The circuits are different and function differently.
If the load is in series with the collector, you have a common-base or common-emitter circuit (usually the latter). If the load is in series with the emitter, you have what's called an emitter follower. Look up Wikipedia articles on these configurations.
With the emitter follower, there's strong negative feedback built into the circuit since the base current flows through the load as well. With high load currents, there's more voltage across the load and the base current drops which reduces the load current - that's the negative feedback.
If you have a common-emitter circuit and just reverse the transistor connections, you'll still have a common-emitter circuit, but the characteristics will be a little different since the emitter and collector, although similar, are built a little differently.
As a switch, a load must always be connected to the collector of a PNP transistor and the emitter connected to the + power rail.
This is known as a "high side switch", meaning the load's connection to the + power rail is being switched on or off.
For a NPN transistor, the load is also connected to the collector but the emitter is connected to the ground. This is known as a "low side switch" as the load's connection to the ground is being switched on or off.
Low side switches are cheaper and the circuit is simpler.
Most loads can use a low side switch but certain loads requires a high side switch.
You can actually run a Bi-Polar transistor the way Mike describes and drive a load. However you only have an Hfe gain of maybe 2, so not much current in the load can be driven ... it is however still considered a load.
You can take a bipolar transistor, either PNP or NPN and connect the collector lead to the point in the circuit where the emitter would normally be connected and the emitter lead to the point for the collector. It will still work, but usually not well since the emitter and collector structures in the transistor are similar, but not identical. The emitter structure is acting like a collector and the collector structure is acting like an emitter.
As Beau mentioned, the gain will usually be terrible in this situation and other characteristics will also be poor, but functional.
In "the old days", bipolar transistors were physically built more symmetrically and you could do this with much less change in the transistor's characteristics.
You can actually run a Bi-Polar transistor the way Mike describes and drive a load. However you only have an Hfe gain of maybe 2, so not much current in the load can be driven ... it is however still considered a load.
Are you showing electron flow there? It's throwing me off...lol.
Also is the positive sign cut off on the left transistors base?
Could someone fill in the arrow heads and mA values assuming .1v drop CE and .7 drop BE on the right transistor.
Thanks for helping me figure this out!
Sorry, not trying to confuse... I was just showing where the current path was, not necessarily the direction.
Anyway, in the reversed transistor model, the "current path" is across the B-C junction , not the B-E junction as it is when the transistor is oriented the way in which it was intended to operate.
The attached image is a PNP bi-polar transistor that is configured as a diode... The Collector is tied to the Base
The Collector connects directly to the substrate(P+), while the Base and Emitter sit in what's called an Nwell (N+). This 'Nwell' creates a diode (PN junction), so assuming that the Collector was not connected to the Base, the diode formed would be ...
C -->|-- B
Another PN junction diode is formed between the Emitter and the Base with a special (P+) doping layer...
E -->|-- B
Notice that the two diodes are back to back...
C -->|-- B -- |<-- E
In this configuration, the transistor is OFF, current across the C-E junction is blocked.
Note: Regardless of the transistor being ON or OFF, the B-E and B-C junctions will always act as a diode.
When you apply power(NPN) or ground(PNP) to the Base
the C-E junction begins to conduct. Without going into more detail, it's like filling an above ground pool with water, and applying an influence to the Base is like pushing down on the edge of the pool. Soon you have an avalanche effect and current flow.
Because of the physical construction, the Collector and Emitter can be exchanged, but one configuration is more optimal than the other.
OK...So I am studing up as mike suggested. I am reading electronics fundamentals by floyd. It was reading about voltage divider bias,,,and base bias etc ........Then a few chapters latter,,,we go into common collector bias,common emitter,,and common base.
what...or how does voltage divider bias and base bias,,,,fit or relate to these other bias configurations?....ie Is voltage divider just another bias configuration?? ANY help pls
we used to learn the current always flow from - to +.
I'm sorry but I have to point out that this is opposite to the convention the rest of the world has been working to since the discovery of electric current.
For example from Wikipedia:
"In metallic solids, electricity flows by means of electrons, from lower to higher electrical potential. In other media, any stream of charged objects may constitute an electric current. To provide a definition of current that is independent of the type of charge carriers flowing, conventional current is defined to flow in the same direction as positive charges. So in metals where the charge carriers (electrons) are negative, conventional current flows in the opposite direction as the electrons. In conductors where the charge carriers are positive, conventional current flows in the same direction as the charge carriers."
In realty is does not matter . as long as you keep the same system....
