3
\$\begingroup\$

I'm just getting somewhat familiar with voltage regulators and am trying to make sure I'm understanding a bunch of parameters correctly.

Scenario:

I have a project that spends 1/3 of its time drawing 20uA, 1/3 drawing 40mA and 1/3 drawing 150mA. The project is powered by a 3 cell LiPo battery, meaning input voltages ranging between 9 and 12.4V. I'm trying to decide whether I should power the whole thing off a single LDO, a single switcher, or try to separate out the parts that push consumption to 150mA, and use an LDO for the low power part and the switcher as necessary for the high power.

I'm looking at an LDO with the following quiescent current (listed as ground current in the datasheet)

  1mA   @ 150mA output current
400uA @ 50mA output current
100uA @ 1mA output current

I'm also looking at a switching regulator with the following efficiency:

90% @ 150mA
90% @ 50mA
85% @ 1mA
60% @ 100uA

If I'm understanding the comparison correctly, then the LDO will pass along (output current + the ground current) * output voltage and waste (output current + the ground current) * voltage difference, which reduces to just input voltage * (output current + ground current), meaning:

11.1V * (100uA + 20uA) = 13.2mW in the first state, and so on (the 11.1 is the nominal voltage of the battery, so it'll spend a fair bit of time there)

For the switcher, the total power consumption as I understand it is Vout * Iout / Eff, so at that first state it would be 3.3V * 20uA / .6 = 1.1mW. My understanding was that for switchers, the quiescent current is factored into the efficiency graph you're given in the data sheet, hence steep fall-off at the start.

Is this right?

\$\endgroup\$
  • \$\begingroup\$ It might be worth pointing out that, since your load seems to be a consistent 3.3 volts, you don't need an LDO at all. Almost any linear regulator will do, since the regulator will drop a minimum of 5.7 volts, and almost anything will do that. From that fact alone, it's clear that a switcher will be much more efficient than a linear, since the efficiency of a linear will be at most 3.3 / 9, or 37%, with an "expected most of the time" efficiency of 3.3 / 11.1, or 30% (ignoring quiescent current, which is reasonable as a starting point). \$\endgroup\$ – WhatRoughBeast Jul 2 '14 at 4:28
  • \$\begingroup\$ 2nd point i have doubt...for switcher...if input is 12V, output is 3.3V, Iout = 100uA and Iquiscent = 20uA..Then if efficiency = 60% ....input power consumption due to load current is 3.3v*100uA/0.6...input power consumption due to quiscent current of IC = 12V * 20uA (Vin_max*Iq).....then u need to sum both....Quiscent is not the converted or switched current to apply efficiency to it....correct me if i am wrong.. \$\endgroup\$ – user19579 Jul 2 '14 at 6:43
3
\$\begingroup\$

Yes, you have it correct in the two equations for total power consumed from the battery.

There's typicals and maximums which are a bit higher than the typicals for Iq and a bit lower for efficiency, of course, and the efficiency could vary if you use a different inductor or other conditions vary.

\$\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.