dakrat

everybody and their momma knows that...

STEALTH

iTrader: (21)

Charlie Suangka of Team Novak - "On some tracks and some conditions the 5800 is faster then others. Just kind of depends on what you are using it for. Most High Traction Onroad Users tell us its like a 15-19 turn.

In and Off-Road Car they say 11-12 turn. This is with the SS ESC that is currently available.

The Timing Chaning Issues, bring up many other problems for usage. Just like it does with regular motors. People have enough to mess with just in gearing and the mechanical timing in the motor. Variable Timing and "auto timing" are all on the horizon for brushless. It'll be like Electric V-Tec one day. Pick a profile for when you want the kick. At this time, what can be done is fairly limited with the technology available. When the technology gets to the point that we can offer these features in an Easy to use streamlined type of package, we'll be all over it. But right now, we're just not there yet.

We have discussed offering some software "update-ability" but usally there is no way for us to add it to the ESC with out it getting bigger, or the ESC needing to be Taken nearly completely apart. This is not something we want customers doing themselves just yet.

As for the phaze ratings, Don't Quote me on this, but basically, a Brushless ESC is like 3 regular ESCs. Each Phaze has a coil to fire. So each phaze has an on resistance rating. When the Novak Brushless ESC goes to Brushed mode, all 3 phazes work together, essentially turning 3 ESCs into 1. So the onresistance is lowered further. This is why all 3 of the wires get soldered to the Negative of the Motor. It just uses them all at the same time like it's one big ESC.

On resistance is the calculation of the ratings of the fets. The more fets in parallel, the more division is done, so the rating is lower and lower. I'll go double check, but I'm pretty sure that's it.

The Velociti Motors are Slightly different then the SS motors, but basically the same for the most part.

Thanks
Charlie"

TEAM-AE-4-EVER

It's too bad neither Novak nor LRP has figured out a way to either get rid of the need for a fan or in some way incorporated it directly into the ESC?!? Oh well... at least we got two giants in the electronics industry of RC going head-to-head yet again on another very useful and fast growing product! The competition will drive down prices and force both to push their pencils in creating a better product for all of us!

THE_REAL_FROBOY

tekin also ahs a brushless in the works and so does mamba now. so that is 4 newer brushless systems

Somerwil

iTrader: (1)

I received my Sphere yesterday. It looks fantistic, but the engine is still to come... *(^*&%^$#*%$&. ^

What's so difficult guys at LRP? I just wanna race my new equipment asap!

Gr,
Mike

Team Duratrax

Don't forget that Team Orion are planning for a BL system to.

Now with all the top manufacturers pumping out BL's it looks like Trinity will have to as well I laugh because they have been the ones dissing BL's.

Sean

shouldbe

Let's explain what turns(5.5T or 6.5T) here mean. For a BL motor, there are three coils, just like the their brushed brothers, although the coils are just inside the motor shell but not in the rotor. Each coil has a turn. This is the real "rounds" the coil winds inside the motor. And each coil has two ends. Thus two ends times three coilds equal 6 ends. However, you can only see three wires outside the BL motor. Why? Because they connect each other inside the motor.

So what? Is there any difference to connect each coil together? Definitely yes. Basically, there are two ways to connect the coils together inside a BL motor: Delta and Y(or star). If you connect each coil end one by one, you just create a triangle. Bind a wire to each vertex of the triangle, you can introduce three wires out of the BL motor. That is called Delta winding. If you connect one end of each coil together and left it untouch inside the BL motor, and then introduce the other ends of the three coil out of the BL motor with three wires, you just form another winding: Y winding.

The real "rounds" for the coils wind inside the motor is usually what we talk about the "turns" of a BL motor. However, simply talking about the "turns" is not enough to describe the performance of a BL motor. With the same turns, but different windings, the effect of the motor could vary greatly. Here below describe the difference:

1. Kv, a way to describe how fast a BL motor could run when adding certain voltage. With Delta winding, the Kv of the BL motor will be higher. Although with Y winding, the Kv will be much lower. There is coefficient between the two windings. Kv_Delta=1.7*Kv_Y
2. Torque per Ampere(or in another word Kt), a way to describe how much torque a BL motor output when draining certain ampere of current. On the contrary to Kv, Kt_Y=1.7*Kv_Delta.

i.e. with Y winding, the torque per ampere is much stronger than Delta winding although the rpm in the same voltage is much lower.

