0
\$\begingroup\$

In the below pictured circuit for the MC33063 by TI, how important is the capacitance of the inpu-output caps? I understand why they're there, I don't know what the recommended capacitance is derived from. I am using the QFN package.
The reason I ask is that I have very little space left on my board and the caps tend to get relatively large, even at 16V, not to mention more. I think I might be able to cram a 100uF/6.3V on the output in a B case, and a 100uF/16V C case on the input, both tantalum, but I doubt they'll be low ESR.
Possibly if the capacitance was slightly lower, I could either use a smaller case or lower ESR. Real input voltage will be 12-14V.
Would that be possible?
Also, what is the smallest package available for the diode?

1

Update: This is what I've come up with so far. If I can find all three in B case, given that the output will be a pair of 220uF/6.3V, than I can fit them.
Please don't mind the dirty schematic with incorrect values for 4.1 output etc.

1

|improve this question
\$\endgroup\$
2
\$\begingroup\$

The input and output caps are very important .The buck convertor like many others draws current in lumps ,basicly when the switching transistor inside your chip is off no significant current is drawn and current is drawn when the switch is on .The current waveform is at the switching frequency and is in a trapizoidal shape rich in harmonics and probably will have a spike on the leading edge ! SO the input cap is selected for low ESR at the switching frequency to encourage the HF crud to flow through the cap and NOT radiate along the input cable .This is why you often see in inductor in the input power wire .Most importantly the cap must be able to handle the ripple current for your product to have a long life .Generaly its easy to substitute a ceremic SMD cap here.On the output the cap must be low enough impedence to keep ripple to an acceptable level for your load AND it must handle ripple current ,not as much as the input on your orthodox chip,AND its part of the control loop in fact its ESR effects the bode plot .If you change the output cap you must check for instability.Have you considered a higher frequency approach that has no chip that reduces switching losses that will run on ceremic caps all round ? Maybe the LF chip isnt so small due to big caps and coil .

|improve this answer
\$\endgroup\$
  • 1
    \$\begingroup\$ Yes, I have made a circuit (4L pcb, no breadboard) with the LT8609 and it is very finicky. Different outputs than I wanted, the waveform was everywhere, etc. Not saying the chip is bad, but I guess the design has to be much more pro. Thus I chose the wellknown 33063 used in chargers etc, cloned by china with horrible parts, hoping that the chip isn't as prone to problems due to cap type etc. I don't want to botch the design, but I have very little board space. I don't mind spending a bit more on better caps, as long as I can make them fit. The price of the 33063 is like 1/10 of the LT anyways. \$\endgroup\$ – Fid Sep 22 '15 at 21:22
  • \$\begingroup\$ Fid you are in a PCB realestate crisis ! one option is to use the "Roman Black ' Buck convertor .It actualy has Eastern origions .Its not reliable but its only got two transistors in its published form .You could customise it for your application and have something reliable and small,but remember there are still capacitive needs on input and output. \$\endgroup\$ – Autistic Sep 26 '15 at 8:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.