I found this Buck converter circuit online, but not so sure what's the purpose of the circuit marked in red box and how it works.
Please come in and discuss if anyone has an idea of it. Thank you
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\$\begingroup\$ The circuit, as completely shown won't do anything useful unless burning the MOSFETs is seen as useful. \$\endgroup\$ – Andy aka Nov 9 '20 at 9:42
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The main purpose is to measure the input current (i.e. the current drawn from the source).
T101 is a current transformer. When the input current (a switching current) flows through T101's primary, a non-zero current proportional to the input current (and, in-phase of course - look at the dots) flows through the secondary:
$$ n=\frac{N_p}{N_s}=\frac{V_p}{V_s}\\ V_p\ I_p=V_s\ I_s\\ \therefore \frac{I_s}{I_p}=n $$
So, for example, for 1A primary current the secondary current can be 1mA.
The output current is then rectified via CR100. And this current drops a non-zero voltage across the burden resistors (R101 and R102. These two resistors also act as a voltage divider). This voltage is fed to the measurement circuit (analog or ADC) through an RC filter formed by R202-C102 pair.
One advantage of this circuit is that the measurement can be done with respect to any reference since the output current is isolated.
answered Nov 9 '20 at 4:24
