UCC28051: UCC28051

Hey Ramesh, 

Ramesh Parnapalli said:

This output is current in nature. How is this voltage converting into voltage ?

The output of error amplifier is current, however there is compensation circuit at COMP pin which is  Zcomp = [XCv1+Rv1] || XCv2], thus as per Ohms law current flowing through impedance results into voltage.

Icomp = gm x (2.5V-Vos_sns), Vcomp = Icomp x Zcomp.

For more detail, kindly look at section 8.2.2.8 Voltage Loop Design from datasheet : 

https://www.ti.com/lit/ds/symlink/ucc28051.pdf?ts=1790611252963&ref_url=https%253A%252F%252Fwww.google.com%252F

Ramesh Parnapalli said:

. Is there any resistor diver circuit connected to the output of gM E/A ?

No.

Ramesh Parnapalli said:

Can you please provide a schematic on how they are connected? and also provide the values if any?

kindly refer below application schematic- 

Hope this helps to answer your question. Let me know if you have any questions for me.

Thanks

Regards

Anita

Thanks for the reply,

True but the voltage is limited to 3.8V and maximum voltage is 5V. Who is restricting the voltage. Ther has to be some limiter.

However, the circuit shows for non-isolated circuit. How about the isolated circuit PMP11155 Schematic

Hey Ramesh, 

Ramesh Parnapalli said:

True but the voltage is limited to 3.8V and maximum voltage is 5V. Who is restricting the voltage. Ther has to be some limiter.

May I know what is the reference of 5V max voltage at COMP pin? 

Kindly refer below app note to understand how compensation circuit work with transconductance amplifier.

https://www.ti.com/lit/an/slva662/slva662.pdf?ts=1790654558968&ref_url=https%253A%252F%252Fwww.google.com%252F

Thanks

Regards

Anita