Hi, I have a 15:1 gear ratio in my permanent magnet DC motor/gear assembly. When i turn the motor manually i can generate up to 40V of regenerative voltage.
My H-bridge circuit is one of those high voltage 3-phase international recifier modules, where i'm only using 2 of the phases to make an h-bridge. I have bootstrap capacitors, which limits my duty cycle to around 95%; anything higher may cause the bootstraps to not properly charge. All is fine here.
My problem happens though when i'm generating some voltage by manually moving the motor in the SAME direction where my H-bridge wants to move the motor. I suspect that this voltage is causing my bootstraps capacitors to not charge, causing me to go to "undervoltage lockout". That's what tech support told me.
How do I get around this? I want to be able to start the motor moving even if someone is already pushing the thing manually in that direction.
Is your H-bridge on (but at zero PWM) when someone starts pushing the motor? Does it act as a brake to the manual pushing? When you turn on the bridge, spending some time with both switches in the LO state should get your caps charged. One thought: you could add a pair of floating gate supplies (9V batteries?) to test the theory, or to get the caps charged to begin with, or to keep them charged during long off periods...
Is your H-bridge on (but at zero PWM) when someone starts pushing the motor?
Yup, i have the H-bridge "armed" in the direction i want to move it. That is, the bottom switch (IGBT) is constantly on, allowing the bootstrap to charge up, while the top switch is at 0 duty cycle (thus off).
Does it act as a brake to the manual pushing?
In the direction that I armed the h-bridge in, it does not act as a brake. As a feature of the product, it is supposed to be easy to move. When i sense this manual movement (via encoder changes), i want to start PWM'ing. Or of course, when someone presses an actuator button, i'll start moving. If i set the h-bridge to be armed in the opposite direction (with the opposite bottom switch turned on), this will act as a brake, but then this is contrary to the intended purpose of wanting an "easy push" feature of the motor.
When you turn on the bridge, spending some time with both switches in the LO state should get your caps charged.
When you say "LO", you mean both of them turned on? I thought that spending some time with the bottom switch turned on while the top is turned off is what charged the bootstrap. Thus too long of an on duty cycle ( >98% e.g.) would starve the bootstrap from charging up in time for the next PWM cycle.
Okay, so I have the cap charged up while keeping the bottom switch low. But like I said, any regenerative voltage manually generated will screw up the role of the bootstrap. I think it takes something like 10V above the motor rail voltage to properly charge the bootstrap cap to have enough power to turn on the top switch. Since the bootstrap is charged to 15V (my power supply to the drive), any manually generated voltage above say, 5V (15 - 5 10V source in series with the motor rail voltage.
This makes little sense to me. If both lower MOSFETs are turned on, which means both H-bridge outputs are LO (that's what I was referring to by LO), then the caps have to be charged. And you should experience motor braking. Then when you detect a little externally-forced motion you can release the brakes, and start PWM, which should also maintain the cap charge. Hmm, as an aside, how would you hold on a slope?
It's a permanent magnet DC motor rated at something like 120VDC. Supply voltage comes from a tap off the primary in my autotransformer, which is 100VAC rectified, thus it's at 100 x 1.414 = 141VDC
I've got a PIC microcontroller optocoupled from all four IGBTs on the drive module. The application is a pedestrian sliding door like you see in a shopping center, hotel, airport, etc.
Okay, so when i press a button to "start" it, the motor is initially stationary, my PWM duty cycle starts off slow, accelerates at a rate so that the operation doesn't look "jerky". This part works okay.
When I am manually pushing it while the motor is still off, then i press the 'start' button, i see my PWMs running on my oscilloscope coming out of the microcontroller, but the motor drive module still isn't kicking in. Once I release my hand, a split second later, the motor will jump as if the drive module finally kicked in. I was told by international rectifier's tech support that i am being inhibited by an undervoltage lockout of the module.
Since the bootstrap/charge pump consists of a cap (and a diode in the module), if the manually generated voltage is negative, then the diode won't conduct.
What? The undervoltage lockout must refer to your H-bridge's chip supply voltage, is that sagging for some reason? What's the part number of your IR module?
The bootstrap should charge independently, as the motor node is externally forced, if the controler's supply is live. The bootstrap circuit is simply a diode and cap reacting to peak output power node voltages. These externally applied voltages cannot exceed the attached driver's rails without being clamped to those rails.
If the externally applied voltage is out of sync with the controller's drive, you might be triggering protective features in the controller, or confusing it's logic to a point where it locks out.