3. equivalent turn compare to brushed motor. Usually, since the magnet used in BL is Neodymium(NeFeB), it is much stronger than normal magnet used in brushed motors, there is no way to compare the turns of BL motor with brushed motor. i.e. there is no direct comparison. However, there could be a relative way to compare them. i.e. if you have a BL motor winding 10 turns act like a 17 turns brushed motor, you can say a 5.5 turns BL motor will act like a 9.35 turns brushed motor!!!

Ok, let come to the point!

1. Without mentioning the winding method, talking about the turn of BL is nonsense. The above mentioned Novak 5.5T could be either Delta or Y winding. But the performance could be different as 1.7 times! So there is no need to be suprised with a BL motor's turn as low as less than 5T! If you check Aveox's webpage, you will find a 2T or even 1.5T BL motor!
2. For NOVAK SS5800(10T) which act as a 15-19T brushed motor, i.e. about 1.5-1.9 times of the nominated turns(10T). Then with the similar magnet, a 5.5T should be 8.25-10.45T!

ttso

iTrader: (1)

Novak and LRP/Reedy new brushless motor are both Y-winding, why? Because IFMAR rule prohibit Delta-winding. It's in rule 4.3.2. As long as it's IFMAR legal brushless motor, it's Y-winding. Why they set the rule like that? "It wasn't me" (tm)

shouldbe

I found you guys has a lot of questions on the explanasion/calculation/effect of internal resistance and thermal consideration of a BL esc. Let's get inside a BL esc and see what is the difference between a brushed competition(forward/brake) and brushed reversible(forward/brake/reverse) esc.

First let's talk about some common ideas.

1. there are many fets inside an esc
2. each fet act as a switch
3. many fets parellel together as a group and act as a one big switch which could pass decades or even hundreds of amperes!
4. the internal resistance is the combined resistance of a group of fets. According to Ohm law, if one fet's internal resistance is R, then the group n pieces of fets' resistance is R/n

Then let's talk about competition esc.

0. to make a competition esc, you need at least two(or two groups of) fets. One is for "brake", the other is "driven"
1. usually within one competition esc, there are 16 fets. Believe me, I've repaired more than 8 "competition" esc. e.g. LRP quantum I, VFS2000, etc. Some vendors add more fets in one esc, e.g. VFS2000 has 20 fets inside. That's why it is a little bit bigger than normal high-end competition esc.
2. within the 16(or 20, it doesn't matter) fets, 1/4 of them are for "brake" purpose. They are just parellel the both ends of the motor. Yes, inside the esc's motor+/battery+ and motor- terminal, there are 1/4 of the 16 fets! And other than "brake" functions, the build-in diode in each fet could protect the "driven" fets as well. I burned more than 5 fets when I first repair a competition type esc just because the "brake" esc is faulty or does not work. With a "Shottkey" diode usually accompanying the esc, which is put between two ends of motor taps, the "driven" esc as well as the "brake" esc could be protected greatly.
3. within the 16 fets, 3/4 of them are for "driven" purpose. Inside the esc, between motor- and battery-/GND terminal, there they are.
4. How many amperes does an esc sustain? It depends on the capability of the "driven" fets. Usually a fet used in RC esc is very high-end fet. e.g. the VFS2000's fet's internal resistance is only 4.8 mohm. It can sustain about 5A current continueously on whatever situation(very high environment temperature, high switching frequency, etc, anyway it is terrible for you if you are a fet ^_^). Then for 3/4 of 16 fets, i.e. 12 fets, they can sustain 12*5A=60A, with only less than 2 watts of power consumption themselves!
5. the internal resistance of the esc. You may calculate it when you read here. Yes 4.8 mohm/12=0.4 mohm! You see that figure quite familiar, right? Yes, usually a esc vendor will show you like this: 0.0004 ohm(or make you more surprised like this: 0.00037ohm, yeah? ^_^)

Ok, then let's talk about reversible esc.

