STMicro Drivers

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Re: STMicro Drivers

Post by bkgable » Wed May 09, 2012 12:05 am

Good info
In other news, I got a response from ST regarding heatsinks:
  • Max Temp 125C
    Ambient 30C
    Rdson 0.75 Ohm
    Iload 3 A
    Heatsink 7 deg C/W
I forgot that the power depends on the current squared.

3A x 0.75 O = 2.25V but that is at DC. High drive voltages allow the inductive current to ramp up faster but have to be turned off at 3A.

I wonder what the thermal resistance would be for a 1x2x1/8 " piece of copper soldered to the power tab? Maybe it could be conductively isolated from a board layer. Are you thinking two layer board?

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Re: STMicro Drivers

Post by MLange » Wed May 09, 2012 1:16 am

bkgable wrote:I wonder what the thermal resistance would be for a 1x2x1/8 " piece of copper soldered to the power tab? Maybe it could be conductively isolated from a board layer. Are you thinking two layer board?
Keep in mind that this is an HTSSOP28, not a PowerSO. Unfortunately, there isn't a PowerSO version of the L6470 as of yet :(

I'm looking at a two-layer board, as four-layer is rather cost-prohibitive (and internal layers can't sink much heat at all anyways). They recommend 12 thermal vias underneath the chip (I'd imagine they mean under the thermal pad). I'm attempting to create as much surface area on the module as possible for heat sinking via PCB to make the heatsink's job easier.

In regards to the second option, I've removed the limit switch resistor and filter cap (to be included on the motherboard instead) and the ADC voltage divider circuit (ditto) to reduce the part count.

Regarding the move of the limit switch resistor and filter from the module to a motherboard, not everyone is going to want or have limit switches, and they would need to be tied high. No sense adding extra parts that are never going to be used, and could easily be added to the motherboard if desired.
Regarding the move of the Power Compenation Voltage Divider to the motherboard, A) This allows the ADC to be used either as an ADC or as Power Compensation (as decided by the user and/or motherboard setup), and B) the Voltage Divider potentiometer needed is much too large to add to each module, and the voltage that the divider outputs would be the same for all modules on the same motherboard. This helps, again, reduce the part count and overall cost, and allows more room for cooling.
aldenhart wrote:You might consider doubling up on the motor outs 1A, 1B, 2A, 2B, and certainly doubling (at least) the motor power and grounds.
Here's a slightly updated version. The connectors I'm looking at (Digikey# S9206-ND for the module, and one of S9198-ND (through hole) or S9190-ND (surface-mount) for the motherboard), and as such, pins are rated at 3A each. I guess I could always make it 2x14 and add another couple grounds if needed.
L6470Module2-Updated.png (45.33 KiB) Viewed 1460 times
I'm thinking that the signal pins will likely take a route around the outside of the bottom layer of the PCB and pop up through vias to reach their respective pins, while power and output will stay on the top layer whenever possible (of course, all this allowing for the largest uninterrupted ground pours possible for heatsinking)
Shapeoko #280 (Inventables Batch #1)

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Re: STMicro Drivers

Post by ukewarrior » Tue Oct 01, 2013 12:19 am

Did this idea die?

I would prefer to build my machine around drivers that can handle more current than the pololu modules.

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Re: STMicro Drivers

Post by aldenhart » Tue Oct 01, 2013 8:10 pm

The gShield will do about 2.5 amps (or slightly more) per winding with cooling, which is somewhat more than the Pololus will do. But really any NEMA17 is OK with (almost) any driver. It's the bigger NEMA23's that require the additional current. I have been playing with the Automation Technologies (Keling) motors like this one - KL23H286-20-8B. It delivers 425 oz-in in a NEMA23 taking only 2.8 amps as a parallel wired motor. Here's a vid:

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Re: STMicro Drivers

Post by hugov » Thu Oct 03, 2013 10:40 am

So you are at or over the edge of what the ST is rated for (3 amps), and well under it's 7 amp peak. I still don't know what that means - what is "peak"? if 3 amps is RMS then the peak is 3 x 1.414... (4.2...amps). What does 7 amps mean, exactly?
RMS is just a way of time-averaging current, and it's used because it handles situations with positive and negative currents such that the two don't cancel out. How it relates to peak current depends heavily on the shape of the current waveform. For DC current, peak = RMS. For sinusoidal current (as in mains-powered appliances etc) the RMS current is 0.707 times (1 / sqrt(2)) the peak current. For a square wave current with 50% duty cycle from 0V to the supply rail (as you would get with a half-bridge or single-ended driver), the RMS current is 50% the peak current. For a square wave (any duty cycle) between positive and negative rails (as you would get with a full-bridge driver) it's 100% of the peak current. For arbitrary waveforms, you can do the maths, as per here:

Imagine a square wave periodic waveform that has a 1% duty cycle, and a peak current (the current during that 1% that it's on) of 100 amps. The RMS (and in this case also other averages like the geometric mean) current is 1 amp. But if you put a (non-slow-blow) 1 amp fuse inline with it, it will blow as the peak is 100 amps.

In this case, with the ST part, there are two different limitations. The first is the IC being limited in time-averaged current handling, and this is the RMS rating. The failure mode here is things like hotspots on the IC heating up to failure while the average temperature is within safe limits - this rating cares about average current because of thermal mass and that temperature changes can't happen instantaneously. The second limitation is the peak current - the instantaneous maximum current seen through the drivers. The failure mode here is things like bond-out wires inside the IC's package fusing.

Hopefully this clears up the distinction between RMS and peak.

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