Emitter Follower Help: Understand Physics & Art of Electronics

[Jony130]

Another question about a BJT, how does a change in Ib produce a change in Vbe (on a microscopic level)?

Also as you increase the forward biased voltage across the Base-Emitter depletion region the depletion width decreases, right? Well, can you increases the voltage so much that the depletion width is effectively gone (is this even a valid question?) ?

I don't know, you must ask some physics. I'm not interested in this subject. And it is irrelevant from the application point of view.

 

[perplexabot]

Thank you for all your help.

Anyone?

 

[perplexabot]

Another question about a BJT, how does a change in Ib produce a change in Vbe (on a microscopic level)?

A change in Ib causes a change of carriers in the Base region, this causes an attraction/repulsion of the carriers in the Emmiter region, which causes the Base-Emitter depletion region to change width, this width correlates to a change in Vbe, but why?

Also as you increase the forward biased voltage across the Base-Emitter depletion region the depletion width decreases, right? Well, can you increases the voltage so much that the depletion width is effectively gone (is this even a valid question?) ?

I still don't know the answer to this?

 

[Ratch]

perplexabot,

Another question about a BJT, how does a change in Ib produce a change in Vbe (on a microscopic level)?

A BJT is a voltage controlled current source. Whatever voltage you put on the base-emitter will determine the current of the collector. The voltage input/current output is exponential, so designers use a lot of resistance in the base to swamp out the exponential variation. That is why they use a current source (high resistance) in the base to drive the transistor. A lot of folks think that a BJT is controlled by the base current, but it is not. It is controlled by Vbe in an exponential way according to Schockley's diode equation. The base current is also controlled by Vbe in a exponential manner. So the base current is related to the collector current is a somewhat linear manner through Vbe, but it is Vbe that is controlling both. So the base current is an indicator of the collector current, but does not control it. The reason for the exponential relationship is that a BJT operates by diffusion, whereas a FET and a vacuum tube operates by electrostatic means.

Also as you increase the forward biased voltage across the Base-Emitter depletion region the depletion width decreases, right? Well, can you increases the voltage so much that the depletion width is effectively gone (is this even a valid question?) ?

The emitter-base junction is just like a diode in that if you put too much voltage across it, it will burn out. That will happen before you can close the depletion region.

Ratch

 

[perplexabot]

perplexabot,

A BJT is a voltage controlled current source. Whatever voltage you put on the base-emitter will determine the current of the collector.

I think I am lost, doesn't the voltage drop across base-emitter have to be .7? Just like a diode? So if you apply a voltage of 3V, .7V will drop across the Base-Emitter, but happens to the other 2.3V?

The emitter-base junction is just like a diode in that if you put too much voltage across it, it will burn out. That will happen before you can close the depletion region.
Ratch

By "burn out," does it dissipate heat until it literally burns itself out?

 

[Ratch]

perplexabot,

Originally Posted by perplexabot
Another question about a BJT, how does a change in Ib produce a change in Vbe (on a microscopic level)?

A change in Ib causes a change of carriers in the Base region, this causes an attraction/repulsion of the carriers in the Emmiter region, which causes the Base-Emitter depletion region to change width, this width correlates to a change in Vbe, but why?

Are you answering your own questions now? The Ib does not cause the depletion region to change, the Vbe does. Ib change is just a consequence of the Vbe change.

Originally Posted by perplexabot
Also as you increase the forward biased voltage across the Base-Emitter depletion region

the depletion width decreases, right? Well, can you increases the voltage so much that the depletion width is effectively gone (is this even a valid question?) ?

I still don't know the answer to this?

I believe that question was answered.

I think I am lost, doesn't the voltage drop across base-emitter have to be .7? Just like a diode? So if you apply a voltage of 3V, .7V will drop across the Base-Emitter, but happens to the other 2.3V?

You will burn out the diode if you apply 2.3 volts to it. If you apply 2.3 volts to a diode in series with a resistor, and do not exceed the current rating of the diode, then the voltage across the diode will be around 0.7 volts for a silicon diode, and 0.4 for a germanium diode. You should study Schlockey's ideal diode equaton.

By "burn out," does it dissipate heat until it literally burns itself out?

Correct.

