boost converter overshoot / undershoot

[justDIY]

I thought all was well with my little boost converter, but then I decided to measure things, to see how the efficiency was doing. What I found was worse than I expected.

Test setup:

Four white leds, wired in series.
Sense resistor 38.3 ohm 1% smt
Requested output ~30mA (32 to be exact)

Voltage source 1:

Two weak AA batteries, approximately 2.5 volts.
Voltage at sense resistor 1.0 volts, output current 26 mA
Voltage across the leds 12.9

Voltage source 2:

USB port, approximately 5 volts.
Voltage at sense resistor 2.05 volts V/R=I = 53 mA output
Voltage across the leds 13.4

So my regulation is terrible for some reason. Aside from a larger sense resistor and pwm input disabled, my design follows the reference design exactly. (Figure 22. White LED Supply)

I've tried different amounts of output capacitance, in case ripple is messing with the feedback. perhaps I need to decouple the feedback pin to ground with a small cap (i've seen that in some designs)? did I miss something?

switcher datasheet:
https://www.ti.com/lit/gpn/tps61040

schematic:
https://projects.dimension-x.net/pictures/smps/dip16n_cc_boost_sch.png

pcb layout:
https://projects.dimension-x.net/pictures/smps/dip16n_cc_boost.png

assembled gizmo (show with 10uf output cap):
https://projects.dimension-x.net/pictures/smps/tps61040_cc_1.jpg

 

[OutToLunch]

Why such a large output cap? Just one of thise 4.7uF caps you have on the input and put it on the output.

Regarding your layout - I would have done some things differently... I would have rotated the sense resistor 90 degrees counterclockwise and then moved it up so that the external connection to it goes directly to the left (when looking at the layout pic you gave a link to). The connection to the FB pin would just go straight down then to the left to the IC pin. The ground connection on both terminals would be connected (currently the output ground is not even tied to input ground on the board). The existing connection between ground copper and the IC ground pin is thin - I would fatten it up as much as possible and then (with the sense resistor being rotated) just continue the ground connection from the ground pin of the IC to the output ground.

I think the biggest problem here is that the output capacitor ground is just floating. Even if you have it tied around externally to the input ground, it is too long of a path - introducing parasitics. Make it more direct - if need be, just solder a wire from the cap ground to the ground trace on the board.

 

[justDIY]

the ground that seems to be floating is connected to the input ground on the breadboard, with a jumper on the breadboard (picture was taken pre-wireup)... revision two takes care of that, however, I just want to make sure everything is the way it should be, before committing to another pcb panel.

I can see now that the weak ground for the chip might be a problem, I just assumed since the trace was so very short that the resistance wouldn't be a problem. I'll try soldering a jumper from that pin to the nearby ground.

I cannot use the 4.7u cap like I have on the input, as it's only rated for 16 volts. When this project is complete, I hope to drive 8 leds in series which will require 26 volts. I have a 35v ceramic on order, waiting to see if it makes it before xmas.

here's a link to the revision two board, the big IC is a pic micro.

https://projects.dimension-x.net/pictures/smps/artlight_2.png

 

[Roff]

Are you using the recommended inductor? (Looks like you probably are). Saturation current is a critical parameter.

 

[justDIY]

I'm using the 10uH inductor specified by the reference design, also listed as "typical" by the datasheet

Since the PFM peak current control scheme is inherently stable, the inductor value does not affect the stability of the regulator. The selection of the inductor together with the nominal load current, input and output voltage of the application determines the switching frequency of the converter. Depending on the application, inductor values between 2.2 µH up to 47 µH are recommended. The maximum inductor value is determined by the maximum on time of the switch, typically 6 µs

I haven't had a chance to improve the grounding yet ... I hope that's all it is.

I'm also going to hook up the scope and make sure the switch is running smoothly and not "bursting".

Some applications require a very tight line regulation and can only allow a small change in output voltage over a certain input voltage range. If no feedforward capacitor CFF is used across the upper resistor of the voltage feedback divider, the device has the best line regulation. Without the feedforward capacitor the output voltage ripple is higher because the TPS61040/41 shows output voltage bursts instead of single pulses on the switch pin (SW), increasing the output voltage ripple. Increasing the output capacitor value reduces the output voltage ripple.

