What about when one diode cathode is at 0V and the other is at Vcc?

Taking the case with both diodes at 0V and Vx=0.7V, if the diode D3 was removed, would the transistor be on or off? Do you need it to be on or off in this case? What does that tell you about the role of D3?

Before I answer to you, let me ask something about my own answer. I said that when both diode's cathodes are to +Vcc they are reverse biased, and therefor they are cut off and that D3 is also reverse biased beacuse voltage at anode and cathode is equal which is 0V. Now I think that this is not correct. If input diodes are cut off, I'll have a more positive voltage at the anode of D3 than at the cathode so it would be forward biased... Now I'm confused again!

I'll reply to your 2 questions after I have solved this last question of mine!

 

If input diodes are cut off, I'll have a more positive voltage at the anode of D3 than at the cathode so it would be forward biased

Exactly.
And remember that if the diode is reverse biased you can replace a diode with a open circuit. Or even remove it from the circuit.

 

Before I answer to you, let me ask something about my own answer. I said that when both diode's cathodes are to +Vcc they are reverse biased, and therefor they are cut off and that D3 is also reverse biased beacuse voltage at anode and cathode is equal which is 0V. Now I think that this is not correct. If input diodes are cut off, I'll have a more positive voltage at the anode of D3 than at the cathode so it would be forward biased... Now I'm confused again!

I'll reply to your 2 questions after I have solved this last question of mine!

That is correct. The way to treat diodes in a very simple way, which is almost (not always) good enough is to replace them with one of two things, either a 0.7V battery (or whatever voltage you want to use for the forward voltage drop) with the positive terminal at the anode and the negative terminal at the cathode if the diode is forward biased or an open circuit if the diode is reverse biased (which included forward biased but at a voltage less than the forward voltage drop). The test to see if you have made the right choice is that if you assume that the diode is forward biased, then the battery you replaced it with must have at least some current flowing through it from positive to negative. If the diode is not forward biased, then the voltage across the open circuit you replaced it with must be less than the forward voltage drop, anode to cathode.

 

Ok, thanks for replying to me and confirming my thoughts. I'll try to answer now the questions WBahn made a couple posts back!

When one of the diode's cathodes is connected to GND, then the same happends when both are grounded. In fact, there are 3 situations when this happends. Both input diodes grounded, D1 grounded (no matter D2) and D2 grounded (no matter D1).
This happends because in any of this 3 situations, Vx will be 0.7V which will be dropped by D3 forcing Vb to be 0V.
Now that all 4 situations are covered, what I don't understand is why only when both diodes are to +Vcc, the transistor is cut off. I just said that at those 3 situations (00, 01, 10, for input diodes) Vb is 0V, why transistor is conducting in this situations?

 

I'm not sure if I understand your previous post properly. But transistor will be TURN-ON only when both input diodes (D1 and D2) are connected to +Vcc. Why? Because there is a path for a base current from +Vcc to GND through +Vcc---> R1--->D3--->Q1 base-emitter --->GND.
And Vx is equal to ??

 

I'm not sure if I understand your previous post properly. But transistor will be TURN-ON only when both input diodes (D1 and D2) are connected to +Vcc. Why? Because there is a path for a base current from +Vcc to GND through +Vcc---> R1--->D3--->Q1 base-emitter --->GND.
And Vx is equal to ??

Sorry I have misspelled the post. Now it's correct. I made corrections in the 2nd line.
But the simulation I made in LTSpice shows that voltage in transistor collector is 5V when both input diodes are connected to +Vcc. Is the simulation wrong?

 

Sorry I have misspelled the post. Now it's correct. I made corrections in the very first line.
But the simulation I made in LTSpice shows that voltage in transistor collector is 5V when both input diodes are connected to +Vcc. Is the simulation wrong?

I already said that was wrong, but you still ignored it.

 

What?? From what I can see the simulation shows correct result. Notice that only at time larger than 1ms and smaller than 2ms both input diodes are connected to +Vcc. And in this time of period Vc is close to 0V

 

I already said that was wrong, but you still ignored it.

What is wrong ?? Are you do not like Rb resistor value ??

 

I already said that was wrong, but you still ignored it.

I'm sorry but I didn't saw that reply...

What?? From what I can see the simulation shows correct result. Notice that only at time larger than 1ms and smaller than 2ms both input diodes are connected to +Vcc. And in this time of period Vc is close to 0V

In that period of time colector voltage shows to be 0V in the green curve, Jony130!!!