0. To make a esc reversible, you need at least 4 fets. Imgine the character "H". The upleft, upright, downleft, downright "|" are the 4 fets, i.e. 4 switch. the "-" in the middle of "H" is the motor. That's why you may hear about "H bridge" when talking about controlling a reversible motor. Each corner of "H" is called "arm" of the bridge.
1. Suppose upleft and downright fet is "switch-on" and upright and downleft fet is "switch-off", then the current will go through the upleft fet, motor, downright fet. Then the motor is turning, let's say, clockwise.
2. On the contrary, if you turn the sequence vice versus, the motor will turn counter-clockwise.
3. if there is no fet is "switch-on", then the motor does not turn around.
4. if all the down fets are switch-on, and all the up fets are switch-off, then the motor brakes!
5. if all the fets or left side or right side of the fets are switch-on, then.....the esc burns!!!
6. ok, talk too much, let's talk about the internal resistance. If the internal resistance of each fet is R, what is the on-resistance of the esc? suppose there is only 4 fets in the esc. The on-resistance is R+R=2R.
7. you may not see the difference between the competition esc and reversible esc. Let's use the above 16 fets as an example. In competition esc, there are 12 fets as "driven" fets, then the internal resistance of the esc is: R/12.
8. However, for the a reversible esc, the 16 fets are divided into 4 groups, each group has 4 fets. And each group's resistance is R/4. And what is on-resistance of the esc? According to point 6, it is R/4+R/4=R/2
9. Now you see the difference: R/12 vs R/2, 6 times diffence of internal resistance, when the escs use the same fets, i.e. in similar volumn, weigh, cost, competition esc is 6 times smaller in internal resistance against reversible esc! i.e. reversible esc usually could support only higher turn motor, although competion esc could support as low as 7T, or even lower(e.g. Novak's GTB "does not has a turn limit")
10. Yes, that's why reversible esc usually much bigger than competion esc, because they have to use much more fets to support similar turn motor!

Ok, finally, let's talk about brushless esc.

0. Within a BL esc, there are "|-|-|-" structure insides the esc. See it is quite similar to the reversible esc. The only difference is that it has two more "arms" and need more fets as switches. Each "|" is called a phase. Be aware that each phase contains a upper arm and lower arm.
1. For one bl esc, there at least are 6 fets as switches. The three "-" are the three coils of motor.
2. similar to reversible esc, to drive the motor, there should be one upper arm and one lower arm to be switched-on. However, different from reversible esc, the switched-on upper arm and lower arm are changed from time to time. The changes of the switch-on status of each upper arm and lower arms are controlled by BL esc!
3. When the motor is running, there is always one upper arm and two lower arms, or two upper arms or one lower arm is switched-on. So the on-resistance of the esc could be: R+R/2=1.5R
4. Suppose we still have 16 fets. Then for each arm, there are 16/6(three phases, 6 arms)=2.667 fets. Then each arms' internal resistance is R/2.667. And the on-resistance of the esc is: R/2.667+R/(2.667*2)=4R!!!!
5. Now you see the diffence of internal resistance among competition, reversible and bl esc: R/12 vs R/2 vs 4R, i.e. 1:6:48!
6. i.e. bl esc is much more easy to be thermal! because they have much more resistance than competition(anyway, be aware, when they have the same size, same weigh, same fets! actually, they don't, so the difference is not that big!)

The final result will really make you astonishing, right? In fact, the above result is based on that, all the three types of esc has the same quantity of fets. Actually, they don't! Competition esc usually has 16 fets, although reversible esc could have more than 20 fets, and bl esc(car bl esc) usually has more than 36 fets! That's part of the reason why bl esc is so expensive, they do have more fets! And usually bl esc will intend to use more cutting-edge fets than competition. Because with normal advanced fets(e.g. the one used in VFS2000, 4.8 mohm, only cost less than 0.5 USD), competition esc could be very excellent! However, for bl esc, they must use the most advanced fets(e.g. a 3.6 mohm fet, it will cost around 1 USD). After all, all these factors make bl esc more expensive, less performance than competition esc(reversible esc has the similar problem, but less important than bl esc).

Ok, let's go deeper into fet's world.

Is there any more factors made bl esc less performed? Yes, there is. What is it? The throttle! You didn't believe that throttle will affect the performance of a bl esc, did you? I bet you don't. However, it really did.

