Dear,
In compensation networks, we use OTA and Opamp as EA in DC-DC converters (also another applications). The different between them is the physical connection of the compensation network with respect to the inverting input of OTA or Opamp. In case of Opamp. In particular, the connection will be realized between the output and inverting input while the connection is between output and ground in case of OTA (please see the attached figure).
I have read the documents and they stated that there is a "local feedback path" in Opamp but not with OTA that results in the difference of compensation network as I have mentioned above.
Any one can help me clarify what is the "local feedback path"? Thanks for your help.
Dear Van Ha,
It is my pleasure that you share this question with me.
At first i would like that you have some skills in analyzing the electronic circuits.
At first i will try to stress out the difference between the two circuits used to realize the compensated low pass filtering in converters.
At first the filter action is accomplished by Zf in the voltage op amp circuit and with H3(s) in the transconductance op amp circuit. They have the same circuit configuration.
In the conventional op amp the filter circuit Zf is connected between the output of the pacifier and its inverting input which means it feedback current from the output to input.
In the other circuit of the gm amplifier no feed back is accomplished between the output and the input and the same filter circuit Zf is directly connected at the output of the amplifier and the total output current flow in this filter impedance.
If we analyse the two circuits we find that both circuits have formally the same transfer characteristics. that is similar H(s).
As an exercise i would like that you derive the transfer function of each circuit and introduce it in a separate post.
The transfer characteristics of the conventional op amp is defined as
H(s)= (Ve- Vref)/ ( (Vo R4/(R4 +R1)) - Vref),
while the transfer characteristics H(s) of the gm amplifier can be defined by:
H(s)= Ve/ ( (Vo R4/(R4 +R1)) - Vref).
Formally the difference between the two circuits is that the output of the conventional op amp is referenced to the reference voltage Vref while the output of the gm circuit is referenced to the ground.
By this definition i gave you the key to solve the problem.
Best wihses
Dear Van Ha,
welcome again!
If you compare the two circuits you see that the filter impedance Zf is connected in the conventional op between the output of the amplifier with a potential Ve and the inverting input of the op amp. Considering the op amp is ideal, then V+=V- = Vref. This result points out that Zf is connected between Ve and Vref.
In the other circuit of the gm amplifier, the filter impedance is connected between connected between the output terminal with a potential Ve and ground.
As for the input the two circuits have a differential input such that
the divided Vo by R4 and R1 is applied on the inverting input and and compared to Vref on the noninverting input.
This is a qualitative explanation why the transfer characteristic definition is different in the two circuits as i mentioned in my first post.
In the next post a complete derivation of H(s) in the two circuits will be given.
Best wishes
- Badges
- Science topic
- Similar topics
- Biological Science
- Bioecology
- Systems Ecology
- Landscape Ecology
- Connectivity