In the thread below, there are different opinions whether output voltage of a boost converter operating in continuous conduction mode (CCM) depends on load current or not.

Boost Converter Output Voltage dependence

Opinion 1: output voltage only depends on input voltage and duty cycle

Vout = Vin/(1-D)

Opinion 2: output voltage depends on input voltage, duty cycle and load current.

For example, as load current increases the duty cycle should also increase to make up the extra energy delivered to load.

So which one is right? If output voltage doesn't depend on load current, then why do we need controller?

When the load current increases, the averaged inductor current should also increases. So how do we do that?

  • \$\begingroup\$ For the DC/DC converter build with the ideal circuit's components and the ideal two-directional switch (diodes are not allowed), the inductor will always work in CCM even widout the load current. And as long as input voltage is pure DC (ideal voltage source and constant) the DC/DC converter behaviors like an ideal DC transformer. The output voltage will be only set by the duty-cycle (if we ignore the transient state). \$\endgroup\$ – G36 Dec 12 '17 at 20:11

With a boost converter you need to transfer energy to the load by charging an inductor with energy and then releasing that stored energy. In CCM, inductor current never falls to zero so the energy transferred is dependant on the peak current and the min current.

What the current peaks at during charging represents the maximum inductor stored energy and, what the current drops to (during load replenishment) represents the energy left in the inductor just as the cycle repeats.

The difference in energy is what is "given" to the load.

For a fixed value inductor and input power supply voltage, the rate at which current linearly rises (di/dt) is constant and depends entirely on V = L di/dt, that well-known formula. I'm assuming perfect lossless components of course!

So, Imax, for a given operating frequency will always end-up at some fixed-value above Iaverage and, Imin will be the same fixed-value below Iaverage. We could call that fixed value Ipeak.

So, energy given to the load, W is: -


Re-arranging using \$I_A\$ and \$I_P\$

W = \$\dfrac{L}{2}[(I_A+I_P)^2-(I_A-I_P)^2]\$

where \$I_A\$ is average current and \$I_P\$ is the peak above (or below) \$I_A\$. You also might be able to get to this formula (hopefully): -

W = \$2L\cdot I_A\cdot I_P\$

This means that both the average current and the peak current determines the energy transfered to the load. But, for a given input supply voltage, operating frequency, duty cycle and inductor value you cannot control \$I_P\$.

So, if the load resistance increases in value BUT you wanted to keep the average output voltage the same, the only option (other than D) is to increase operating frequency to reduce the peak current attained by the inductor during charging. This of course reduces average current gradually over a few cycles and what you find, in some controllers, is that the average current becomes lower and the frequency slews back to the original value.

More load current means lower frequency, less load current means higher frequency.

At the end of the day, I believe you still need a "clever" controller (to alter frequency) so I'm not sure that this question and the optional answers have much bearing on the practical world.

That's the way I see it anyway. Good (but flawed) question!!

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  • \$\begingroup\$ Yes, your answer already includes transient period of load step. This is where I was confused. If there is no way to change duty cycle or switching frequency temporarily during this time then the converter can not work with varying load. When we say that output voltage only depends only input voltage and duty cycle we are talking about steady state not transient state. \$\endgroup\$ – anhnha Dec 13 '17 at 8:13
  • \$\begingroup\$ Insight on this problem is difficult to find on the internet. I'd encourage you to use a sim tool to set up a simple scenario. However, to re-affirm, if the load current changes and you cannot alter the cycle time or duty cycle then you are are screwed with respect to regulation. \$\endgroup\$ – Andy aka Dec 13 '17 at 10:01
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    \$\begingroup\$ I already simulated it in psim and confirmed that. That is the reason I was confused by some people saying that output voltage doesn't depend on load current. That statement is only correct for steady state operation with a specific load current. When load current changes, during the transient period we need to change cycle time or duty cycle to change the average inductor current. \$\endgroup\$ – anhnha Dec 13 '17 at 10:59
  • \$\begingroup\$ @anhnha Cool. I'm glad you did a simulation to confirm my thoughts - I was thinking about simulating it myself but the math seems so simple that it's hard to see how it could be wrong. \$\endgroup\$ – Andy aka Dec 13 '17 at 12:54
  • \$\begingroup\$ @anhnha in response to this question and the general lack of clarity on this subject to be found on the internet, I've added a section on DCM and CCM operation to my website that you may be interested in if you find the time. It's a work in progress so don't expect perfection at this stage. My work underwrites my answer above Stades website - the CCM flyback design will be added in a few minutes. \$\endgroup\$ – Andy aka Apr 3 '18 at 9:52

The answer is: it all depends. Here is an easy way to determine everything. We will assume perfection (no resistive or diode switching losses).

When you close the switch, it is easy to calculate the amount of current increase in the inductor. Because V = L di/dt then di/dt = V/L. di/dt is in amperes/second (dt is the amount of time the switch is open each cycle). Similarly, assuming the output is a stable voltage, you can calculate the current decrease when you close the switch in the same manner, except this time the voltage is (Vin - Vout), and the time is the "off" time.