I can't explain it too clearly. However, you can see that in this way. Let's explain with competition esc first. With same continueous current, the more the throttle is, the less peak current is. e.g. if the continueous current is I, and the throttle is 50%, then the peak current will be I/50%=2I! And the power consumption of the fet will be (2I)^2*R*50%=2I^2R. When the throttle is 100%, the power consumption is I^2R! The power consumption in 50% throttle is twice of the 100% throttle! If the throttel is n%, then the power consumption on that throttle is 1/n% of the full throttle!

That's the figure of competition esc. Fortunately, when the throttle is low, the contineous current is low as well. However, for bl esc, things are much worse than competition. Why? because when 100% throttel, each phase/arm of the bl esc only switch-on for 100%*1/3. i.e. when 100% throttle, for each phase/arm of bl esc, the throttle is 33%, then the power consumption could be 3 times of a competition esc! If the throttle is not 100%, things will be much worse!

Ok, after reading these characters, you must be frighten and dare not to use a bl esc on car. However, to save you confidence on bl esc, let me show you some good aspect of this issue:

1. For a car esc, the continueous current usually less than 35A. That will not much lower the power consumption of your bl esc.
2. A bl esc usually has much more fets than competition esc. usually for car bl esc, 36 fets is at least. Some bl esc like Schulze or Hacker, or LMT, has more than 50 fets(LMT Micro 1895 has 60 fets, HACKER Master Car Competition has 54 fets).
3. A bl esc usually uses much more advanced(and more expensive) fets, which have much lower internal resistance(I don't know if LRP has used or not, as I knew, the most advanced esc has as low as 2 mohm, IR(International Rectifier) product). If the bl esc vendor could use most advanced fets, bl esc could performance as cool and high performance as competition esc did!

AI2

shouldbe: interesting discussion there! thanks.

but I have to disagree regarding your calculations for BL controller's partial throttle FET operation. Firstly you must be talking of an average current, because that is different to a "continuous" current you mention when you are talking about switching FETs at partial throttle. Clearly a 50% duty cycle chopped current cannot produce a continuous current.

I can see in your calculations that 2I peak current at 50% duty cycle would produce double the power dissipation compared to just I amps at 100%. However the power delivery is different. The 2I Amps at 50% will also deliver more power than I Amps at 100%, just by the same calculations you used to calculate the power dissipation, so no wonder dissipation is also higher... how can you have more power dissipation/output for the same current? just doesn't make logic sense to me..

For the BL controller, suppose at 100% duty cycle if we are producing I continuous Amps of current, then clearly this would mean that each phase is producing only I amps for 1/3 of the time (not something like 3I amps as by your thinking above) because the sum of these each 33.3% duty cycles then gives you the total continous current of I amps running through the motor. if anything the fact that each group of the FETs per phase only operates 1/3 of the time helps to reduce the total power dissipation in the FETs, eg compared to if they were active for 100% of the time. Just common sense to me...

So guys, as far as BL tecnholcogy is concerned its a very good thing that at full power any of the FETs only switch-on for 1/3 of the time! Helps keep down the size of the controller!

cheers

shouldbe

Yes, you correct the mistake of mine: it is average current, but not continueous current. "Average" is what in my mind when I type "countinueous", hehe.

about the power consumption and power output, you are wrong. With the same average current50% and 100% duty cycle, the output power is the same, because the same average current I times the same voltage U equal the same output power. However, the power consumption of the fets are different. That is because of the "square" in the equation P=I^2*R. Due to the "^" when peak I is double, the power consumption is four times of origin, and after "average" the result by 50%, the final power consumption is twice as origin. See the inpact of "^"? You can make a calculation youself with different duty cycle. You will find my conclusion is correct.

In fact, according to the above analysis, the partial throttle could be a problem for a competition esc, especially when the esc support not low enough turn's motor and run on its limit. e.g. a 13T limit esc will not be happy working with 13T motor under partial throttle in bad environment(high temperature, etc.) for long time. Fortunately, most of the competition esc has developed themselves far beyond the limit. i.e. their power consumption is too small when fully loaded in the worst environment, the extra power consumption made by partial throttle could be omitted at all as well. Unfortunately, bl esc has not reach that far, the full throttle power consumption still count in some bad case. It could be worse in partial throttle as well.

