# Buck Converter Error Amplifier Input Polarity Confusion (TL598)

I'm trying to design a buck converter using TL598. I'm studying the technical article Designing Switch Mode Power Supplies With the TL598 - Application Report for this purpose. This article references the paper The K Factor - A new mathematical tool for stability analysis and synthesis (Dean Venable) for designing the error amplifier compensation circuit and does the all calculations according to it.

The application report gives the polarity like this: (Page 29) (Page 32)

However, Dean's paper gives it like this: TI Article:

Inverting Input     : Fixed Reference Voltage over the Compensation Network
Non-Inverting Input : Buck Converter Output Voltage


Dean's Document:

Inverting Input     : Buck Converter Output Voltage over the Compensation Network
Non-Inverting Input : Fixed Reference Voltage


What is the reason for this contradiction? Texas Instruments clearly states that their calculations are based on the Dean Venable's article. They are completely consistent with each other except for this input polarity detail. Which polarity is the correct one? And what is the cause for this ambiguity?

## 2 Answers If you look at just the opamp circuit, both articles have the same opamp function
OUT = function(I+ - I-)
using the labels that I tagged on.

For TI, the opamp circuit is hooked up such that
OUT_TI = function(5V_OUT - VREF)

For Dean's, the opamp is hooked up such that
OUT_DEAN = function(VREF - 5V_OUT)

Therefore, at OUT, the polarities are opposite
OUT_TI = -(minus) OUT_DEAN
but otherwise equivalent.

Now look at the TI circuit diagram, the bubble (circled in blue) implies inverted level. Even without the bubble, the function of "PWM Comp" block (or the "CONTROL CIRCUIT" in Dean's) is not completely spelled out, so one may assume the control level is designed to work as intended regardless.

On one circuit you have a "block" described as "gate drive" and on another circuit there is a block called "control circuit". Who is to say that one block doesn't have an inversion in gain compared to the other. Swapping the inputs on an error amp introduces a numerical or mathematical inversion.

Even if one didn't have this inversion, a high output on one circuit turns a mosfet on which starts to put energy into the LC filter whereas a high from the control circuit (bottom diagram) will turn off the BJT - this in itself is an inversion.

You are taking this all too literally. If I were to be literal I'd want to know why a NPN BJT is shown in Dean's paper when it looks like it should be PNP.