Struggling to understand transistors

MCU88

Joined Mar 12, 2015
358
Trans-istor. Trans-Resistor. Geddit? Current controlled voltage source. A typical BJT (Bipolar Junction Transistor) -- has about 4MOHM of DC resistance between the emitter and the collector. By applying (n) current at the base junction of the BJT, this in turn changes the resistance of the emitter to collector junctions. So it could be perceived as an current controlled variable resistor. So we take (n) small amount of current at the base to control a much larger amount of current at the collector OR the emitter. If an voltage is applied to the collector, the emitter voltage tends to follow the base voltage less 0.6V drop. This is called an emitter follower...
 
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WBahn

Joined Mar 31, 2012
29,979
The operation of a transistor, particularly a BJT, is far better modeled as a controlled current source and not as a controlled voltage source.
 

Glenn Holland

Joined Dec 26, 2014
703
While we're on the subject, is the barrier voltage at the B-E junction universally 0.7 volts or does it depend on the specific transistor?

My impression is that the diodes in transistors may be doped differently than a plain discrete diode and the barrier may be different.
 

MCU88

Joined Mar 12, 2015
358
While we're on the subject, is the barrier voltage at the B-E junction universally 0.7 volts or does it depend on the specific transistor?

My impression is that the diodes in transistors may be doped differently than a plain discrete diode and the barrier may be different.
Textbooks will tell you 0.6V, but in my experience it is slightly higher than this after probing around circuits with an multimeter...

I do know for certain that the best way to test an transistor on a digital multimeter is to use the diode test function. The DUT (Device Under Test) -- Perceived as two diodes back to back. But for testing leakage and punch through a lot of people will swear by an analogue meter with the needle movement and set on ohms.
 

WBahn

Joined Mar 31, 2012
29,979
There's nothing magical about 0.7V. You have an exponential relationship between base-emitter voltage and collector current. For most silicon transistors, the base-emitter voltage increases by about 60mV (at room temperature) for every factor of ten increase in collector current.

The transistor can be designed to have just about any desired base-emitter voltage and any particular collector current. If you want a lower voltage you simply make the junction areas larger in order to reduce the current density - and sometimes this is done and is why many power transistors typically operate at lower base-emitter voltages than small-signal transistors. But because this involves more device area, the usual design is to match the junction area to the current density needed for effective thermal management and this results in the base-emitter voltages of the same general type of BJT transistor being roughly the same. The same is true of other materials but the resulting voltage range is different.
 
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#12

Joined Nov 30, 2010
18,224
While we're on the subject, is the barrier voltage at the B-E junction universally 0.7 volts or does it depend on the specific transistor?

My impression is that the diodes in transistors may be doped differently than a plain discrete diode and the barrier may be different.

Here's a partial answer: http://forum.allaboutcircuits.com/blog/turn-on-voltage-of-a-bipolar-transistor-vbe-ic.571/

Yes, different uses require different doping, but there is nothing fixed about the Vbe.

Oops. WBahn already said that.:oops:
 

blocco a spirale

Joined Jun 18, 2008
1,546
Copy the simple amplifier schematic below.
Ignore the capacitors, make the supply = 10V and all resistors =1k and hFE=100

Knowing what you (now) know, calculate the voltages and currents and write them on the schematic, then, Make R4 = 2k and do the calculations again, try different values for R3...and observe what happens. Try different values for R1 etc....

 
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MCU88

Joined Mar 12, 2015
358
[QUOTE="blocco a spirale, post: 835395, member: 24400"]Copy the simple amplifier schematic below.
Ignore the capacitors, make the supply = 10V and all resistors =1k and hFE=100

Knowing what you know, calculate the voltages and currents and write them on the schematic, then, Make R4 = 2k and do the calculations again, try different values for R3...and observe what happens. Try different values for R1 etc....

[/QUOTE]

That circuit is called an common emitter amplifier. It is an AC amplifier and capacitor C1 is for input coupling, C2 for output coupling and C3 for roll off at (n) frequency. Just thought I'd mention it. And it is an good exercise to do; play around with different resistor values and pull out an calculator, paper and pen and do the OHMs law.
 

Jony130

Joined Feb 17, 2009
5,487
For me the simplest BJT model is a water analogy.
A water valve is always used as a device to control the flow of water. Similarly, always think of a bipolar transistor as a device used to control electric current flow by assistance of a base current.
This figure will help you understand how BJT work

03.png

To open (Turn - ON) a BJT transistor you need the base current to flow.
So the battery supplies this base current( Ib) and this base current allows BJT to open (Turn-ON).
Base current flow in this path:
+battery --->base resistor --->base-emitter junction---> -battery.
This base current Turn-ON the BJT and this allows the flow of a much large current between collector and emitter.
Ic = β * Ib and Ie = Ib + Ic is a basic principle of a transistor "action".

