I believe whats happening is when you reverse bias the pn junction of the transistor, and when the electrical field around the depletion region of the pn junction increases to about 7 volts, what happens is a charge carrier movement across the pn junction allowing small current to flow, this current which flows through the resistor at the emitter causes a linear voltage drop across this resistor, and since its a small resistor compared to the resistance of the depletion region of the pn junction, the resistor drops a small amount of the voltage of Vb, and the rest is dropped over the transistor and that is what u r measuring, its not negative resistance, its basically when u increase the voltage, the stronger the electric field around the depletion region of the pn junction the more charge carriers cross the junction in form of current, the more current through the emitter resistor, the more voltage drop across it, the LESS voltage is left to be dropped over the transistor, and thats why it looks like negative resistance but its not. This is how i see it.
It's not the reduction in Vec that needs to be explained though. As you say, the voltage drop across R as the current begins to flow results in the reduction to Vec, but when Vec then drops below the threshold voltage the current still flows and actually increases. So I'm wondering what happens to the current when Vb is gradually reduced below the threshold AFTER the current starts flowing and it's messing with my head so I had to do this experiment myself. After bringing Vb up above the threshold and registering the current, which increased as expected even though Vec was below the threshold, I then slowly reduced Vb again until it was below the threshold and Vec increased accordingly and the current dropped. I can't say I was surprised by this, it would be naive to expect anything different, but intuitively I would expect the current flow to be dependent on Vec (not Vb) being above the threshold. I think this demonstrates that it is actually "negative resistance" even though strictly speaking I would consider resistance to be simply V/I, not dV/DI, as someone else has pointed out. It's actually called "negative differential resistance" according to the Wikipedia article.
I'm over-thinking this obviously, because it's clear that once the reverse-bias threshold has been exceeded, the current is simply determined (pretty much) by the voltage drop across R. So you are probably right. I'm glad I did the experiment myself though. [EDIT] Damn it, I must be confused. Is it the current that determines the voltage drop, or vice versa?
Great work. May I recommend that we need to use constant voltage source to get more accurate results for such experiments - the reason for saying so is off the shelves batteries may not exhibit a "Constant Voltage" current source power supply. It is always great idea to have a bench type power supply with adjustable voltage and current sources at the output, this way we will be more professional at doing this. You may google the definition of Constant Voltage Current Source (CVCS) or CCVS. You have great collection of great videos and I thank you for all of your valuable work...
I am about to finish the physical chemistry series. While it’s challenging, I think you could early make videos on all of the topics as an additional playlist. The videos out there aren’t the best and they aren’t as nearly good as your videos. Plus you will make tons of likes and subscribers especially those who struggled with it. I think you will enjoy it since you’re perfect at math. Just an advice from one of your fans.
Here is a semiconductor physics explanation of this phenomenon. You basically created a tunnel diode in the Base region. As you increase the reverse bias you increase size of the depletion region to a point where the un-inverted region becomes so small that because of the quantum mechanical properties of electrons, and thin un-inverted region they can tunnel through the barrier. Do a search on tunnel diode oscillators ( reverse diode oscillators).
in a simpler way, simply voltage is negative resistance or "inverse resistance" in the way that if u have 2 batteries in series it cancels out the resistor inbetween and shorts get shorted away from!!!
I love this guy and his one and only intro “in this video”
That's relatable
Although I neither study nor like electrical science, I really love your videos. THANK YOU SIR !! ❤
I believe whats happening is when you reverse bias the pn junction of the transistor, and when the electrical field around the depletion region of the pn junction increases to about 7 volts, what happens is a charge carrier movement across the pn junction allowing small current to flow, this current which flows through the resistor at the emitter causes a linear voltage drop across this resistor, and since its a small resistor compared to the resistance of the depletion region of the pn junction, the resistor drops a small amount of the voltage of Vb, and the rest is dropped over the transistor and that is what u r measuring, its not negative resistance, its basically when u increase the voltage, the stronger the electric field around the depletion region of the pn junction the more charge carriers cross the junction in form of current, the more current through the emitter resistor, the more voltage drop across it, the LESS voltage is left to be dropped over the transistor, and thats why it looks like negative resistance but its not. This is how i see it.
It's not the reduction in Vec that needs to be explained though. As you say, the voltage drop across R as the current begins to flow results in the reduction to Vec, but when Vec then drops below the threshold voltage the current still flows and actually increases. So I'm wondering what happens to the current when Vb is gradually reduced below the threshold AFTER the current starts flowing and it's messing with my head so I had to do this experiment myself. After bringing Vb up above the threshold and registering the current, which increased as expected even though Vec was below the threshold, I then slowly reduced Vb again until it was below the threshold and Vec increased accordingly and the current dropped. I can't say I was surprised by this, it would be naive to expect anything different, but intuitively I would expect the current flow to be dependent on Vec (not Vb) being above the threshold. I think this demonstrates that it is actually "negative resistance" even though strictly speaking I would consider resistance to be simply V/I, not dV/DI, as someone else has pointed out. It's actually called "negative differential resistance" according to the Wikipedia article.
I'm over-thinking this obviously, because it's clear that once the reverse-bias threshold has been exceeded, the current is simply determined (pretty much) by the voltage drop across R. So you are probably right. I'm glad I did the experiment myself though.
[EDIT] Damn it, I must be confused. Is it the current that determines the voltage drop, or vice versa?
Have a good day man, needed this
thank you so much sir! you are so helpful I am so grateful that you keep making videos
Great work. May I recommend that we need to use constant voltage source to get more accurate results for such experiments - the reason for saying so is off the shelves batteries may not exhibit a "Constant Voltage" current source power supply. It is always great idea to have a bench type power supply with adjustable voltage and current sources at the output, this way we will be more professional at doing this. You may google the definition of Constant Voltage Current Source (CVCS) or CCVS. You have great collection of great videos and I thank you for all of your valuable work...
no. running it this way makes sure it wont blow up if for some reason it decides to drop its resistance to 20 ohms for whatever reason
Dude do more of these . With experiments.. this video is fantastic
I am about to finish the physical chemistry series. While it’s challenging, I think you could early make videos on all of the topics as an additional playlist. The videos out there aren’t the best and they aren’t as nearly good as your videos. Plus you will make tons of likes and subscribers especially those who struggled with it. I think you will enjoy it since you’re perfect at math. Just an advice from one of your fans.
heyman
Here is a semiconductor physics explanation of this phenomenon. You basically created a tunnel diode in the Base region. As you increase the reverse bias you increase size of the depletion region to a point where the un-inverted region becomes so small that because of the quantum mechanical properties of electrons, and thin un-inverted region they can tunnel through the barrier. Do a search on tunnel diode oscillators ( reverse diode oscillators).
So is negative resistance used in any practical application or is it simply a phenomenon that must be contended with when building circuits?
in a simpler way, simply voltage is negative resistance or "inverse resistance" in the way that if u have 2 batteries in series it cancels out the resistor inbetween and shorts get shorted away from!!!
Thanks ❤
Nice video ........
........
do anyone know what blackboard app are ''the organic chemistry tutor'' use
i think he uses a tablet. If you go on his website (link in description), on "My youtube experiment", it shows what he uses
Pls what software did you use to make this video
You may use Easy Video (ezvid) - it is free to use but you need to familiarise on its controls
I would still consider that positive resistance. V/I is still a positive number, even if dV/dI is a negative number.
Try made Oleg Lossev crystadine receiver with this transistor.
Bro actually I want to knw give detail explanation about negative resistance
Ca ne marche pas