This is one of those things that I have always had a really hard time wrapping my head around.

In the horrendously simplified ideal step-up SMPS model below, I have set it up so when at 50% duty cycle, the output voltage is double the input voltage. For the sake of easy math, the current in the inductor is maintained at ~1A.

simulate this circuit – Schematic created using CircuitLab

As you can see, the power transfer is 1W so it operates at 100% efficiency.

But here is the kicker... only half the current taken from the supply ever reaches the load...

Intuitively one's head is telling one it can only be 50% efficient.

Granted an ideal SMPS would not be 100% efficient, but numbers in the 90 - 98% efficiencies in a real world SMPS are not unusual...

What is going on?

• The current is being converted to magnetic flux and back again. Think of it like chopping a bar of chocolate in half and putting the two pieces end to end. You've still got a bar of chocolate of the same area but it's longer and thinner. Nov 3, 2017 at 15:41
• You might be wondering about current flowing without voltage resulting in zero watts... Nov 3, 2017 at 15:41
• @Finbarr sigh.. now I have to go buy a bar of chocolate.... thanks Nov 3, 2017 at 16:14
• Me too. I'm abandoning plans for my series of electronic engineering - confectionery crossover articles. Nov 3, 2017 at 16:20
• I'm glad you asked this question! This bothered me for a few years, and the few people I asked (this was pre-stack!) didn't know. I finally asked my EE father-in-law, who answered like it was obvious :) Nov 3, 2017 at 16:28

Current != energy.

While current is flowing to ground through the switch, the energy represented by that current flow is being stored in the inductor.

When the switch opens, that energy is delivered to the load.

• +1 "Current != energy" Good man, I knew you would see right through that. It is a common misconception though. Nov 3, 2017 at 16:11

The current flowing from switch into ground for the first half a cycle has all the voltage across the inductor. The energy supplied by the source is used to build a magnetic field in the inductor. In other words, for this first half cycle your inductor is the load.

For the next the inductor acts as a second source. It will use the energy stored in the magnetic field to output that current. So all the energy "taken" in the first half cycle by the inductor is "given back" during the next half cycle. Hence, it is 100% efficient.

• I like this answer because it approaches the problem intuitively, simplifying the circuit and making sense of it. Nov 4, 2017 at 4:34

The answer of course was put most succinctly by Dave Tweed.

Current != Energy

The trouble is we all grew up and started out with this model...

simulate this circuit – Schematic created using CircuitLab

We learned that I1 = I2 + I3, and we work out that if I2, the current in the first path, = 1A, and I3, the current in the second branch, = 0.5A that the power dissipated in the first path must be twice the power dissipated in the second path. Which, for this purely resistive design, is in fact true.

That got stuck in our heads, that the power gets distributed in the ratio of the currents. However, in reality that is only true for a linear systems and purely resistive loads.

In truth the actual equation dictating power transfer is

P1 = P2 + P3

That is the power entering the circuit must equal the power consumed by each path through that circuit.

In our switching circuit..

.. the fact that half the current is going down that "ideal" charge path does not matter. Since, with an ideal inductor, the path has no resistance, the path dissipates no power, no Joule Heating, despite the current. Meanwhile, energy is absorbed by the inductor.

Or to put it another way, and as others have pointed out, while the charge path is active, and although the current is taking that path, the inclusion of that lossless storage mechanism effectively removes all the energy from the current to be released later into the second path to get you up to the new voltage.

The fact that half the current from the supply never makes it to the load is irrelevant.

Of course, if that charge path has some resistance, as it would in a real SMPS, there will be some power lost during the charge cycle. Whatever is lost, will no longer be available at the output side, and your efficiency will drop accordingly.

• I truly don't get where you are going with this 'half the current' story. The output energy will always be less than the energy stored in the inductor .....the output current will ALWAYS instantaneously equal the input current ...and the output voltage (peak) will depend on the R values. Nov 4, 2017 at 23:23
• @JackCreasey I guess you either see it or you don't. No big deal. Nov 5, 2017 at 0:54

Your reasoning is correct, you just got the current through the inductor wrong. The current through the inductor stays roughly constant over the time of one cycle if you switch fast enough. Therefore, the resistor current is the same as the input current half of the time, and zero for the other half.

