Arlo's power stage Leaf controller runs and drives page 103
- Thread starter Arlo1
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Lebowski
10 MW
ok but you just calculated yourself 6 FET´s take 0.11 A to drive, how do you get to 2A or even 10A from there ?Arlo1 said:
I agree with Lebowski. The current consumption may be caused by shoot through in the extra FET's of the gate driver. In a previous post I suggested using bipolar transistors because then there won't be any problems with shoot through. Put a 1ohm resistor between the PMOS and NMOS and measure it with a scope, to see if there is shoot through.
I'm not talking about shoot through in the IRFP4468, I'm talking about shoot through in the FET's you added to the gate driver, to make them drive the IRFB4468 gates harder. If there is shoot through in those FET's, they will drain your 3s battery.
Is your little scope capable of measuring such short events?
Is your little scope capable of measuring such short events?
Bearing.
The only reason I have not sorted this is because we just did 2500km in my srt4 to visit a good friend/pro motocross-er to see his track and hang out and race some bikes/minibikes/indoor electric go karts as kind of a holiday! Once I'm caught up here at home again i will fix it all up and run some more tests.
My scope is 2.5Mhz so not great but I'm looking at a 100Mhz from Rigol or Owon... Just got to watch the penny's because my slow season is just starting.
There is a reason I listen to Dave!bigmoose said:
The only reason I have not sorted this is because we just did 2500km in my srt4 to visit a good friend/pro motocross-er to see his track and hang out and race some bikes/minibikes/indoor electric go karts as kind of a holiday! Once I'm caught up here at home again i will fix it all up and run some more tests.
My scope is 2.5Mhz so not great but I'm looking at a 100Mhz from Rigol or Owon... Just got to watch the penny's because my slow season is just starting.
Sounds like great fun!
To follow an application note, rather than taking advice from some random dude on the Internet, is probably good advice, generally. But now, when you have unsolved problems, it might be time to question things like an application note. Make some measurements of the FETs of the gate driver.
I have read a few application notes, and judging by the quality of some of them, I think they are sometimes written by young engineers, as the first assignment on their new job, when they are fresh from school and doesn't understand that a circuits behavior isn't as simple as the schematic.
To follow an application note, rather than taking advice from some random dude on the Internet, is probably good advice, generally. But now, when you have unsolved problems, it might be time to question things like an application note. Make some measurements of the FETs of the gate driver.
I have read a few application notes, and judging by the quality of some of them, I think they are sometimes written by young engineers, as the first assignment on their new job, when they are fresh from school and doesn't understand that a circuits behavior isn't as simple as the schematic.
Lebowski
10 MW
So very true.... especially now that you have unexplainable issues, measure everything, don't guess.bearing said:
And I, sir, do not like to be called `some random dude on the internet'
Ok bearing i read though what i could and its almost impossible to have a shoot through in the buffer fets the only way is if the first two in the buffer stage are off at the same time this meens 12v to the hi side gate and 0v to the low side gate at the same time as well r100 will help make sure the on fet is shut off first then the other will get turned on in the second stage. I will test them but i am sure my three driver boards with all the consumptions are using to much power for the ribbon cable. Not to mention nobody has steped in and said "this is how you properly calculte gate charge amperage"!!!!
No, it's the other way around. It's hard to not get shoot through with this circuit. Only if the gate thresholds of both P- and N-MOS are higher than VDD/2, you are free from shoot through. The 100 ohm resistor will of course help a lot, but there is probably always a short shoot through event. Since the treshold level of a FET gets lower as the FET heats up, shoot through gets worse and worse as the circuit heats up.
I can make a try. The unit of Ampere is Coulomb per second. In the datasheet of the FET you can find how many Coulomb it takes to charge the gate. Lets use Total Gate Charge (Max.) Qg = 227 nC. Since you are using f = 40kHz, total charge moved to the gate in one second is Q = Qg * f = 227nC * 40kHz = 9mC. This means that the average current is I = Q / t = 9mC / 1s = 9mA per FET + losses. Since you have 6 FETs, the total average current is 9mA*6 = 54mA + losses. So that's what drains the 3s battery.
The heat in your wires comes from the RMS current. In this case, the RMS current is a lot higher than the average, since the current is pulsed into the gates with high peaks. Lets say the gate current is a 1A rectangular pulse. This means it has a duty cycle of 0,009. Irms = 1A * SQRT(0,009) = ~0,1A to charge the gate. Since the current that discharges the gate passes through the same wire, then Irms_tot = 0,2A. If the ground wire is shared, it has to pass the RMS current from the 3 lower FETs gates + the RMS current which charges the bootstrap caps. This means the RMS current in a shared ground wire is probably in the order of 1A RMS.
