# What is the purpose of the op-amp feedback circuit?

This is an existing circuit of buck regulator circuit I have been reviewing with.(Here I have not mentioned the part number of the due to the confidentiality issue) Following is the screenshot of the feedback resistor design. Reference voltage of the buck regulator is 1 V and output is designed for 3.3 V . Feedback regulator resistors (51 kΩ , 22.1 kΩ) combination design I have reviewed and verified the values.

Opamp feedback path and its functions are confusing sections for me. Following is my understanding, current feedback across the 100 Ω resistor is taken across to the amplifier input and voltage output comes in parallel with the feedback path ( 22.1 kΩ) through a buffer.

Can anyone suggest the purpose of the opamp-feedback circuit? And what could be possible output voltages at the buffer (op-amp) output.

• My first guess is that it's a current limiter. The right opamp measures the voltage across a $100\Omega$ resistor, amplifies it, and then pulls up the FB pin if it is too large. A large FB will make the regulator think that the output voltage is too high, causing it to source less current. – Sven B Nov 7 '18 at 8:31
• @SvenB, can you please just give a reference I can read to understand how buck will source less current? – vt673 Nov 7 '18 at 8:44
• I'd refer you to any introduction about buck regulators if you want to know how they can source less current. YouTube has video's explaining the principles. – Sven B Nov 7 '18 at 9:07
• 100Ω current sense resistor? You sure that's not a 1Ω or 0.1Ω? It should be. It looks like an attempt to regulate both voltage and current. – Misunderstood Nov 17 '18 at 17:54
• @Misunderstood, good catch. Initially there was a schematic error, Actually it is 0.1 Ω – vt673 Nov 19 '18 at 6:54

## 1 Answer

The OPAMP circuit does the current limiting. It is not a real current feedback, because the feedback path is alive only during over-current situation, and it can only reduce the output current.

The first stage is a difference amplifier - other call it as differential amplifier as well, but the latter is used more broadly - what converts the differential voltage across the 100Ohm series resistor into a single-ended one, then another OPAMP buffers it and feeds this error voltage back to the FB pin, if the diode opens. The diode rectifies this feedback, or with other words, it disconnects the OPAMP circuit from the FB pin under normal output current range. Not clear why is a zener diode drawn in the schematic. Any simple diode would do the job.

Why is it a current limit? The converter will push current into the output as long as the voltage at the output does not reach the desired value, which is set by the feedback resistors and a voltage reference. Note that they can not reduce the output voltage by sucking current back from the load. In the latter case, they would be called a generator or a driver. The output voltage is reduced by the load itself. Once the voltage on the FB pin is greater than the reference, the regulator thinks that the output is already too high, and it will not increase the output voltage, hence the output current decreases.

• From Ltspice simulation, it is noted that opamp feedback voltage after the buffer is equal to 936.84 mV at load current is 3.5 A. I was wondering how this can be calculated using equation, any suggestion? From the voltage drop across the 100 mΩ sense resistor at 3.5 A, it is noted that voltage to the amplifier input will be 350 mV. Using simple opamp equation Vin * ( 1 +R1/R2) = 350 * ( 1 + 22.4 kΩ / 22.4 kΩ) = 350*2 = 700 mV. Please suggest where am missing the point. – vt673 Nov 7 '18 at 11:35
• The opamp output voltage is Vo = Iload * 100mΩ (22.4kΩ/1kΩ). So, you will get 1V at (1V/22.4)/100mΩ = 0.446A – G36 Nov 7 '18 at 17:22