I have a basic question about measuring input current. The buck regulator I have chosen is rated to deliver 3.3V/2A from +12V input. My question is -

Is the input current 2A (load current) + few mA (needs for operation)?

Or because of stepdown from 12V to 3.3V, based on load current, the input current is 500mA + few mA?

Is the 2A output current taken entirely from the input or step down conversion delivers 2A after consuming few mA.


4 Answers 4


It's closer to your second option, but not quite.

A buck regulator will take a few mA for its operation, which could be considered to be constant. Say 15mA at 12V. Let's ignore that for the moment.

If perfect, it would output the same power as input. Your output of 3.3V at 2A is 6.6W, so a perfect buck regulator would need 550mA from the 12V supply (6.6W/12V). (Image from the Wikipedia page)

enter image description here

To see this, consider that a perfect switch has no losses, a perfect diode either conducts like a wire (no power is lost) or blocks perfectly, and a perfect inductor with zero resistance and zero eddy current losses dissipates no power (it simply stores energy and barfs it up again). So from conservation of energy, input power (averaged over a cycle in steady state) must equal output power.

In a real buck regulator implementation, there are, however, losses that are at least roughly proportional to the output power due to resistance in the inductor and switch, voltage drop in the diode and so on. Say the efficiency is 80%. It will then require 6.6W/0.80 = 7.5W input power, or 625mA, plus the 15mA it needs just sitting there, so 640mA total (usually the quiescent current will be folded into the overall efficiency figure so we don't need to add it).

Because of the 15mA (or whatever it is for your real regulator) the efficiency of the regulator (output power divided by input power) will decrease for lower output currents.


I just to add to the other answers, which cover the basics fairly well, but ignore the input current waveform, which can be important in some applications.

The current is drawn from the 12V supply in pulses, and the average current during the pulse is indeed the same as (or greater than) the load current. The duty cycle of the pulses is about 3.3V/12V = 27.5%, so the overall average current (ignoring the efficiency issues) is on the order of 2A × 27.5% = 0.55A.

Some power sources, such as solar panels, are current-limited; they cannot supply more than their short-circuit current under any circumstances, even for short periods of time. For example, suppose you wanted to power your regulator from a 10W, 12V solar panel. Such a panel will supply up to about 0.833A to a resistive load, and will have a short-circuit current of about 1A. If you try to draw more than 1A at the 3.3V output of your buck regulator, the voltage will start to droop. The solar panel simply cannot "charge" the inductor with more current than this.

With this sort of source, you must have an input decoupling capacitor (in parallel with the panel) that can supply the pulse current needed by the buck regulator without excessive voltage droop. This allows the solar panel to provide just the average current required, with a small amount of "ripple" caused by the switching of the regulator. To continue the example, let's say your regulator operates at 100kHz, and you want to limit the ripple to 1V. The capacitor needs to supply the difference in current between what the panel can supply and what the regulator reuqires:

$$I_{CAP} = 2 A - 0.833 A = 1.166 A (max)$$

It needs to do this for the on-time of the buck regulator:

$$t_{ON} = t_{PERIOD} \cdot Duty Cycle = 10 \mu s \cdot 27.5\% = 2.75 \mu s$$

The capacitor required is:

$$C = \frac{I_{CAP} \cdot t_{ON}}{\Delta V} = \frac{1.166 A \cdot 2.75 \mu s}{1 V} = 3.2 \mu F (min)$$

  • \$\begingroup\$ For peak-current-limited applications, could an inductor between the supply and the filter cap be helpful? In the absence of an inductor, taking energy out of the cap would drop the supply voltage; the amount of power the supply could put into the cap would be decrease with the cap voltage. If one added an inductor, the amount of power the supply would put into the cap would still decrease with cap voltage, but the power which didn't go into the cap would go into the inductor, and could be reclaimed later in the cycle. \$\endgroup\$
    – supercat
    Mar 26, 2014 at 16:28
  • \$\begingroup\$ @supercat: Yes, putting an inductor between the source and the capacitor would definitely help reduce the amount of ripple current that the source sees. But once you're past the "knee" and into the current-limited regime, the current doesn't vary all that much anyway. The main purpose of the capacitor is to reduce the ripple voltage so that the assumptions that went into the design of the buck regulator aren't violated. \$\endgroup\$
    – Dave Tweed
    Mar 26, 2014 at 16:42
  • \$\begingroup\$ The panel's current is going to be constant in either case (limited by the panel itself); the issue is voltage. Every 1% sag in the panel voltage represents a 1% loss of harvestable power, so if one can have the cap sag without the panel voltage sagging I would think that would be a "win". \$\endgroup\$
    – supercat
    Mar 26, 2014 at 17:11

Consider a simple synchronous buck regulator made from two switching transistors and an inductor-capacitor low pass filter. This one converts 12V to 5V: -

enter image description here

The switches never operate together meaning one is closed while the other is open circuit. The waveform at the junction is as shown and the output from the LC filter is shown in red. In reality the triangular shape of the voltage is much less exaggerated than I've shown it.

Given that the switches (operating at frequencies between 10kHz and 10MHz typically) are theoretically either open circuit or short circuit, no power is dissipated in them. Inductors and capacitors don't dissipate power either so the circuit is 100% efficient at power conversion from a high voltage to a lower voltage.

OK that's the simple story. Obviously switching losses and imperfect on-resistance in the power transistors make efficiency more like 90 to 95% at full load. Quiescent currents to drive op-amps and other stuff in the buck convertor add a few mA too.


Consider an ideal universal switching converter:


simulate this circuit – Schematic created using CircuitLab

If SW1 is in the upward position x% of the time, and SW2 is in the upward position y% of the time, then x% of the electrons that flow through x will flow through the left side supply/load, and y% of the electrons that flow through the inductor will flow through the right side. The average voltage through the inductor will be V1*x% - V2*y%. In order for the amount of current flowing through the inductor on each cycle to be the same, the average voltage must be zero, implying that v1*x% must equal v2*y%. Otherwise, if v1*x% exceeds v2*y%, current will accelerate rightward through the coil; if v2*y% exceeds v1*x%, current will accelerate leftward.

Thus, for any combination of non-zero duty cycles, assuming ideal components, a cycle-to-cycle equilibrium state will exist when the voltage ratio of the two voltages will be equal to the ratio of duty cycles, and the current is equal to the reverse ratio. Power can flow in either direction as needed to maintain this equilibrium. As a practical matter, converters are generally operated with one of the switches in a fixed "upward" position while the other one cycles up and down (for buck-mode conversion, SW2 would be left "up"), so that all of the electrons going through the inductor pass through either the supply or the load, but that's not strictly necessary. The way the behaviors are affected by the duty cycles of sw1 and sw2 will be the same regardless of whether one (or even both) duty cycles are equal to 100%, provided only that neither duty cycle is zero, and that the frequency is high enough that the coil current doesn't change too much during each cycle [the ideal model assumes coil current doesn't change at all during each cycle; the higher the frequency, the less the current will change during each cycle, and the more accurate the model will be].


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