Electronic – DC current through diodes in AB amplifier


I can't figure out how the DC current through the diodes is calculated in eq. 8.22 on this PDF from University of North Carolina.

I_D = \frac{V_{CC}/2-V_{BE}}{R_f+R_2}

\$I_D\$ should be equal to the current through the resistor \$R_2\$ minus the current going through the base of the transistor \$Q_1\$. I am able to obtain this equation only supposing \$I_B = 0\$, which is not true, specially because they are matching the diode to the transistor to avoid thermal problems.

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Best Answer

Yes, that equation assumes Ib=0. We all know that can't be true IRL, but if the transistors aren't in saturation, it's likely that Ib is far lower than the bias currents used in the resistor/diode network. So the estimate isn't a perfect solution in the real world, but it's close enough...and engineering is never about perfection, it's all about being close enough.