Boost converter - can someone please explain the phenomenon

[newtekuser]

Use something like MP3437 for example :

~2.3$ SOT-23-8 / TSOT-23-8 : https://www.digikey.com/en/products/detail/monolithic-power-systems-inc/MP3437GJ-P/18090393

10 QFN : https://www.digikey.com/en/products/detail/monolithic-power-systems-inc/MP3437GRP-P/18089144

Cheaper but a bit less efficient :

~1$ PAM2423 : https://www.digikey.com/en/products/detail/diodes-incorporated/PAM2423AECADJR/4033258

Nice specs, many thanks!

[Mechatrommer]

Attached my revised design. I have resized the inductor footprint and diode and re-arranged components on the board. However, I can't come up with a layout where both traces from the inductor to the IC are short due how IC and inductor pins are placed, so I have to choose between making 5V or 12V traces long. In this case, I positioned components so that 12V traces are shorter.
With my limited experience routing PCBs the only way I can see to make traces shorter is to go to a 4 layer board.

imho, inductor/diode placement is less critical compared to feedback resistor as long as you dont place them unreasonably far away. but then feedback resistor layout will only affect transient/switching respond, this is a later stage to fine tune to get minimal output noise. but i guess your problem in OP has nothing to do with pcb layout, its overloaded / current limited issue ymmv.

There's a beefier version MC33167 / MC34167 that has a 5.5A switch current limit so it would handle higher output but it's more expensive and at this point, there's better cheaper switching regulators.

thats why when i cant find suitable IC (those smps ICs that i found are pathetic when it comes to max current) i will go will external switching mosfet. with external mosfet we are free to choose for tens or hundreds of amperage output, but for smaller app and lower noise, built-in mosfet in smps IC is recommended. ymmv.

[shapirus]

I made a boost converter using the equations provided by Ti for calculating the power stage of a boost converter.
My requirements are: 5Vin, 12V out with a power draw of 400mA from the load.

You won't get more than ~250 mA on the output at 12 V when boosting from 5 V input with the MC34063A, unless you add an external switcher, and by then you'll be better off using a different chip altogether (take a look at e.g. MT3608, which requires an even simpler external circuit to operate).

The MC34063A has a maximum allowed current of 1.5A for the internal switcher. It is also rather picky about the inductor you use. But once you get the external components right, it works quite well for what it is.

My boost converter outputs 12.2V without a load but once I connect a 12VDC fan that takes 0.300mA, my output drops to about 6V. Input current also measures s 1.2Amp which is the max limited by my bench top power supply setting.

What is the value of Rsc? What is Isat of your inductor?

You may find it helpful to refer to MC34063A calculators such as http://www.nomad.ee/micros/mc34063a/

[newtekuser]

I made a boost converter using the equations provided by Ti for calculating the power stage of a boost converter.
My requirements are: 5Vin, 12V out with a power draw of 400mA from the load.

You won't get more than ~250 mA on the output at 12 V when boosting from 5 V input with the MC34063A, unless you add an external switcher, and by then you'll be better off using a different chip altogether (take a look at e.g. MT3608, which requires an even simpler external circuit to operate).

The MC34063A has a maximum allowed current of 1.5A for the internal switcher. It is also rather picky about the inductor you use. But once you get the external components right, it works quite well for what it is.

My boost converter outputs 12.2V without a load but once I connect a 12VDC fan that takes 0.300mA, my output drops to about 6V. Input current also measures s 1.2Amp which is the max limited by my bench top power supply setting.

What is the value of Rsc? What is Isat of your inductor?

You may find it helpful to refer to MC34063A calculators such as http://www.nomad.ee/micros/mc34063a/

I'm using this 220uH inductor with Isat of 1.7A: https://www.digikey.com/en/products/detail/bourns-inc/SRR1280-221K/1970008
Rsense value is 0.051 ohm, don't ask me how I calculated it bc I can't find my notes on it, or the formula in the IC data sheet

I searched for MT3608 but Digikey and Mouser don't have it in stock. I actually purchased this part off eBay long time ago, but the number written on the IC I got is 7217P not MT3608 as in the auction listing. The package is also very small - will this be able to handle the heat dissipation?

[shapirus]

I'm using this 220uH inductor with Isat of 1.7A: https://www.digikey.com/en/products/detail/bourns-inc/SRR1280-221K/1970008
Rsense value is 0.051 ohm, don't ask me how I calculated it bc I can't find my notes on it, or the formula in the IC data sheet

Well I'd suggest using values provided by the online calculators as a starting point.

I searched for MT3608 but Digikey and Mouser don't have it in stock. I actually purchased this part off eBay long time ago, but the number written on the IC I got is 7217P not MT3608 as in the auction listing.

