Engineering Help on current flow/direction (DTL circuit)

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The discussion focuses on understanding current flow in a DTL circuit involving transistors T1, T2, and T3. The "blue arrow" indicates that current flows only through path 1 because T2 is OFF, which is determined by the voltage at its base being insufficient to turn it on. T3 is ON, allowing current to flow through path 2, but T2 remains inactive due to its base voltage being lower than the required threshold. The analysis highlights that for T1 to pull the output low, T3 must be conducting, confirming the circuit's logic behavior. The conversation emphasizes the importance of understanding transistor states and their impact on circuit functionality.
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Homework Statement


Hello everyone,

In the circuit below I don't understand why the "blue arrow" only goes through path 1, instead of path 1, 2 and 3?

Homework Equations


The Attempt at a Solution



dtllogic.jpg
 
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It looks as though the diagram is taken out of some context where the "blue arrows" are part of a description of the circuit's operation, and no doubt referred to in that text.

The diagram indicates that transistor T3 is "ON", so it will conduct current via path 2 although it's not shown, probably because its not relevant to the point the descriptive text is making. T2, on the other hand is indicated to be "OFF", so no current should flow in path 3.
 
Hi gneill.

It really got too much out of context. It is like you said, it is a text that describes a circuit using transistors and diodes and what happens if both inputs (e1 and e2) are '1' (5V).

I don't understand why T2 is OFF.
 
frank1 said:
I don't understand why T2 is OFF.

Start by working out what is needed to produce a logic 0 at the output...

T1 must be ON pulling the output down so the Base of T1 must be about Vbe = 0.7V.

Current must be flowing...
a) through the resistor connected between the base of T1 and ground.
b) A little into the base of T1.

That current must come from the emitter of T3 so T3 must be ON. If T3 is ON then it's collector voltage will be about 0.3V (Vcesat) above the emitter. The emitter is at the same voltage (0.7V) as the base of T1 so the collector of T3 is about 0.7 + 0.3 = 1V.

Now look at T2. For T2 to be ON the base of T2 must be greater than about...

Vcesat of T1 + Vd + Vbe = 0.3 + 0.7 + 0.7 = 1.7V say

However the base of T2 is at same voltage as the collector of T3 which is 1V so T2 must be OFF.

Good job T2 is OFF when T1 is ON or the smoke will get out.
 
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