What do you mean when you say the supply collapses? I see a voltage drop across the supply when I turn on initially. Is this it? I am running with a bench supply but tried a large car battery and found the dip decreased a lot. Adding 1200 uF of capacitance across the bridge supply helped a lot too.


The ringing I see seems to be after the dip and recovery of the voltage.

 

The cap needs to be close to the bridge not near the psu.

 

I've put the caps as close to the bridge as I can at the moment. You can see the caps at the bottom of my board:



I am planning to make a PCB so will try and position them as close as possible then. As I have a dev board I just wanted to try and debug as much as possible with it first.

What is the function of C1 and C2. Would they just couple spikes to ground?

 

Ok, looking at layout you may need a cap across the PSU on the micro side, if driven from same PSU then noise could could be getting here and effecting the FET drivers and bootstrap voltage.

 

C1 and C2 on your schematic are part of the oscillator circuit. Is that what you meant? Read here: http://en.wikipedia.org/wiki/Crystal_oscillator#Modeling

If you meant the big bridge bypass capacitors (the 680uFs), they provide peak current to your motor and make it possible to shorten the stray inductances that carry high load current.

When you work with switching power supplies, converters etc you have to provide a short path from power supply to the bridge and load and you have to provide a path back to the power supply too.

All wire/trace stray inductances on the path from these bypass capacitors to the bridge should be as short as possible. The WHY and HOW is quite well explained here: http://www.irf.com/technical-info/appnotes/an-978.pdf page 11 (6. LAYOUT AND OTHER GENERAL GUIDELINES).

Just put those capacitors physically as near as possible to the H-bridge.

The power supply voltage measured with an oscilloscope at the capacitor terminals should not be very low or long.
Read also page 9 of http://www.hvlabs.com/files/HIP4081application.pdf

You say, "it overlaps with the turn-on of the other side...". Can you draw a quick diagram how the sequence of your four gate signals looks like?

 

I think Mark was referring to C1 and C2 on the HIP4081 dev board (p11 of http://www.hvlabs.com/files/HIP4081application.pdf). These couple each side of the motor to ground.

I found using the large caps across the supply to the bridge reduced the voltage dip and when I run off a battery rather than bench supply the dip disappears almost entirely (I presume as the battery can provide a high peak current whilst the bench psu current limits).

I have attached a picture of the PWM scheme below. The A signals are the inverse of the B signals. The high signal duty is the inverse of the low side duty. An increase in duty of the high side signal is coupled with a decrease in duty of the low side signal which increases the conduction time. In the diagram I have shaded the areas when conduction occurs:



Going back to my captures of the gate source voltages, would you expect to see a step. I presume this is a function of the bootstrap? I am thinking there is a crossover between the high side FET turning off and low side turning on, causing some shoot through due to this step.

 

Regarding your PWM picture: If ALO and AHO are the lower and upper FETs of the same leg I don't understand how the gate signals could be ON at the same time.

(refering to the diagram on page 11)

 

What is the function of C1 and C2. Would they just couple spikes to ground?

They would act together with L1 and L2 as a low-pass filter. They are not needed in a simple motor application.

 

I just realised that I labeled the signals incorrectly! Alo and Bho labels should be swapped, I've updated the image, sorry!

 

Is it perhaps worthwhile adding some Schottky diodes accross the gate resistors to further reduce the turn-off time of the FETs? This is mentioned in the app note page 7 http://www2.vmi.edu/Faculty/squirejc/Research/IC_Datasheets/motor_drivers/HIP4081 app note 9405.pdf

 

Is it perhaps worthwhile adding some Schottky diodes accross the gate resistors to further reduce the turn-off time of the FETs? This is mentioned in the app note page 7 http://www2.vmi.edu/Faculty/squirejc/Research/IC_Datasheets/motor_drivers/HIP4081 app note 9405.pdf

Your problem starts right at the moment where there is no current flow from + through the motor to - , i.e. both upper FETs are ON and both lower FETs are OFF.

The problem may not occur with a different PWM scheme, but you should be able to solve it using this too.

Adding diodes for faster discharge of the gates seems counterintuitive since turning the FET off faster and therefore interrupting the current faster will facilitate the oscillations even more.

I would try to turn the FETs off less rapidly by increasing the gate resistor values (if heat dissipation of the FETs at the given switching frequency allows it) or the put a snubber (as simple as a RC) over the low-side FETs. See this app note for snubber design: http://space.dianyuan.com/blog/u/50/1170381240.pdf

 

Why is having a fast turn-off a bad thing? I would have thought it would be beneficial to turn off the gates as quickly as possible to prevent shoot through?

I am currently using 10 ohm resistors on the gates. I'll boost them to 100 ohm and see how it performs.

I tried a simple RC snubber across the motor (based on values used by OSMC http://www.robotpower.com/osmc_info/ who use the HIP4081) but it didn't seem to make much difference. I'll look into designing the snubber and seeing how it performs.

Colin

 

Why is having a fast turn-off a bad thing? I would have thought it would be beneficial to turn off the gates as quickly as possible to prevent shoot through?

I am currently using 10 ohm resistors on the gates. I'll boost them to 100 ohm and see how it performs.

The deadtime should be determined by the two resistors connected to the HIP4081 (pin 8 and 9), not the gate resistors.

Slower turn-off maybe beneficial to avoid/decrease ringing but will increase switching losses in the MOSFET and therefore heat it up more.

A snubber does exactly this, it either clamps voltage or decreases the rate of rise of voltage or current.