Not sure what 7217P would be. The MT3608, per its datasheet, is supposed to be marked as "B628DC". I'm sure there are many alternatives that are similarly easy to use and can provide a decent output current at good efficiency, but I don't know them without searching. Maybe someone else can suggest something.

The package is also very small - will this be able to handle the heat dissipation?

Small package is fine, if the resistance of the switching element is low enough, which is the case with MT3608. Besides it uses a MOSFET, unlike MC34063A which uses a BJT for switching. MT3608 is rated for 4A max switching current (not the same as continuous output current) and 0.6 W absolute maximum power dissipation. Let's say we want to run it at no higher than 0.4 W, which, at an efficiency of 90% (taken from the datasheet curves), would mean 4 W input = 0.4 W loss + 3.6 W output, and 3.6 W output at 12 V is 300 mA output. With 400 mA at 12 V on the output the power loss in the IC is going to be about 530 mW, still below the absolute maximum rating, but you do need to design the PCB with heat dissipation considerations in mind.

[newtekuser]

I'm using this 220uH inductor with Isat of 1.7A: https://www.digikey.com/en/products/detail/bourns-inc/SRR1280-221K/1970008
Rsense value is 0.051 ohm, don't ask me how I calculated it bc I can't find my notes on it, or the formula in the IC data sheet

Well I'd suggest using values provided by the online calculators as a starting point.

I searched for MT3608 but Digikey and Mouser don't have it in stock. I actually purchased this part off eBay long time ago, but the number written on the IC I got is 7217P not MT3608 as in the auction listing.

Not sure what 7217P would be. The MT3608, per its datasheet, is supposed to be marked as "B628DC". I'm sure there are many alternatives that are similarly easy to use and can provide a decent output current at good efficiency, but I don't know them without searching. Maybe someone else can suggest something.

The package is also very small - will this be able to handle the heat dissipation?

Small package is fine, if the resistance of the switching element is low enough, which is the case with MT3608. Besides it uses a MOSFET, unlike MC34063A which uses a BJT for switching. MT3608 is rated for 4A max switching current (not the same as continuous output current) and 0.6 W absolute maximum power dissipation. Let's say we want to run it at no higher than 0.4 W, which, at an efficiency of 90% (taken from the datasheet curves), would mean 4 W input = 0.4 W loss + 3.6 W output, and 3.6 W output at 12 V is 300 mA output. With 400 mA at 12 V on the output the power loss in the IC is going to be about 530 mW, still below the absolute maximum rating, but you do need to design the PCB with heat dissipation considerations in mind.

I got in touch with the seller and they confirmed the number on my MT3608 IC is the production code. 

[newtekuser]

I took another stab at this design, this time using the more efficient MT3608 IC. When routing the PCB I am having trouble keeping the switching loop (IND, IC SW pin1, D) farther away from the feedback resistor network (R1, R2) to avoid coupling. I double checked the schematic and it looks in line with the example from the data sheet, but the way the rats nests are set up I must connect output of D as well as R2 to 12V output, so can't figure out how I can "isolate" them.

Any suggestions?

[Ian.M]

That's fairly easy - Flip C2 to bring its Gnd pin down the board and move its ground return current.  Move R1 and R2 up next to R3, side by side to minimise the trace length to pin 3.  The feedback must be tapped off at or after the capacitor pad, not at the diode, to reduce its coupling to the pulsing current in the switching loop, so connect +12V to the top end of the repositioned R1, from the C2 pad or output connector pad, running the trace up then parallel to the board edge, with a very well via stitched Gnd guard trace or fill right next to it to screen it from the switching node and diode.

More via stitching for the MT308 Gnd pin would also help - if you cant fit a via under the IC (in addition to not instead of the one right of the pin), run a trace under the IC down towards the inductor with the via as close to the IC as possible. 

However you should be using wide traces, lands and fills for all the high current connections.  Your other topic https://www.eevblog.com/forum/beginners/mp3437-boost-converter-ic-vdd-pin-voltage/ had a datasheet for the MP3437 with a suggested layout showing how to do this sort of stuff right.  Study it and try to apply the style and concepts to this MT3608 layout!

[shapirus]

BTW, while you're at it, take a look at https://embeetle.com/#blog/return-currents. Useful!

[newtekuser]

That's fairly easy - Flip C2 to bring its Gnd pin down the board and move its ground return current.  Move R1 and R2 up next to R3, side by side to minimise the trace length to pin 3.  The feedback must be tapped off at or after the capacitor pad, not at the diode, to reduce its coupling to the pulsing current in the switching loop, so connect +12V to the top end of the repositioned R1, from the C2 pad or output connector pad, running the trace up then parallel to the board edge, with a very well via stitched Gnd guard trace or fill right next to it to screen it from the switching node and diode.

