LMR62421: Boost Converter to drive Motor Inrush Current Properly

In case the inrush current is a red herring, I did put a 10 ohm resistor in series with the motor and all that happened was the total output voltage to the motor (averaged on a multimeter) dropped. 3V into 10 ohms is well under the power rating for this chip so I think the issue might be unrelated to the inrush current (though when it's output is open circuit the voltage is exactly correct).

I also tried connecting pin 1 of Q3 to the input voltage of the boost converter as shown in some boost converter application notes to achieve turn off. The thought was that perhaps the gate voltage wasn't enough to turn on. However, the output voltage continued to be low, and the difference between output and input voltage wasn't enough to turn the PFET fully on. I instead simply shorted across those leads to see if the motor would activate when the enable pin was toggled.

Shorting across the PFET and testing EN alone gave the same result of a 4-5V output instead of 6V, which sometimes kicks up to 6V as it should if the input voltage is high enough. Naturally a 5V supply to the boost converter works perfectly.

Hi Brian,

Thanks for providing the detailed information. I'll check and get the feedback to you next Monday.

Regards

Lei

Hi Brian,

Could you take a measurement of the Vin voltage waveform based on the current setup shown in the attached schematic? Please measure as close as possible to the device Vin pin.

Then, try to remove the filter of FL1 and check whether the regulated 6V can be achieved correctly.

In your description, a 5V supply can make the boost converter works perfectly. Does this mean that the boost converter can drive the motor correctly? If so, please also provide the current load profile of the motor when Vin=5V.

Regards

Lei

Thank you Lei!

I will check the VIN waveform as close as I can, but when I checked it further from the chip (at FL1) it was pretty clean with a small dip when the chip starts up but only 80mV or so. It's probably the most suspect signal though, I'll see if I can get a better measurement triggered to when the chip tries to kick on.

• I have tried shorting across FL1 a few times in case it was limiting current to the converter, there was no change.
• I should have said "when I plug in the battery charger chip", since the voltage going out to the motor is not 5V (it is 2.8-4.2V always). But when I plug in the USB charger the battery charger chip puts out a somewhat higher charge than the battery itself. It kicks the system on when VIN gets extra support.

The battery charger chip is rated to 1A, the BQ25060. I totally buy that the supply impedance is somehow too high, but shorting the battery directly to the input of the boost converter didn't change anything, so I don't think that's it (or at least not all of it). I'd believe that the 18650 has more internal resistance than a USB cable, but certainly enough for steady state.

I saw one suggestion to bootstrap VIN using the boost converter output, but that seems unwise to me. I saw another that suggested supplying VIN from a separate supply as the inductor -- I could do that here by giving VIN the pre-FL1 system voltage.

Hi Brian,

Thank you for your checking. The most suspicious reason that the device Vin pin may be pulled down below the UVLO should not be correct based on your experiments. 

To further check, I'd like to start from the condition that the whole circuity can work perfectly, that is, a 5V supply. Will you be available to measure the load current at this condition? And then decrease the supply from 5V to 3.3V, with the Vin, SW and Iout being monitored.

Just let me know if you have any new findings!

Regards

Lei  

Dear Lei,

Thank you so much for the support in figuring this out!

I used a USB oscilloscope to capture many pieces of information that maybe will help us.

These are measurements of Vin under different conditions. They are measured across the capacitor unfortunately so it's not perfect.

Battery Plugged in, EN disabling Boost Converter.

Battery Plugged in, when EN is first enabled. It dips to 2.633V according to my oscilloscope which should be sampling fast enough to see a faster drop but again I'm on the capacitor so it could be worse at the chip.

Steady state failing to start with battery only.

After plugging in the USB charger (increases Vin as it would be charging the battery).

After unplugging the USB charger while leaving it on (the battery is sufficient for steady state). A little more ripple

I also looked at the output voltage to compare. I took these measurements across the leads of my motor so they are somewhat far from the boost converter (and after the output MOSFET).

