XOR Gate Circuit From Diodes

[sxy]

My answer to mine question:
I=V/R
So if the resistence exists the current must go through the resistence;
and from there to the ground.
And because of the parallel connection with the LED;
the same voltge drop will be on the LED.
Am I right?
Thanks

 

[sxy]

Anyway you agree with me that the current do exist and goes to the ground in the case of A=B=1?
Thanks a lot

 

[Langsdonspingel]

Hi, I have little effort to answer your question.

The diodes are used in a bridge configuration (Rectifier) to sort out the input logic into the positive level, means if there is a High state input it will always flow to the base of NPN to switch it on. And the Low state will always flow out to the emitter of NPN transistor. For more information also see https://www.electro-tech-online.com/10eloto/xor-gate-circuit-from-diodes.163375/

Thanks

 

[sxy]

Thanks for your answer; but in that circuit there are no transistors at all.
All the best

 

[sxy]

In the file atached you can see the circuit.
How the current can get to the Base of the transistor.
There is no path to the Base.
Thanks

 

[rjenkinsgb]

In that variation, whichever input is held to ground (not open circuit) keeps the base feed resistor low, while the other input being high is routed to the emitter and provides power.

If both inputs are high, there is no "low" to the base resistor so the transistor does not conduct, and if both inputs are low, no power to the emitter.

ps. Zener diodes are not appropriate, as too high a voltage between the two inputs will cause them to all conduct. 1N414 are not zeners anyway; the common signal diodes are 1N914 or 1N4148.

 

[rjenkinsgb]

To reduce the questions/arguments..
Current paths to the transistor.

If the input voltages are reversed, it will route through the opposite (left<>right) diodes and positive will still be to the emitter, negative to the base resistor.

 

[sxy]

Thanks for your answer.
What about the previous question
Am I right about the reason why the current splits to the ground and to the LED
in case of A=1 B=0.
All the best

 

[sxy]

My answer to mine question:
I=V/R
So if the resistence exists the current must go through the resistence;
and from there to the ground.
And because of the parallel connection with the LED;
the same voltge drop will be on the LED.
Am I right?
Thanks

 

[sxy]

XOR-Transistor circuit: in case of: A=0 ;B=1 what is the current path to the Base?
thanks

 

[rjenkinsgb]

in case of: A=0 ;B=1 what is the current path to the Base?

The opposite two diodes, as already stated: Base ground via upper right and power to emitter via lower left.

The previous two questions do not make sense [non-transistor circuit].
If one switch is on and the other off, one resistor is connected directly between power and ground; that will take some current but has no bearing on the LED current.

The other resistor is in a series chain from power via a diode, the LED, another diode and finally the resistor to ground. There is at no point a resistor in parallel with the LED alone.

 

[crutschow]

Since you seem interested in discrete XOR circuits, below is a two-transistor one for you to chew on:

 

[sxy]

Diode D2 and the LED are conected in parallel to the left resistor so the voltage drop on the resistor
equals to the voltage drop on both the diode D2 and the LED together.
thanks

 

[rjenkinsgb]

Diode D2 and the LED are conected in parallel to the left resistor so the voltage drop on the resistor
equals to the voltage drop on both the diode D2 and the LED together.

No.
Follow the current path through all the components.

The right hand resistor connects directly to the supply, via the switch. So, the full 9V across it and it has no effect on or relevance to the LED with the switches that way.

The current path through the LED consists of the upper right diode, the LED, the lower left diode and the resistor.

The voltage on the left hand resistor will be somewhere around 9V - (0.6 + 1.8 + 0.6) = 6V
(Assuming it's a red LED).

If B was on and A was off, just exchange left<>right for diodes and resistors & the same otherwise applies.

 

[sxy]

Thank you very much for your answer.
This explaination with the diagram was excellent;
but I didn't understand your previous explanation about the XOR gate with the transistor at all;
how exactly the current goes to the Base of the transistor if A=0 and B=1?
all the best

 

[sxy]

What do you mean in the sentence:
"If the input voltages are reversed..."
thank you

 

[rjenkinsgb]

What do you mean in the sentence:
"If the input voltages are reversed..."

A at 0V and B at V+.

In that case the resistor is pulled to 0.6V by the upper right diode and the emitter is connected to V+ via the lower left diode.

A bridge rectifier routes positive and negative to the appropriate output from either (or any, with multi-phase rectifiers) input.

 

[sxy]

Thanks for your answer.
How the current can get to the upper right diode; it has no path to get there.
I understand that input B has a route to the emitter; via the lower left diode;
but from input B you can not get to the upper right diode.
Maybe you can show it on the diagram please.
The lower right diode prevents from the current to get to Base resistor.
Thanks a lot

 

[rjenkinsgb]

How the current can get to the upper right diode; it has no path to get there.

With input A grounded, exactly the same way it gets to the lower right if input B is grounded...

Remember the base is going to be at roughly 0.6V less than the emitter, which is in turn 0.6V less than A or B, if either is connected to power.

The base resistor is positive and the grounded (0V) input negative, so the appropriate right side diode conducts and pulls the resistor down to around 0.6V above the 0V or low level input.

 

[sxy]

Thanks for the answer.
How exactly the current from input B can pass the low right diode that is reverse biased?
Can you please draw the current path input B=1 on diagram?
thanks a lot