You got 1V input voltage, 50% on-time and 4 Ohms at the output. The voltage across the resistor will be 2V like you already calculated. So, for half of a cycle, 500mA will flow through the resistor and through the inductor. During the other half, that same 500mA will flow through the inductor to ground since the current has to stay constant. The inductor current is only 500mA and not 1A like you assumed.

The circuit draws 500mA constant current from its input and delivers 500mA to its output, at double the voltage, for half of the time. That's 100% efficient.

Edit: SPICE simulation showing that the inductor current is indeed 500mA.

Edit 2: If you add a capacitor on the output of the thing, the situation becomes dramatically different.

Same as before: 1V input voltage, 50% on-time and 4 Ohms at the output. This still results in 2V across the resistor. But now that voltage is constant, so the resistor current is 500mA continuously, not just 50% of the time like it was without the cap. That results in the resistor dissipating 1W. The inductor has to deliver the same energy in one unit of time that the resistor uses up in two units of time, therefore the inductor current is twice that of the resistor current: 1A. This gives us 1W input power. Any "excess" energy that the resistor doesn't use immediately flows into the capacitor for later use.

So with the capacitor added, the circuit suddenly consumes double the power than without the cap, but it's still 100% efficient.

• This answer just describes the operation and does not really answer the question. Nov 3, 2017 at 16:12
• @Trevor I think it does - sorry if my answer wasn't clear enough. Your assumed 1A inductor current is wrong, so I did the calculation again for you and showed that there's in fact just 500mA constant current flowing through the inductor so the circuit only draws half a watt. I'll edit the answer and try to make it clearer. Nov 3, 2017 at 16:20
• Nope its 1A continuous in the coil, 1A 50% of the time to the load = 500mA. Nov 3, 2017 at 16:22
• @Trevor I threw it into LTSpice. It's 500mA, really. The current through the inductor has to be the same as the current through the resistor when the switch is in the "on" position. Kirchhoff's circuit laws dictate this. Since the voltage across the resistor is 2V, the current through it is 500mA and so is the current through the inductor. Nov 3, 2017 at 16:30
• Sorry, not sure what's up with your simulation, maybe your Ron is too high... I get this i.sstatic.net/WxBp3.png Nov 3, 2017 at 16:58

Assuming that your switching rate allows the current in the inductor to build to 1 A before switching .....the current flowing in the secondary will also instantaneously be 1 A when the switch opens, so there is no paradox.

Run the simulation below. The input has an LR time constant of 1 Ohm * 1 mH when the switch (FET RDS(on) ignored) is on. I set the inductor ESR to 1 Ohm in this example.
The output has an LR time constant of 3 Ohms * 1 mH when the switch is off.

If Vin is 1 V, Vout will be instantaneously 1 A * 2 Ohms --> 2 V and will drop as defined by the LR time constant while the switch is open. I leave it to you to discover the RMS voltage for Vout.
Steady state conditions would be: 1A for the switch closed. 0.3 A for the switch open. So the current from the 1 V supply will vary from 1 A maximum to 0.3 A minimum.

simulate this circuit – Schematic created using CircuitLab

Vin in Blue, Vout in Orange:

In subsequent cycles the input current peak drops due to the time constant.
I leave you to play with the values.

• Yes, it's the half the current goes south that throws your mind through a loop though.. Dave hit it on the head. Most folks tend to think of current and energy as being the same thing, including me most of the time I guess, but it is not that simple. Nov 3, 2017 at 17:04
• @Trevor. But half the current does not go south. If the current in the input circuit is 1 A, then the output current will (instantaneously) be 1 A, no if's or buts. The current then drops as the inductor energy is dissipated. Equally the 2 V is not set in stone, if you increase the resistor value to say 100 Ohms, the same rule applies. Now you have 1 A * 100 Ohms --> 100 V peak. The rule is therefore, the input current (peak or DC) will equal the output current, at least instantaneously. The output energy will always be less that the inductor stored energy. Nov 4, 2017 at 23:20
• But if the current in the inductor is 1A only half an amp goes to the load, or more accurately 1A for half the time, the other half amp goes straight back to the source. Nov 5, 2017 at 2:27
• @Trevor ...1 A for half the time does not equal 0.5 A....it is 1 A for half the time. Nov 5, 2017 at 2:44

As @Jonathan S. stated, the input current is not 1A.

Transient avoided in the discussion.

Here is the behavior in a steady-state condition.

It is so better seen that :

The inductance current is ALWAYS 0.5A. The resistor current is also 0.5A, but for HALF time ...