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I am trying to build a PWM controller as shown in the figure below. Considering the example that my input is 5V and output is 10V and I need a duty cycle of 50%. enter image description here

Ref for image: https://i.sstatic.net/0kkWg3nC.png

My doubt was what happens once the output is 10V and the error is zero? Would the output of the error amplifier be zero? In that case then the comparator output would also be zero and there would be no regulation right? What am I missing here?

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  • \$\begingroup\$ aaab - Hi, Where did the image come from? To comply with the site rule on referencing, details of the original source of copied / adapted material must be provided by you, next to each copied / adapted item. If the original source is online & public, please edit the question & add the webpage/PDF/etc. name & its link (URL) (e.g. website name + webpage title + its URL). If the source is offline (e.g. printed book / private intranet) then edit the question & add full source details e.g. title, author, page, publisher, edition. TY (More rules in tour & help center.) \$\endgroup\$
    – SamGibson
    Commented Nov 15 at 11:16
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    \$\begingroup\$ aaab - Hi, There's a misunderstanding. You added a link below the unreferenced image, which just links back to that image. However what is needed is a link back to where you originally copied the image from. Please update the reference link to point to that other website's source webpage for the image. TY \$\endgroup\$
    – SamGibson
    Commented Nov 15 at 12:24

2 Answers 2

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My doubt was what happens once the output is 10V and the error is zero?

It won't happen; the error won't be zero because it needs to be finite to drive the right level of PWM. It's exactly the same as an op-amp; we say that the two inputs (inverting and non-inverting) are at the same voltage when negative feedback is used but, because the op-amp has an open-loop gain of many tens of thousands, if we did the math we would see that the two inputs might be milli-volts or microvolts different to each other.

Short story: a finite but acceptably small error drives the system.

So, depending on your circuit's open-loop gain, the output might stabilize at 9.99 volts or 9.999 volts. How bad might that be?

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  • \$\begingroup\$ This is incorrect, the error can go to "0" the gain is "infinite" at DC \$\endgroup\$
    – Stiddily
    Commented Nov 15 at 13:06
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    \$\begingroup\$ @Stiddily the error cannot go to zero without infinite gain and, even an integrator is limited to the open-loop gain of the op-amp it uses. Hence, it is a finite error that drives the system to an acceptable measure of goal accomplishment (unlike what your answer appears to say). I suggest you look at what a practical and non-ideal integrator can deliver AND also consider the effects of op-amp offset voltages. \$\endgroup\$
    – Andy aka
    Commented Nov 15 at 13:09
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It will help to look at what an error amplifier actually looks like, and what that "compensation" block is: https://www.ti.com/lit/an/slva662/slva662.pdf.

The above link is for the TI article "Demystifying Type 2 and Type 3 Compensators".

I won't go into the differences between, type 1,2 and 3, but you'll see a common theme. There is no DC path in the compensation. It is all capacitors. The goal of any op-amp is to get it's inputs to be equal. In the case of an SMPS Error Amplifier, this means that the scaled representation of the output voltage is equal to the reference voltage shown in your image.

Like you pointed out: different loads, different inputs, different outputs, etc, all require different duty cycles to maintain. Duty Cycle is set by the output of your EA. The EA output is compared against a sawtooth, so a higher EA gives a higher duty cycle and a lower EA value gives a lower duty cycle.

To understand how EAs work, lets start with the steady state case AKA the output is what we command. Let's say that this occurs at a DC of 50%. If I change nothing, the SMPS will happily stay in this state. This is an error of 0, this is normal.

If I increase the load on the output, the amount of energy required to keep the output stable will increase. More current out = more current in. The EA will see this decrease in output voltage and start to increase it's output to increase duty. The rate at which this occurs is set by the Rs and Cs in the "compensation" block.

Similarly, if we then remove this added load, we are pushing too much current into the output and the output votlage will start to rise (i = C dv/dt). The EA sees this and starts to decrease it's output level.

Suffice to say, the EA has 3 states. It either increases, decrease, or maintains its output to maintain the duty cycle. The "error" may be 0, but that doesn't mean that duty cycle goes to 0!

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