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Since i started my training about design of DCDC I'm really confuse with control part.

At school i learned that DutyDycle=Vout/Vin for Buck, and that's it.

But now, i see that, Duty cycle is set by the output of error Amplifier and can indeed change from 0 to 100 according the output Voltage state. Is not at all a constant :/.

My second confusion and is my main question is about the picture joined.

If the result of EA is 0V (Line i drawn in red)enter image description here. What happen? duty cycle will be zero? This is not a problem ? because we should also have DutyDycle=Vout/Vin. Is there any constant define for the PWM modul for exemple if EA is 0V -> give a fixe dutycycle ?

I'm so sorry my question is not totally clear because is not clear in my mind too. If someone could give me a clear view about this part it will be a big party for me.

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2 Answers 2

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The formula describing the relationship linking \$V_{out}\$ to \$V_{in}\$ shows that the duty ratio \$D\$ controls the output voltage for a converter operated in continuous conduction mode or CCM: \$V_{out}=DV_{in}\$. That means that based on the input conditions, the control loop will permanently adjust the duty ratio to maintain \$V_{out}\$ on target. For instance, assume you want to deliver 5 V from a 10 V source, then the duty ratio will be 50% and the loop will maintain this value. The input source increases to 12 V, then the loop will reduce the duty ratio to 5/12 = 42% or so.

For a buck converter, the loop can freely adjust the duty ratio between 0 and 100%. 100% means the switch is permanently turned on. For instance, the source drops to 4.9 V then the loop will push the duty ratio to the maximum value, maintaining the switch on in an attempt to meet the 5-V target. Now if for some reason the load current drops to a very small value or disappears (unloaded converter), the converter enters the discontinuous conduction mode (DCM) and the relationship linking \$V_{out}\$ to \$V_{in}\$ is no longer \$D\$ alone but involves the switching frequency and other variables. To maintain \$V_{out}\$ on target, the only option the loop has is to reduce the duty ratio down to a small value and eventually 0%: the converter skips cycles. This is a very common type of regulation. The below picture shows how the duty ratio is elaborated in voltage-mode control:

enter image description here

Please note that if many converters can accept 0% operation, most need an upper limit for the duty ratio as a main power switch in a boost converter or a buck-boost converter cannot stay on too long otherwise saturation of the magnetic element eventually occurs and the switch may be destroyed.

As a final note, whether a converter is operated in voltage- or current-mode control, it is still the duty ratio which sets the operating point. However, in voltage mode, the loop directly drives the duty ratio via the pulse width modulator (PWM) while in current mode, it indirectly sets it by setting the inductor peak current. You will find many details about the buck converter in a seminar I gave at APEC in 2019.

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  • \$\begingroup\$ thank you verry much @verbal Kint for your explanation, i will check the seminar , because i want to understand this since while . Tell me if i'm wrong. Let reduce the subject to CCM and voltage-mode control, Vout=D.C*Vin it's a general relation between the Input Voltage and the output Voltage as you said "shows that the duty ratio D controls the output voltage ". But in real life we never have or mesured it since the dutycycle is set by the loop? \$\endgroup\$
    – kem
    Commented Jun 3, 2020 at 21:17
  • \$\begingroup\$ Here you said "For instance, assume you want to deliver 5 V from a 10 V source, then the duty ratio will be 50% and the loop will maintain this value. The input source increases to 12 V, then the loop will reduce the duty ratio to 5/12 = 42% or so." Can i suppose that , the Pin Vref(V+) of the EA is the desired output and the Vout mesured( V-) is input Voltage . Hence when the Vout mesured( V-) change , the dutycycle adjust in order to have the good Vref. \$\endgroup\$
    – kem
    Commented Jun 3, 2020 at 21:18
  • \$\begingroup\$ if all my previous exemple are ok ,( and to be honest i hope. It make more sence for me), Will i really mesure 0% of duty cycle when Vref(V+) equal to Vout mesured( V-) ? \$\endgroup\$
    – kem
    Commented Jun 3, 2020 at 21:20
  • \$\begingroup\$ or we define a standard Dutycycle, accorrding Vout=D.C*Vin and we define a small signal d.c with what i said before . \$\endgroup\$
    – kem
    Commented Jun 3, 2020 at 22:00
  • \$\begingroup\$ "that means that based on the input conditions, the control loop will permanently adjust the duty ratio to maintain Vout on target. " But the EA do not see the input condition is connected only to vout and Vref. \$\endgroup\$
    – kem
    Commented Jun 3, 2020 at 22:22
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Only in an ideal buck circuit will \$\frac{V_{out}}{V_{in}}={\rm d.c.}\$. In a real circuit there will be resistive loss in the switch and inductor, and other factors that reduce the output voltage slightly compared to the ideal value. Or (as pointed out in comments) if the load draws very little current, repeated on cycles could push the output voltage above the ideal value.

Your circuit uses feedback to adjust the duty cycle as needed to achieve the desired output voltage.

duty cycle will be zero?

Suppose the d.c. goes to zero. Then the feedback voltage (\$V_{FB}\$ in your diagram) will go to zero. Then the v-error amp output will increase, leading to an increase in d.c., so it won't stay at zero.

As long as the feedback voltage is below the target value, the error amp will keep increasing the d.c. until the target is reached (up to the maximum d.c. the circuit is capable of), and vice versa.

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  • \$\begingroup\$ What is EA supposed to be? Is the OP asking about when the output voltage is zero? Or when the error amplifier output is zero? \$\endgroup\$
    – DKNguyen
    Commented Jun 3, 2020 at 20:37
  • \$\begingroup\$ The EA is the error amplifier. I'm asking when the output of the Op is zéro, because we have vout =Vref. So that mean i can forget Vout/Vin=d.c in real life we never have this ? \$\endgroup\$
    – kem
    Commented Jun 3, 2020 at 20:54
  • \$\begingroup\$ @kem, it's an approximation. Say you want Vout/Vin = 0.65. Then a d.c. of 0.65 is a close guess to what it should be. Maybe it will actually be 0.67. Maybe it will actually be 0.62. It won't very likely be 0.33 or 0.98 unless something else is very wrong with your circuit. From this approximation you can make reasonable design choices about your circuit, like the required inductor and capacitor values, estimate current and voltage ripple, etc. \$\endgroup\$
    – The Photon
    Commented Jun 3, 2020 at 21:34
  • \$\begingroup\$ ThePhoton thank you for you explanation, But i already know this , since i have basis and there is lot of general explanation on google. But My problem is about specific thing in regulation loop that i do not find a clear explanation on the internet. I fill like the Input voltage do not play a big role there and only the output voltage variation and Vref of EA set the dutycycle , i'm right ? .@DKNguyen and ThePhoton \$\endgroup\$
    – kem
    Commented Jun 3, 2020 at 21:54
  • \$\begingroup\$ @kem, you'll put attenuation (a voltage divider) in the feedback path, so that the Vfb is equal to Vref when Vout is at the target voltage. But the input voltage is very important because it determines how quickly the inductor charges when the switch is closed. \$\endgroup\$
    – The Photon
    Commented Jun 3, 2020 at 21:57

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