Does output voltage of a boost converter depend on load current in CCM?

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: -

\$\dfrac{L}{2}[I_{MAX}^2-I_{MIN}^2]\$

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!!

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.

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.