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Boost converter - can someone please explain the phenomenon
Posted by
newtekuser
on 07 Mar, 2024 06:08
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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.
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.
If I bump the current limit on the 5V input to 1.4A, then my output voltage gets to 12V and the fan rotates at full speed. However, the output voltage continues to climb past 12V and things get awfully hot.
What I don't get is why it's drawing so much power.
From what I understand the formula to calculate the input current is In = (Iout * Vout)/Vin = 0.96A
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#1 Reply
Posted by
Ian.M
on 07 Mar, 2024 06:21
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Our crystal balls have gone cloudy, and its too much trouble to hack your CCTV so we've got no idea whatsoever what you are doing wrong!
So far it appears you *may* be using a TI switcher chip, and have a problem with unstable output voltage. Without a full schematic, your PCB layout and closeup photos of the assembled board, we simply cant help significantly, though my first suspicion is you may have FUBARed the layout and noise is swamping the feedback signal.
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#2 Reply
Posted by
newtekuser
on 07 Mar, 2024 06:41
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Our crystal balls have gone cloudy, and its too much trouble to hack your CCTV so we've got no idea whatsoever what you are doing wrong!
So far it appears you *may* be using a TI switcher chip, and have a problem with unstable output voltage. Without a full schematic, your PCB layout and closeup photos of the assembled board, we simply cant help significantly, though my first suspicion is you may have FUBARed the layout and noise is swamping the feedback signal.
Attached. I could not fit the 220uH inductor on the existing pad so I Frankenstein'ed it
The booster IC is an MC34063AD
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#3 Reply
Posted by
Ian.M
on 07 Mar, 2024 07:20
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It looks like you've got a massive loop area for the input cap => current sense => inductor => IC switching transistor => ground, and when the transistor cuts off, its extension from the inductor => diode => output cap, with the feedback divider in the middle of it, thus inductively coupled to it.
The capacitors should be right next to R5 and the diode, with their ground legs towards the pin 1,8 end of the IC to minimise the loop area. If you've got a full ground plane, you may be able to fix it by drilling the board next to the cathode of the diode and the input side of R5 and bodging in more decoupling. Also try reducing the divider resistors by an order of magnitude to reduce the impedance at the feedback pin. It may also help to flip the inductor over so its leads cross to reduce the loop area.
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From what I understand the formula to calculate the input current is In = (Iout * Vout)/Vin = 0.96A
You forgot to include efficiency. Only under ideal circumstances is the above true. The chip you choose was already discussed
here, as being not so good due to the use of BJT's for the switching.
The efficiency found in the datasheet is at best 89% for a step up converter and at worst 58% for an inverting converter shown in the different test cases. So assume 80% which already brings up the input current to 1.2A.
The datasheet also shows sample PCB layouts which are wise to follow.
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#5 Reply
Posted by
Berni
on 07 Mar, 2024 07:55
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There is so much wrong with the PCB layout indeed. That diode also looks undersized and doesn't have any copper area around it to aid in cooling.
The chip used is also a old obsolete switchmode chip that has poor efficiency.
For these kinds of things it is best to just buy ready made modules from Aliexpress. They cost less than what it would cost you to even just buy the components, let alone make a PCB and cost of your time to assemble it. The efficiency of those cheap modules is not quite world changing (since they are built down to a price), but it is good enough to give the power levels the listing says it does. You can still pimp it out for higher efficiency by giving it a better low mOhm inductor or better diode.
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#6 Reply
Posted by
Ian.M
on 07 Mar, 2024 08:10
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For these kinds of things it is best to just buy ready made modules from Aliexpress. They cost less than what it would cost you to even just buy the components, let alone make a PCB and cost of your time to assemble it. The efficiency of those cheap modules is not quite world changing (since they are built down to a price), but it is good enough to give the power levels the listing says it does. You can still pimp it out for higher efficiency by giving it a better low mOhm inductor or better diode.
That very much depends on the module. Typically smaller Chinese DC-DC modules are good for a continuous load of under 50% of their nominal current or power ratings, and maybe 90% of their max. voltage rating. Try to push them anywhere near their ratings and they overheat within seconds.
