The timestamps for the different topics covered in the video: 1:11 What is Common Base Configuration 2:55 Input Characteristics of the Common Base Configuration 5:45 Output Characteristics of the Common Base Configuration 6:33 Three Different regions of operation (Active, Saturation, and Cut-off) 10:52 Signal Amplification in CB Configuration
all videos are just awesome unbelievable that that area vailiable in utube thank god i got it some how . thanks a lot to u carrry one for other topic also .
I might've missed it, but introducing the shottkey equation would help. That way, base emitter resistance can be introduced without hindrance, not to mention even that output curve.
Valuable information thank you so much you made it this topic easy and now i also able to explain this topic to others once again thanking you sir thank you so much
great video. However I have aquestion. We can see that in output, collector current is given by alpha times emitter current. I want to know why does the output voltage (Vc) not have an effect on that. I know the mathematical relation but whats the actual reason?
The discussion of three connection modes are all mixed up in three videos. Looks like providing the same confusion poorly organised text books are providing. 1. The fact that base current turns on the transistor is common thing, why emphasise that in Common base case configuration? 2. Input and output impedance should be called out in all the configurations. Did you omit that in common emitter configuration, I am not sure? 3. Why common collector discuss alpha beta and gamma. Isn't that a general transistor characteristics? I mean it is good for exam. In other words why a circuit designer should consider these values is more important than jumbled mathematical relationships between them. Beta is the only useful stuff according to me. I am wrong, but your lecutre doesn't point out where I am wrong. 4. Practical help on designing a rudimentary amplifier or switch using a transistor and real available resistors would make these sessions more interesting. My englinnering lectures and labs didn't teach me that, had to learn that myself after playing with some real components and breadboard. 5. Actually the stuff is simple and should be taught in high schools without the scary charts.
Hello Anand, this is a pretty old comment but I loved this comment. The fact that the whole idea of a transistor is so poorly organized in almost all books is notable. Also the idea of what is common is simply the idea of what factor are you measuring. No lecture and book emphasize on this.
@ 5:01 You mentioned the Input impedance [is the slope of the curve; i.e TanX. But here, TanX=Per/Base= Ie/Vbe; which doesn't seem to be the Impedance from the curve. Please correct it.
Impedance is resistance here (R = V/I). For input characteristics, changes in I is very big compared to the changes in V. That's why, impudance is low.
Hello Sir, I have a question for the constant Ic when Vcb increase. As what you said in common emitter transistor, as Vcb incrases, the depletion region will increase so that the effective width between base and emitter decreases. Thus, more electrons will flux into collector region. That explanation could also be applied here. Then Ic should not be constant. Instead it should also increase. But why does the current keep constant here? Thank you very much for your video!
great explanation sir,i have following questions when i watched this video 1).why you don,t use output impedance in your calculations?is that not effect for output voltage and current? 2).same current flow through the circuit from input side to output side . but voltage is increasing without using any voltage source.so the power is increased .how that happened?
the thing is we are also applying the DC voltage to the BJT. AC input signal is the one which we want to amplify. But to do so, we need to bias the BJT (we need to apply the proper DC voltage to the BJT). This external DC biasing helps in the amplification. There is a conservation of energy. For more info, I would recommend you to watch the video on BJT- Large signal model. Here is the link: ruclips.net/video/ME41uPsDxFo/видео.html
During the whole video you have explained using an NPN transistor. Why couldn't you use NPN while showing the amplification example at last ? Why did you have to switch to PNP ? Can you please demonstrate the amplification using an NPN transistor too ?
At 4:55 , u said input impedance as slope of the graph... And we all know slope is always defined as change in y divided by change in x....but then u define Ri= change in Vbe / change in Ie.. Can you explain?
Ri is the inverse of the slope. The slope gives you the conductance. That's why it is change in the Vbe to the change in Ie. I hope it will clear your doubt.
Sir I found out one mistake in your video as per my knowledge. In your video you told that the output characteristics curve for common base configuration is linear but I verified from Sedra Smith that the curve is not the liner it has a small positive slope which indicates that the Ic depends upon Vcb in the active mode.
