A Deep Dive into Transistors

Transistors - NPN & PNP - Basic Introduction

AI-generated summary and notes. Check quotations, numbers, and important claims against the source video. Captions may contain errors.

Watch the source video on YouTube

Estimated reading time: 24 minutes for the text on this page.

Summary

In this video by The Organic Chemistry Tutor, viewers are treated to a comprehensive introduction to transistors, specifically focusing on NPN and PNP types. These components act as electronic switches and amplifiers, capable of utilizing small currents to manage larger ones. The video delves into the specifics of their structure and functioning, unraveling how these bipolar junction transistors operate in various modes, namely active, cutoff, and saturation. With detailed explanations, the tutor breaks down the roles of emitter, base, and collector, alongside practical examples like using transistors as switches to control current flow through light-emitting diodes.

Highlights

Transistors can switch and amplify currents in electronic devices like LEDs. 💡
NPN transistors have an n-type semiconductor structure for emitter and collector, with a p-type base. 🔀
PNP transistors have two p-type materials and one n-type base, reversing the NPN structure. 🔄
Correct biasing (forward or reverse) is crucial for the transistor's operation in circuits. ➖➕
Calculating current flows and potential differences in transistors involves Ohm's Law and understanding transistor parameters like beta (current gain). 📐

Key Takeaways

Transistors are crucial components that can act as switches and amplifiers in electronic circuits. 🔌
NPN and PNP transistors function similarly but with key differences in current direction and biasing. ⚡
Understanding forward and reverse bias is essential for troubleshooting and designing circuits involving transistors. 🔄
Current flow in transistors is based on the conventional current, contradicting the natural flow of electrons. 🔄
To avoid damaging transistors, it's important to properly manage the base current and resistor values. 🛡️

Overview

Transistors, particularly NPN and PNP types, are pivotal in electronic circuits, functioning primarily as switches or amplifiers. The video by The Organic Chemistry Tutor walks viewers through the fundamental concepts by explaining how small currents can control larger currents, thus powering devices like LEDs.

The nuances of NPN and PNP transistors are unfolded with detailed descriptions of how current flows differently in these devices. From understanding their structure, composed of different semiconductor materials, to exploring current flows, the video provides a vivid picture of transistor operations.

With informative examples, the video explains transistor applications and how they operate in various modes such as active, cutoff, and saturation. Practical insights into role players like the emitter, base, and collector make the video an essential guide for understanding and applying transistors in electronic circuits.