 

[justDIY]

alrighty, using that left over copper strip along the top of the pch, I bonded the 'floating' ground pin to the input ground. I used three bridges, one at the input, one where the ground for the chip connects, and one at the far end, where the output capacitor is.

I removed the 1uf electrolytic, and replaced it with a 0.1 uf ceramic

at first, I was excited, my meter showed 1.20 volts across the sense resistor ... but then it started dropping, and was hovering around 1.00 when I shut it down. figuring my batteries might be sagging, I connected the circuit to a 5v supply. Sense resistor voltage was over 2 volts!

the switch output looks clean on the scope, all the pulses are clear and evenly spaced, although I'm not exactly sure what output voltage and multiple pulses would look like.

here's some photos on what the scope says, first is the switch output, second is the feedback pin
https://projects.dimension-x.net/pictures/smps/tps61040_waves_1.jpg
https://projects.dimension-x.net/pictures/smps/tps61040_waves_2.jpg

 

[Styx]

why havn't you setup the feedback with a potential-divider? the cct has it just tied to GND

Also what are you hoping in trying to "create" a differential power supply via taking the +VE at the output of the switcher and the -VE at the feedback - anything the load does will mess up the switcher since it will put a voltage on that pin?

this is a single-ended booster chip

 

[Roff]

why havn't you setup the feedback with a potential-divider? the cct has it just tied to GND

Also what are you hoping in trying to "create" a differential power supply via taking the +VE at the output of the switcher and the -VE at the feedback - anything the load does will mess up the switcher since it will put a voltage on that pin?

this is a single-ended booster chip

I think the idea is to create a current source for the LEDs. Look at fig. 22 in the datasheet.

 

[justDIY]

i tried soldering a jumper from the ground pin on the device directly to the ground plain, giving it a very strong ground ... no change ... when the input voltage is under 3 volts, the output current is below the set-point ... input voltage over 3 volts, output current is way above the set-point.

 

[justDIY]

I found the design manual for their constant current led driver evaluation module, and was looking at the waveforms they have in there.

https://www.ti.com/litv/pdf/slvu068a

my led current waveform looks very similar to theirs (figure 2-2), perhaps not as exaggerated ... hard to hold the probe, camera and fiddle with dials at the same time

but my switching waveform looks nothing like theirs (figure2-3), unless it's a fake ... mine doesn't ring like that at all ... isn't ringing generally a bad thing anyway?

 

[Roff]

Your layout looks like ENABLE is grounded through RPD (I'm assuming the PIC is not installed). I guess I'm missing something.

 

[hjames]

The "ringing" in 2-3 of the app note is indicative that the switching regulator is operating at less than full capacity - Essenially every cycle ends with the feedback voltage above 1.2V, so the chip doesn't try to immediately begin the next cycle. Instead it waits, and what you see is the end of the inductor flapping around with it's parasitics. The next cycle starts when the fb voltage drops below 1.2V, and the chip "charges" the inductor up to 400mA.

Also, according to the app note (pg 2-6) there's ~8mA of ripple at 4.2v. In your circuit this would end up being 300mV worth of ripple, which is what you're seeing. BTW, in their circuit, it would end up being .8V worth of bouncing up and down...

 

[Ocelaris]

Sorry to ask such a simple question, but what and where did you get that copper trace? Is it a etching kit or just tape?

Looks good to me none-the-less! SMD wonder works! I'd just be happy to get one of those going.

 

[justDIY]

Your layout looks like ENABLE is grounded through RPD (I'm assuming the PIC is not installed). I guess I'm missing something.

Ron, you are correct. However, I'm not using the revision two pcb yet, still working with the one in the first post.

My thoughts with RPD (pull-down) is a 'safety' measure ... if the PIC pin driving enable goes floating for some reason, RPD will hold it low until the PIC drives it high.

 

[justDIY]

Sorry to ask such a simple question, but what and where did you get that copper trace? Is it a etching kit or just tape?

Looks good to me none-the-less! SMD wonder works! I'd just be happy to get one of those going.