Increasing Vce implies that the depletion region increases, implying that less voltage would be required to forward bias Vbe. How does that make any sense? Shouldn't it have been the other way around
Minority charge carriers are responsible for current if the diode is connected in reverse bias is that mean minority charge carriers in both p side and n side??plz answer...
when we say, the common base, the base is common for the AC signal. That means the AC input signal is applied between the emitter and the base, while the output is measured between the collector and the base. For the DC biasing perspective, the base terminal may not be at the ground potential. But to use the BJT as an amplifier, the base-emitter junction has to be forward biased and the collector-base junction has to be reverse biased. (To ensure the operation in the active region) Actually w.r.t to ground there will be some finite DC voltage at the base. Here to simplify the analysis it shown such way. I hope it will clear your doubt. If you still have any doubt then do let me know here.
I have a doubt, why in output characteristics we consider Current of Collector, since current is going inside the transistor so shudnt it be input characteristics? Or do we consider the direction of electron?
sir is it supposed that when we increased VCB then we make the junction between base and collector more reversed then the collector current reduced not to be constant ?
I want to thank you again. How did the high output resistance help this circuit? As the output voltage was dependent on load resistant. I figured from the output characteristics, that the high o/p resistance is helping for a current mirror as the current does not change much but does it help for voltage amplification purpose?
Because it is PNP transistor. So, its direction will be opposite to NPN transistor. The transistor shown earlier was NPN transistor. I hope, it will clear your doubt.
Yes, exactly. I was wondering this comment. He used PNP transistor here, but in both NPN and PNP, the i/p will be along VBE side and o/p will be on VCB side. So there's no problem with the equations though.
I have posted a few Quiz related to BJT on the second channel. Here is the link: ruclips.net/channel/UCGA2TO8ylVqFHpucwn_6Jlw And soon many more will be posted.
I have one doubt in my mind ..if collector Base junction is in reverse bias then why electrons from emitter move from reverse bias depletion region ? Why dont electron pushed back to emitter by repeltion of negative ions of P-region of base?Is there breakdown occur at reverse bias depletion region? Please answar and clear my doubt
Here, emitter base junction is forward baised. That means electrons will be move towards the base region (Due to the repulsive force by the negative terminal of the voltage Vee ) and holes will get atteracted towards the negatvie terminal of the emitter voltage (Vee). At the same time collector base junction is reverse biased. So, once the electrons enters the base region, they will behave like a minority charge carriers. And they will get pushed by the electric field of the reverse bias voltage at the collecor base terminals. So, as a minority carrier they will reach into the collector region . And then by the force of the positive terminal of the Vcc, they will get attracted towards it. I hope, I am able to clear your doubt. If you still have any doubt, please let me know here.
Thank you very much..i understood this concept its due to electric field..Can u clear one more doubt ..how BJT amplified signal means increase output power if we know that power is fixed and we can't generate it ..i understand mathematical reason of voltage and current amplification but what is physical reason?..Can we call BJT mixing input power with battery power so output power become high?
Typically the output impedance in this configuration is very high. It will come in parallel with RL. And if we take the parallel combination of those two resistors then it is close to RL. That’s why here it is not considered. For exact calculations you can watch the video on common base amplifier. Here is the link : ruclips.net/video/MckBbhfxdkk/видео.htmlsi=dYJmsP0UI30uVH_q
I have a doubt, you have written in properties of CB that it has low power gain, but if current gain is 1 then power gain is same as the voltage gain and so it is high. Right?
I have a dought sir, what is the difference between Vbb, Vee and Veb when the base is common. And the volatege is always apllied between base and emitter then why three different notations are used
How is the voltage from the Base to Emitter considered to forward biased for both the PNP and NPN when they are on? Is this not a mistake? If you're talking about Vbe in both cases then they both can not be considered forward biased. The only way it makes sense is if in regards to the PNP you are saying Veb is forward biased. But why do it like that?
Its really a good video ,but I have one doubt , voltage gain is actually based on load resistance, which we can change according to our convenience than why its rage is fixed?
In common base amplifier, along with load resistor, the gain also depends on the collector resistor and the transconductance. That's why just increasing the load resistance alone, you will not see much increase in the gain after some point. For more information, about the gain of the common base amplifier, you may refer this video: ruclips.net/video/MckBbhfxdkk/видео.htmlsi=1KOQ9EApVd8GTiZy
y in active region, Vcb didn't affect Ie, but at input characteristic there Vcb increase, Ie also will increase?!!!!!can someone explain to me, urgent!!!
It depends how the amplifier is configured. Some amplifier amplifies current, some voltage, while some amplifiers are power amplifier and amplifies both current and voltage.
You will get the output between the collector and the base terminal. But usually, the output is connected to the load. So, if you connect the load resistor RL across the two-terminal then you will get the same output across the load resistor. I hope it will clear your doubt.