Transistors - NPN & PNP - Basic Introduction Transcription

00:00 - 00:30 in this video we're going to talk about transistors transistors are electronic devices that can be used as switches where you could use a small amount of current to drive a larger current to power something let's say like a light emitting diode or you could use them to amplify a small AC signal we're going to talk about two types of transistors the NPN transistors and a PNP transistor
00:30 - 01:00 these transistors are known as bipolar Junction transistors and that they have two junctions so here's one way to describe the NPN transistor the first Junction is right here is between the P and the n-type semiconductor material and here is the second Junction so by means two so we have a bipolar Junction transistor the emitter and a collector
01:00 - 01:30 of an NPN transistor are made up with the n-type semiconductor the base is made up of the p-type semiconductor an NPN transistor is essentially two diodes connected to each other the diode always points towards the n-type material so here's a typical diode and this is the PN Junction of that diode the p-type
01:30 - 02:00 part of the diode is known as the anode and this is the cathode when you connect the anode to a positive potential and the cathode to the negative potential current will flow through the diode and the diode will be in forward bias mode so it's going to be active with a voltage drop of about point 6 to point 7 volts for a silicon diode now if you reverse the polarity let's say you
02:00 - 02:30 connect the positive terminal of a battery to the cathode part of the diode and the negative terminal to the anode in this case the diode will be in reverse bias mode it's going to be off unless you exceed the maximum reverse voltage that I can handle so it's important to understand the terms Ford bias and reverse bias because when solving problems associated with transistors you need to know if you're dealing with the active region the cutoff region or the
02:30 - 03:00 saturation region now let's spend a minute talking about the PNP transistor so everything is in Reverse for this particular transistor so instead of having two n-type materials in one p-type material we have two p-type materials and an n-type material so the base is made up of the n-type semiconductor the emitter and the collector is made up of a p-type material so the equivalent symbol for the PNP transistor using diodes looks
03:00 - 03:30 like this now if you purchase let's say one of the small NPN or PNP transistors they would look like this let's use the 2 n 2 - 2 - NPN transistor this is the emitter this is the base and that is the collector now let's talk about how we can draw the electrical symbols for an NPN and a PNP transistor especially when
03:30 - 04:00 drawing circuits so let's start with the NPN transistor first draw a circle and then draw the base the emitter points away from the Sun of the transistor when we use a different color to represent that so that is the emitter whenever you see the arrow that is the emitter part of the transistor this is the base and that is the collector the base always is
04:00 - 04:30 in the middle so now for the PNP transistor the base is going to be in the same location but this time the emitter is going to be pointing towards the center as opposed to away from the center of the transistor so this is the emitter that's the collector and this is the base now for an NPN transistor the emitter is simply it's usually connected to the ground for a PNP transistor the collector is typically connected to the ground the
04:30 - 05:00 emitter of the PNP transistor typically goes to the positive voltage source and the collector of the NPN transistor goes towards the positive voltage source now for the base the base of NPN transistor also goes to the positive voltage source but for a PNP transistor the base goes towards the ground now you definitely want to connect a resistor to it because you want the base current to be very
05:00 - 05:30 very small to much a base current can destroy the transistor it can allow a large collector current to flow through it and that could cause it to overheat damaging the component now let's talk about some other things that you need to know regarding these two transistors current flows into the base of an NPN transistor whereas for PNP transistor it flows out from the base keep in mind this is opposite to the direction of electron
05:30 - 06:00 flow conventional current flows from a high potential to a low potential but electrons flow from a low potential to a high potential so just keep that in mind one of the many reasons why I like to use conventional current when analyzing circuits involving transistors is that the direction of the emitter arrow tells you the direction of the emitter current the collector current flows towards the NPN transistor and the emitter current flows away from it in the direction of
06:00 - 06:30 the emitter arrow now for the PNP transistor current is going to flow from the positive potential to ground and so this is going to be the emitter current and as you can see it's in a direction of the arrow the collector current is going to be flowing away from the PNP transistor and the base current flows away from it so these two transistors are completely opposite to each other in an NPN transistor the base current flows into
06:30 - 07:00 it the transistor or into the base terminal whereas in a PNP transistor it flows away from the base terminal as you can see the collector current flows towards the transistor in an NPN type device when the PNP type divides it flows away from it but the direction of the arrow always tells you the direction of the emitter current whether you're dealing with an NPN or a PNP transistor so those are some basic things you want to know about these two types of