Ratch

 

[perplexabot]

Thank you for your help.

perplexabot,
Are you answering your own questions now?

No need for ridicule.

Since Ib doesn't control Vbe then I am assuming my previous post is wrong:

Here is my reasoning of why feedback occurs in a transistor, please tell me if I am right/wrong or am missing something:

Vout changes due to noise, or a change in Re or some other reason and so Ie changes and since Ib = Ie + Ic, Ib will change as well, the change in Ib will change Vbe which in turn increases/decreases the Base-Emitter depletion region width causing more/less carries to pass through which changes Vout back to what it was.

Please help me fix my reasoning!

 

[perplexabot]

I just edited my above post, I think i should have made a new post. Sorry for the pointless post, can anyone help?

 

[sophiecentaur]

Another question about a BJT, how does a change in Ib produce a change in Vbe (on a microscopic level)?

Also as you increase the forward biased voltage across the Base-Emitter depletion region the depletion width decreases, right? Well, can you increases the voltage so much that the depletion width is effectively gone (is this even a valid question?) ?

Your original question about the emitter follower is covering three areas. It covers the (Physics of the) internal operation of the transistor, it covers the use of the current amplification in a circuit (circuit theory) and it also covers Feedback. All three of these are pretty vast subjects and best discussed separately. You may get on better if you can break your thinking up into black boxes, which is what Engineers have to do all the time. It is lucky that electronics is very suitable for this approach but also it can be a problem when people know nothing of one particular area.

 

[perplexabot]

Your original question about the emitter follower is covering three areas. It covers the (Physics of the) internal operation of the transistor, it covers the use of the current amplification in a circuit (circuit theory) and it also covers Feedback. All three of these are pretty vast subjects and best discussed separately. You may get on better if you can break your thinking up into black boxes, which is what Engineers have to do all the time. It is lucky that electronics is very suitable for this approach but also it can be a problem when people know nothing of one particular area.

Hi, thanks for the reply. I am assuming you are suggesting that I treat the transistor as a black box and not think of the internal operation due to my lack of knowledge of the physics behind it. Now I am not disagreeing with you, I do not have the knowledge a physicist has, but I would definitely like to have an idea of how Feedback works, specifically for a transistor, since it is the building block of almost everything else. If the discussion of this topic is too vast, please refer me to a link.

Thanks again.

 

[dlgoff]

Another question about a BJT, how does a change in Ib produce a change in Vbe (on a microscopic level)?

Here's a link that I think you'll find interesting.

http://hyperphysics.phy-astr.gsu.edu/hbase/solids/trans2.html

A base emitter voltage of about 0.6 v will "turn on" the base-emitter diode and that voltage changes very little, < +/- 0.1v throughout the active range of the transistor...

 

[perplexabot]

Here's a link that I think you'll find interesting.

http://hyperphysics.phy-astr.gsu.edu/hbase/solids/trans2.html

Thank you, that is something that will keep me busy for a while.

 

[sophiecentaur]

Hi, thanks for the reply. I am assuming you are suggesting that I treat the transistor as a black box and not think of the internal operation due to my lack of knowledge of the physics behind it. Now I am not disagreeing with you, I do not have the knowledge a physicist has, but I would definitely like to have an idea of how Feedback works, specifically for a transistor, since it is the building block of almost everything else. If the discussion of this topic is too vast, please refer me to a link.

Thanks again.

No, not at all because I don't know how much you know. What I am suggesting is that you can choose not to bother with the Physics when designing a circuit (except for very clever stuff that is pushing the envelope). You can use a whole hierarchy of approximations, starting with a very few parameters to describe how the transistor actually behaves in a circuit. When you are concerned with detailed frequency response and linearity, you may find it worth while but that is way down the line.
Also, where feedback is concerned, the basic feedback formulae often consider very high gain devices (Op Amps are often assumed to have infinite gain) and are very simple. But even a single, modern BJT exhibits high enough gain to treat it as 'infinite' when making a first stab at circuit design. Basically, in many circuits with feedback, the performance is 'defined' by (some of) the surrounding Resistors and Capacitors rather than the performance of the active devices.
Just splitting the problem up into three, will make the whole thing much easier to grasp. The whole of Engineering is based on 'just enough' or 'appropriate' accuracy. (Same thing applies to cost, too.)
Life's hard enough!