If collector base junction width is increased then base current will decrease and collector current increases but i dont understand why emitter current increased if the collector base juction width is increased?
@@aamir99204 Yes, just to explain the signal amplification in CB configuration, the PNP transistor was shown. (since the direction of input and output currents are more relatable in PNP CB configuration)
In PNP transistor, EB junction is forward-biased. (Emitter is P, and base is N), while CB junction is reverse-biased. (Collector is P and base is N). In the NPN transistor, BE junction is forward biased (Base is P and Emiiter is N), while BC junction is reverse biased. (Collector is N and base is P, so higher voltage is applied at Collector side). I hope, it will clear your doubt.
@@ALLABOUTELECTRONICS but if base-emitter in NPN is foward biased, that means current flows into emittor(N) and out of the base(P), because that is how a [NP] diode would behave in foward bias. But current is flowing out of the emitter in your NPN diagram, which makes the emitter-base reverse bias.
@@chibuzordesmond3937 I would recommend you to watch the basic video on BJT. If you will watch it, then your doubts will get clear. Here is the link: ruclips.net/video/-VwPSDQmdjM/видео.htmlsi=B9kH68XWb2-uVTNo
@@ALLABOUTELECTRONICSI've watched it and my understanding now is that current always flows from + to - in the circuit, I've been taking the convention for granted but let me know if that is correct.
The timestamps for the different topics covered in the video:
1:11 What is Common Base Configuration
2:55 Input Characteristics of the Common Base Configuration
5:45 Output Characteristics of the Common Base Configuration
6:33 Three Different regions of operation (Active, Saturation, and Cut-off)
10:52 Signal Amplification in CB Configuration
Some the most valuable 15 minutes of my life
🌚🌝
Of...
Your videos are such treasure for people wanting to learn electronics, thank you!
You are like a gift to people like us who want to learn something. It was more productive than my engineering education. Thank you.
I was sufferd from this concept in my yesterday class .
It is very helpful for me
Thank you bro
Bhai tu cricket pe dhyan de
@@parthsharma7146 haha
Awesome
😮
This will help me to get good marks in my sem exam...thanks sir😍
Transfer + Resistor = Transistor. This was my wow moment!
Best class. I have ever done!
Hat's off🌻, Sir..
From Bangladesh... 🙂
also from bd.
all videos are just awesome unbelievable that that area vailiable in utube thank god i got it some how . thanks a lot to u carrry one for other topic also .
Neso academy and this channel are surviving my life
May god bless you sir, you are like a hero to us
Wow this video waited 3 years for me....
I'm so grateful, thank you sir😍
I might've missed it, but introducing the shottkey equation would help. That way, base emitter resistance can be introduced without hindrance, not to mention even that output curve.
Yes, thats true. But I just wanted to keep it as simple as possible.
Sir a 100 thanks for clearing the basics of transistor
Valuable information thank you so much you made it this topic easy and now i also able to explain this topic to others once again thanking you sir thank you so much
Excellent teaching than our polytechnic college faculty
vidyalankar?
Crystal clear explanation👍
great video. However I have aquestion. We can see that in output, collector current is given by alpha times emitter current. I want to know why does the output voltage (Vc) not have an effect on that. I know the mathematical relation but whats the actual reason?
Very nice every point is covered
The discussion of three connection modes are all mixed up in three videos. Looks like providing the same confusion poorly organised text books are providing.
1. The fact that base current turns on the transistor is common thing, why emphasise that in Common base case configuration?
2. Input and output impedance should be called out in all the configurations. Did you omit that in common emitter configuration, I am not sure?
3. Why common collector discuss alpha beta and gamma. Isn't that a general transistor characteristics? I mean it is good for exam. In other words why a circuit designer should consider these values is more important than jumbled mathematical relationships between them. Beta is the only useful stuff according to me. I am wrong, but your lecutre doesn't point out where I am wrong.
4. Practical help on designing a rudimentary amplifier or switch using a transistor and real available resistors would make these sessions more interesting. My englinnering lectures and labs didn't teach me that, had to learn that myself after playing with some real components and breadboard.
5. Actually the stuff is simple and should be taught in high schools without the scary charts.
Yes, that’s true :(
Hello Anand, this is a pretty old comment but I loved this comment. The fact that the whole idea of a transistor is so poorly organized in almost all books is notable. Also the idea of what is common is simply the idea of what factor are you measuring. No lecture and book emphasize on this.
excellent tutorial. Thank you!
Beautiful explanation 💟
Very nice explanation.