So the rules are: the current increase has to equal the current decrease, so input voltage, output load, and duty cycle all affect the output voltage at the point the load changes. Consider that in discontinuous mode we are closing the switch and filling a bucket with energy, then opening the switch and pouring it all out into the load. However, in continuous mode, we are filling up the bucket when the switch closes, then pouring out only a portion of it. In continuous mode, there is always a partially full bucket (a partially charged inductor.)

To increase the load current, the switch will have to stay closed longer, so that the bucket will rise to a new level. Once this occurs, the duty cycle will return to the original steady state value regardless of the load.

So in answer to your two questions:

Question 1. Opinion 1 is correct if you are considering the steady state condition with a theoretical device; Opinion 2 is correct during startup, varying load, varying input voltage, and the effects of losses.

Question 2. Make sure you have a big enough bucket! The inductor current will rise and fall as the switch opens and closes. Remember that at the peak current, you must not saturate, and if the inductance sized incorrectly (too small value), the current will rise too quickly and you will go into discontinuous mode, or conversely (too large value), the system will not regulate well.

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  • \$\begingroup\$ Yes, your answer makes sense. I didn't think about the transient state. \$\endgroup\$ – anhnha Dec 12 '17 at 7:09

Ideally, the output voltage of a buck converter running in continuous mode is a multiple of the input voltage, with that multiple dependent only on the duty cycle.

However, back here in the real world it's hard to find those ideal diodes with no forward voltage drop, inductors with no series resistance, and capacitors with no losses. The various non-idealities cause some voltage drop with higher current.

The usual solution is to close the loop to control the duty cycle to whatever it takes to achieve the desired output voltage. With good parts, that duty cycle will remain largely constant for a particular combination of input and output voltage. However, it will go up some with higher output current. That's because the control loop has to push the circuit a little harder to compensate for the inevitable losses.

For a boost converter, it's more complicated. Unlike with a buck converter, the duty cycle is a tradeoff between enough time to store energy in the inductor, and enough time to deliver the stored energy to the load. 100% duty cycle, for example, continuously charges up the inductor but never delivers anything to the output.

However, there is still a fixed voltage ratio between input and output that is only a function of the duty cycle with ideal components. For a buck converter, if D is the fraction of the time the inductor is connected to the input voltage and the inductor is connected to ground the remaining time (1-D), then the ratio of output voltage to input voltage is simply D.

    Vout / Vin = D

Now think of a boost converter as the buck converter run in reverse. That means Vout and Vin are swapped. It also means with consider the "on" time of the inductor when it is connected to ground, not Vout. Therefore D of a boost converter is 1-D of the same thing viewed as a buck converter.

Applying all this flipping from buck to boost to the equation above, yields the equation for a boost converter:

    Vin / Vout = 1 - D

Rearranging this to tell us what the output to input voltage ratio is yields:

    Vout / Vin = 1 / (1 - D)

This is easier to see by analyzing a simplified switching converter:

First let's consider this a buck converter. You stipulated continuous mode, so the switch always either connects the left side of the inductor to VA or ground. The result is a simple low pass filter.

However, this same circuit works in reverse as a boost converter. With the switch always connected to one of the two choices and no diode, this is really a DC transformer. It works the same way for either buck (input is VA, output is VB) or boost (input is VB, output is VA). If we consider the duty cycle to be the fraction of time that the switch is connected to VA, then VB is simply the duty cycle times VA. That's the buck converter view.

The relationship works identically in reverse. VA is VB divided by the duty cycle. The only difference for typical boost converter analysis is that we usually consider the duty cycle the fraction of the time the switch is connected to ground instead of VA. In other words, we use 1-D relative to what we call "duty cycle" for a buck converter.

Now before you complain that this is unfair because the diode is missing and that current can run backwards thru the inductor, remember that you stipulated the converter was running in continuous mode. The duty cycle, input and output voltage ratios, and output current demand are such that the current is always flowing in the inductor. If you knew this was always true, you could remove the diode.

The circuit as shown, without a diode, does actually work both ways, and inductor current can flow either direction. This is basically a "DC transformer", with the voltage ratio strictly a function of the duty cycle, no matter which way you define it.

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    \$\begingroup\$ But even with ideal components, the boost converter operating in CCM will also depend on load current in transient stage. If there is no controller, then the converter can only operate at a specific load current. \$\endgroup\$ – anhnha Dec 12 '17 at 17:18
  • \$\begingroup\$ @anh: I just realized that you asked about a boost converter and I answered for a buck converter. I have update the answer to talk about both, and how they are really two sides of the same coin, at least in continuous mode. The answer is still that the voltage ratio is only a function of duty cycle with ideal components in continuous mode. \$\endgroup\$ – Olin Lathrop Dec 12 '17 at 18:18

I had a similar doubt regarding buck converter i.e., whether the output voltage is determined by the duty ratio alone. The theory says, for an ideal buck converter under steady state conditions and in CCM mode, the output voltage is governed only by duty ratio and not on the load resistor (or load current).

