As Q 2 becomes more conducting, its collector current increases. Q 3 also conducts more heavily and eventually reaches the point of saturation also. V O = V CC - V R1 - V BE4 - V D1 V O = 5 - 0.94 - 0.65 - 0.6 = 2.81V Point 3: V i = 1.2V, V O = 2.81VĪs the input voltage is further increased, Q 2 conducts more heavily, eventually saturating. With Q 3 beginning to conduct there is a conduction path for current through Q 4 and the diode, D 1, which then turns fully ON. This confirms that Q 2 is still operating in the forward active mode. Under this condition the collector to emitter voltage drop across Q 2 is: The voltage drop across R 2 is then V R2 = 1.23mA × 2.2 kΩ = 2.7V.
With operation in the linear active region, the collector current in Q 2 is 0.97 × 1.27mA = 1.23mA. Note that with transistor Q 3 just at turn on, V BE3 = 0.6V which means that the current through R 3 is 0.6V/470Ω = 1.27mA. Eventually, Q 3 reaches the point of conduction. Base current to Q 2 is supplied by the now forward biased base-collector junction of Q 1 which is still in saturation. V IN = V BE2 - V CE1(SAT) = 0.6 - 0.1 = 0.5 Point P2: V IN = 0.5, V OUT = 3.8VĪs the input voltage is further increased, Q 2 becomes more conducting, turning fully ON. V OUT = V CC - V BE4 - V D1 V OUT = 5 - 0.6 - 0.6 = 3.8V Point P1: V IN = 0.5, V OUT = 3.8VĪs the input voltage is slightly increased, the above state continues until, with Q 1 on and in saturation, the voltage at the base of Q 2 rises to the point of conduction. While there is no load present, there are leakage currents flowing in the output stage which allow the transistor Q 4 and the diode D 1 to be barely conducting in the ON state. This ensures that Q 2 is off which, in turn, means that Q 3 is off. Since the only source of collector current is the leakage of Q 2, Q 1 will be driven into saturation. With the input near 0 volts and the base current supplied to Q 1, this transistor can conduct in the forward mode. Figure 6 TTL inverter input vs output transfer curve
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