'LM2596' modules are a special case - most Chinese market LM2596 ICs are fakes, and seem to be clones of the older LM2576 IC. Unfortunately that operates at approx. 1/3 the switching frequency so if the design equations in the LM2596 datasheet are used (which is typically the case), the result is the inductor and decoupling caps are only one third of the value they should be. The current must thus be derated by a factor of three, (in addition to the 50% 'Chinese' derating) so a '3A' module is typically only good for 500mA.
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#7 Reply
Posted by
Berni
on 07 Mar, 2024 08:16
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I have bought a bunch of difference cheap Aliexpess DC/DC boards and they mostly seamed to have domestically designed Chinese chips on them.
Maybe i was lucky but they did perform to what the listing says they do. That being said running a 3A rated module at 3A continuously will make it get ridiculously hot, the efficiency wont be great..etc but it does work. But when you run them at 50% load they run at a reasonable temperature. The PCB designs of the modules also looked fairly decent with minimized loop area and such.
But yeah Aliexpress is a bit of a lottery pick at times, you do run into crappy fakes here and there. I just make sure to test stuff i get from there before using it.
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#8 Reply
Posted by
Ian.M
on 07 Mar, 2024 08:20
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I'd *rather* have a Chinese IC with a Chinese language datasheet, on any such module because it improves the odds that the design engineer understood the chip's design equations and ratings properly!
If you've got the fake 'LM2596' modules and decent rework skills, it may well be worth fitting a genuine LM2596, which should get you up above 1A continuous, as you cant get the board and other components for the price of the module. However the diode is typically under-specced with insufficient trace area to handle its dissipation, and the inductor is likely to have a higher DC resistance and lower saturation current than desirable, so odds of getting anywhere near 3A are poor, even with forced air cooling.
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A module like
this one is certainly enough for the OP's purpose. According to the
datasheet it can switch at max 4A.
I have several of these modules but not used them with high loads in boost mode yet, but for buck purposes I have used the XL40xx versions and they are working fine on average load. I use one in my CV system for powering high torque brushless 6V servos from a 24V supply. Have not measured the currents but the module is lasting for several winter seasons without bringing up the temperature in the box up to high. ~32 degrees Celsius after a full day of operation.
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#10 Reply
Posted by
newtekuser
on 07 Mar, 2024 15:42
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It looks like you've got a massive loop area for the input cap => current sense => inductor => IC switching transistor => ground, and when the transistor cuts off, its extension from the inductor => diode => output cap, with the feedback divider in the middle of it, thus inductively coupled to it.
The capacitors should be right next to R5 and the diode, with their ground legs towards the pin 1,8 end of the IC to minimise the loop area. If you've got a full ground plane, you may be able to fix it by drilling the board next to the cathode of the diode and the input side of R5 and bodging in more decoupling. Also try reducing the divider resistors by an order of magnitude to reduce the impedance at the feedback pin. It may also help to flip the inductor over so its leads cross to reduce the loop area.
Thanks for the suggestions! I am using a full ground plane on the bottom layer btw.
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It is not specified in the datasheet, but it might help to put a 100nF capacitor between pin 6 and ground as close as possible to pin 6. Might be that the ic is oscillating on the supply at high frequencies due to not being properly decoupled.
I have seen this solve problems multiple times with different types of circuits.
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#12 Reply
Posted by
newtekuser
on 08 Mar, 2024 02:14
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It is not specified in the datasheet, but it might help to put a 100nF capacitor between pin 6 and ground as close as possible to pin 6. Might be that the ic is oscillating on the supply at high frequencies due to not being properly decoupled.
I have seen this solve problems multiple times with different types of circuits.
Thanks, I'll try that! Btw, I put the board under a thermal camera and the only component that gets hot is the IC.
I will re-route traces to put components closer to each other and also downsize the divider resistors as Ian.M suggested. If that doesn't work still, I'll try a different IC that is more efficient.
My understanding of loops is that at least I should have no ground loops in my design because I'm using a solid ground plane and the return path for GND is "instantaneous" since the path between the component and the ground is direct.
Based on this my options are either going four layer board and route power via a dedicated power plane, or as suggested, minimize the loop by re-arranging components closer together. Is this correct?