I etched it using using the photolithography method

 

[justDIY]

Also, according to the app note (pg 2-6) there's ~8mA of ripple at 4.2v. In your circuit this would end up being 300mV worth of ripple, which is what you're seeing. BTW, in their circuit, it would end up being .8V worth of bouncing up and down...

so then, it's the non-linear voltage-current relationship of leds that's screwing things up ... the small increase in voltage say from 3.1 to 3.4v (from the ripple) is causing the large increase in current?

I still don't understand why the regulator doesn't skip pulses or reduce its frequency when it sees over 2 volts on the feedback in.

 

[OutToLunch]

You're waveforms are somewhat convoluted - took me a sec to figure it out. The pic you show of the SW pin is really depicting the OFF time of the internal FET, which is about 400nsec. The voltage on the SW pin is about 15V and the input is 5V, so the voltage across the 10uH inductor is about 10V. This yields a peak inductor current of about 400mA - which is in line with what the datasheet says - the switch turns off when it hits 400mA.

The switch turns on when the voltage on the FB pin has fallen below 1.233V. With the input at 5V, the switch will be on for about 900nsec (800nsec plus 100nsec prop delay) for it to reach the limit of 400mA. Looking at the other waveform you show of the voltage on the FB pin, this seems to fit also - the switch turns on when the voltage jits 1.233V. You don't see anything happen to the voltage on the FB pin until the switch releases 900nsec later - which with the rate of fall of the voltage on the FB pin (approx -180mV/usec), the bottom of the FB voltage should be at about 1.07V - which is fairly close to where it is. It appears that the IC is doing what it is supposed to be doing. I guess now we need to figure out how to make it do exactly what you want it to do.

Oh - if you look at the waveform for the voltage on the FB pin, the rate of fall is about 182mV/usec. With a 0.1uF capacitor and the voltage falling fairly linearly, you can assume that the current coming out of the cap is all due to the LEDs and that calculates out to be about 18mA.

I think what you're going to have to keep in mind is that 1.233V divided by the sense resistor is not going to be the average current through the LEDs. I don't really have the time to figure out the average voltage across that resistor, but it is going to be a function of the input voltage, max inductor current (400mA), the output capacitance and the current through the LEDs. When the switch turns on, there will be a delay which will cause the FB voltage to fall below 1.233V before the switch turns off and the inductor dumps current to the output cap. That delay will be: tdelay=400mA*L/Vin. (tdelay is really just the ON time of the FET)

When the FET turns off, the inductor will dump the current into the output capacitor. We know what the peak inductor current is (400mA) but we really don't know what the voltage on the SW pin will be - we can assume 15V from the scope shot you supplied, though. So while not entirely accurate:
Toff=400ma*L/(15V-Vin)

This time can approximate the peak voltage the output capacitor will charge to. We know the inductor current will decrease linearly from 400mA to zero in the time Toff. That current is put directly into the output cap (ignoring the constant current draw of the LEDs). This can be put into equation form with :
iL=400mA-(15-Vin)*t/L
With iL=Cout*dv/dt, solve for dv in terms of dt, then integrate from 0 to Toff and you should get a rough approximation of the ripple voltage on the output capacitor. If I did it right it should be:
Vp-p=Cout*L*(400mA^2)/(15-Vin)-C*((15-Vin)/(2*L))*(400mA*L/(15-Vin))^2

With all that info, you should be able to reconstruct the ripple voltage across the current sense resistor, transform that into an average current calculation and then use that equation to solve for the value of the current sense resistor that you want for a particular average LED current. Keep in mind that as the input voltage varies, so will the output current.

Hope that wasn't too confusing, I wrote in in a stream of consciousness type manner.

 

[hjames]

This is a very simple regulator chip - more advanced ones have all sorts of compensation mechanisms inside them, in comparison to this one being just a flipflop and a pair of comparators. It doesn't know how to skip pulses, and it's pulse width is pretty much defined by how fast the inductor charges up - which is directly related to the input voltage. Plus, the faster the inductor charges up, the timing delay built into the chip will cause the inductor current to exceed 400mA more, causing even more ripple.

In more advanced regulators, there is another control loop around the inductor current limit - by the time the inductor is fully "charged" you can't do anything else, it's going to cause the output voltage to bump up by some given quanta. However, if you can change the current lmiit of the inductor, then you can control how much it's going to overshoot - and keep the output voltage in tighter regulation.