Transistors

[Dylan Wisdom]

What does a common base circuit do?

 

[BobK]

Voltage gain or current follower. I have never used that configuration.

Bob

 

[Dylan Wisdom]

Is there any good way to learn all of this? Since obviously I have a lot to learn.

 

[BobK]

Yes, check out the first entry in the thread "Book Reviews"

Bob

 

[Dylan Wisdom]

Thank you very much!

 

[KrisBlueNZ]

Common emitter and common base configurations amplify voltage.

See http://en.wikipedia.org/wiki/Bipolar_junction_transistor#Applications

Under "Amplifiers" there are links to articles on the three configurations, which have reasonable descriptions. If you have any specific questions after reading those sections, feel free to ask.

 

[dcac]

Bob, is right partly, if you want to amplify the voltage you will need to use common emitter in case you do not care about 180 shift which is present at your output signal, if you want to have no phase shift for your signal then you will need common collector circuit. All this circuit act as current amplifiers. If you want amplify voltage then you are off looking at FET tranbsistors.

 

[KrisBlueNZ]

Bob, is right partly, if you want to amplify the voltage you will need to use common emitter

Both common emitter and common base configurations have voltage gain.

in case you do not care about 180 shift which is present at your output signal, if you want to have no phase shift for your signal then you will need common collector circuit.

Wrong again. Common base configuration is also non-inverting.

All this circuit act as current amplifiers. If you want amplify voltage then you are off looking at FET tranbsistors.

Also wrong. BJTs are current-driven, but they also provide voltage gain. Input signal voltage is converted into a signal current by series resistance and/or the transistor's own incremental resistance. FETs are voltage-controlled but they can also provide current gain. It depends on the configuration.

Common emitter / common source: voltage gain; current gain; inverting.
Common collector / common drain: no voltage gain; current gain; non-inverting.
Common base / common gate: voltage gain; no current gain; non-inverting.

 

[LvW]

. BJTs are current-driven, but they also provide voltage gain. Input signal voltage is converted into a signal current by series resistance and/or the transistor's own incremental resistance.

May I add a short comment for clarification?
During design of a transistor amplifier we very often make use of the relation Ic=Ib*beta.
However, this does not mean that the BJT would be "current-driven".
In fact, the BJT is a voltage-controlled device (Ic=Io*exp(Vbe/Vt).
A voltage Vbe between base and emitter causes two currents Ib and Ic, and both are related by a factor called "current gain beta".
More than that, the graphical verification of the above mentioned exponential relationship between Vbe and Ic gives a curve (transfer characteristics) with a SLOPE that is called "transconductance gm".
(By the way: The name "transistor" - transfer resistor - also points to this voltage-current relation).

 

[Harald Kapp]

In fact, the BJT is a voltage-controlled device (Ic=Io*exp(Vbe/Vt).

But: in a typical amplifier application you do not use this equation to design the amplifier. Typically you're looking for a linear relation between Input and Output of an amplifier, not an exponential one. Therefore you use a base resistor and/or feedback to essentially convert the Input voltage to a proportional base current which is amplified and converted back to an Output voltage by a resistor.

Whether an element is voltage driven or current driven is often a matter of your (personal) point of view. The characteristics can be used to judge which view is better suited to an application. Take for example a simple diode:
You could say a diode is voltage driven. Apply a voltage across the diode and current will flow. The Problem is that due to the exponential characteristic of the diode a small change in voltage will create a huge Change in current. This is mostly inconvenient.
If, on the contrary, you consider the diode as current driven, a small change in current will result in an even smaller change in voltage. This can be handled much easier in most cases.

 

[LvW]

But: in a typical amplifier application you do not use this equation to design the amplifier. Typically you're looking for a linear relation between Input and Output of an amplifier, not an exponential one. Therefore you use a base resistor and/or feedback to essentially convert the Input voltage to a proportional base current which is amplified and converted back to an Output voltage by a resistor.
.

Yes - I agree. Therefore, I did point to the fact that during design we assume that Ib controls Ic. It was only my intention to remind the interested readers to the physical facts.
By the way - the well known feedback using an emitter resistor works only because of the voltage-control feature of the BJT. This resistor linearizes the Ic=f(Vbe) characteristic drastically.

 

[KrisBlueNZ]

Agreed. You could say that any semiconductor is voltage-driven, because you need voltage to make current flow! From that point of view, LEDs are voltage-driven! But LEDs are characterised as current-driven, and BJTs are characterised as current amplifiers, because that's the simpler and clearer way to explain them. As Harald says, it depends on how you want to look at it, and which way is the most useful for explaining a particular aspect of their behaviour.

Even your (LvW) point about the emitter degeneration resistor only working because the BJT is voltage-controlled can be described in terms of current control, although I agree it's more sensible to describe it as negative voltage feedback.

 

[Fishstickilicious]

This thread is an interesting read. When you guys talk about "inverting", do you mean simply that the phase is shifted? I.e. If I had a square wave with a 30% duty cycle driving the base of a BJT (assuming the proper resistor is connected), does that mean that my load would experience a 70% on/off duty cycle during operation, or would it experience a 30% duty cycle with the phase shifted relative to the signal driving the base?

 

[Arouse1973]

I am with Harald on this one. I think as he mentioned it is all about which aspect of the circuit your interested in. All BJT transistors have the ability to amplify current don't they. But you don't have to use that feature for voltage amplification. I find It's simpler to think of voltage if that's what you are amplifying and think of base current if you are interested in current gain.
Adam

 

[Harald Kapp]

Inverting means that a high voltage at the input of a circuit generates a low voltage at the output and vice versa.
In terms of a sinusoidal voltage this is equivalent to a 180 ° phase shift.