@ 5:01 You mentioned the Input impedance [is the slope of the curve; i.e TanX.
But here, TanX=Per/Base= Ie/Vbe; which doesn't seem to be the Impedance from the curve.
Please correct it.
Impedance is resistance here (R = V/I). For input characteristics, changes in I is very big compared to the changes in V. That's why, impudance is low.
Hello Sir, I have a question for the constant Ic when Vcb increase. As what you said in common emitter transistor, as Vcb incrases, the depletion region will increase so that the effective width between base and emitter decreases. Thus, more electrons will flux into collector region. That explanation could also be applied here. Then Ic should not be constant. Instead it should also increase. But why does the current keep constant here? Thank you very much for your video!
great explanation sir,i have following questions when i watched this video
1).why you don,t use output impedance in your calculations?is that not effect for output voltage and current?
2).same current flow through the circuit from input side to output side . but voltage is increasing without using any voltage source.so the power is increased .how that happened?
the thing is we are also applying the DC voltage to the BJT. AC input signal is the one which we want to amplify. But to do so, we need to bias the BJT (we need to apply the proper DC voltage to the BJT). This external DC biasing helps in the amplification. There is a conservation of energy.
For more info, I would recommend you to watch the video on BJT- Large signal model.
Here is the link: ruclips.net/video/ME41uPsDxFo/видео.html
@@ALLABOUTELECTRONICS thank you sir ,i am following your tutorials and those are very helpful to me for develop my knowledge
During the whole video you have explained using an NPN transistor. Why couldn't you use NPN while showing the amplification example at last ? Why did you have to switch to PNP ? Can you please demonstrate the amplification using an NPN transistor too ?
4:32 “decrease”.
Excellent explaination. Thanks so much
At 4:55 , u said input impedance as slope of the graph... And we all know slope is always defined as change in y divided by change in x....but then u define Ri= change in Vbe / change in Ie.. Can you explain?
Ri is the inverse of the slope. The slope gives you the conductance. That's why it is change in the Vbe to the change in Ie.
I hope it will clear your doubt.
Sir I found out one mistake in your video as per my knowledge. In your video you told that the output characteristics curve for common base configuration is linear but I verified from Sedra Smith that the curve is not the liner it has a small positive slope which indicates that the Ic depends upon Vcb in the active mode.
No bro,he told that from ideal case.What you have said is the practical case.In practical case Ic=(alpha×Ie)+Icbo
Yes it is not linear, but for approximation we take it for linear
Thanksssss time saveddd❤❤♥♥🔥
Great video as usual but I think you made quite a few mistakes confusing reversed bias and forward bias.
great explanation thank u
Increasing Vce implies that the depletion region increases, implying that less voltage would be required to forward bias Vbe. How does that make any sense? Shouldn't it have been the other way around
Minority charge carriers are responsible for current if the diode is connected in reverse bias is that mean minority charge carriers in both p side and n side??plz answer...
In common base mode, what is the potential at the base? (The positive terminal of one battery is connected to the negative terminal of the other)
when we say, the common base, the base is common for the AC signal. That means the AC input signal is applied between the emitter and the base, while the output is measured between the collector and the base. For the DC biasing perspective, the base terminal may not be at the ground potential. But to use the BJT as an amplifier, the base-emitter junction has to be forward biased and the collector-base junction has to be reverse biased. (To ensure the operation in the active region)
Actually w.r.t to ground there will be some finite DC voltage at the base.
Here to simplify the analysis it shown such way.
I hope it will clear your doubt.
If you still have any doubt then do let me know here.
@@ALLABOUTELECTRONICS Thank you!!
Everything is explained so perfectly. But The Properties of Common Base Configuration are not mentioned??
Very clear consept
I have a doubt, why in output characteristics we consider Current of Collector, since current is going inside the transistor so shudnt it be input characteristics? Or do we consider the direction of electron?
sir is it supposed that when we increased VCB then we make the junction between base and collector more reversed then the collector current reduced not to be constant ?
When reverse voltage increased depletion region gets wider thus base width decreases which allows Ieto pass Vcb junction.Hence Ic almost equal to Ie
Thank you so much!!
Thanks for this video!
13:18 but how it possible that we give low potential and we get high potential energy conservation is valid here?
Thank you for this video
Thank you so much sir ❤🙏👍
4:32 why if depletion region is wider then base is narrower??
Well done
I want to thank you again. How did the high output resistance help this circuit? As the output voltage was dependent on load resistant. I figured from the output characteristics, that the high o/p resistance is helping for a current mirror as the current does not change much but does it help for voltage amplification purpose?