 

In that period of time colector voltage shows to be 0V in the green curve, Jony130!!!

Exactly, this is how it should work. Are you trying to say that this is wrong ??

 

I'm sorry but I didn't saw that reply...

My mentioned was in #32.
If you don't know what I mean then to check the post in #31 by ericgibbs, check the circuits and comparing with yours maybe you will find the difference.

 

hi psy,
Its a 2 Input NAND circuit, look at this Truth table.

 

hi psy,
Its a 2 Input NAND circuit, look at this Truth table.

I'm a bit confused about the last posts in this page 3.

I'll try to resume:

Situation 1
D1 -- GND
D2 -- GND
------//-----
D1 -- GND
D2 -- +5Vcc
-----//-----
D1 -- +5Vcc
D2 -- GND

Vx = 0.7V
Vy = 0V
Vb = 0V

Transistor NOT conducting


Situation 2
D1 -- +5Vcc
D2 -- +5Vcc

Vx = Vγ3+Rb*Ib+Vbe..... (or Vx = Vcc + R1*Ib or Vx = Vcc + R1*I1)
Vy=Rb*Ib + Vbe ............. (or Vy = Vcc + R1*Ib + Vγ3 or Vy = Vcc + R1*I1 + Vγ3)
Vb = Vbe........................... (or Vb = Vcc + R1*I1 + Vγ3 + Rb*Ib or Vb = Vcc + R1*Ib + Vγ3 + Rb*Ib)

Note: the alternatives within curly braces are just to say that there are at least 2 ways to go to GND and also that I1 is the same as Ib.

So, for this situation, transistor should BE conducting, as D3 is forward biased.


Is this correct?


Now, Jony130 said that

I'm not sure if I understand your previous post properly. But transistor will be TURN-ON only when both input diodes (D1 and D2) are connected to +Vcc. Why? Because there is a path for a base current from +Vcc to GND through +Vcc---> R1--->D3--->Q1 base-emitter --->GND.
And Vx is equal to ??

But in the simulation, when both input diodes are connected to +Vcc, the collector voltage is 0V, so transistor is NOT conducting... That is what I'm understanding... But there is something I'm surelly understanding the wrong way!

 

But in the simulation, when both input diodes are connected to +Vcc, the collector voltage is 0V, so transistor is NOT conducting... That is what I'm understanding... But there is something I'm surelly understanding the wrong way!

The transistor is Conducting, all the +5V supply is dropped across the collector resistor, so the collector is close to 0V.!!

 

Ok, I got it... I was assuming that Vce = 0V transstor was not conducting... Wrong assumption...

Thank you all for one more hard journey on explaining things to a guy which is of hard understanding! I'm sorry for this!

Psy

 

But you still don't seem to have answered the question you had about what the role of D3 is in the circuit.

 

But you still don't seem to have answered the question you had about what the role of D3 is in the circuit.

WBahn, I forgot to answer to that here. Or better, I already knew the answer but didn't understood why. Now I think I understood. D3 is important to ensure that when any of the input diode's cathodes (or both cathodes) are grounded, that Base-Emitter junction is not forward biased and so that Vce is maximum giving us the logic level 1. If it wasn't there for these 3 situations, Base-Emitter junction would be forward biased not allowing the transistor to give us the logic level 1. I don't know if these are the best words to explain it... Maybe not.

 

WBahn, I forgot to answer to that here. Or better, I already knew the answer but didn't understood why. Now I think I understood. D3 is important to ensure that when any of the input diode's cathodes (or both cathodes) are grounded, that Base-Emitter junction is not forward biased and so that Vce is maximum giving us the logic level 1. If it wasn't there for these 3 situations, Base-Emitter junction would be forward biased not allowing the transistor to give us the logic level 1. I don't know if these are the best words to explain it... Maybe not.

I think you have the main idea. If any of the input values are LO, then the voltage at the node you've been calling X would be one diode voltage drop above ground. Depending on the diode, the transistor, the resistor values, the temperature, etc., etc., that might (or might not) be enough voltage at that node to cause the transistor to conduct enough such that the output voltage would fall out of spec for being a Logic HI. Since the input could be as high as 1V and still be called LO, this is almost certainly enough to turn on the transistor enough to bring the output voltage out of spec (it would be good for you to remove the diode and do the simulation). With the diode in place, the questionable behavior doesn't come into play until near the top of the allowed voltage at the input for a logic LO.

 

Thanks once more!
Psy