologies where you give the duty cycle (D) of the _active_ side (the one that is a transistor, not a diode). This causes confusion with the reversible topology as low device and high device take turns being the active device depending whether you are driving or regenning.

Always using the duty cycle of the _high_ device (H) eliminates some annoying "(1-D)"s from the design formulas, at least in the common cases where input and output voltages are positive.

[Edit: Changed last sentence to refer to "H" for high-side duty cycle]

[weber]

Coulomb, I believe that Thingstodo is using only the DC-DC stages from two off-the-shelf 12 Vdc to 110 Vac single-phase inverters. He connects the input of each to a separate 12 V battery and connects their 150 Vdc outputs in series to get 300 Vdc. He then feeds this 300 Vdc to the DC bus of an off-the-shelf 208 Vac 3 phase VFD which then drives a 3 phase AC motor.

Do you expect to enable the DC/DC step-down based on a minimum voltage rise above the nominal bus voltage?

Personally I wouldn't use separate drive and regen DC-DCs at all. But if I did, I would want them to both be permanently connected, i.e. not have to detect regen and switch between them.

In my previous message I was assuming they would be dumb DC-DC converters with a fixed step up or step down ratio, but I realise now that off-the-shelf types will almost certainly try to regulate their outputs so they would have to be switched based on VFD DC bus voltage, as you say.

I have no plans to do anything of this kind. It still seems easier and cheaper to just wire lots of small batteries in series to get the high voltage I want for my VFD and motor. But the insulation and isolation (ELV segmentation) safety issues are fairly onerous, so if someone designed a VFD that incorporated a 1:2 boost and its additional cost and space requirements were not too much greater than those of the avoided segmentation contactors and BMS units, then I'd be using it.

But it looks to me that a 1:2 boost VFD couldn't help but be around twice the cost of a standard VFD. And a 1:10 boost just looks completely out of the question, to me.

[coulomb]

Its 3 terminals can be treated as input, output, and common in any of the 6 possible permutations, whereupon...

Ok, I'll bite. These would be the six permutations:



Will somebody pleeeeease check my equations   

Here, D represents the PWM ratio (as a fraction of one) for the switch in the single direction case (i.e. no regen). This is so that the equations agree with the Wikipedia articles (for Buck, Boost, and Buck-boost converters). Where the other IGBT's PWM simplifies the maths, I've used U for the Upper transistor's PWM fraction, or B for Bottom transistor's PWM fraction.

I'm also not sure that the capacitor should stay were it is for the two buck-boost variants.

[ Edit: Added paragraph explaining D, U, and B. ]
[ Edit2: Fixed Alt Buck-boost variant ]
[ Edit3: Fixed equation; renamed variants with Pos and Neg ]
[ Edit: This is still not right, at least the sign is wrong in the last variant. I'm working on a new set, with great help from Weber, but I just need a bit more time. ]

[PStechPaul]

This is an interesting thread. I had a similar idea, and I built a converter that takes 12V or 24V and steps it up to 320VDC with which I'm powering a 2HP VFD and a 2HP 3phase motor on a small tractor. I've been discussing my project on: mytractorforum

and also on:
diyelectriccar

You can see my most recent demonstration of it at:
http://www.youtube.com/watch?v=j5TyhdY-cHQ

The main difference in my approach is using a step-up transformer rather than an inductive boost topology. I'll supply more details if interested, but first I have to see if the forum will let me post, since I just joined.