07:00 - 07:30 transistors but for the most part we're going to be focusing on the NPN transistor as part of our circuit analysis now let's go over some formulas that you want to be familiar with when dealing with transistors the first one is this one the collector current is equal to beta times the base current the base current IB the collector current IC beta is also represented by an H Fe when
07:30 - 08:00 you're buying a transistor look for this number H Fe that's your beta that's your current gain now ie is the sum of IC plus IB that's another equation that you want to write down and if we replace IC with beta times IB we could represent ie in terms of IB so factoring out IB ie is
08:00 - 08:30 equal to IB plus I mean times beta plus 1 so you want to make sure that you know these two equations and then once you have it you can come up with this equation if needed so here's a problem for you let's say we have an NPN transistor and let's say that this transistor has a base current of 50 micro amps and the
08:30 - 09:00 transistor has an HF II or a beta of 200 calculate IC and ie in this example calculate the collector and the emitter currents feel free to pause the video so IC is going to be beta times IB beta is 200 and so if we multiply that by 50
09:00 - 09:30 micro amps 2 times 5 is 10 carry over the 3 zeros we get 10,000 micro amps now keep in mind 1000 micro amps is equal to 1 milliamp so 10,000 micro amps that's 10 milli amps so that is the collector current it's 10 milliamps now the emitter current is the sum of the base current plus the
09:30 - 10:00 collector current so the base current is 50 micro amps and then the collector current is 10,000 micro amps so the emitter current is 10 thousand 50 micro amps which is 10 point zero 5 milliamps but you can say that's approximately 10 milliamps so when HFE is large let's say if it's greater than 100 ie
10:00 - 10:30 is approximately equal to IC 10 milliamps or 10 point zero five or even ten point one that's a still about 10 milliamps so sometimes you could use this approximation to simplify calculations in the future especially when you have difficult transistor problems are you trying to solve now let's talk about some other things that you need to know with regard to NPN transistors in circuits so that's the
10:30 - 11:00 base that's the collector this is the emitter by the way if you're wondering how to connect the battery for this circuit the negative terminal of the battery you want to connect to ground the positive term of the battery goes towards where the +9 voltage or + 9 V is so the original circuit is equivalent to this circuit with the positive terminal connected at that point just in case you're wondering so this is the base resistor RB this is RC the collector
11:00 - 11:30 resistor and this is the resistor that's associated with the emitter ar e now this is going to be ve that is the potential at point e with respect to ground the potential at Point C with respect to ground is known as VC and the potential at point B with respect to ground is VB so if you were to connect a multimeter to point E and the ground the
11:30 - 12:00 voltage that you would read is ve VCE is the difference between potentials VC and ve so if you connect a multimeter between Point C and E the voltage that you'll get is VCE
12:00 - 12:30 now this here if you connect the multimeter between those two points that's gonna be vbe so vbe is the difference between the potential at B and the potential at E now for most NPN transistors this is going to be somewhere between point six and point seven volts so this value is pretty constant therefore you could use it when
12:30 - 13:00 solving circuits involving transistors the last one you need to be familiar with is V CB this is the difference between the potential at Point C and the potential at point B now another one which is quite different from the previous tree is VCC this is the collector supply voltage in this case it's going to be positive nine volts it's basically what you see here Vee
13:00 - 13:30 let me use capital e is the emitter supply voltage and in this example that's the ground voltage so that's gonna be zero which we really don't have to worry about for that circuit now let's talk about how the transistor can be used as a switch in circuits so what we have here is a 12 volt battery connected to the base and the emitter of the transistor by means of a base resistor RB and the switch s1 which is
13:30 - 14:00 currently open the 12 volt battery is also connected to the collector and the emitter region of the transistor through R C and D 1 this is going to be a green LED light emitting diode now when a switch is open there's no current flow into the base in order for the transistor to be active vbe has to be point six or greater so right now because there's no current flow into the
14:00 - 14:30 base the transistor is off which means that there's no current flowing from the collector to the emitter of the transistor which means that d1 the light emitting diode is also off so s1 is a switch that controls a small amount of current which will also control a larger current flown through the LED and that's the purpose of the transistor at least one of its many
14:30 - 15:00 operations it's using a small current to control a larger current now let's see what's gonna happen when the switch is turned on so now let's close a switch at this point current will flow from the positive terminal of the battery through the base so that's gonna be IB and then current will flow from the emitter now once we