More via stitching for the MT308 Gnd pin would also help - if you cant fit a via under the IC (in addition to not instead of the one right of the pin), run a trace under the IC down towards the inductor with the via as close to the IC as possible. 

However you should be using wide traces, lands and fills for all the high current connections.  Your other topic https://www.eevblog.com/forum/beginners/mp3437-boost-converter-ic-vdd-pin-voltage/ had a datasheet for the MP3437 with a suggested layout showing how to do this sort of stuff right.  Study it and try to apply the style and concepts to this MT3608 layout!

Thank you very much for the suggestions! Re-attaching the revised layout. The only DRC I get now are about the new unconnected vias. I have concerns about thermal imbalances from using those fills but I'm not able to use thermal reliefs because some of the spaces are too narrow and copper won't fill through.
Hope I didn't misunderstand anything.

[Ian.M]

Looks a lot better but there are a few small improvements left.  Move C1 left a bit and put its Gnd via just down from it to make space to widen the trace to the inductor.  Also widen the trace from the inductor to the diode.  The 12V trace to R1 can (and probably should) be narrow as it carries negligible current.

N.B. the new vias should all be connected topside by a ground track or fill, which you might as well join to the IC's Gnd pin for better heat dissipation.

[shapirus]

Why not rotate the inductor 90 degrees and run the traces alongside each other to reduce the loop area?

[newtekuser]

Why not rotate the inductor 90 degrees and run the traces alongside each other to reduce the loop area?

Couldn't see the forrest for the trees Is this what you had in mind?

[newtekuser]

Looks a lot better but there are a few small improvements left.  Move C1 left a bit and put its Gnd via just down from it to make space to widen the trace to the inductor.  Also widen the trace from the inductor to the diode.  The 12V trace to R1 can (and probably should) be narrow as it carries negligible current.

N.B. the new vias should all be connected topside by a ground track or fill, which you might as well join to the IC's Gnd pin for better heat dissipation.

Thank you very much!!! I have updated the layout in my previous post. The DRC errors are also gone after connecting the vias topside and hooking up to the IC GND pin. 

[shapirus]

Couldn't see the forrest for the trees :) Is this what you had in mind?

Yes, more like in the attached picture. Cut corners where you can. I also see no reason (there may be!) why not to increase the width of high-current tracks where you practically can.

I'd also turn J1 and C1 180 degrees to avoid that narrowing of the power carrying track between C1 and the IC, and dragged C1 a little lower to put its ground via in line with GND of J1 and the IC.

Then, I'd drag J2 a bit lower as well, again to improve the current return path, and, I think, rotate it 180 degrees too, so that its terminals are placed very close to the respective ones of C2.

Draw your return current traces *before* pouring copper on the ground plane :).

C2 and J2 could be moved closer to D1.

Also, routing using copper filled polygons is inconvenient, as it makes it difficult to rework when you move parts. It's easier to use traces (of whatever width is suitable), as they are dragged along with the respective footprints, but mainly because the layout looks much tidier with them. Polygons are fine, but they I think should be added when everything else is ready and only where you really need them.

[newtekuser]

Another failure 

My output is only 4V something and it draws 200mA just idling and getting very hot. I wonder if the IC I got is indeed a MT3608.

I chose a 6uH inductor (per calculations, I should have used a 3.3uH but the IC data sheet says 4.7uH is minimum), R1 = 27K, R2 = 1.4K (0.6 x (1+27K/1.4K)) = 12.1V

Duty cycle was calculated based off 92% efficiency (1 - (5x0.92)/12) = 0.61

I checked for shorts and there are none.


Inductor - https://www.mouser.com/ProductDetail/652-SRP1265A-6R0M
Diode - https://www.mouser.com/ProductDetail/757-CMS04TE12L%2cQ%2cM
27K resistor - https://www.mouser.com/ProductDetail/71-RCS080527K0FKEA
1.4K resistor - https://www.mouser.com/ProductDetail/667-ERJ-P06F1401V
10K resistor - https://www.mouser.com/ProductDetail/710-560112120004
22uF cap - https://www.mouser.com/ProductDetail/963-MSASE21GBB5226MT

[newtekuser]

dont expect anywhere near 100% efficiency for boost ... my rule of thumb for boost is 50% efficiency ... next thing is check your inductor value, too small and it will into saturation quickly and IC will smoke ... etc ...

The inductor value was chosen based on calculations. However, I swapped it with a 220uH inductor and it behaves the same.

[shapirus]

Do you have an oscilloscope? Check if there's any switching activity going on at all.

[newtekuser]

Do you have an oscilloscope? Check if there's any switching activity going on at all.

I probed the switching node at pin 1 with the scope and when I measure frequency it only reads around 60Hz. Datasheet says it should oscillate at 1.2MHz. Did I get a dud?

