Switching Power Supply Study Group

1nnocent

Switching Power Supply Study Group [Copy link]

 

In the book Switching Power Supply Design (3rd Edition),

The logical relationship behind this is that the EA output voltage decreases, which corresponds to Figure 1.4. This does not make sense. Do I need to swap the in-phase and inverting terminals of the EA amplifier to make sense? Or is there something wrong with my understanding?

As Vdc increases, the voltage divider of R2 increases. Shouldn't Vea increase? If the intersection point moves backward, the duty cycle should increase, right? The on-time should increase, which is a positive feedback. I think that exchanging the ± of EA will form a negative feedback, which will advance the intersection point and reduce the duty cycle. The following is part of Figure 1.4

maychang

This is not necessarily a mistake.

The original diagram does not show the polarity of the "current amplifier". If the "current amplifier" is an inverting amplifier, then when Vea increases, the duty cycle of PWM decreases. Of course, if the "current amplifier" is a non-inverting amplifier, then the two input terminals of EA should be reversed.

maychang

It is not easy to see that there are doubts in this picture.

可欣

I just reviewed analog electronics. The in-phase input terminal means: the output terminal polarity is the same as the in-phase input terminal polarity; the reverse input terminal means: the output terminal polarity is opposite to the reverse input terminal polarity; the book says it is negative feedback, of course, the input increases and the output decreases. In-phase input or reverse input is only related to the output terminal polarity, not the output terminal value. Positive and negative feedback are related to the output terminal value and the input terminal value (the difference between the two input terminals).

I think the original poster is confused about the concepts of non-inverting input terminal, inverting input terminal, and positive and negative feedback.

1nnocent

maychang posted on 2020-8-13 19:19 This is not necessarily an error. The original image does not show the polarity of the "current amplifier". If the "current amplifier" ...

I haven't been exposed to current amplifiers before, so I have some questions when I see this.

1nnocent

Ke Xin published on 2020-8-13 20:39 I just reviewed analog electronics. The in-phase input terminal means: the output polarity is the same as the in-phase input polarity; the reverse input terminal means: the output terminal is the same as the reverse input terminal...

I regard EA as a comparator, not an amplifier.

PowerAnts

Q1 is a P tube, what's the problem?

maychang

PowerAnts posted on 2020-8-14 11:42 Q1 is a P tube, what's the problem?

This book uses this symbol to represent a bipolar transistor.

maychang

1nnocent posted on 2020-8-14 09:26 Ke Xin posted on 2020-8-13 20:39 I just reviewed analog electronics. The in-phase input terminal means: the output polarity is the same as the in-phase input polarity; the reverse...

EA is not a comparator, but a linear amplifier with finite gain.

maychang

Ke Xin published on 2020-8-13 20:39 I just reviewed analog electronics. The in-phase input terminal means: the output polarity is the same as the in-phase input polarity; the reverse input terminal means: the output terminal is the same as the reverse input terminal...

The feedback loop here includes Buck circuit, LoCo filter circuit, EA amplifier, PWM generation circuit, and "current amplifier". This large feedback loop must be negative feedback, but the polarity of each local link in it is not necessarily negative feedback.

maychang

Continuing with my reply on the 2nd floor.

The first post is not complete. If you also post the waveform below this picture in the book, you can see that the switch tube starts to conduct at the lowest point of the sawtooth wave and turns off at the intersection of the sawtooth wave and Vea. So, if the two input terminals of EA and the two input terminals of the PWM generation part in this circuit are correct, then the "current amplifier" must be an inverting amplifier.