If there is shoot through in the buffers, or some other unwanted losses, then both average and RMS gets higher, of course.
Ok so for a few nano seconds there is shoot through in q1 - q5 and also q5 - q7 but big BUT the r100 or 470 ohm resistor in my case (which i have to change back to 100 ohms) limits the shoot through to ultra low current! This makes sure its almost impossible to have both of the second fets on at the same time meening its almost impossible to have shootthrough on the fets without a resistor inbetween them!
Yes but what im getting at is there is no way q6/q8 or q2/q4 can have shoot through!!! Because it always turns one fet off before turning the other one. As well in your rms calcs a couple posts ago you forgot to add the draw of 3 2113 drivers 3 current sensors and 24 buffer fets and all the 10k pull down resistors!!! Not to mention 1amp on the ribbon ground is overload anyway!!bearing said:
Ok another thing the fets are ixfk230n20t they have a gate charge of 378nc so re-work the math and 378 with 50 khz = 18.9mC so 1.89mA then x 6 =113.4mA then look at the RMS for that ~2 x what you estimated then add 3 ir2113 fet drivers and 3 alegro current sensors and all the buffer fets and all the pull down resistrors and continious amps are likely to be in the 2-3 amp range and RMS.... with a wire rated for .226 amps !bearing said:
Arlo1 said:
That depends on the capacitance of the gate. Let's say the capacitance is 0, then R1 won't help at all, there will always be shoot through. With capacitance on the gate, then Cg and R1 will delay the turn on, which hopefully happens after turn off, of the FET on the other side. The gate capacitance of these IRF110/-9110 is very small, so I'm not convinced that the R1*Cg delay is enough. That's why I suggest measuring them.
Arlo1 said:
I left them out because you asked how to calculate gate charge current, so I wrote as simple as possible on that subject. My gut says the IC's and pull downs are negligible. Buffer FET's hasn't been measured yet. I have never questioned that the ribbon is overloaded. If it melts, it's clearly overloaded.
Remember that the RMS current isn't draining the 3s battery, the average current is. An average current of 2A is just insane. If it's really that high, something must be wrong, IMHO. Could be anything, but I'm sure the problem isn't solved by replacing the ribbon cable with a bigger one.
Ok so I finaly got to start pulling more of it apart to fix it all. The negative trace on all three boards is burnt and one of them looks like a section exploded off the board which I have a feeling was the first pop in the video. I am going back and fourth through all the data sheets to find what was out of its limits. I was at 4500-5000 rpm and 50khz pwm then runs prior were up to 40khz and ~4000-4500 with out a bit of heat from the fet drivers so before I blame them I need proff.
THERE IS A BIG CHANCE i have stray inductance and floating grounds causing weird shit. I know better then to have things as far apart but my driver layout was designed in a hurry without the fets in my hands.
But having to thin of wire for the negative was not the problem although I am changing my design to have the 12 volt come in to the driver boards on a seprate set of wires as well I am going to put the reg on one of the driver boards because the brain doesnt need 12v for anything.
I will also put the 3w 10k resistors on the driver boards to reduce the voltage at the driver so then header pins are not having two pins <.1 inches apart with a difference of up to 340v on them!
I am working on some new designs to get the board closer to the fet legs ATM as well.
THERE IS A BIG CHANCE i have stray inductance and floating grounds causing weird shit. I know better then to have things as far apart but my driver layout was designed in a hurry without the fets in my hands.
But having to thin of wire for the negative was not the problem although I am changing my design to have the 12 volt come in to the driver boards on a seprate set of wires as well I am going to put the reg on one of the driver boards because the brain doesnt need 12v for anything.
I will also put the 3w 10k resistors on the driver boards to reduce the voltage at the driver so then header pins are not having two pins <.1 inches apart with a difference of up to 340v on them!
I am working on some new designs to get the board closer to the fet legs ATM as well.
I see you write about having wires with a voltage difference of twice the battery voltage a lot, so I think it's time to bust that "myth". The biggest potential difference you can have between two points with a 170V battery, is 170V, unless you are transforming the battery voltage in some way.
No because if one phase is -170 and the other is 170+ then you have 340 peak to peak. I understand it will be off from that because when on phase is at a peak the next on will not so maybe 170v on one while the other 2 are ~ -80v. This is still added together for a total voltage difference and potential of an arc.bearing said:
To get -170V, you need another battery of 170V connected with it's positive wire to ground.
The PWM:ing in a normal system with one battery of 170V is between 0 and 170V, so the max the difference is 170V-0V. The motor sees 170V peak-peak, or +-85V from the neutral point. With pure sine wave, max phase-phase would be 85V + 85V*SQRT(3)/2.
Anyway, it doesn't matter that much. I think it's good that you are moving drivers closer to the FET's. To minimize inductance in wires between FET's and gate drivers, you can twist the wires.