Same doubt . Now u have an idea if so kindly share
Thank you sir 🌟
Sir Aap bohathi achhi video banatehe
Thank you
Very well explained.. keep it up
At 12:04, why suddenly the current directions have reversed when compared to what u had been showing in first 12 minutes of the video?
Because it is PNP transistor. So, its direction will be opposite to NPN transistor. The transistor shown earlier was NPN transistor. I hope, it will clear your doubt.
Thank you sir 💜🙏😇
Sir at 11:06 you have taken pnp transistor but earlier you took NPS transistor it’s a bit confusing .
Yes, exactly. I was wondering this comment. He used PNP transistor here, but in both NPN and PNP, the i/p will be along VBE side and o/p will be on VCB side. So there's no problem with the equations though.
Well explained .. tqs sir
could you mak e a video about solving sample problems in three BJT configuration the CB,CE,CC...plsss
I have posted a few Quiz related to BJT on the second channel.
Here is the link:
ruclips.net/channel/UCGA2TO8ylVqFHpucwn_6Jlw
And soon many more will be posted.
What happens to collector current if base current increases tell the reason plz...
Thanx and very helpful
Why electrons didn't cross the collector base junction while operating in saturation regions
The Ro || RL is neglecting and assuming the parallel is almost 1K, right? 13:16
Yes.
I have a doubt on that
As we see that r0 is 1000 kohms and how rl becomes 1kohms
Nice video
Keep going on
Very good 👌
Timestap: 11:02 dc biasing vokrages are not shown means... Both ac abd dc were applied simultabeously.. Kindly clear.. M from electrical background
Yes, both are applied simultaneously. Just for simplicity, DC voltages are not shown.
I have one doubt in my mind ..if collector Base junction is in reverse bias then why electrons from emitter move from reverse bias depletion region ? Why dont electron pushed back to emitter by repeltion of negative ions of P-region of base?Is there breakdown occur at reverse bias depletion region? Please answar and clear my doubt
Here, emitter base junction is forward baised. That means electrons will be move towards the base region (Due to the repulsive force by the negative terminal of the voltage Vee ) and holes will get atteracted towards the negatvie terminal of the emitter voltage (Vee).
At the same time collector base junction is reverse biased.
So, once the electrons enters the base region, they will behave like a minority charge carriers. And they will get pushed by the electric field of the reverse bias voltage at the collecor base terminals.
So, as a minority carrier they will reach into the collector region . And then by the force of the positive terminal of the Vcc, they will get attracted towards it.
I hope, I am able to clear your doubt.
If you still have any doubt, please let me know here.
Thank you very much..i understood this concept its due to electric field..Can u clear one more doubt ..how BJT amplified signal means increase output power if we know that power is fixed and we can't generate it ..i understand mathematical reason of voltage and current amplification but what is physical reason?..Can we call BJT mixing input power with battery power so output power become high?
Why minority charge carriers are responsible for current in reverse bias
Great video
I didn't waste my 15 mins
🙏
Respected sir, Why we don't include output impedance while calculating output current?
Typically the output impedance in this configuration is very high. It will come in parallel with RL. And if we take the parallel combination of those two resistors then it is close to RL. That’s why here it is not considered. For exact calculations you can watch the video on common base amplifier.
Here is the link : ruclips.net/video/MckBbhfxdkk/видео.htmlsi=dYJmsP0UI30uVH_q
okay sir, thankyou
Sir, at 12:59 Ro and RL in parallel?
Yes.
You are a thop sir
I have a doubt, you have written in properties of CB that it has low power gain, but if current gain is 1 then power gain is same as the voltage gain and so it is high. Right?
Nice one ☺️
I have a dought sir, what is the difference between Vbb, Vee and Veb when the base is common. And the volatege is always apllied between base and emitter then why three different notations are used
The output characteristic. How does a output current Ic flow when it's reverse bias
Time 8.30
Are you talking about cut-off region. In the cut-off region both junctions are reverse biased.
@@ALLABOUTELECTRONICS No, in active and saturated regions the collector base is reversed but how is there a Current flow Ic?
In the introductory video on BJT, I have already explained it. If you go through it, you will get it.
@@ALLABOUTELECTRONICS thx for the help bro. This help me with exams
9:24 it is reverse biased then Ic must be due to majority charge carriers....
How can the base terminal of a transistor be both positive and negative at the same time in a common base configuration?
