0
\$\begingroup\$

I am going through Texas instrument web bench design suit and get 40% efficiency for some booster circuits

if booster can handle the load (Vout and Iout) according to design given by TI what is the disadvantage of having 40% efficiency over 70% or 85% efficiency design?

is the booster unstable?

will bolster fail at some time (within calculated load range can there be a voltage drop after some time?)

will booster waste energy as heat?

Energy loss and heating is OK for my application.

\$\endgroup\$
  • \$\begingroup\$ "I don't bother energy waste and heating is" Do you mean energy loss and heating is OK, or do you mean it is intolerable? \$\endgroup\$ – TimWescott Nov 21 '19 at 15:41
  • \$\begingroup\$ @TimWescott I meant energy loss and heating is OK for my application \$\endgroup\$ – oppo Nov 21 '19 at 15:42
4
\$\begingroup\$

if booster can handle the load (Vout and Iout) according to design given by TI what is the disadvantage of having 40% efficiency over 70% or 85% efficiency design?

It'll get hot, and waste heat.

will booster waste energy as heat?

Yes. You can figure this out using the first law of thermodynamics: you have a system into which electrical energy is going, and out of which is coming electrical energy. There's 2.5 times as much electrical energy going in as coming out -- that means the other 1.5 times is coming out as heat, because there's really no other significant energy-shedding mechanism.

is the booster unstable?

Not necessarily. Efficiency and stability aren't necessarily related.

will bolster fail at some time (within calculated load range can there be a voltage drop after some time?)

That depends. For every watt you put in, you'll get out 400mW, and 600mW will get burnt up as heat. If your booster's thermal design (meaning -- if your circuit isn't big enough) isn't up to shedding that heat, it'll get too hot and burn up.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ 40% efficiency is terrible; unless the boost circuit is consuming a small fraction of the overall board power its size will be dictated by its efficiency. You may find that even though you don't care about the overall efficiency or board heating, you do care about size you need to make the boost circuit just for thermal considerations. If that ends up being the case ask a separate question spelling out your input voltage, output voltage, and power requirements, and asking what can be done. I suspect you're looking at simple boost converters where you need multiple stages, or flyback. \$\endgroup\$ – TimWescott Nov 21 '19 at 16:03
  • \$\begingroup\$ its single stage boost converter 70V 20mA output with powered from a single-cell lipo battery, when I reduce the output current to 10mA @ 70V , from design I got 40% efficiency, at 70V 20mA it is 79% \$\endgroup\$ – oppo Nov 21 '19 at 16:08
  • \$\begingroup\$ I'm not sure I believe those numbers -- I suspect that you've gone outside of the parameters that the web bench can handle easily and it's giving you garbage. This is based on the fact that, first, it's showing the same power in for two different power outputs, and second, I find 80% efficiency in a simple 3.7 to 70V boost hard to believe (unless it's a flyback and you're not telling us). Will the chip or pass transistor that you're contemplating safely dissipate 1W? \$\endgroup\$ – TimWescott Nov 21 '19 at 16:22
  • \$\begingroup\$ TPS61390 gives 70V 20mA 79% efficiency on the web bench. also, my load won't take 20mA it will consume around 1mA - 5mA, that's why I reduce load current and checked efficiency went down to 40% \$\endgroup\$ – oppo Nov 21 '19 at 16:30
  • \$\begingroup\$ 70V x 1mA = 70mW. At 40% efficiency the input power is 175mW and booster loss is 105mW. The TPS61390 can easily handle this, and the efficiency is well within its normal operating range. So if you don't care about battery drain it's not a problem. \$\endgroup\$ – Bruce Abbott Nov 21 '19 at 17:00
1
\$\begingroup\$

The main disadvantage of low efficiency is energy loss. If you don't care about that, the booster can handle it, and the power supply can deliver the extra power required, then it's not important - to you.

All boosters have low efficiency at low power because they consume a fixed amount of 'quiescent' power which is not proportional to output power. Operating the booster in this region does not necessarily make it unstable, but it may have higher ripple currents and increased EMI due to running in 'discontinuous' mode.

A booster with low efficiency may also produce more EMI due to the higher current it requires. A 40% efficient booster needs double the current of an 80% efficient booster, and this current is drawn in pulses which tend to produce EMI. The higher current may also make the power supply voltage drop, causing the booster to draw even more current.

The energy consumed by the booster is wasted (mostly) as heat. If it can handle this then by definition it will not fail. However if running at high power it will get hotter than a more efficient booster, and so may have lower reliability and service life. You may not care about the waste heat, but the booster could care about the temperature rise it causes.

| improve this answer | |
\$\endgroup\$

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.