Just a small point, noted you are using an autotransformer. Do you have an isolating transformer in front of that? If not be VERY VERY VERY CAREFUL. If do, just be VERY CAREFUL
which is just a snapshot of the IRAMS10UP60B data manual.
By the way, i only use two of the phases and make it into an H-bridge, while what is shown is a 3phase circuit.
In the circuit, the motor's return is also the 15V ground. When the motor is manually turned, then i have a hunch it causes the diodes connected to the bootstrap capacitors (CB1, CB2 in the diagram) to get reverse-biased, and not conduct. Thus the bootstrap capacitors never get enough juice to fire the top switch.
So i'm PWM'ing from my micro, but the drive's output is not responding until this generated voltage subsides. It's gotta be that the bootstrap capacitors are not charging up because the diode connected to it gets reverse-biased.
Perhaps it's not an undervoltage lockout after all. UVLO is said to occur at 11.1 volts for this drive module. I don't think this is the case.
So not being too device-specific, has anyone in general ever had a problem in starting up a DC motor that is already moving from some external force?
I've examined the datasheets and I still don't buy your argument.
formatting link
BTW, the interior IC is their IR21363, similar to the ir2131, see
formatting link
You've got to show how the diodes get reverse biased, which implies the lower IGBTs somehow fail despite their gates being driven on. These IGBTs are capable of sinking or clamping more than 15A with no more than 3.5V Vce drop (e.g. see irg4bc20k datasheet, fig 1,
formatting link
).
Furthermore, you should be able to observe the actual situation with your scope. For example, you can monitor the bridge current across module pins 12 and 22. The module has an internal 33-milliohm sense resistor, which means iTrip = 0.5/0.033 = 15A. You can monitor the sense voltage with a pair of probes in differential fashion.
The bootstraps are definitely getting discharged when I push the motor manually. I measured the voltage across the bootstrap capacitor with a scope, and found that it gets fully discharged.
One kludge would be to put a diode and relay in parallel with each other in series with the motor. If the relay is open, then the diode will block the discharge path of the bootstrap capacitor. However, the motor will now not move in the opposite direction unless the relay is closed. So if anything, this is just a kludge, and in fact, the mechanical relay will eventually fail.
That said, the method of turning on the bottom FET/IGBT to charge the bootstrap capacitor in an h-bridge circuit is fundamentally flawed.
The advised method would be to alternately clock each diagonal pair in the h-bridge at all times. So i guess at 50/50 duty cycle, you then have two opposing equal forces.
BTW, we assume that you've got a so called "freewheeling" diode across each of your IGBTs, as shown in the irams10up60b datasheet, right?
Not to be too stern, but unless you explain some aspect we are missing, or make the relevant measurements, we have to discount your conclusion, and your dramatic "fundamentally-flawed" assertion. If instead you want to assert there's something fundamentally flawed within the irams10up60b module, that may be.
You're saying the bottom IGBT is turned on, the driver-chip's Vdd power supply is present, yet under this circumstance the associated high-side driver cap can't become charged or if charged becomes discharged? As I pointed out** - for the IGBTs in your machine, it shouldn't be possible for the motor, acting as a generator, to overwhelm the turned-on IGBT.
Please tell us what happens when your push the motor: (1) Does the driver IC's Vdd / Vcc power supply remain AOK? (2) Is the bottom IGBT's gate voltage high, and the IGBT turned on? (3) Does the IGBT's collector voltage stay near ground (+/- 1 volt)? (4) Yet somehow the high-side driver capacitor tied to this same IGBT's collector/drain becomes discharged? Nah, can't be.
Perhaps you can also tell is what's going on with the other IGBTs in your bridge. We're keeping in mind it's your uP program that decides which of the IGBT gates in the module to activate.
** copy of the post, the details of which you didn't address:
Author: W>
I've examined the datasheets and I still don't buy your argument.
formatting link
BTW, the interior IC is their IR21363, similar to the ir2131, see
formatting link
You've got to show how the diodes get reverse biased, which implies the lower IGBTs somehow fail despite their gates being driven on. These IGBTs are capable of sinking or clamping more than 15A with no more than 3.5V Vce drop (e.g. see irg4bc20k datasheet, fig 2,
formatting link
).
Furthermore, you should be able to observe the actual situation with your scope. For example, you can monitor the bridge current across module pins 12 and 22. The module has an internal 33-milliohm sense resistor, which means iTrip = 0.5/0.033 = 15A. You can monitor the sense voltage with a pair of probes in differential fashion.
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.