 

[Ratch]

perplexabot,

Since Ib doesn't control Vbe then I am assuming my previous post is wrong:

Correct, Ib doesn't control Vbe, but it tracks Vbe in a one to one exponential relationship according to Schockley's diode equaton. If you drive a diode with a current, the voltage across it will automatically set itself to the correct value to conform to Schockley's equation.

Here is my reasoning of why feedback occurs in a transistor, please tell me if I am right/wrong or am missing something:

Vout changes due to noise, or a change in Re or some other reason and so Ie changes and since Ib = Ie + Ic, Ib will change as well, the change in Ib will change Vbe which in turn increases/decreases the Base-Emitter depletion region width causing more/less carries to pass through which changes Vout back to what it was.

Please help me fix my reasoning!

Feedback occurs because the output (emitter) has some voltage and current in common with the input (base). If you want to get involved further, you should start a new thread called "feedback" or something like that. Ie = Ic + Ib, not the way you have it written.

Ratch

 

[perplexabot]

Thank you all for your wonderful help. I will need to read up on transistor feedback or maybe start a new thread in the near future as Ratch suggested.

 

[sophiecentaur]

Here is my reasoning of why feedback occurs in a transistor

You are right to say that feedback occurs in a transistor (the Capacity between Collector and Base, for instance, will limit the high frequency gain, for instance). BUT the feedback that is at work in a basic emitter follower is not "in the transistor". The feedback is there because of the presence of the emitter resistor. It would be there for a totally ideal BJT, with a 'step function' for the junction characteristic and infinite current gain. The feedback is due to the position of the transistor in the circuit and is due to the volts that develop across the emitter load. This distinction is very relevant and I'm not sure you are taking it on board (please correct me if I'm wrong). Think of the transistor as a not-very-good version of an ideal device. The way the circuit behaves is affected very little by the details of the transistor's performance. Whenever possible, circuit design is based on the values of the passive components and assumes that they will compensate for the details of the inadequacies of the active components. Just look at a dozen designs of (linear) transistor amplifier design. You will see that there is nearly always a load resistor (or current source) in the emitter circuit. This is to introduce feedback and to remove the effect of the non-linearity in the base current. I strongly urge you to revisit Op Amp circuits with feedback and to figure out how the various forms of feedback work. That all segues nicely into the way discrete 'real' transistors are operated and the way feedback is applied in non-ideal devices.

There is a parallel with the way we tend to treat passive devices like resistors and capacitors. Every resistor you pull out of the drawer is, in fact, a complicated combination of resistance, inductance and capacitance (revealed in its very high frequency performance) but we ignore that in nearly every case and blithely rely on V=IR to tell us how it will work. We don't get down to the smartarse level until circumstances drive us to it - or when the circuit starts playing us up!

 

[perplexabot]

The feedback is due to the position of the transistor in the circuit and is due to the volts that develop across the emitter load.

I need to give that more thought.

Thank you for your correction and the great info. I will be reading about BJTs, OP-AMPs and feedback for the next couple of weeks. Hopefully by then I will be in a better state than I am now.

 

[sophiecentaur]

I need to give that more thought.

This borders on Philosophy - so beware of brain ache!

 

[perplexabot]

This borders on Philosophy - so beware of brain ache!

Let the brain ache begin.

 

[perplexabot]

Hello again. I'm not sure If I should post my Emitter Follower question here or make a new thread. I will try posting it here first. So in "The Art of Electronics" it says the gain for an emitter follower is Gv = RL/(RL + re), where RL is the resistance of the load and re is the impedance looking back into the emitter. Isn't the gain Vout/Vin where Vout is the voltage across the Load and Vin is the voltage applied at the base? It seems as though that is not what is going on here. Please tell me how that equation they got that equation.

My approach was (which is obviously wrong) Vo/Vi = [Rs/(hfe + 1) + re] / [RL(hfe + 1) + re]
Where Rs is the Impedance of the source.

Thank you for your help.

