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I'm having a lot of trouble trying to design a boost converter using the TI MC3306 IC and I am wondering if I mathed this out correctly.

The chip gets extremly hot to touch and the voltage sags.

The input is 3.5v. However, when I add any sort of load of around 0.8-1 amps reported on my power supply, the IC gets really hot, the voltage sags to about 8V. Smaller loads sag less and dont heat up as much. I have calculated the values for 1.5A maximum draw (the chips max rating)

I have tried a few different inductor sizes, all with similar results, 1uh, 3uh, 10uh and 20uh, none of which have solved the problem. The inductor stays cool though. Orignal calculated result is 23uH.

Our load pulls 0.5A straight from the PSU, when attached to the boost, it pulls between 0.8-1.3A. Something in the circut is causing a large inefficiency. I am not sure why this is happening. I had another engineer take a look and they are also not sure. This is my last hope! Thanks greatly for your help. Please let me know if I'm missing info. :)

Required Specs: Vin:3-4V Vout: 11V Max I: 1.5A F: 100KHZ Inductor I saturation: 2.2A

IC Datasheet: http://www.ti.com/lit/ds/symlink/mc33063a.pdf?HQS=TI-null-null-digikeymode-df-pf-null-wwe&ts=1590606565705

Pics: 1. Schematic, 2. No load, 3. With load (1.2A reported by PSU)

schematic With No Load With Load (1.2A reported by PSU

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    \$\begingroup\$ There's a lot here to address. To deliver an average 1.5 A at 11 V, you have to draw an average of about 5 A at 3.5 V. Have you bothered to read this? Would you please go to section 4.3 there and work out your peak inductor current? I'm curious how it compares to the maximum your inductor can handle. And even then, I'm curious how all this works out with respect to your sole use of the internal switch. I'm pretty sure the device is working fine and protecting itself as it should. \$\endgroup\$
    – jonk
    Jun 1, 2020 at 22:32
  • \$\begingroup\$ Did you start your design using the TI Webench? I suspect not because that should have given you a decent starting point. \$\endgroup\$
    – user16324
    Jun 1, 2020 at 22:36

2 Answers 2

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Main & fatal problem: You are trying to use the IC at vastly above its rated power level.

The internal switch 'just can't do it' and power dissipation would be immense.

The datasheet shows switch current of 1.5 A absolute maximum.
At 1A the switch (a Darlington pair) saturates at 1 to 1.3 V.
On a 3.5V supply the switch loss with 1.3V drop will be Vsw/Vcc = 1.3/3.5 = 37%, and total losses will be higher.
At 11V, 1.5A out Pout = 16.5W so loss in the switch (if it was capable of the load, which it isn't) would be about 0.4/0.6 x 16.5W = 11 Watts. Which would not be viable.

This is a very olde IC indeed. (I have used several hundred thousand of them :-) ).
If you MUST use this IC then using it to drive an external transistor - probably a FET, would make it work.

Secondary (but fatal) problem: The inductor saturation current is too low.

If efficiency (with an external transistor) was say 80% then
Power_in = Power_out / efficiency
= 16.5W/0.80 = 20.6W ~= 20 Watts.
Iin average = P/V = 20/3.5 = 5.7A.
I peak is "higher again" due to current in the inductor being limited to the on period.
A current rating of closer to 10A would be wise.

___________________________________________

From long long ago and far away (unless you live in China) - this will do a better job for you. I cut and pasted this from a larger circuit and removed extra functionality that you do not need. The NPN/PNP pair provide gate drive to the FET. The transistors used (or their smd equivalents) are very good for many purposes but many other transistors could be used here.

Use a FET to suit your application - the CES 2310 is a very nice part but under rated for your power level. See comment below for where this circuit was used. Far better and more modern ICs are available if cost is not a major driver.

enter image description here

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  • \$\begingroup\$ That smps was used in some of these (roll don't click) - cost was a factor - the IC cost a few cents in China in high volume. \$\endgroup\$
    – Russell McMahon
    Jun 2, 2020 at 11:57
  • \$\begingroup\$ Thank you so much for the detailed explanation! I definitely learned a lot here :) \$\endgroup\$
    – Michael
    Jun 4, 2020 at 21:01
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If you have 11V out with 1.5A load current, that's 16.5W. Allowing for reasonable efficiency of say 80% that's 20.6W of input power.

So with 3V in, you will draw 6.88 A from the input supply, and that will be the average inductor current. The peak inductor current will be higher by 1/2 of the ripple current. (In a boost, the inductor carries the input current.)

So an inductor with a 2.2A saturation rating is well undersized for your application.

If you had 30% ripple current, then the average inductor current could be 1.87A before saturation. This would give you 5.61W of input power capability at Vin minimum of 3V, and 400mA of output current at 11V Vout. (80% efficiency)

You can probably get more out of your converter if the inductor soft-saturates, but as you noted things will start to get warm....

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    \$\begingroup\$ This puppy is designed to operate discontinuous, from what I gather. With the internal switch drop lowering the voltage across the inductor as much as it does, the peak inductor current is pretty high. Well over 10 A, I think. \$\endgroup\$
    – jonk
    Jun 1, 2020 at 22:39
  • \$\begingroup\$ @jonk At 100kHz with 23uH it should be continuous for input current > about 1A. I think, I ran the numbers pretty quickly. I don't think there's anything in the MC33063 that forces a discontinuous design, but it's an ancient part that I probably used last about 15 years ago, so I may be wrong. It is a bipolar switch, so 100kHz is pushing the limits. \$\endgroup\$
    – John D
    Jun 1, 2020 at 22:50
  • \$\begingroup\$ If you read the BOOST discussion here, it presumes discontinuous mode and they do NOT present a continuous mode design so far as I can see. (See section 4.2 and Figure 4.9, for example.) I take that as suggesting a recommended mode for this part. But as you say, I may be also not spending enough time to work out details. \$\endgroup\$
    – jonk
    Jun 1, 2020 at 22:52
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    \$\begingroup\$ I think the OP should design according to this application note. I'm not at all surprised at the OP's experiences, though. There's an eval board that retails at about $60 for it. The OP should review that schematic for the eval, too, if not actually buy the board. Plus, take careful consideration of operating it either Darlington or not and using an external BJT. The inductor will have to be different, too. If they expect that much current at the output. All in all, it doesn't look as though they've done their due diligence. \$\endgroup\$
    – jonk
    Jun 1, 2020 at 23:02
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    \$\begingroup\$ Plus the 1.5A abs max switch is 'a little sad' at 7A+ :-) \$\endgroup\$
    – Russell McMahon
    Jun 2, 2020 at 11:29

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