This is the output voltage when the battery is failing to start it up.

This is the output voltage after I plug in the USB charger (so increases Vin a bit). Beautiful.

And after I unplug the USB charger the battery is completely adequate to maintain the power and voltage output.

I can set up a more complex test to look at voltage thresholds with a power supply, but let me know if the above gives any clues. To do this test I would connect the power supply in place of the battery so that I can simulate a battery with different charge levels, if that's a good enough simulation I can give it a try. However, I watched it go through about 3.6-3.7V at Vin before it started failing (with 570uF of extra electrolytic capacitance at Vin). Without the extra capacitance at Vin I couldn't get it to reliably start even at 4V battery voltage.

It would be great if you have any advice on simulating these circuits better, is there any tool that can simulate these parts (both this and the battery charger are after all from TI) in a generic enough way that I can approximate the rest of my circuit it would come in awfully handy.

Hi Brian,

Thank you for providing these waveforms. Based on your measurements, I believe if connect pin1 of FL1 to a "strong" Vsource, the circuit you attached should work correctly. So I still think the problem is the output current capability of the battery or BQ device(when the USB charger is not plugging in).

Sorry I'm not familiar with the BQ device and need to check first. I'll give you the feedback late today.

Regards

Lei

Thank you, I appreciate it. It is supposed to provide 1A of capacity, but the battery itself is essentially connected directly to the input of FL1. I disconnected it entirely from the circuit and just connected the battery without a change in behavior, I don't think it is doing anything problematic.

If we assume that the battery impedance is the issue, is there a way to get this chip to survive being connected to a motor load with a clever circuit? 570uF is too much capacitance to add at the input, and it seems that it may need many times that to work.

Hi Brian,

What's the voltage of the battery? Maybe it's near the threshold since plugging in the USB charger to charge the battery, and then unplugging in the USB charge, the circuit can work. Could you provide a waveform of the voltage at the Vin pin of LMR62421 when the pin1 of FL1 is connected directly to the battery? 

In your original circuit, is the pin1 of FL1 connected to the OUT pin or BAT pin of BQ25060? Could you also attached the schematic of the BQ device and the battery?

The performed experiments show that with "strong" Vsource, the boost circuit can work, but with the "weak" battery, the boost circuit failed to regulate the Vout. Is it possible to try a different battery? 

If confirmed that the battery impedance is the issue, and no other options of battery with lower impedance, I think increasing the Cin is the only available option.

Regards

Lei

This is a lithium ion 18650 that is newly purchased, and it has a charge voltage of 3.7V during the test. It has had one charge cycle, when I charged it up to 4.2V (the motor worked fine at that point) and once the charge voltage dropped below 3.6-3.7V the input voltage wasn't enough. The battery is designed to discharge to 2.8V and I think the boost converter is supposed to be able to handle this voltage input level.

Sure, I can do that (connecting the battery to FL1). I do also have a current probe.

I can also try a different battery; they are brand new though.The charge isn't even low, they're at their nominal voltage. If they were discharged they'd be all the way down near 2.8V and I assume definitely not work to start the boost converter anymore.

Is there another boost converter that might stand a better chance?

What about a bigger output capacitor and delaying the output PFET for a substantial time (so as to get the boost starter bootstrapped)?

Is there a circuit that people use with these to slowly connect the load to the circuit to avoid Vin crashing (if that's what's happening)? A colleague suggested a capacitor between Vout and the gate of the PFET but I'd love to find an application note about driving brushed DC motors with a boost converter.

I have a hard time guessing how big of a capacitor would be needed here, but it seems like 10,000uF would be required to be safe... that doesn't seem right for a small DC motor, this thing only uses 80mA average. I'll see if I can get a current probe onto the motor wire to see the inrush current. Maybe the few hundred milliamps I saw before was in error.

Dear Lei,

I captured the startup current to the motor from switching it on with the USB cable plugged in, so this should represent normal motor current during startup.

