# TTL NOR Gate analysis

I'm looking at the design below. I am reading this webpage where the picture came from.

What I'm aiming at is the wisdom behind the design. Looking at the two transistors facing downward, I assume that the voltage at the collector region is amplified from the base since the emitters, given that A and B are receiving low input signal, are grounded. That'd make the transistors facing each other to turn on, and have the output spot perform the NOR logic. Does that make sense?

• Yes, your analysis is correct. Mar 19, 2021 at 18:14
• romeo, Please do note that the "facing down" transistors actually do have a forward conducting diode between base and collector when A or B is high (5 V). Don't forget that conducting path. So if A is high, then the 4.7 k conducts via base to collector, then base to emitter of the second BJT, so that BJT pulls its collector down and the LED is off. Similarly, if B is high -- the LED is off. But if both A and B are low, then the downward facing BJTs are ON, pulling their collectors tightly to their "low" emitters, causing both inner BJTs to be off, allowing the LED to be on.
– jonk
Mar 19, 2021 at 18:29

In TTL logic, the input transistor works in a manner similar to the two diodes in your DTL inverter. If the input, (emitter) is high, current will flow from the base through the collector. If the emitter is low, current will not flow out of the collector. Unlike most uses of a transistor, this use in not particularly related to amplification. It replaced the double diodes in DTL logic because it switches faster.

Yes, there is a lot of wisdom here; "current steering" is such a wisdom. But what is this?

Input LOW. When the input A/B is LOW (grounded), the base current of the input transistor is completely diverted (steered) through its forward-biased base-emitter junction although the base-collector junction is also forward biased. Why?

In this arrangement, two "diodes" are connected in parallel. The first is the base-emitter junction of the input transistor with a forward voltage of 0.7 V. The second "diode" is composite; it consists of two forward-biased PN junctions (the base-collector junction of the input transistor and the base-emitter junction of the second transistor) in series. The total forward voltage of this network is 0.7 + 0.7 = 1.4 V. As a result, the base current flows through the "diode" with less forward voltage, and the second transistor is off.

This trick can be visualized by connecting two LEDs with different threshold voltages in parallel as can be seen in this video.

Input HIGH. Another "wisdom" in this TTL input stage is that, when the input A/B is HIGH (connected to Vcc), the input transistor operates in "reverse mode", where its emitter and collector are swapped. As a result, its gain is very low, and a very small current is drawn from the previous stage.

Actually we do not need any gain here. We do not need a transistor; a humble backward-biased diode (like in DTL) will work better. So, in this state, the simpler DTL works better than TTL.