For the assumption of bl esc when partial throttle, I can't say my conclusion is totally correct. It may be not that terrible. However, at least bl esc will not work better than competition esc on partial throttle. That is definitely correct. I agree with you that 1/3 of working time will decrease the power consumption of each phase, however, if you have a same number fet competition esc, you will see there is no difference for the 1/3 of working time, you just add more fets to share the power consumption. That is.

The power consumption in partial throttle for bl esc confused me for a long time. I just make some "draft" calculation above because I loss the old calculation I made before. Anyway, whatever the theoretical calculation or real application, it shows that bl esc usually much easier to be thermal. In fact, I've two car bl esc(LMT micro 1895 car and Hacker Master Car competition) in hand. the latter one has had burned due to bad environement. Bl esc for car has a long way to go.

Finally, if you go to IR's web site, you can find their most advanced fet DirectFET. One of those fets has as low as 2 mohm internal resistance. It usually is used on notebook computer, and may cost a lot. Let's say 1.5USD/fet. However, if NOVAK or LRP or Hacker or LMT could use those fets, bl esc could run as cool as competition esc!

shouldbe

Let's talk about the motor. What makes a motor better than the others? The turns, the efficiency affect the performanc of a motor most. For car application, a lower turn and higher efficiency motor is better than the others.

Why shall I say that? Let's talk about what we want a motor be and some basic theories of motors.

For car application, we'd like to have a fast and full of punch power motor. What decides the speed of the motor? The voltage and the turns. Since the number of batteries are fixed, then only the turns decide the speed. The lower turn, the faster speed. However, you can not ask for a too low turn motor because of: 1) lower turn motor is not efficient in low speed, you can not run it all in high speed, can you? 2)lower turn draws too much current, which will drain your batteries too soon.

And what decide of the punch? The electron-magnetic force does! What decide the electron-magnetic force? The ampere draw and structure of the motor does. Be aware that there is a limit of the electron-magnetic force generated by current going through the coils, beyond that limit, although the ampere draw increases very fast, the electron-magnetic force will increase very slow. i.e. the energy does not change from electronics form into mechnical form, but change to heat! Heat is lost of energy or in another word, efficiency. Low efficiency, low performance. How to generate electron-magnetic power as much as possible, lower the turns, although we come to the conflict discussed above, too lower turn is not efficient.

Ok, how to choose a powerful motor? Choose a lower turn but efficient enough motor.

A motor with less coil resistance is more efficient.
A motor with appropriate magnetic design which could generate very high electron-magnetic force lineary along with amper draw is more efficient.

Haha, I talk too much, many of which I am not for so sure. Hehe, just for discussion, or even for your bedtime story. Hehe.

AI2

Hey shouldbe,

good that some people are thinking about fascinating things like this on the forum... However i still think you're wrong about that and would like to explain again why this is the case so that we don't confuse the readers here unnecessarily.

I agree the square term in P=I^2.R doubles the power as we double the current and run at 50% duty cycle. clearly this means the power dissipation in the FET is doubled, but so is the power being delivered by the load (motor in this case). It's obvious isn't it? after all the same curren tis flowin ghtrough the load (motor) and the FETs. Let's just look at it using the exact same formula:

P = I^2.R, where now R is the resistance of the motor windings. If we have I amps at 100% duty cycle then the theoretical power delivered by the windings is: P= I^2.R.

OK now we double the motor current and switch at 50% duty cycle, then the power delivered by the motor windings is now:
P = (2.I)^2. R * 50% = 4. I^2 . R / 2 = 2.I^2.R

ie. the power delivered by the motor is doubled, the same way the total FET dissipation is doubled! So when you are talking about double heat dissipation in FETs, this is actully occurring when the power delivered by the motor is also double. I cannot explain it any better to you. Just think about it again logically, I'm sure you'll agree this is correct.

I agree with you that speedo efficiency is less during partial throttle due to switching losses, this is worse for brushless than brushed controllers, because the brushless FETs always switch even at 100% duty cycle.

cheers

burito

Wow... interesting discussion, but I think you guys are a little off-topic here!
Would certainly belong in a more "open thread" and not necessarily the thread about the LRP brushless system.

ProMaX

But hey they dont and they run hot and need big heatsink! They maximise profits by use cheap FET and/or less number of them and charge high price also!!