The reason I'm not going with a device from Aliexpress is that I want to integrate this into a bigger project of mine that I'm designing from scratch and do not wish to use pre-made parts for it.
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#13 Reply
Posted by
Ian.M
on 08 Mar, 2024 02:40
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Before giving up on this revision, drill the board (not on any track) next to the outer ends of R5 and D1 as I suggested above and next to pin 6 as Pcprogrammer suggested to get access to the ground plane and patch in 100nF thru-hole ceramic caps for more and better placed decoupling to see if it helps. Also, try to lower the divider resistances. If you don't have the exact values, do what you can and see if it then holds the correct output voltage for the new divider. With a 1K R2, R3 of 8K2 would be 0.5V low, 9K1 would be 0.63V high and putting the just removed 10K on top of the existing R3 in parallel would be 0.47V high, plenty close enough for a 12V fan.
If my suggestions help, its then a good idea to respin the board to get the inductor on and improve the layout with fair confidence it will work as expected, though I'd still place pads for the extra decoupling so you can add them again if it doesn't behave with just the electrolytics.
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#14 Reply
Posted by
ArdWar
on 08 Mar, 2024 03:33
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The IC getting hot isn't unexpected, it isn't exactly efficient switcher and not particularly good at heat dissipation either (~90C/W is surprisingly bad tbh).
The "loop problem" isn't (necessarily) about ground loop. Your switching loop is uncomfortably large and "loose". For boost converter, the critical loop is between switch transistor, output diode and output capacitor. at least try to make that loop as small and tight as possible. Granted your IC isn't exactly the fastest switcher so layout isn't as critical, but this is a good practice nonetheless.
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#15 Reply
Posted by
newtekuser
on 08 Mar, 2024 03:57
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Before giving up on this revision, drill the board (not on any track) next to the outer ends of R5 and D1 as I suggested above and next to pin 6 as Pcprogrammer suggested to get access to the ground plane and patch in 100nF thru-hole ceramic caps for more and better placed decoupling to see if it helps. Also, try to lower the divider resistances. If you don't have the exact values, do what you can and see if it then holds the correct output voltage for the new divider. With a 1K R2, R3 of 8K2 would be 0.5V low, 9K1 would be 0.63V high and putting the just removed 10K on top of the existing R3 in parallel would be 0.47V high, plenty close enough for a 12V fan.
If my suggestions help, its then a good idea to respin the board to get the inductor on and improve the layout with fair confidence it will work as expected, though I'd still place pads for the extra decoupling so you can add them again if it doesn't behave with just the electrolytics.
You were not kidding, I thought it was a figure of speech for actually adding the cap to the revised board
I’ll drill it, sure!
As for the resistors I was thinking a 5.1K and a 44.2k for 12.08v Would these do?
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#16 Reply
Posted by
newtekuser
on 08 Mar, 2024 04:00
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The IC getting hot isn't unexpected, it isn't exactly efficient switcher and not particularly good at heat dissipation either (~90C/W is surprisingly bad tbh).
The "loop problem" isn't (necessarily) about ground loop. Your switching loop is uncomfortably large and "loose". For boost converter, the critical loop is between switch transistor, output diode and output capacitor. at least try to make that loop as small and tight as possible. Granted your IC isn't exactly the fastest switcher so layout isn't as critical, but this is a good practice nonetheless.
(Attachment Link)
Thank you very much for the explanation on the switching loop! Next time I’ll respin the board I’ll correct that.
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#17 Reply
Posted by
Ian.M
on 08 Mar, 2024 04:14
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That's only a factor of two lower divider resistance, which may not be enough to swamp the noise we suspect is messing up the feedback. Go with my suggestion of ten times less, and if it improves stability, you can then get the quiescent current back down by removing R4, the 10K load.
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#18 Reply
Posted by
MrAl
on 08 Mar, 2024 07:08
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Hello to all,
This has already been said, but I'd like to stress the efficiency issue a bit more.
With an ideal converter, the input and output power relationship is simply:
Pin=Pout
and in terms of voltages and currents:
Vin*Iin=Vout*Iout
However, no converter is ideal which would mean 100 percent efficient. Instead, there is an efficiency factor that is always less than 1.