 

[sophiecentaur]

Gv = RL/(RL + re) is almost unity, which is what you'd expect. This, of course, is Voltage Gain; (the Power gain is much greater - relating to the current gain). I think the value of Vi, used for that formula just relates to the Voltage on the base. If you include a finite series resistance between input volts and the base then this will affect the gain (potential divider effect) Perhaps this is where your problem lies - but you are right to consider the source resistance for many practical applications. Let's face it, you use a buffer when the source impedance is high, dontcha?

 

[NascentOxygen]

Hello again. I'm not sure If I should post my Emitter Follower question here or make a new thread. I will try posting it here first. So in "The Art of Electronics" it says the gain for an emitter follower is Gv = RL/(RL + re), where RL is the resistance of the load and re is the impedance looking back into the emitter.

That's what theory predicts, from the small signal equivalent circuit.

Isn't the gain Vout/Vin where Vout is the voltage across the Load and Vin is the voltage applied at the base?

That's the universal definition of voltage gain, sure. That's how you measure it, but gives no inkling of what theoretically it should be.

My approach was (which is obviously wrong) Vo/Vi = [Rs/(hfe + 1) + re] / [RL(hfe + 1) + re]
Where Rs is the Impedance of the source.

Rs can't enter into the equation for gain, if Vi is the voltage at the base. Whatever voltage is dropped across Rs is immaterial. The loss due to source resistance is excluded by the decision to measure Vi at the base, and not as Vs, for the gain calculation. (So set Rs=0 here.)

[PLAIN]https://www.physicsforums.com/images/icons/icon2.gif Vi and Vo are the small signal voltages—the alternating component sitting atop the DC bias level.

 

[perplexabot]

Thanks for the replies. Such great help.

Gv = RL/(RL + re) is almost unity, which is what you'd expect. This, of course, is Voltage Gain; (the Power gain is much greater - relating to the current gain). I think the value of Vi, used for that formula just relates to the Voltage on the base. If you include a finite series resistance between input volts and the base then this will affect the gain (potential divider effect) Perhaps this is where your problem lies - but you are right to consider the source resistance for many practical applications. Let's face it, you use a buffer when the source impedance is high, dontcha?

Your right. I think I didn't take into consideration the voltage divider effect.

That's what theory predicts, from the small signal equivalent circuit.

That's the universal definition of voltage gain, sure. That's how you measure it, but gives no inkling of what theoretically it should be.

Rs can't enter into the equation for gain, if Vi is the voltage at the base. Whatever voltage is dropped across Rs is immaterial. The loss due to source resistance is excluded by the decision to measure Vi at the base, and not as Vs, for the gain calculation.

[PLAIN]https://www.physicsforums.com/images/icons/icon2.gif Vi and Vo are the small signal voltages—the alternating component sitting atop the DC bias level.

Ok, now I know why Rs can't enter the equation, because the voltage at the base is the voltage that is left after voltage division with Rs.

 

[Jony130]

The AC voltage gain of a Emitter follower we can find using small-signal analysis.
The small-signal T model of emitter follower look like this:

And I hope that you see that Vout = Ie*Re and Vin = Ie *(re + Re)

So the voltage gain

Vout/Vin = Ie*Re/ (Ie*(re + Re)) = Re/(re +Re) (voltage divider equation).

And when we connect the load resistor parallel to Re the gain will change to

Vout/Vin = Re||RL /(re + Re||RL) and if RL<<Re we have RL/(re+RL)

 

[perplexabot]

The AC voltage gain of a Emitter follower we can find using small-signal analysis.
The small-signal T model of emitter follower look like this:

And I hope that you see that Vout = Ie*Re and Vin = Ie *(re + Re)

So the voltage gain

Vout/Vin = Ie*Re/ (Ie*(re + Re)) = Re/(re +Re) (voltage divider equation).

And when we connect the load resistor parallel to Re the gain will change to

Vout/Vin = Re||RL /(re + Re||RL) and if RL<<Re we have RL/(re+RL)

Hmmm. Interesting model. I have not seen it before. Very easy to understand. Thanks for the full analysis.

 

[Jony130]

If you want included Rs resistor we modify the diagram

And we can find overall voltage gain of this circuit

Av = Vb/Vin * Vout/Vb = Rin/(Rs + Rin) * Re/(re+Re)

where Rin is a Emitter follower input resistance.

Rin = (β+1)*(re + Re)

 

[berkeman]

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