I did a simulation to confirm this. The load is switched from 5 ohm to 1 ohm at 0.5 ms. Duty ratio is maintained constant at 0.5 throughout the simulation.

The simulation results are as follows: Simulation results

It clearly proves that the inductor current gets modified as per the load current demand and reaches a new steady state accordingly. The output voltage also goes through a transient, but comes back to the original steady state values

The simulation model used is as follows. Please note there is no feedback controller employed and the variable loading is simulated using a voltage controlled switch having on and off resistance as desired.

LTSpice Simulation model

When the load current increases, the averaged inductor current should also increases. So how do we do that?

All I can say is every power converter has a unique equilibrium condition ( also called steady state condition), where the energy inflow (in either Ton or Toff duration) becomes equal to energy outflow (in the other duration). As long as it is not in equilibrium, the system naturally drifts towards the equilibrium state. This is what happens when a load change happens. During transient time, the energy intake in an inductor is not matching with energy outflow, which results in a build of inductor energy.

I just want to caution, if you try to do the opposite action , ie, reduce the load current (increase load resistance) the above conclusion (constant o/p voltage independent of load current) need not hold good. The reason being as the load current decreases the converter may enter Discontinuous conduction mode (DCM) and o/p voltage is dependent on load current in DCM.

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I see that most of the people have answered your question in a very satisfactory way, at least for me. But, I can add here a more simplification toward your ambiguity. IN CCCM, it is assumed that in a lossless circuit, the relation between input/output is given by: Vout = Vin/(1-D)

Now, run any circuit simulator with an ideal circuit components (means that whatever the components you are using, switches, diodes, capacitors or even inductors are all ideal) and adjust your duty cycle for the given Vin and the required Vout. No mater how much you are loading your boost converter, the output will be more or less the same. Your hard day starts once you include a lossy component in your circuit, then the output load is becoming involved. That's simply because the relation Vout = Vin/(1-D) has originally been developed based on the assumption of a lossless circuit components, which means that the equation doesn't count for the losses. More current means more conduction loss (also known as: copper loss), it counts as (I·I·R) the square of the of your average output current. Here comes the importance of a controller to correct the duty-cycle to ensure a constant output voltage at any loading condition, provided that the controller is able to switch the converter from near 0% to near 100% of duty-cycle.

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Clarification: "CCM" stands for Continous Conduction Mode, meaning that the current through the inductor of the boost converter never has a chance to decay to zero. This is in contrast with DCM (Discontinuous Conduction Mode), where there is no current flowing for a part of the cycle. Boost (and buck) converters with diode rectification operate in DCM under light loads, and transition into CCM as the load increases.

Opinion 1: output voltage only depends on input voltage and duty cycle

Vout = Vin/(1-D)

This is true, assuming an ideal boost converter in CCM. The output voltage (Vout) of an ideal boost converter operating in CCM is only dependant on the duty cycle (d) and input voltage(Vin), not current:

Vin / Vout = 1 - d

That said, reducing the load current enough will eventually cause the converter to operate in DCM (unless it employs synchronous rectification), where the load current has a great effect on the output voltage. Also, the output voltage of any practical boost converter will be slightly affected by the load current even in CCM, mainly due to the parasitic resistances in the circuit.

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  • \$\begingroup\$ "your equation is that of an ideal buck converter": no, it is for boost converter. By the way, the equation you rewrite Vout = Vin(1 + d / (1 - d)) is exactly the same as mine. I don't think you answer my question. You just stated the fact based on the formula and didn't explain why. \$\endgroup\$ – anhnha Dec 12 '17 at 7:07
  • \$\begingroup\$ I am wondering how can you misread my formula as such. It is Vin /(1-D) not Vin/D as you said. With Vin = 10V and D=0.8 then Vout = 10/(1-0.8) = 50V. This is the same result as your formula. \$\endgroup\$ – anhnha Dec 12 '17 at 9:50
  • \$\begingroup\$ Wonder no more: I'm writing this on a smartphone, far from a PC (or pen and paper for the that matter), which makes everything slow and difficult. Yeah the two formula (yours and mine) are equivalent when you don't accidentally turn a - into a / \$\endgroup\$ – jms Dec 12 '17 at 10:25
  • \$\begingroup\$ jms: what do you mean by this, "you don't accidentally turn a - into a /"? My formula has nothing wrong at all. Maybe because you are viewing it in a phone so it got all these problems. \$\endgroup\$ – anhnha Dec 12 '17 at 10:56
  • \$\begingroup\$ I already acknowledged that the two formulae are identical (both are correct) in the previous comment. By "you don't accidentally turn a - into a /" I refer to how I calculated Vout = Vin/(1-D) as Vout = Vin/(1/D) because I wrote / instead of - \$\endgroup\$ – jms Dec 12 '17 at 11:05

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