A Problem with TTL Inverter Circuit

wanneng12345

A Problem with TTL Inverter Circuit [Copy link]

In the following typical TTL inverter circuit, when the input Vi is high, T1 is saturated, and the collector potential VC1 of T1 is high, causing T2 to turn on. T1 is an NPN tube. When the NPN tube is saturated, the internal current flows from the collector to the emitter, that is, it flows into the collector. So, where does the base current of T2 come from? At this time, the collector current of T1 is from C to E, so how can there be current flowing into the base of T2?

maychang

The third screenshot posted by the original poster (page 106 of the original book) has made it very clear: when the input terminal is high level, and T2 and T5 exist, T2 and T5 must be turned on at the same time. This is definitely not what the first post said; "When the input Vi is high level, T1 is saturated, and the collector potential VC1 of T1 is high level", but T1 emitter junction is reverse biased and T1 collector junction is forward biased.

maychang

"So, where does the base current of T2 come from?"

Since the emitter junction of T1 is reverse biased and the collector junction of T1 is forward biased, the base current of T2 flows from the positive terminal of the power supply (Vcc) through the resistor R1, the collector junction of T1, the emitter junction of T2, and the emitter junction of T5 to the negative terminal of the power supply (GND).

maychang

Because the current passes through the resistor R1, T1 collector junction, T2 emitter junction, T5 emitter junction to the negative terminal of the power supply (GND), and passes through three forward PN junctions, the book says that the base voltage of T1 is 2.1V, which is the forward voltage drop of the three PN junctions.

ZXopen

A Problem with TTL Inverter Circuit

maychang

In digital circuits (such as the inverter 7404 mentioned in the first post), each tube does not work in a linear amplification state, but in a switching state, so the operation of the tube in the linear amplification state cannot be applied to the digital circuit.

In fact, when the input terminal of T1 in the inverter is low, it is in saturation. But when the input terminal of T1 is high, T1 is no longer a triode, nor is it in saturation. Instead, the emitter junction is reverse biased and the collector junction is forward biased. The emitter junction has no effect on the circuit as a whole, and the collector junction is equivalent to a diode.

chunyang

The original poster's analysis is wrong, and the basic working principle of transistors is not fully understood. For the original poster's diagram, when the input is high level, the working point of T1 cannot be met. At this time, it can only be regarded as two diodes, with the base as the positive pole and the collector and emitter as the negative poles. Therefore, the power supply provides a strong base current to T2 through R1 and the collector junction of T1, causing T2 to saturate, and then T4 is cut off and T3 is saturated, so the output is low, forming an inversion. When the input is low level, the working point of T1 is established, T1 is turned on, T2 loses the source of base current and is cut off, and then T3 is cut off, and T4 is saturated due to the base current provided by R2 and then outputs a high level, forming an inversion.