Hmm, its now 30 years since I went to that class studying electronics and I think I still keeps the books we used. But I do think the teacher said it was a very common to say the electrons flew from + to - but the laws of physics said it was the opposite way. So thats was what we shold learn and since that days I always think of the current flow from - through the cathode and then the anode, opposite to the arrow in the chematics, to +.
I was just testing a Zetex ZXT1049 NPN transistor for gain, and found a normal gain (hFE) of 615 but was surprised to find an inverted gain of 270. Wow, it is indeed speced to be an especially high gain transistor, but it turns out that even its inverted gain is higher than most generic transistors! I pulled out a couple of generic 2N3904, 2N3906 and 2N2222 transistors, and the forward gains ranged from 120 to 255, but the inverted gains were as Beau said, from only 2 to 6. Sorry, still sort of off topic diversion.
Youngbill asked, <OK...So I am studing up as mike suggested. I am reading electronics fundamentals by floyd. It was reading about voltage divider bias,,,and base bias etc ........Then a few chapters latter,,,we go into common collector bias,common emitter,,and common base. what...or how does voltage divider bias and base bias,,,,fit or relate to these other bias configurations?....ie Is voltage divider just another bias configuration?? ANY help pls >
That covers a lot of ground, and there are so many things you can do with transistors. The book does explain what all those circuits are good for right? There should be a distinction between analog (amplifier) and digital (switch) circuits. Yes, a voltage divider is a common means to bias a transistor. You can have a PNP transistor used as a high side switch, in which case a voltage divider to the base determines the point at which it makes the transition from OFF to ON and back. The normal switching point for a transistor is around 0.6 V, but with a resistor divider you can move that to 1.7 V for a Propeller or 2.5 V for a Stamp, half the power supply in either case.
Could you explain,How to test and find the gain. I was trying to test for saturation,,,and I was watching V-ce and watching collector amps,,and base amps,,,but was not sure exactly when I was at saturation,,,(so that I could do the math,,, or ..to know .. what size of base resistor was needed.)
I had a 100 ma load,,,,at 5 volts..using a pn2222.
What I found is ...I just was not sure,,when adjusting the R,,base,,,when I was exactly at saturation,,,,Thanks for everyones replys
A transistor is in saturation when the voltage on its collector is less than the voltage on the base. How much base current is required will depend on the load you have on the collector. It takes more base current to drive the transistor into saturation when the collector load is 100Ω, than it does when the collector load is a higher resistance.
You are on the right track, if you are watching Vce and Ic. If you measure or calculate the base current {Ib = (Vdd - Vbe)/Rb), then divide that by Ic when Vce=Vbe. That is the large signal gain as the transistor enters saturation.
You can make a graph of Ic versus Ib as you vary the base resistor. That is the large signal gain and it is a curve that starts high at small currents and becomes low as it enters saturation. The slope of that line at any point is called the small signal gain.
youngbill
Posts: 54
Comments
If the load is in series with the collector, you have a common-base or common-emitter circuit (usually the latter). If the load is in series with the emitter, you have what's called an emitter follower. Look up Wikipedia articles on these configurations.
With the emitter follower, there's strong negative feedback built into the circuit since the base current flows through the load as well. With high load currents, there's more voltage across the load and the base current drops which reduces the load current - that's the negative feedback.
If you have a common-emitter circuit and just reverse the transistor connections, you'll still have a common-emitter circuit, but the characteristics will be a little different since the emitter and collector, although similar, are built a little differently.
This is known as a "high side switch", meaning the load's connection to the + power rail is being switched on or off.
For a NPN transistor, the load is also connected to the collector but the emitter is connected to the ground. This is known as a "low side switch" as the load's connection to the ground is being switched on or off.
Low side switches are cheaper and the circuit is simpler.
Most loads can use a low side switch but certain loads requires a high side switch.
You can actually run a Bi-Polar transistor the way Mike describes and drive a load. However you only have an Hfe gain of maybe 2, so not much current in the load can be driven ... it is however still considered a load.
You can take a bipolar transistor, either PNP or NPN and connect the collector lead to the point in the circuit where the emitter would normally be connected and the emitter lead to the point for the collector. It will still work, but usually not well since the emitter and collector structures in the transistor are similar, but not identical. The emitter structure is acting like a collector and the collector structure is acting like an emitter.
As Beau mentioned, the gain will usually be terrible in this situation and other characteristics will also be poor, but functional.
In "the old days", bipolar transistors were physically built more symmetrically and you could do this with much less change in the transistor's characteristics.
Are you showing electron flow there? It's throwing me off...lol.
Also is the positive sign cut off on the left transistors base?