The PWM:ing in a normal system with one battery of 170V is between 0 and 170V, so the max the difference is 170V-0V. The motor sees 170V peak-peak, or +-85V from the neutral point. With pure sine wave, max phase-phase would be 85V + 85V*SQRT(3)/2.
Anyway, it doesn't matter that much. I think it's good that you are moving drivers closer to the FET's. To minimize inductance in wires between FET's and gate drivers, you can twist the wires.
Ok im not sure you understand what im saying??? If the battery is 170v then you apply 170v to the phase then when the next magnet comes by the same phase its the oposite polarity so you will apply 170v negative to that phase which gives you 340v peak to peak. I proved this with my oscilloscope in my videos. The center of your battery is not newtral newtral is negative from your battery. Now what im worried about is having the voltage from the phase wires running all the way to the brain board and this gives the difference of any two phase wires on the headder pins and inside ribbon wire it self. If you are at max rpm with 100% phase amplification you have the potential of max of one phase to about 2/3 of max in reverse (negative 2/3 max) on the other two phase wires! I will put this in a 3 phase graph if needed later. I know you are thinking it can only power one phase positive or negative at a time but the motor is spining and the back emf is always there.bearing said:
Thank you, toolman, I appreciated that.
Arlo, I don't know what else to do than showing this picture.
The FET's contains diodes, so the electric potential of a phase wire can never get higher than Vbatt + Vf, and it can never go lower than 0-Vf. So, the maximum difference between two points in the circuit is Vbatt + 2*Vf. The EMF grows from the neutral point and outwards, symmetrically, towards the rails, as the RPM increases. When EMF reaches the rail, the motor can't spin any faster. You can "fool" it by advancing the timing, and use the inductance of the motor as part of a "3-phase boost converter". The boosted voltage stays within the motor, and can not be measured from the phase wires, unless you disconnect them from the controller while the motor is spinning.
It seems like you are getting upset by my posts. It was never my intention from the beginning to upset anyone, but I think that, as the last days has tested my patience, I made these recent posts about voltages somewhat out of pride or prestige, in addition to the good intention of sharing knowledge. I don't like where this went, and I think I'm only hindering progress at this point. You lack some knowledge and/or education, but I think you are a really smart guy, and can read up on what you need. Wish you the best, and good luck with this.
Arlo, I don't know what else to do than showing this picture.
The FET's contains diodes, so the electric potential of a phase wire can never get higher than Vbatt + Vf, and it can never go lower than 0-Vf. So, the maximum difference between two points in the circuit is Vbatt + 2*Vf. The EMF grows from the neutral point and outwards, symmetrically, towards the rails, as the RPM increases. When EMF reaches the rail, the motor can't spin any faster. You can "fool" it by advancing the timing, and use the inductance of the motor as part of a "3-phase boost converter". The boosted voltage stays within the motor, and can not be measured from the phase wires, unless you disconnect them from the controller while the motor is spinning.
It seems like you are getting upset by my posts. It was never my intention from the beginning to upset anyone, but I think that, as the last days has tested my patience, I made these recent posts about voltages somewhat out of pride or prestige, in addition to the good intention of sharing knowledge. I don't like where this went, and I think I'm only hindering progress at this point. You lack some knowledge and/or education, but I think you are a really smart guy, and can read up on what you need. Wish you the best, and good luck with this.
Bearing i want all of the people who read my threads to know what is true and what is false. I am not trained in this feild but i am learning from the best. If you are offended that im trying to point something out then im sorry but this is a public forum which needs acurate info.
Now having said that i can visualize what is going on very well. Here is what im trying to say.
Look only at 2 phase wires phase wire A and phase wire B
Step 1 H bridge A has the positive fet on and H bridge B has the negative fet on this produces the positve side of the sine wave and if the motor rpm is maxed out unloaded the top of the sine wave will be = battery voltage 170v in my case then step 2 we reverse and H bridge A turns on the negative fet and H bridge B turns on te positive fet this will be the negative side of the sine wave and again unloaded at max rpm it will be = to the battery voltage but in reverse so in my case 40s lipo -170v giving you a peak to peak of 340v
Now having said that i can visualize what is going on very well. Here is what im trying to say.
Look only at 2 phase wires phase wire A and phase wire B
Step 1 H bridge A has the positive fet on and H bridge B has the negative fet on this produces the positve side of the sine wave and if the motor rpm is maxed out unloaded the top of the sine wave will be = battery voltage 170v in my case then step 2 we reverse and H bridge A turns on the negative fet and H bridge B turns on te positive fet this will be the negative side of the sine wave and again unloaded at max rpm it will be = to the battery voltage but in reverse so in my case 40s lipo -170v giving you a peak to peak of 340v
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