BJT Class AB Complementary Push-Pull Output Stage Bias/Quiescent Current Relation

This circuit helps in the analysis and/or verification of the mathematical relation between the bias currents in a BJT class AB complementary push-pull output stage amplifier. At quiescent the load current is zero, the sum of the voltage drops across the diodes is equal to the sum of the base to emitter voltage drop of the NPN transistor and the emitter to base voltage drop of the PNP transistor Vd1 + Vd2 = Vbenpn + Vbepnp nVt · ln (Id1/Isd1) + nVt · ln (Id2/Isd2) = nVt · ln (Icnpn/Isnpn) + nVt · ln (Icpnp/Ispnp) ln (Id1/Isd1) + ln (Id2/Isd2) = ln (Icnpn/Isnpn) + ln (Icpnp/Ispnp) ln (Id1·Id2/Isd1·Isd2) = ln (Icnpn·Icpnp/Isnpn·Ispnp) Id1·Id2/Isd1·Isd2 = Icnpn·Icpnp/Isnpn·Ispnp Icnpn·Icpnp = (Isnpn·Ispnp / Isd1·Isd2)(Id1·Id2) The diodes have the same current Id1 = Id = Id2 also, at quiescent the load current is zero and the current through the two transistors are equal Icnpn = Ic = Icpnp Ic² = (Isnpn·Ispnp / Isd1·Isd2)(Id²) Ic = [√(Isnpn·Ispnp / Isd1·Isd2)] · Id This is the general equation. In this particular circuit diodes D1 and D2 are of the same type Isd1 = Isd = Isd2 and the transistors are matched Isnpn = Istr = Ispnp Ic = [√(Istr² / Isd²)] · Id Ic = (Istr / Isd) · Id The transistor saturation current (Is) was modified from 1e-16 A to 5e-15 A and the the diode saturation current was modified from 1e-14 A to 1e-15 A. Substituting Ic = (5e-15 A / 1e-15 A) · Id Ic = 5 · Id The bias/quiescent current of the transistors will be 5 times that of the diodes. Note: The current through the transistors will be unequal when the load current is nonzero.