I have a question about Bandgap Reference circuits.

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How did he get Vt*ln(n)?? This is how I did it: enter image description here


1 Answer 1


\$A\$ is the area. \$I_S\$ is the device's saturation current, which is proportional to \$A\$. The device with area \$nA\$ would have saturation current \$n I_S\$. So your second line should have been:

$$ = V_T \ln \left( {{I_C}\over{I_S}} \right) - V_T \ln \left( {{I_C}\over{n I_S}} \right) $$

(Note \$n\$ in denominator instead of numerator) The rest then goes through to what you were trying to prove.

  • \$\begingroup\$ Ah. That makes much more sense now. So there's two ways of doing this bandgap. I can either drive in uneuqal currents into each BJT or I can drive the same current, but change the BJT size ratios. Thank You! \$\endgroup\$ Apr 1, 2020 at 17:12

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