# Question for 74LS00

#### georgiosSkod

Joined Nov 20, 2021
5
From the last few posts, I am able to understand the circuit with floating inputs. However I am not sure on how the absence of an input doesn't allow current flow from the base to the emitter.
In the inverter example, the impression I got is that the diagram would look like this:

In that case current from the emitter to ground should still be allowed, right?

#### LesJones

Joined Jan 8, 2017
3,702
With nothing connected to the input there is nowhere for emitter current to flow. The diode on the input will not conduct with the cathode connected to positive. (Via the 4k resistor and base emitter junction of the transistor.) Current from the 4K resistor flows through the base collector junction (Which behaves like a diode. ) and causes the second transistor to conduct.

Les.

#### MrChips

Joined Oct 2, 2009
25,950
A simple way of looking at it is to analyze a diode-resistor logic AND gate.

Logic output is LOW when A or B is tied to ground (OR gate for NEGATIVE LOGIC).
Logic output is HIGH when both A and B have HIGH inputs (or not connected).

#### KeithWalker

Joined Jul 10, 2017
2,318
From the last few posts, I am able to understand the circuit with floating inputs. However I am not sure on how the absence of an input doesn't allow current flow from the base to the emitter.
In the inverter example, the impression I got is that the diagram would look like this:
View attachment 253306
In that case current from the emitter to ground should still be allowed, right?
Papabravo answered this question in post #9. Why do you keep asking the same question?

#### crutschow

Joined Mar 14, 2008
29,537
In that case current from the emitter to ground should still be allowed, right?
Wrong.
Don't you know that a diode can conduct in only one direction?

#### KeithWalker

Joined Jul 10, 2017
2,318
Wrong.
Don't you know that a diode can conduct in only one direction?
Except for the reverse leakage current.

#### crutschow

Joined Mar 14, 2008
29,537
Except for the reverse leakage current.
Don't think that's pertinent to the TS's question.