In the circuit above, I don't understand how to close and open the BJT that I circled.
To be specific, I think that when I give input 1 the circled BJT should be opened, but how? What is the voltage on the base of the circled BJT?
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2 Answers
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When the emitter-base junction of the input transistor is reverse biased, then the base-collector diode conducts from base to collector, thus applying current to the base of the circled transistor, turning it on.
When the input is high the input transistor is no longer operating as a transistor. When the input is low the input transistor operates as a common base transistor, pulling the base of the circled transistor low thus turning it off.
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\$\begingroup\$ When the input is high, the input transistor is still a transistor but it operates in the so-called "reverse active mode" with a very small beta. \$\endgroup\$ Commented Sep 18, 2022 at 20:40
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\$\begingroup\$ I will just add that this is an undesirable phenomenon in TTL because it causes a small input current to flow. For comparison, there is no such problem in DTL, because there is a backward-biased PN junction at the input. This is not explained in textbooks and students are left with the impression that this is something good and desirable. \$\endgroup\$ Commented Sep 19, 2022 at 7:14
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The input part of the TTL logic gate above can be easily explained at low input voltage (logic 0) by the phenomenon of current steering, the essence of which is: If we connect two diodes with different forward voltages in parallel, the current will flow through the diode with lower forward voltage.
The input transistor can be presented as two "diodes" (base-emitter and base-collector junctions) with common anodes (the base). The base-emitter junctions of the circled transistor and the output transistor Q2 can be also considered as "diodes".
Thus, at low input voltage (logical 0), there are two "diodes" in parallel. The left diode is the base-emitter junction of the input transistor. The right diode is the string of three "diodes" in series: the base-collector junction of the input transistor, the base-emitter junction of the circled transistor and the base-emitter junction of Q2. As a result, the base current of the input transistor is diverted ("steered") through the lower output part of the previous stage to ground and the circled transistor is off.
The high input case (logic 1) is trivial - the left "diode" is off and the current is diverted (steered) entirely through the right "diode" string. The circled transistor is on.
answered Sep 19, 2022 at 14:40