Engineering Need help understanding the transistor circuit of an Boolean NOT gate

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The discussion focuses on understanding the transistor circuit of a Boolean NOT gate. A user seeks clarification on their circuit design, which is deemed incorrect due to missing components and potential short circuits. Key corrections include the necessity of a base resistor (Rb) to limit current and a collector resistor (Rc) to prevent shorting when the transistor is on. The correct operation of the circuit is explained: when the input is logic 1, the output is logic 0, and vice versa, effectively demonstrating the NOT gate function. Additionally, the current flow direction is clarified based on the transistor's state.
x64bob
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need help understanding the transistor circuit of a Boolean "NOT gate"

can some one please link or upload a scheme or would this one i attached work? and how do you display in which direction the current flows? that's what I'm trying to show with the battery halves...
 

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No, what you have drawn will not work as an inverter. Just look at the simplified schematic of some inverter gates in their datasheets:

74S04 -- http://rocky.digikey.com/scripts/ProductInfo.dll?Site=US&V=296&M=SN74S04N

74HCU04 (unbuffered inverter) -- http://rocky.digikey.com/scripts/ProductInfo.dll?Site=US&V=296&M=SN74HCU04N
 
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What I believe is: you want to let the transistor act like an inverter.
The circuit you attached won't work because:
1 - you must put a resistor Rb on the base of the transistor.
2 - when the input is logic 0, the transistor is turned ON and it will behaves like a wire, therefore , a short is going to happen because the ON state transistor will short the 6V to ground.

I attached a circuit that will work properly... and here you are:

[PLAIN]http://www.kpsec.freeuk.com/images/trinvert.gif

As you can see, there are 2 resistors in this circuit: Rc and Rb...

Rc is put so there will be no short circuit when the transistor is ON.
Rb is put to limit the base current ib.

This circuit will work properly as a Boolean "NOT Gate"

The logic 1 input signal will turn the transistor ON, so it will act like a wire, and the output is shorted to the ground, which means the output is logic 0.
And vice versa, when the input signal is logic 0, the transistor turns OFF, and there will be no current passing through the resistor, therefore the whole +Vs voltage will drop between the collector and the ground as output voltage, so the output voltage is logic 1 now !

As a summary, when the input signal is logic 1 , the output signal is logic 0...but when the input voltage is logic 0 , the output voltage is logic 1...so the circuit is acting just like a "NOT GATE" !

I hope this helps...
 
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I forgot about the current...
When the transistor is on (i.e. the input signal is logic 1) the current will pass from the +Vs to ground, in other words from the collector toward the emitter.
But when the transistor in off (i.e. the input signal is logic 0) no current will flow through the transistor because it's OFF.
If the output signal is applied to a load (e.g. Resistor RL) the current will flow from +Vs through Rc and then through the load RL to the ground.
 

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