Could someone fill in the arrow heads and mA values assuming .1v drop CE and .7 drop BE on the right transistor.
Thanks for helping me figure this out!
Sorry, not trying to confuse... I was just showing where the current path was, not necessarily the direction.
Anyway, in the reversed transistor model, the "current path" is across the B-C junction , not the B-E junction as it is when the transistor is oriented the way in which it was intended to operate.
so what about the current values?
The Collector connects directly to the substrate(P+), while the Base and Emitter sit in what's called an Nwell (N+). This 'Nwell' creates a diode (PN junction), so assuming that the Collector was not connected to the Base, the diode formed would be ...
C -->|-- B
Another PN junction diode is formed between the Emitter and the Base with a special (P+) doping layer...
E -->|-- B
Notice that the two diodes are back to back...
C -->|-- B -- |<-- E
In this configuration, the transistor is OFF, current across the C-E junction is blocked.
Note: Regardless of the transistor being ON or OFF, the B-E and B-C junctions will always act as a diode.
When you apply power(NPN) or ground(PNP) to the Base
the C-E junction begins to conduct. Without going into more detail, it's like filling an above ground pool with water, and applying an influence to the Base is like pushing down on the edge of the pool. Soon you have an avalanche effect and current flow.
Because of the physical construction, the Collector and Emitter can be exchanged, but one configuration is more optimal than the other.
what...or how does voltage divider bias and base bias,,,,fit or relate to these other bias configurations?....ie Is voltage divider just another bias configuration?? ANY help pls
I do agree with the way you show the direction of the current path, we used to learn the current always flow from - to +.
Campeck's left drawing looks very confusing for me as the arrows looks backwards.
I'm sorry but I have to point out that this is opposite to the convention the rest of the world has been working to since the discovery of electric current.
For example from Wikipedia:
"In metallic solids, electricity flows by means of electrons, from lower to higher electrical potential. In other media, any stream of charged objects may constitute an electric current. To provide a definition of current that is independent of the type of charge carriers flowing, conventional current is defined to flow in the same direction as positive charges. So in metals where the charge carriers (electrons) are negative, conventional current flows in the opposite direction as the electrons. In conductors where the charge carriers are positive, conventional current flows in the same direction as the charge carriers."
Or see here: http://mste.illinois.edu/murphy/HoleFlow/ElectricFluid.html
hole vs electron .
In realty is does not matter . as long as you keep the same system
..
That just cannot be, it's just plain backwards to every other physics class in the world.
Exactly. That's why we have the convention.
Hmm, its now 30 years since I went to that class studying electronics and I think I still keeps the books we used. But I do think the teacher said it was a very common to say the electrons flew from + to - but the laws of physics said it was the opposite way. So thats was what we shold learn and since that days I always think of the current flow from - through the cathode and then the anode, opposite to the arrow in the chematics, to +.
But it really does not matter, I do agree.
Which one is more optimal?
I apologize for inadvertently hijacking this thread with a transposition error on my part.
We're just trying to get the basics sorted out. Else there is going to be a lot of electric motors running backwards:)
Youngbill asked, <OK...So I am studing up as mike suggested. I am reading electronics fundamentals by floyd. It was reading about voltage divider bias,,,and base bias etc ........Then a few chapters latter,,,we go into common collector bias,common emitter,,and common base. what...or how does voltage divider bias and base bias,,,,fit or relate to these other bias configurations?....ie Is voltage divider just another bias configuration?? ANY help pls >
That covers a lot of ground, and there are so many things you can do with transistors. The book does explain what all those circuits are good for right? There should be a distinction between analog (amplifier) and digital (switch) circuits. Yes, a voltage divider is a common means to bias a transistor. You can have a PNP transistor used as a high side switch, in which case a voltage divider to the base determines the point at which it makes the transition from OFF to ON and back. The normal switching point for a transistor is around 0.6 V, but with a resistor divider you can move that to 1.7 V for a Propeller or 2.5 V for a Stamp, half the power supply in either case.
I had a 100 ma load,,,,at 5 volts..using a pn2222.
What I found is ...I just was not sure,,when adjusting the R,,base,,,when I was exactly at saturation,,,,Thanks for everyones replys
You are on the right track, if you are watching Vce and Ic. If you measure or calculate the base current {Ib = (Vdd - Vbe)/Rb), then divide that by Ic when Vce=Vbe. That is the large signal gain as the transistor enters saturation.
You can make a graph of Ic versus Ib as you vary the base resistor. That is the large signal gain and it is a curve that starts high at small currents and becomes low as it enters saturation. The slope of that line at any point is called the small signal gain.