Help on current flow/direction (DTL circuit)

  • Context: Engineering 
  • Thread starter Thread starter frank1
  • Start date Start date
  • Tags Tags
    Circuit Current
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 2K views
frank1
Messages
25
Reaction score
0

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
 
Last edited:
Physics news on Phys.org
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.