Sir may I know best and simple reference book for BASIC ELECTRONICS ...VTU BOARD
Electronics Principle by Albert Malvino and David Bates
Tq sir
So nice
Super nice
Thanks ☺️
How is the voltage from the Base to Emitter considered to forward biased for both the PNP and NPN when they are on? Is this not a mistake? If you're talking about Vbe in both cases then they both can not be considered forward biased. The only way it makes sense is if in regards to the PNP you are saying Veb is forward biased. But why do it like that?
Its really a good video ,but I have one doubt , voltage gain is actually based on load resistance, which we can change according to our convenience than why its rage is fixed?
In common base amplifier, along with load resistor, the gain also depends on the collector resistor and the transconductance. That's why just increasing the load resistance alone, you will not see much increase in the gain after some point. For more information, about the gain of the common base amplifier, you may refer this video: ruclips.net/video/MckBbhfxdkk/видео.htmlsi=1KOQ9EApVd8GTiZy
Don't forget to see the description, there is a short note, which is very helpful❤
y in active region, Vcb didn't affect Ie, but at input characteristic there Vcb increase, Ie also will increase?!!!!!can someone explain to me, urgent!!!
Wonderful vedio....it was very helpful ..want some vedio on physical electronics and semiconductor theory
👍🙂😁nice
I amplifier we amplify voltage or current ? Please answer
It depends how the amplifier is configured. Some amplifier amplifies current, some voltage, while some amplifiers are power amplifier and amplifies both current and voltage.
@@ALLABOUTELECTRONICS in general bjt as a amplifier which circuit us used ?
Usually, for amplification BJT is configured in common emitter configuration as it provides both voltage and current gain.
@@ALLABOUTELECTRONICS there are three types of config cb cc & ce right ? Then what is fixed bias voltage divider bias ? It so confusing please help
Can you post videos for heat engine in thermodynamics
Bro i reffered some books too, in that output is taken at VCB and in some it is taken at resistor Rl which one is right please clear this bro......
You will get the output between the collector and the base terminal. But usually, the output is connected to the load.
So, if you connect the load resistor RL across the two-terminal then you will get the same output across the load resistor.
I hope it will clear your doubt.
I didnt get why on reducing effective base width, less VBE is required, why will the curve shift to left?
Well explained! There are three cows tho 🐄
If collector base junction width is increased then base current will decrease and collector current increases but i dont understand why emitter current increased if the collector base juction width is increased?
Tell me the applications of cb configuration
The common Base amplifier has low input impedance and high output impedance, so it is used for the impedance matching at RF frequency.
Aunty Google it
How can I send questions problems involving calculations
sir u have shown symbol for pnp transistor in video, but you are refering to npn so which is right?
If you see the symbol closely, then it is the symbol of NPN transistor. The arrow is going from base to emitter.
Sir at 11:34
@@aamir99204 Yes, just to explain the signal amplification in CB configuration, the PNP transistor was shown. (since the direction of input and output currents are more relatable in PNP CB configuration)
input current, how it is decided whether we take the conventional direction or the direction of electron?
Conventional current
I didn't get that last line,transfer of resistance
9:40 ICBO, what is that stands for
It is the reverse current flowing between collector and base when the emitter is open circuited.
@@ALLABOUTELECTRONICS thank you.
You mixed up which half is foward bias and which is reverse bias, in both the pnp and npn transistor diagrams. 2:55 and 2:32
In PNP transistor, EB junction is forward-biased. (Emitter is P, and base is N), while CB junction is reverse-biased. (Collector is P and base is N).
In the NPN transistor, BE junction is forward biased (Base is P and Emiiter is N), while BC junction is reverse biased. (Collector is N and base is P, so higher voltage is applied at Collector side).
I hope, it will clear your doubt.
@@ALLABOUTELECTRONICS but if base-emitter in NPN is foward biased, that means current flows into emittor(N) and out of the base(P), because that is how a [NP] diode would behave in foward bias.
But current is flowing out of the emitter in your NPN diagram, which makes the emitter-base reverse bias.
@@chibuzordesmond3937 I would recommend you to watch the basic video on BJT. If you will watch it, then your doubts will get clear.
Here is the link: ruclips.net/video/-VwPSDQmdjM/видео.htmlsi=B9kH68XWb2-uVTNo
@@ALLABOUTELECTRONICS will do
@@ALLABOUTELECTRONICSI've watched it and my understanding now is that current always flows from + to -
in the circuit, I've been taking the convention for granted but let me know if that is correct.