I'm not super familiar with brushed DC motors, does this profile mean anything to you? I see it reaching a peak current of ~1.4A for 1ms or so, though I know that's not supposed to cause any issue with this boost converter chip.

Another colleague suggested perhaps a thermister in series would help.

Hi Brian,

Thank you for providing the current waveform, it's interesting and provides much useful information.

Just curious about one question, although may not be related with this topic: in stable status, is the motor 36000RPM?

From the motor current waveform, I think the start-up process of the motor is:

1. Starting from the rest status, 6V is forced and the initial current increases to a big value of ~1.4A. Then the motor starts to rotate;
2. The rotate speed gradually increases. In this phase, big current is still needed. After around 100ms, the motor enters into the stable status;
3. In the stable status, small current is needed, and the battery can make the motor work. The picture of "After unplugging the USB charger while leaving it on (the battery is sufficient for steady state)" can prove this.

The problem is that, if power the circuit with battery only, the motor will be trapped in the phase2 of ~12000RPM, one third of the speed in the stable status, which is circled in the following picture.

In this trapped status, big "pulse" current of 200Hz is needed for the motor. That's why you can observe the 200Hz ripples in the Vin pin. 

From the following picture you provided, we can see that with a 0.8A peak current "eaten" by the motor in the trapped status, the boost Vout will be pulled down to 2.2V. That should be one of the reasons that the motor can't enter to the next stable phase. 

For LMR62421, support Iout=0.8A, Vin=3V, Vout=6V, and efficiency=80%, then the needed input current will be 2A. Although it's close to the current limit, it should not be the reason that make a low Vout of ~2.2V. So I think the reason is the limit of the maximum output current of the battery. 

Could you help check what's the maximum output current of the  18650 Li-ion battery you used? Is it possible to try a 18650 Li-ion battery with maximum output current >=2A;

I also want to try to increase the Cout. No need as big as 10,000uF, but three or four times of C11, such as 47uF, although you've tried a 100uF capacitor and got lower Vout.  

Hope we are closing to get the motor work normally.

Regards

Lei

Thanks Lei, I had not heard of this Stage 2 of the motor starting.

I took an impedance spectra from the battery while it was at a normal voltage. The response includes the wiring, so the high frequency response might be from inductance in wiring rather than from the battery, but fair enough to include.

It sounds like I should try putting a bigger capacitor at the output, possibly with a ferrite to make sure I don't break the boost converter trying to charge it in the first place when EN turns it on (after all, a low ESR ceramic is a dead short too...), and then use a delayed turn on to activate the output FET after a little while so that the boost converter can charge up the output capacitor fully.

I am going to try that next based on what we've discussed, but I admit I am worried that once the capacitor is charged the boost converter will drop down into a low duty cycle idle mode and not increase the duty cycle in the feedback loop fast enough to keep up, I was not sure if it was better to start the converter with the motor connected or to let the regulator reach max voltage (and zero current) and hope it can ramp up the current quickly enough.

Do you know how the converter behaves with large output capacitors? I can easily attach 100uF ceramic, but I can put a small resistor in series with it if a direct connection would confuse the regulator.

Hi Brian,

The "stage 2", or "phase 2", is not a classical concept defined in books. I just used it based on the motor behavior from rest to stable status of fast spin, and the current waveforms you provided.

I don't think the Cout is the problem, because with all the same setup, the motor can be driven correctly when you plug in the USB charger. The only difference is that the previous stage(BQ device) can provide higher output current to LMR62421 than the battery.

My suggestion is to use a 18650 battery with higher maximum current output or make the Cout two or three times bigger, with the purpose to prevent the Vout being pulled too low. The "phase 2" is a special status, as long as the output energy can be a little higher to make the motor rotate faster, then the needed current of the motor will decrease significantly. 

I'm new to the properties of a DC motor, please correct me if I have any wrong understandings.

Regarding the Cout, if it's extremely large, then the output pole and cross frequency will be very small, and the phase margin may be small as well.  

Regards

Lei