15:00 - 15:30 get a small amount of current flowing through the base that is when the potential of B is 0.6 or more the transistor will now be on so the switch between the collector and the emitter of the transistor is now active thus current is going to flow from the battery through the collector I mean through RC so this is gonna be IC it's gonna flow through the light emitting diode so now this diode is on and then
15:30 - 16:00 to the collector so that's IC and then to the emitter which will be IE so that's how this the transistor can act as a switch you could use a small current such as IB to control a larger current IC now let's talk about solving a circuit such as this so let's put some numbers to this problem let's say that
16:00 - 16:30 RB is a hundred colognes you want a large resistor because you don't want too much current flowing to the base of this transistor and let's set our C to one Cologne we don't want too much current flowing to the Lyman and diode such that it burns out most Lightman dials can handle about 20 to 30 milliamps of current so a 1 kilo ohm resistor won't prevent more than 12 milliamps of current flowing through d1 if you do 12 divided by a thousand you
16:30 - 17:00 get point 0 1 2 amps or 12 milliamps so that is the maximum IC current that we can get in a circuit it's VCC divided by RC if we had Ari we would add that as well but we don't have that this is also known as the saturation current now for this problem I'm gonna change RC from 1 kilo ohms to
17:00 - 17:30 220 ohms and we're gonna say that beta is let's make it a hundred that's gonna be our in cheffy value for the transistor and this time I put the transistor in a circle which is what it should be rs for resistors D is for diodes in this case light emitting diode asses for switches and Q that's for a transistor so in this problem go ahead and
17:30 - 18:00 calculate IB IC and ie and calculate the potentials at Point C B II as well as point D and we'll call this point a so go ahead and solve this problem the first thing I would do is set the ground potential to 0 and then we could determine the potential at point B keep in mind vbe is between point six and point seven volts so we're gonna say
18:00 - 18:30 that vbe is 0.6 volts for this problem as the base current increases vbe increases slightly so the base current is very very very small vbe is gonna be close to 0.6 volts if the base current is relatively large let's say like a hundred micro amps or one milliamp and it really depends on the transistor but if it's relatively large vbe might be close to 0.7 volts but for this problem we're gonna say that it's 0.6 volts to
18:30 - 19:00 keep it simple so that means the potential at point B is point 6 volts now that we know that we can now calculate the current flowing through our B which is IB the potential here is at 0.6 volts and the potential on the other side of the resistor when s1 is closed that's going to be the potential at Point a which is 12 volts according to
19:00 - 19:30 Ohm's law in order to calculate the current flowing through a resistor it's going to be the voltage across that resistor divided by the resistance so in this case IB is going to be the voltage across the resistor which is VA minus VB is the difference between the potential at ay and a potential IB divided by RV
19:30 - 20:00 and so that's going to be 12 volts minus 0.6 volts divided by a hundred kilo ohms so eleven point four volts divided by a hundred kilo ohms now something I do want to mention is that when you divide votes by ohms you're gonna get the current in amps if you divide volts by kilo ohms you're gonna get the current in milliamps so eleven point four
20:00 - 20:30 divided by 100 will give us a current value of 0.1 one for milliamps so when you understand that you don't need to convert amps into milliamps and nor do you need to convert kilo ohms into ohms so now that we have the value of IB we can calculate the value of IC so this is
20:30 - 21:00 point 1 1 4 milliamps I see is gonna be beta times IB and beta is a hundred IB is point 1 1 4 milliamps so multiplying it by a hundred and we need to move the decimal 2 units to the right so we get a current of 11 point 4 milliamps for IC so that is the current that is flowing through RC it's eleven
21:00 - 21:30 point four milliamps so now that we know the current flowing through RC we can calculate some potentials around that point we know the potential at a is 12 volts so using Ohm's law we can calculate the potential at point D so the voltage is going to equal the current times the resistance the voltage is going to be the difference between the potentials at a and D this is going
21:30 - 22:00 to equal IC times RC so rearranging the equation to calculate the potential at D is going to be the potential at a minus the voltage drop across RC which is IC times RC so the potential at a is 12 volts and keep in mind VD is gonna be lower than V a because the current flows from a high potential to a low potential so it's gonna be 12 volts minus the current of now since this is an ohm so
22:00 - 22:30 we want this to be in amps eleven point four milliamps if you divide that by a thousand you'll get the current in amps which is point zero one one four amps times the resistance of 220 ohms so twelve minus point zero one one four times 220 that gives us a potential at point D which is nine point four nine
22:30 - 23:00 two votes now that we have the potential at point d we can now calculate the potential at point c now for this we can get an estimate because we know the voltage drop of a typical green LED for the most part is approximately two volts it can vary it can be 1.8 volts it could be 2.2 but we're going to go