When discussing square waves, it is strictly not correct to use the term "phase" since a rectangular waveform is made up of a (theoretically) infinite number of sine waves. Nevertheless it is common to refer to the phase of the fundamental wave as the phase of the rectangular waveform.
In your example inverting a rectangular waveform with a dutx cycle other than 50% is not identical to phase shifting it by 180 °! Shifting such a signal in time preserves the ducty cycle, whereas inverting the signal changes the duty cycle.

 

[KrisBlueNZ]

This thread is an interesting read. When you guys talk about "inverting", do you mean simply that the phase is shifted?

It's clearer to describe it as inversion. The signal comes out upside down. For many waveforms, such as a sinewave, a square wave with 50% duty cycle, etc, this corresponds to a 180 degree phase shift, but it's clearer to define inversion as inversion, not phase shift.

I.e. If I had a square wave with a 30% duty cycle driving the base of a BJT (assuming the proper resistor is connected), does that mean that my load would experience a 70% on/off duty cycle during operation, or would it experience a 30% duty cycle with the phase shifted relative to the signal driving the base?

Inversion refers to the output voltage vs. input voltage. For a common emitter stage, when the base is driven higher, the collector voltage becomes lower; that's what inversion means.

In your case you're comparing the input signal with the voltage across, or current through, the load. As you know, when you provide base bias, the transistor conducts, and voltage appears across the load, causing current to flow in the load. In this sense there is no inversion - when the input is driven active (high), the load is activated.

So if you drive a common emitter switch with a 30% duty cycle, the load in the collector circuit will be energised for 30% of the time. The voltage across the load will have a duty cycle of 30%. The VOLTAGE AT the collector of the transistor will have a duty cycle of 70%.

 

[Arouse1973]

Yes I have also wondered why people refer to an inversion as a 180 degree phase shift. As a phase shift by definition involves a shift in time doesn't it?.
Adam

 

[LvW]

.......and BJTs are characterised as current amplifiers, because that's the simpler and clearer way to explain them. As Harald says, it depends on how you want to look at it, and which way is the most useful for explaining a particular aspect of their behaviour.

Yes - of course, everybody can try to find the best explanation of the BJT´s behaviour. I think, this applies, In particular, during the process of designng a circuit.
Nevertheless, there are some applications which - for my opinion - cannot be explained based on current-control, for example: Current mirrors and Barry Gilbert´s translinear loops.
Furthermore, I am not sure if it is possible to derive the gain expressions for the classical differential amplifier (long-tailed pair) based on the current-control model only.
More than that, for my opinion there is even a kind of contradiction when we say:
* We need a certain voltage Vbe (0.6...0.7)V to "open" the transistor, which means to allow a certain quiescent current Ic;
* However, the variation of this current (which allows output voltage variation across the collector resistor) is caused by Ib variation rather than by a corresponding change of Vbe.

 

[KrisBlueNZ]

Well, given that Vbe and Ib are clearly related, any circuit that can be explained in terms of one can be explained in terms of the other; it just depends which way is more helpful for understanding the component or circuit. In the case of a current mirror, the two transistors have the same Vbe, therefore the same Ib flows in each, and this causes the (ideally) equal collector currents. Do you agree with that explanation? Would you say that explanation paints transistors as voltage-controlled, or current-controlled? Or both?

Re the "kind of contradiction", that description of Vbe and Ib could just as well be applied to an LED; an LED is characterised as current-controlled because the brightness is roughly related to the current (at least, more closely related to current than to voltage), and the brightness is the quantity we're interested in. Similarly, with a transistor, the base current determines the collector current, which is the quantity we're interested in, which is why transistors are normally described as current-controlled.

BTW the use of the word "open" in analogy to a door or a gate, to mean "cause current to flow in" or "forward-bias", could be confused with "open" in the phrase "open circuit", which has the opposite meaning, where "open" means "a break". I mentioned this to Harald because he was doing it too. I guess it's to do with the German language. Could I suggest "forward-bias", "turn ON", or "cause conduction"? Anything other than that specific word "open".

I agree with Fishstickilicious and Adam R. This is an interesting thread

 

[(*steve*)]

I'm enjoying your contributions LvW. (Although I think we may have skipped ahead a little too fast for Dylan -- assuming he still has an unanswered question).

Nevertheless, there are some applications which - for my opinion - cannot be explained based on current-control, for example: Current mirrors and Barry Gilbert´s translinear loops.

I think you may have answered a question I posed here. (Thanks)

Furthermore, I am not sure if it is possible to derive the gain expressions for the classical differential amplifier (long-tailed pair) based on the current-control model only.

That may be another place, but most people will be in the business of using the equations, not deriving them.

More than that, for my opinion there is even a kind of contradiction when we say:
* We need a certain voltage Vbe (0.6...0.7)V to "open" the transistor, which means to allow a certain quiescent current Ic;
* However, the variation of this current (which allows output voltage variation across the collector resistor) is caused by Ib variation rather than by a corresponding change of Vbe.

I think that the voltage control of transistors should not be foisted on people until they learn the practical heuristics of current control.

It's a layer of complexity that isn't needed for anything that a beginner does (aside from needing to overcome Vbe), and only rarely for more experienced applications.

I come from a practical first, then theoretical approach to electronics. It's an approach that works well for people who just want to learn how to make a transistor switch a relay on and off, possibly not so well for someone learning electronic engineering.