[shapirus]

I probed the switching node at pin 1 with the scope and when I measure frequency it only reads around 60Hz. Datasheet says it should oscillate at 1.2MHz. Did I get a dud?

Quite possibly, as long as everything is wired correctly, input voltage is correct etc. I think we can rule out the board layout, as it should work, albeit poorly, even on a solderless breadboard.

So if everything else is done correctly, then it must be the IC that's wrong. But before this can be concluded, you need to make damn sure that everything else is done right and you're not seeing some normal operation under certain conditions.

I'd also suggest looking for switch mode controller ICs at a reputable distributor that's comfortable for you to buy from, whatever it might be. Narrow the search down to what you can obtain, then look at the datasheets to narrow it further down to what will be comfortable to use. Maybe get a few different ICs. Try them on a solderless breadboard, or maybe on a solderful (meh) breadboard to get at least some results that'll tell you that you're on the right path.

[newtekuser]

I probed the switching node at pin 1 with the scope and when I measure frequency it only reads around 60Hz. Datasheet says it should oscillate at 1.2MHz. Did I get a dud?

Quite possibly, as long as everything is wired correctly, input voltage is correct etc. I think we can rule out the board layout, as it should work, albeit poorly, even on a solderless breadboard.

So if everything else is done correctly, then it must be the IC that's wrong. But before this can be concluded, you need to make damn sure that everything else is done right and you're not seeing some normal operation under certain conditions.

I'd also suggest looking for switch mode controller ICs at a reputable distributor that's comfortable for you to buy from, whatever it might be. Narrow the search down to what you can obtain, then look at the datasheets to narrow it further down to what will be comfortable to use. Maybe get a few different ICs. Try them on a solderless breadboard, or maybe on a solderful (meh) breadboard to get at least some results that'll tell you that you're on the right path.

I did notice one issue in that I routed the connection to the EN pin incorrectly. Should have been tied it to supply voltage for startup but I tied it to ground via 10K resistor (R3). Maybe the 10K resistor wasn't necessary anyways.
However, even after removing the resistor, there is no change in behavior, maybe I had damaged the IC at this point.
My other suspicion was improper diode orientation, but I measured it with a multimeter in diode mode and it only conducts to the output and not backwards to the switching node.

Unfortunately suppliers like Digikey and Mouser don't have this IC in stock until late April.

[MrAl]

dont expect anywhere near 100% efficiency for boost ... my rule of thumb for boost is 50% efficiency ... next thing is check your inductor value, too small and it will into saturation quickly and IC will smoke ... etc ...

The inductor value was chosen based on calculations. However, I swapped it with a 220uH inductor and it behaves the same.

One of the things about a switching converter that is a little different than other stuff is that the inductor has to have the right inductance and also the proper operating current specification.  Some inductors just won't work in some applications because they saturate with DC current too easily.
This could be the problem, and this ties in with the inductance and the switching frequency.

[pcprogrammer]

And that is why I like to use Aliexpress modules. For my remote control project I'm using small buck converter boards, that I take the components of and solder them to my own board. Can also be cheaper if it is only for a small series. These complete boards are so cheap that buying loose components costs more.

I'm using one of these in a satellite receiver to power the VFD from 12V to 35V. Not a lot of current draw for this, but these modules can do about 2A on the output.

Edit: Fixed a link

[radiolistener]

My boost converter outputs 12.2V without a load but once I connect a 12VDC fan that takes 0.300mA, my output drops to about 6V. Input current also measures s 1.2Amp which is the max limited by my bench top power supply setting.

300mA at 6V means that the fan has resistance about R = 6 V / 0.3 A = 20Ω.
If you want to keep 12V at 20Ω load, then you're needs to provide P = 12^2 / 20 Ω  = 7.2 W power for your fan.
DC/DC has efficiency about 80%, so you're needs to provide at least 7.2 W / 0.80 = 9 W on your DC/DC input.

Since your DC/DC input is 5V, you're needs to provide at least minimum 9 W / 5 V = 1.8 Amps.
But as you said, your lab PSU max current is limited with 1.2 Amps which is just P = 1.2 A * 5 V = 6 W power.
While your load wants to get 7.2 W power and your DC/DC needs some power loss.

DC/DC cannot produce additional 3 W power from nothing and this is the reason why output voltage drops down.

As you can see your lab PSU just unable to provide required 9 W power.

[shapirus]

I did notice one issue in that I routed the connection to the EN pin incorrectly. Should have been tied it to supply voltage for startup but I tied it to ground via 10K resistor (R3). Maybe the 10K resistor wasn't necessary anyways.
However, even after removing the resistor, there is no change in behavior, maybe I had damaged the IC at this point.

But is the EN pin connected to VCC now? "EN" means enable, active high, and the datasheet says:

A high input at EN turns on the converter, and a low input turns it off.