Hello support-forum,
with the BQ25700A we try to reduce the power dissipation of the input shunt (Rac) by reducing the value to 4 milliohms. Now the input current regulation is unstable.
Is it possible to reduce the input shunt to such a low value or will the compensation not be sufficient? How can the compensation components be calculated for the low shunt value?
It the datasheet the shunt is just given in two values (10 milliohms and 20 milliohms), as I understand this is because of the parameter range in the RSNS_RAC register.
For reduced power dissipation we increased the vlaue of the inductor to 6.8 millihenry. The ripple-current and therefore the power dissipation in the inductor will be reduced.
The datasheet recommends values for the inductor in the range 2.2 micro henry to 3.3 microhenry for a switching frequency of 800 kilohertz. Is it possible to increase the inductor value to 6.8 microhenry and how must the compensation be adjusted?
The resistor at the pin IADPT is used to "read the inductor value through the ADPT pin" (datasheet 8.3.4.1). Is it correct, to set the resistance to 240 kiloohm to set the inductance to 6.8 microhenry (extrapolation of the values in table 3)?
Regards,
Volker