23:00 - 23:30 with an estimate of 2 volts so that means that the potential at C will be seven point four nine volts approximately and so VCE which is the difference between these two that's going to be seven point four nine volts as well ideally speaking when create an transistor circuit like this especially when using it as an amplifier you want VCE to be half of the collector supply
23:30 - 24:00 voltage so in this example the collector supply voltage is 12 volts when designed and the amplifier you want VCE to be half of that six volts this is known as midpoint bias if you want to gain the maximum benefit of the amplifier now there are three important regions to be familiar with when dealing with transistors the first is the cutoff region next we have the active region and the third operating region is the
24:00 - 24:30 saturation region now during the cutoff region the emitter has a higher potential than the base so in this case we can say that VB is less than V e or we can say that vbe is less than 0.6 if that's the case the transistor is off in this case the emitter base Junction is
24:30 - 25:00 in reverse bias mode notice that the direction of the current is opposite to the direction of the arrow so that's reverse bias mode now the potential of the collector is going to be higher than the potential of the base so the collector base action is also in reverse biased mode so when those two junctions are in reverse bias mode you're operating in the cutoff region the transistor is off then we
25:00 - 25:30 draw a line to separate these regions during the cutoff region VCE is approximately equal to the collector supply voltage and the collector current is approximately zero amps you might get a small collector current depend on the voltage but if it is there it's gonna be nano amps it's very very close to zero next we have the active region now in the active region the potential at the
25:30 - 26:00 base is going to be greater than the potential at the emitter and it has to be vbe has to be more than 0.6 volts or should be equal to or greater than 0.6 volts to be more specific when that happens the transistor is going to be in its active state and now the base emitter Junction is in forward bias mode now the collector still has a higher potential with respect to the base so
26:00 - 26:30 the collector base region is still in reverse bias mode in forward bias mode the current is in the direction of the diode arrow in reverse bias mode you can see that the current is in the opposite direction of the arrow of the diode now let's talk about the saturation region in the saturation region the potential of the base is still greater than the potential of the emitter so vbe is still
26:30 - 27:00 equal to a greater than 0.6 volts so this is in forward bias mode now in the saturation region the base has a higher potential than the collector so current is going to flow from a high potential to a low potential so therefore the collector base region is also in forward bias mode now for the cutoff region we said that VCE is equal
27:00 - 27:30 to the collector supply voltage of VCC in the active region VCE is going to be between zero and VCC the collector supply voltage under saturation conditions VCE is zero so let's say if the collector supply voltage is nine volts VCE is going to be between nine volts and zero volts if it equals nine
27:30 - 28:00 volts it's in the cutoff region if it equals zero it's in the saturation region so that's why it's important to know what the VCE value is because by knowing it you could determine if the transistor is operating in the cutoff region in the active region or in the saturation region so VCE knowing that value is very important now the second thing you could use to determine if the transistor is operated in the cutoff octave or
28:00 - 28:30 saturation region is the IC value if the IC value is close to zero its operating in the cutoff region if the IC value is between zero and it's a maximum value which I'm gonna call is that's the saturation current then it's in the active region the maximum IC value or the saturation value is going to equal the collector supply voltage divided by
28:30 - 29:00 RC plus re if R is there if re is not there if there's no emitter resistor it's just VCC over RC assume we don't have an LED or alignment in diode if there is a light emitting diode like in the example problem that we had the saturation current will be VCC minus the voltage drop of the diode divided by RC we didn't had our in that example so we could use that to calculate the maximum saturation current now in the active
29:00 - 29:30 region IC can be calculated by using this formula beta times IB this means that as beta increases IC will increase proportionally in the saturation region if you increase the beta current I see will not change it'll be at its maximum value so that's how you know if you're in the saturation region if your increase in the beta current by decrease in RB and if you realize that IC is not going up then the transistor is
29:30 - 30:00 operating in the saturation region and then you can also tell by measuring VCE if you connect a meter between the collector and the emitter pin of the transistor if it's approximately zero the transistor is in saturation mode so those are some ways in which you can determine if the transistor is in the cutoff region the active region or the saturation region now if you want to design an amplifier if you want to make the most ideal amplifier you want to
30:00 - 30:30 create a midpoint bias type amplifier that is when VCE is 1/2 of VCC and when IC is 1/2 of the saturation current