BJT act more like control by base current current source
http://forum.allaboutcircuits.com/threads/common-emitter-amp-doubts-and-questions.91084/#post-663195
 

Mike33

Joined Feb 4, 2005
349
When starting out, I got completely confused by the relationships at work here, and couldn't find an easy to read text or what have you that dealt with the issues. Even AFTER I 'knew' what Ib and Ie WERE, I didn't really "get" what to DO with that info. I just cared about Ic!! I really cared about driving things, thought a lot about VOLTAGE, and didn't consider that unless you set things up right you will get inverted output, etc.

It wasn't until I learned to think about CURRENT that things made more sense....perhaps watching some stuff like this tutorial will help the OP; I know that coming across simple tut's like this sure helped me out, more than one can imagine! These tut's tend to make "Ib" and "Ie" make more sense, IMO.

Keep in mind, it's ALL about current with BJT's, with the exception that you need a TURN ON VOLTAGE at the base (Vb) of about .7V, which you have to account for. After that, you can think just about your load, as long as you've chosen the right BJT for the job and whatnot (power ratings...).

There are tons of good videos on just just this subject - some beginner, and some quite advanced as we progress. Enjoy!
 

Thread Starter

tjohnson

Joined Dec 23, 2014
611
I suppose my question really is: What exactly is the relationship between the value of a resistor at the base junction and the current emitted by the transistor?
After reading further in the AAC textbook, I came across the answer in the section about common-emitter amplifiers:
For light exposure levels somewhere between zero and maximum solar cell output, the transistor will be in its active mode, and the output voltage will be somewhere between zero and full battery voltage. An important quality to note here about the common-emitter configuration is that the output voltage is inverted with respect to the input signal. That is, the output voltage decreases as the input signal increases. For this reason, the common-emitter amplifier configuration is referred to as an inverting amplifier.
So the larger the value of the resistor is, the smaller the input signal will be, which means that the output voltage will be greater. Correct?

To the last paragraph,you have that incorrect. A larger resistor in the base leg will reduce the base current, and thus also the collector current, reducing the LED brightness.
It is correct if the transistor is used as a common-emitter amplifier, as it was in the circuit shown in my physics book. Here's an example of the same type of circuit from the AAC textbook (which is what the paragraph I quoted above was describing):
 
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crutschow

Joined Mar 14, 2008
34,285
........................

It is correct if the transistor is used as a common-emitter amplifier, as it was in the circuit shown in my physics book. It is also used that way in the following circuit in the AAC textbook (which is what the paragraph I quoted above was describing):
True. My comment was true if the LED is connected in series with R, which would be the usual connection for controlling an LED with a transistor.
 

WBahn

Joined Mar 31, 2012
29,979
So the larger the value of the resistor is, the smaller the input signal will be, which means that the output voltage will be greater. Correct?
The larger the value WHAT resistor is? Why won't you provide a schematic of the circuit YOU are talking about?
 

WBahn

Joined Mar 31, 2012
29,979
THANK YOU!!

So what is happening here is that the current through R2 splits between the transistor and the LED. The more current that is allowed to go through the transistor, the less current that will go through the LED (and vice-versa). As R3 increases in resistance, that results in less base current, which results in less collector current, which means that more current goes through the LED and it gets brighter.
 

Thread Starter

tjohnson

Joined Dec 23, 2014
611
THANK YOU!!

So what is happening here is that the current through R2 splits between the transistor and the LED. The more current that is allowed to go through the transistor, the less current that will go through the LED (and vice-versa). As R3 increases in resistance, that results in less base current, which results in less collector current, which means that more current goes through the LED and it gets brighter.
And thank you! I think I finally understand this now. As usual, I am amazed that the explanation is so simple. Sorry for not posting a schematic sooner.
 

WBahn

Joined Mar 31, 2012
29,979
And thank you! I think I finally understand this now. As usual, I am amazed that the explanation is so simple.
You're welcome. The explanation for how a concrete example of a circuit works is almost always easier than an esoteric discussion of a general type of circuit (though those discussions definitely have their place).

Sorry for not posting a schematic sooner.
Me, too. But live and learn.
 

Mike33

Joined Feb 4, 2005
349
Ah, that "inverted output" thing I mentioned, lol. See, there's the 'current mindedness' aspect at work here....we have to think about what current flowing in the resistors is DOING.....voltage DROPS. More current, greater voltage drop. Less current, vice versa. Like pulling a spring back and forth, sorta.

Once you're past that, it becomes MUCH easier!

If you were driving the LED right in the collector (or emitter) path as in the video, you wouldn't have that inversion due to voltage drop, as it would BE the load (well, part of it). I think of this as the transistor driving something as opposed to simply switching. Glad you got your answer.
 

Thread Starter

tjohnson

Joined Dec 23, 2014
611
Textbooks will tell you 0.6V, but in my experience it is slightly higher than this after probing around circuits with an multimeter...
That's what I would expect. At 12:31 in the video posted by Mike33, the instructor says that the intrinsic value is 0.7V for transistors and 0.6V for diodes.
 
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