BJTs and Logic Gates (basic stuff)

  • #26
3,003
2
When the emitter is connected directly to 0V it is at that potential... 0V ... it cant be at anything else cuz of the direct connection to the 0V line.
Now when you turn the transistor on you are now directly connecting the +6V line directly to 0V via the collector - emitter junction now you have a short circuit actoss the 6V power supply and as others said earlier the transistor is going to get very hot and probably die quickly.
The 4k7 resistor in the emitter line (leg) to 0V "lifts" the emitter / Vout point above the 0V potential.

Na... no hijack guys :) ya all chimed in whilst I was sleeping through sunday nite and working through monday morn :) thanks for the added input.

I'm not sure how many other ways I can explain this so Saladsamurai finally grasps what is happening :)

Dave


I know, I know guys ... I suck. Haha! And FYI, as infuriating as this is (my lack of understanding), I really am enjoying this too! This is how I learn something new. I struggle through it. I am not gifted by any means (hence the high post count consisting of all questions like this :smile: ). But in the course of the struggle, I pick up so much more information through these interactions than I ever could through a text. I could just as easily accept all of what you are saying at face value and go ahead and build some circuits in the lab and be done with it. But I think there is something more to be had from the learning process.

Ok ... back on topic. Let's discuss this part, since this exact image has crossed my mind during our discourse (added some labels to the image):

ok... forget about the transistor being there for a minute, consider these 3 ccts (A, B, and C) below ....

what do you think the voltage in cct A will be at point 1
what do you think the voltage in cct B will be at point 2
what do you think the voltage in cct C will be at point 3 and point 4 (Vout)

imagine in cct C the first resistor @ 0.5 Ohms is the transistor junction resistance

explain why you will get the voltage readings you do

NEW3Batts.jpg
In Image A:
At point 1 I would have to say that the voltage is 6V relative to the negative terminal. If we start at point d and travel upward through the loop, there is a voltage rise of 6V and then no sink. So at every point in the wire, the voltage is 6V.

In Image B:
At point 2, again the voltage is 6V. If we start at point h and travel upward through the loop, there is a voltage rise of 6V and then no sink before point 2. So any point in the wire from the + terminal to the resistor has a voltage of 6V.

In Image C:
At point 3, same deal of 6V for the same reason, start at point m and travel upward through the loop, there is a voltage rise of 6V and then no sink before point 3. At point 4 (Vout) the voltage would be 6V - I1R1 or from Ohm's Law Vout = 6V - 6V*R1/(R1+R2).
 
  • #27
davenn
Science Advisor
Gold Member
2019 Award
9,237
7,506
Ok ... back on topic. Let's discuss this part, since this exact image has crossed my mind during our discourse (added some labels to the image):

In Image A:
At point 1 I would have to say that the voltage is 6V relative to the negative terminal. If we start at point d and travel upward through the loop, there is a voltage rise of 6V and then no sink. So at every point in the wire, the voltage is 6V.
Ahhh No :) --- its a wire straight from the + to the - terminal of the battery, a short circuit, Voltage will be 0V, current will be maximum.
try it with a really small battery say a 1.5V AA battery or a 6V lantern battery
put your multimeter across the terminals read the voltage 1.5V or 6V whatever the battery
now VERY briefly short out the battery terminals and watch the voltage drop to 0V.
Please dont do it on a car battery, even briefly it can supply massive short cct current

This is the bit you need to understand to see the picture in what is happening in the transistor cct if the 4k7 resistor isnt in the circuit between the Emitter and the 0V line
The transistor junction, with only a VERY tiny C-E resistance is basically a short cct across the +6V power supply.

In Image B:
At point 2, again the voltage is 6V. If we start at point h and travel upward through the loop, there is a voltage rise of 6V and then no sink before point 2. So any point in the wire from the + terminal to the resistor has a voltage of 6V.
correct

In Image C:
At point 3, same deal of 6V for the same reason, start at point m and travel upward through the loop, there is a voltage rise of 6V and then no sink before point 3. At point 4 (Vout) the voltage would be 6V - I1R1 or from Ohm's Law Vout = 6V - 6V*R1/(R1+R2).
Correct
Total current flow in cct It = V / R1 + R2 = 6V / 4700 Ohms + 0.5 Ohms =
6V / 4700.5 = 1mA
So voltage drop across 0.5 Ohm resistor is V = I x R = 0.001A x 0.5 Ohms = 0.0005 V

But that low V drop is only cuz of the presence of the 4k7 resistor, which is limiting the current flow through the cct to 1mA

Dave
 

Related Threads on BJTs and Logic Gates (basic stuff)

Replies
8
Views
5K
  • Last Post
Replies
3
Views
705
  • Last Post
Replies
13
Views
3K
  • Last Post
Replies
4
Views
782
  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
10
Views
589
Replies
9
Views
1K
  • Last Post
2
Replies
25
Views
9K
Top