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I am using a camera module of our custom application. The camera module started consuming more current when compared to a previous board which has all the same settings, chipsets and modules.

In our conversation with a support engineer, we got this answer:

VCAMD power supply in previous board is driven by a 1.27V DCDC, in present board it is driven by LDO. In a dark environment, DCDC will save about 14mA and in a light environment, DCDC will save about 25mA. So test results of both are different.

How can using a DCDC save power, the module would consume the power that it needs?

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It's not that a DCDC (Buck Regulator) saves power, it's that an LDO wastes power.

In effect a buck regulator converts the voltage difference to more available current.
An LDO converts the voltage difference to heat, and heat is a waste product you don't really want.

An LDO regulating, say, 12V down to 5V has to drop 7V and dissipate that power as heat. The more current you draw the more heat is produced. If you draw 1A through that example (5W) it in turn draws 1A from the power source (12W), so it has to lose 7W of power to the atmosphere.

A perfect (they don't exist, but for illustration) buck regulator going from 12V to 5V, with 1A output (5W) would draw in turn 5W from the power source, which at 12V would be 417mA.

Of course, as I say, perfect buck regulators don't exist and there are still losses, so it would actually draw slightly more from the source - more like maybe 6W, or 500mA. Still considerably less than an LDO.

There are down sides to buck regulators though:

  • The are noisy. They work by rapidly switching the power on and off, and that makes for higher radiated and conducted emissions.
  • They are harder to lay out. To keep the EMI emissions low and get them to pass compliance testing, careful layout on the PCB has to be considered.
  • They use more components. An LDO typically is one chip and a couple of capacitors. Buck regulators also need (normally) at least a diode and an inductor as well.

All that adds up to buck regulators being more expensive than LDOs.

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  • 2
    \$\begingroup\$ superb ! thank you for the lines "It's not that a DCDC (Buck Regulator) saves power, it's that an LDO wastes power." \$\endgroup\$ – kakeh Mar 17 '15 at 9:26
  • \$\begingroup\$ Why does mount wasted depend on how light the environment is? \$\endgroup\$ – sharptooth Mar 17 '15 at 10:19
  • \$\begingroup\$ I guess in a dark environment the amplifiers in the system have to work harder (higher gain) to bring the exposure level up? \$\endgroup\$ – Majenko Mar 17 '15 at 11:36
  • \$\begingroup\$ It should be noted that there exist drop in replacements for LDOs that are bucks in a package. Given the small amounts of energy saved by using one vs LDO I would speculate that this could be such a case here. \$\endgroup\$ – PlasmaHH Mar 17 '15 at 13:38
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LDO is an irrelevant term here - the key differences are "Linear" versus "Switched".
LDO means that Vin-Vout can be very small if desired BUT efficiency is governed by what Vin IS, not what it can be worst case.

I'll use SMPS for "Switched mode power supply" and LPS for Linear Power Supply.

For a LPS
Iin = Iout. Vout is set to suit the load.
So Power in = Vin x Iin = Vin x Iout
Power Out = Vout x Iout
So efficiency
= Power out/Power in
= (Vout/Iout) / (Vin/Iout)
Efficiency = Vout / Vin
So, for a fixed Vout, efficiency falls linearly as Vin rises

For a SMPS
Power_out = Power_in x Z
Z is the conversion efficiency and varies with converter type, voltages in and out and their ratio, power levels and more. But as a guide: A non isolated "buck" down-converter can achieve:

  • 98% super space rated no expense barred ultra optimised + luck

  • 95% industry best practice, careful design and manufacture, probably over limited range

  • 90%+ Good design in best part of range
  • 80% - 90% Most designs across much of range. Not usually at very high or very low loads or high voltage ratios
  • < 80% Extreme conditions, end of battery, very light or heavy load etc.

As a rule of thumb most SMPS will provide 85% to 95% efficiency in most cases.


So -

A LPS operating with Vin = 9V and Vout = 3.3V will be 3.3/9 = 37% efficient.
Except in ultra extreme cases any SMPS that cannot do better should be buried at a crossroads with a stake in its heart.

  • A system may use a nominal 9V PP3 Alkaline battery with
    Vnew of about 6 x 1.65V/cell = 9.9V (not for long) and
    Vdead of about 6 x 0.9V 5.4V -
    so the efficiency of a 3V3 linear regulator operating from this battery will vary widely with state of charge.

A LPS operating with Vin = 5V and Vout = 3.3V will be 3.3/5 = 66% efficient.
Most SMPS will be better except in extreme cases.

A LPS using a LiFePO4 battery operating from 3.5V down to 3.1V and operating an LED at 3.0V will be from 3/3.5 to 3.1/3.5 = ~= 86% to 97% efficient - ie efficiency rises as Vin approaches Vout.
In this case the average efficiency will be around 91% to 94% across much of the battery range.
Only the best SMPS buck regulators would have higher efficiencies and a linear regulator may well be a good choice here.

  • Note: I designed a range of products which operate one or more low Vf white LED(s) using a single LiFePO4 cell - guess what sort of regulator I used :-).
    [The LEDs used are carefully chosen to operate from 3V or less under all desired operating conditions. ]
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  • \$\begingroup\$ respect your effort in explaining in detail, thank you very much, i hope you have used an ldo in you case as the operating voltage is near to nominal cell voltage, using a dcdc would have costed you more, but for me it looks like dcdc with an LDO is a good design in case of my scenarios like an SOC \$\endgroup\$ – kakeh Mar 18 '15 at 4:29
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A regulator is used in situations where the load needs electrons with less energy (per electron) than those pushed out by the source. A linear regulator takes electrons from the source, wastes some energy from each, and then feeds them to the load. Every time an electron comes from the load, an electron has to come from the source.

A buck regulator passes electrons through an inductor which can take energy from some and give energy to others. It starts out initially behaving like a linear regulator--passing electrons through the inductor (which takes out some of their energy) and then the load--but once the inductor has stored up a little energy a switch disconnects the source and starts feeding in electrons which are returning from the load. Those electrons won't have enough energy to drive the load again, but since the inductor has just stored up some energy, the inductor can use its stored energy to re-energize those electrons and send them back through the load.

If one is e.g. starting with 10 volts and connects the inductor to the source 1/3 of the time and to the load return 2/3 of the time, then only about 1/3 of the current going through the load will have to come from the source. Since an inductor will add or remove flux at a rate proportional to the applied voltage (minus losses), and the average integrated flux needs to be zero, that implies the average voltage (minus losses) needs to be zero. Since the inductor will be connected to the load twice as much as it's connected to the supply, that will mean that it must drop twice as much voltage when connected to the supply as when connected to the load. Thus, the inductor will drop about 6.7 volts when connected to the supply and produce 3.3 when connected to the load.

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