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I want to build a simple boost converter for learning purposes that outputs 12V/1A from a 5V/2A source. A simple boost converter has an inductor, diode, capacitor and some transistor that can be switched off/on by a PWM signal:

boos converter

I think the formula for output voltage is:

$$ V_{out}=V_{src}/(1-D) $$

Here are my questions:

  • why the formula does not contain which values need L and C to be. Or, L and C are dependent on required current output?
  • is D the period of PWM signal?
  • Wy the formula does not include voltage drop over the diode?
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3 Answers 3

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I think the formula for output voltage is: \$V_{out}=V_{src}/(1−D)\$

The output voltage formula is correct when the load current is sufficient to keep the converter operating in continuous conduction mode (CCM). CCM occurs when the inductor still has current flowing in it (i.e. it has stored energy) when the switching cycle re-starts: -

enter image description here

Image from my basic website and, if the load current falls to a lowish level, the transfer of energy completes before the next charge cycle begins and, you enter what is known as discontinuous conduction mode (DCM): -

enter image description here

In this mode, the voltage transfer formula is more complex: -

$$\dfrac{V_{OUT}}{V_{IN}}=\dfrac{1}{2}+\sqrt{\dfrac{1}{4}+\dfrac{D^2\cdot R_{LOAD}}{2\cdot L\cdot F_{SW}}}$$

Now, the output voltage is also dependent on the effective load resistance, the inductance and, the switching frequency.

Why the formula does not contain which values need L and C to be. Or, L and C are dependent on required current output?

In CCM, the values of L and C affect the ripple voltage. Regard L and C as a low-pass filter; you want a cut-off frequency that is much lower than the switching frequency to avoid significant output ripple. This is a good motivation to find appropriate values. Then, if there is a possibility that DCM can be entered, you need to also factor in the switching frequency, inductance and load current when setting the duty cycle.

is D the period of PWM signal?

D is the duty cycle of the switching cycle. Hopefully the pictures above help you understand this.

Why the formula does not include voltage drop over the diode?

It's an idealized circuit where the volt drop is assumed to be small compared to the output voltage. You can factor it in by increasing the output voltage 0.6 volts if you wish.

Here's a calculator from the above link: -

enter image description here

If you input the same values you will see that it runs in CCM until the load resistance increases to about 20 Ω. I assumed a 10 μH inductor and switching frequency of 100 kHz for this example.

Please also note that for an input power of 10 watts (5 volts and 2 amps) the very best continuous output current at 12 volts is 0.833 amps (see the above calculator picture). If you take power transfer efficiency into account and diode drop you might be able to produce a continuous output current of about 0.7 amps and not 1 amp.

That's an overall 84% power efficiency. But, on a good day with good component selection, you might get nearly 90% efficiency.

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  • \$\begingroup\$ Thank you very much! From your answer I take that this conditions assume the load is constant. If the load is dynamic, lets say it can vary between 15 to 20 ohm at different timepoints, to keep the output voltage stable the duty cycle should be adjusted? that is kind of a PID conroller would be necessary? \$\endgroup\$
    – DEKKER
    Commented May 24, 2023 at 14:38
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    \$\begingroup\$ @DEKKER it doesn't usually need to be as sophisticated as that; usually negative feedback around an op-amp that delivers a voltage that gets converted to duty cycle can suffice. However, there can be complications when moving between modes. It can be a serious subject. \$\endgroup\$
    – Andy aka
    Commented May 24, 2023 at 14:41
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I want to build a simple boost converter for learning purposes that outputs 12V/1A from a 5V/2A source.

You can't get 12V 1A = 12 W from a 5V 2A = 10 W supply. You can get 12V and less current. You could get maybe 0.8A if your converter is efficient

why the formula does not contain which values need L and C to be. Or, L and C are dependent on required current output?

When a boost converter is operated in continuous conduction mode, the steady state output voltage does not depend upon L or C. Thus, they are not included in the formula. A boost converter is in continuous conduction mode when the current through the inductor does not drop to zero during a cycle. The inductor is continuously conducting.

is D the period of PWM signal?

No, D is what is called the duty cycle, or what I like to call the duty ratio. It is the fraction of the time that the switch is "on". It is always a value between 0 and 1.

Wy the formula does not include voltage drop over the diode?

To be precise the formula should include the voltage drop over the diode. However, the author you are reading probably didn't feel that level of accuracy was necessary for their purpose, so they ignored it. Approximations are common both when explaining concepts, and in practical engineering.

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  • \$\begingroup\$ Thanks! So I can just randomly drop in L and C components? I think it is not that simple and one should have a ration which values to use? \$\endgroup\$
    – DEKKER
    Commented May 24, 2023 at 13:06
  • \$\begingroup\$ No, you cannot randomly drop in L and C components. There is a whole design procedure. Or, more correctly, there are a number of design procedures, each with different starting points. \$\endgroup\$ Commented May 24, 2023 at 13:12
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D is duty cycle and since it is unitless number 0..100% or 0..1 there is no time period and all time related things such as frequency or period or L or C or current are reduced away from the formulas between input and output voltage.

D is not the period, it's duty cycle, which is just on-time divided by whole period of time.

Diode is assumed ideal so no voltage drop.

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