Heat problem with a custom 20V boost converter

Thread Starter

jipristt

Joined Jun 4, 2017
20
Hello,

I designed a 14V-16V to 20V/2A boost converter on a printed circuit and after testing the PCB I get a lot of heat out of it.
This is the first time I am building this kind of circuit and dealing with heat issues, so I like to ask for some feedback and
advice.

Below you can see my circuit and some information about it:
1576255447072.png

The areas of heat is U7, Q10 and L4.
I've tested the circuit with 15V constant supply voltage and 1A output.
The System is meant to be connected to a 4s Li-Po battery charger circuit (max 1.5A) that is tested and verified.

One thing that I changed from the schematic is that instead of MTP3055VL I put an IRFP250MPbF because the order was delayed,
I couldn't wait and that was the closest to it that I had for a mosfet.

Regarding this circuit I have the following questions:
1) Is there something that I can change in the design to reduce the heat generation
2) What would be a proper heat sink for the heating components
3) Is it possible to add a heat sink on a toroidal coil and how you can deal with the generated heat

I also include a few pictures about the circuit that might be helpful.

IMG_20191213_175148.jpg IMG_20191213_175242.jpg

Any thoughts or comments are welcome.

Thanks!
 

SteveSh

Joined Nov 5, 2019
105
1) Is the MC33167 attached to any sort of heat sink?
2) For switching regulators, the big heat producers tend to be the switching transistor(s), Q10 in this case, and the magnetics. Do you have power dissipation estimates for those components?

Not knowing more details about your design make it hard to assess. But, the faster you can switch Q10, and the lower its Vds, the lower its dissipation.
 

pmd34

Joined Feb 22, 2014
518
Hi! Erm the circuit is a bit complicated with your choice of switching regulator, however.. if your R29 is really 680 Ohms, this could be the problem:
You need the FET to switch between off and on (and visa versa) very quickly.. the time it is in between these states it is behaving as a resistor.
The gate of a fet behaves a bit like a capacitor, so if you "charge" it though a resistor the voltage increases only "slowly".. so it switches on slowly.. it prevents noisy high frequency switching, but looses a lot of power in the fet during the transition.

So really you want to dump power into that gate quickly.. you could almost emit R29 in fact, but something like a 10 Ohm might be a good compromise. (Fast but not too fast - and noisy).

If you have a scope, have a look at what is happening at the gate and drain of the fet.. you want to see a nice clean quick switch on and off, with minimum of ringing.
 

ronsimpson

Joined Oct 7, 2019
910
I think you are copying this:
1576347402557.png
First thing you have 10uH not 190uH. The current will ramp up very fast.

Second: You are using a buck-boost when a boost only should do the job.
Ron Simpson
----edited----
A boost buck supply, stores power on the inductor. For about 50% you need to store 4.5A from the +16V then for 50% you need to move the power out of the inductor and into a +20V 2A supply.
A boost supply is like building a 4V 2A supply that sits on top of the 16V supply.
 
Last edited:

Thread Starter

jipristt

Joined Jun 4, 2017
20
Hi! Erm the circuit is a bit complicated with your choice of switching regulator, however.. if your R29 is really 680 Ohms, this could be the problem:
You need the FET to switch between off and on (and visa versa) very quickly.. the time it is in between these states it is behaving as a resistor.
The gate of a fet behaves a bit like a capacitor, so if you "charge" it though a resistor the voltage increases only "slowly".. so it switches on slowly.. it prevents noisy high frequency switching, but looses a lot of power in the fet during the transition.

So really you want to dump power into that gate quickly.. you could almost emit R29 in fact, but something like a 10 Ohm might be a good compromise. (Fast but not too fast - and noisy).

If you have a scope, have a look at what is happening at the gate and drain of the fet.. you want to see a nice clean quick switch on and off, with minimum of ringing.
pmd34 that was a good point! I replaced R29 with a 0 Ohm resistor and Q10 is not getting hot anymore. Thank you for that!

I think you are copying this:
View attachment 194520
First thing you have 10uH not 190uH. The current will ramp up very fast.

Second: You are using a buck-boost when a boost only should do the job.
Ron Simpson
----edited----
A boost buck supply, stores power on the inductor. For about 50% you need to store 4.5A from the +16V then for 50% you need to move the power out of the inductor and into a +20V 2A supply.
A boost supply is like building a 4V 2A supply that sits on top of the 16V supply.
Ron, indeed I used the reference design for that application but in order to get 20V/2A I had to change some values including the L. For that I used the formulas provided by the datasheet again and regarding L value I calculate it to be 10uH instead of 190uH. The formula is the one below:
1576500078932.png

However, after reading your post and checking again the datasheet I realized that I messed up with the delta I_L part. For my case I used delta I_L = 2 * I_Lavg and that was indeed taking too much current fast. I changed the coil to a 47uF one that I had laying around (~2A Ipp ripple) and head is reduced significantly. A 190uH coil will give round 0.5A Ipp ripple which is still an acceptable value so I'm ordering some parts and going to try it soon!

Finally, Ron I really liked your suggestion about the boost supply and I think I'll be using it for my next designs! ;)

Thank you guys for your help!
 

ronsimpson

Joined Oct 7, 2019
910
about the boost supply
You will need a different IC for a boost.
One thing about boost, it does not have good short circuit protection.
Example: Vin=16, Vout=20, When you pull too much power the 4V supply will shutdown and stop but the underlying 16V supply will still supply power.
 
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