For a real converter, the power relationship changes to (Eff is the efficiency factor less than 1):
Pin=Pout/Eff
and since Eff is always less than 1, the power input is always greater than the power output.
In terms of voltages and currents again this would be:
Vin*Iin=Vout*Iout/Eff
Unfortunately, the efficiency Eff can be kind of low. To start, 80 percent, which means the efficiency factor Eff would be 0.80, and that means the input power will be a lot more than output power. This is easy to calculate too.
Let's start with an input of 5 volts and an output of 10 volts at 1 amp.
Since the input and output voltages are considered constant, and the output current also constant, the power relationship can then be written as:
5*Iin=10*1 ( where again 5v input, 10v output at 1 amp output).
That would make Iin equal to 2 amps in an ideal converter. Considering the efficiency factor Eff this would become:
5*Iin=10*1/0.80
or simply:
5*Iin=12.5
and so now that means Iin is higher than before by 25 percent:
5*2.5=12.5
so now the input current Iin is equal to 2.5 amps when in the ideal converter it was only 2.0 amps.
Now let's drop the efficiency a little more, down to 70 percent. We then have:
5*Iin=10/0.7
or:
5*Iin=14.29
and this means the input current went up to about 2.86 amps.
We can start to see how this messes up the original calculation with an ideal converter, and how much more current that we might see on the input than in the ideal case. Thus, the efficiency always has to be taken into account.
With 5 volts input and 12 volts output and 1 amp output and 75 percent efficiency the input current would be:
Iin=3.2 amps.
With only 0.5 amps out the input current would be:
Iin=1.6 amps.
With only 0.3 amps out the input current would be:
Iin=0.96 amps.
All of these Iin figures are the average current levels though. The peak levels depend on the inductor value and the pulse width.
If the inductor value is too low this will mean higher peak currents and that could mess things up also. If that is the case, the inductor value has to be increased. Because the inductor value is in the denominator of the current calculation, a larger inductor reduces the peak current. Because the pulse width is in the numerator, the shorter the pulse time the lower the peak current. Unfortunately, it is possible that this chip can skip a pulse which means the next pulse might be longer than usual. This would cause the inductor current to go up higher than usual, so the inductor value gets more critical.
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dont expect anywhere near 100% efficiency for boost, think of current wasted to ground. my rule of thumb for boost is 50% efficiency. so to drive 12V 0.3A (4W) will need 8W from input ie with 5Vin equates about 1.6A thats already exceeding the top of IC spec, not really a good idea as well. next thing is check your inductor value, too small and it will into saturation quickly and IC will smoke. oscilloscope and a little bit experience to see switching node is golden at this stage. i tend to go with larger than recommended inductance, less stress to the IC, although ripple and noises may suffer, but its for later tuning as cooler switching IC is much more critical imho. this is what a lazy math stuff laymen guys like me will do fwiw...
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#20 Reply
Posted by
newtekuser
on 08 Mar, 2024 18:39
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Thank you all for taking the time to explain this! I’m not writing this IC just yet as I only need 12V briefly (less than 30 seconds) but are there any drop in replacements that are more efficient?
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#21 Reply
Posted by
newtekuser
on 09 Mar, 2024 02:54
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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.
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#22 Reply
Posted by
mariush
on 09 Mar, 2024 05:37
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The maximum switch current of a MC34063 is 1.5A ... with 5v input, 12v output you're gonna exceed that with more than 250mA output current.
You can use any online calculator and it will tell you that, for example try
http://www.nomad.ee/micros/mc34063a/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.
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#23 Reply
Posted by
newtekuser
on 09 Mar, 2024 06:03
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The maximum switch current of a MC34063 is 1.5A ... with 5v input, 12v output you're gonna exceed that with more than 250mA output current.
You can use any online calculator and it will tell you that, for example try http://www.nomad.ee/micros/mc34063a/
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.
I looked at those but they have a min input of 7.5v and my supply voltage is 5v.
Thanks for sharing the calculator btw!
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#24 Reply
Posted by
mariush
on 09 Mar, 2024 06:29
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