- #1

perplexabot

Gold Member

- 329

- 5

http://upload.wikimedia.org/wikipedia/commons/b/b8/NPN_emitter_follower.svg

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter perplexabot
- Start date

- #1

perplexabot

Gold Member

- 329

- 5

http://upload.wikimedia.org/wikipedia/commons/b/b8/NPN_emitter_follower.svg

- #2

sophiecentaur

Science Advisor

Gold Member

2020 Award

- 25,797

- 5,197

Without any Maths, you can say that current will pour into the base and, hence, through the transistor (current amplification factor of at least several tens) from C to E, until Vbe is around 0.7V. When that happens, Vout is 0.7V below Vin. If you vary Re, the current through the transistor will change to re-establish the 0.7V Vbe. There is so-called Voltage Feedback because any small change in Vout will cause a change in Ib and hence a large change in Ie - and this will maintain Vout to be what it was. This gives an effectively low output impedance - as the Volts are independent of the load resistance. If Vin is raised, the current that will be drawn through the base will not change appreciably because Ve 'chases Vb', remaining 0.7V below. So the current into the base of the Emitter follower will not change as Vin increases - which is what you would get with a very high resistance - hence the high input resistance.

Why does it 'limit'? If you try to take Vb too high then the transistor will saturate and no more current can pass through Re, so the volts cannot increase. The feedback mechanism runs out. Vbe need only increase by a miniscule amount beyond this and the forward biased be diode will pass loads of current. The input impedance will drop as Vb increases until it approaches that of a forward biased diode in series with Re.

There you are - no Maths and plenty of arm waving. Does it help?

- #3

- 315

- 0

But I can't seem to grasp the physics behind it.

The physics are the same as any BJT. Do you mean the topology of the common collector?

I know it is used as a buffer, it has high input impedance and low output impedance.

Its use won't help you understand how it works.

I don't know what properties cause the output voltage to follow the input voltage.

The property of any transistor in the active region. The emitter terminal voltage is around 0.7 volts different than the base terminal. So the emitter follows the base or vice versa voltage wise.

Also if you have "The Art of Electronics,

I don't, but clipping occurs whenever the input is overdriven or the operating point is chosen badly.

Ratch

- #4

sophiecentaur

Science Advisor

Gold Member

2020 Award

- 25,797

- 5,197

- #5

- 625

- 138

Emitter follower is a very simply circuit. The output voltage is always 0.6V lower the the input voltage.

See some examples

Also notice that the base current is**(β+1)** smaller then emitter current (load current).

So our base current source**B1 ** see our load ( Re resistor) not as** 100Ω** resistor. But** B1 **see **(β+1)*Re** load.

Here you have anther example.

Spouse we have a 1K resistor voltage divider supply from 10V battery.

Without any load connect to the output terminal of our voltage divider the output voltage is equal 5V. Now we connect a 100Ω load resistor across the output terminal.

And now our voltage divider output voltage drops from 5V to 0.83V. So we ruin our circuit.

To fix this issue we add a buffer (emitter follower) see the diagram

Now I hope that you see why we say that emitter follower has a high input impedance and low output impedance. It's all thanks to BJT and his current gain.

See some examples

Also notice that the base current is

So our base current source

Here you have anther example.

Spouse we have a 1K resistor voltage divider supply from 10V battery.

Without any load connect to the output terminal of our voltage divider the output voltage is equal 5V. Now we connect a 100Ω load resistor across the output terminal.

And now our voltage divider output voltage drops from 5V to 0.83V. So we ruin our circuit.

To fix this issue we add a buffer (emitter follower) see the diagram

Now I hope that you see why we say that emitter follower has a high input impedance and low output impedance. It's all thanks to BJT and his current gain.

- #6

perplexabot

Gold Member

- 329

- 5

Why will a change in Vout cause a change in Ib? Why would Vout change in the first place? Isn't Ib the one that is being "controlled" and so is subject to change?There is so-called Voltage Feedback because any small change in Vout will cause a change in Ib and hence a large change in Ie - and this will maintain Vout to be what it was. This gives an effectively low output impedance - as the Volts are independent of the load resistance.

May you please clarify this phrase? I don't understand what you mean by "overdriven or the operating point is chosen badly." Do you mean if Ib is low enough to cause Vbe to be lower than .7V then clipping shall occur?I don't, but clipping occurs whenever the input is overdriven or the operating point is chosen badly.

For the first image, second figure, the current through Re, should it be 54mA?See some examples

- #7

perplexabot

Gold Member

- 329

- 5

- #8

- 625

- 138

Yes, my mistake.For the first image, second figure, the current through Re, should it be 54mA?

As for the clipping. The clipping can occur when transistor change his state eg. conduction to cut-off. See the example form The Art of Electronics

I slightly change the diagram

And ask yourself one question. What voltage you need to provide to open BJT in this circuit. Also look at figure 2.9 in Art of Electronics.

Vout can change for example if you change load resistance.Why will a change in Vout cause a change in Ib? Why would Vout change in the first place?

- #9

perplexabot

Gold Member

- 329

- 5

Cut-off happens at voltages less than .7v?Yes, my mistake.

As for the clipping. The clipping can occur when transistor change his state eg. conduction to cut-off.

Is the answer .7V ?And ask yourself one question. What voltage you need to provide to open BJT in this circuit.

That was what I was thinking, but I wasn't sure. Thanks.Vout can change for example if you change load resistance.

On page 67 of "The Art of Electronics" it says:

The output can swing to within a transistor saturation voltage drop of Vcc, but it cannot go more negative than -5 volts. That is because on the extreme negative swing, the transistor can do no more than turn off, which it does at -4.4 volts input (-5V output).

I don't understand why clipping occurs at -5V (or -4.4V input) when cut-off happens at .7V. In other words shouldn't clipping occur at .7V?

- #10

- 625

- 138

No 0.7V Is a Vbe voltage not the base voltage or load voltage.Thank you for your help.

Cut-off happens at voltages less than .7v?

Is the answer .7V ?

That was what I was thinking, but I wasn't sure. Thanks.

On page 67 of "The Art of Electronics" it says:

The output can swing to within a transistor saturation voltage drop of Vcc, but it cannot go more negative than -5 volts. That is because on the extreme negative swing, the transistor can do no more than turn off, which it does at -4.4 volts input (-5V output).

I don't understand why clipping occurs at -5V (or -4.4V input) when cut-off happens at .7V. In other words shouldn't clipping occur at .7V?

In this imagine without BJT the Ve voltage is equal 5V so to open the BJT Vb voltage must be higher then 5.6V. So If input voltage is greater than 5.6V the BJT is active mode and the voltage across the load resistor is Vout = Vin - Vbe. But if Vb is smaller than the 5.6V the BJT is in cut-off stage. And the Ve voltage is constant 5V. Input voltage don't affect the VE voltage any more because BJT is cut-off.

Last edited:

- #11

perplexabot

Gold Member

- 329

- 5

No 0.7V Is a Vbe voltage not the base voltage or load voltage.

In this imagine without BJT the Ve voltage is equal 5V so to open the BJT Vb voltage must be higher then 5.6V. So If input voltage is greater then 5.6V the BJT is active mode and the voltage across the load resistor is Vout = Vin - Vbe. But if Vb is smaller than the 5.6V the BJT is in cut-off stage. And the Ve voltage is constant 5V. Input voltage don't affect the VE voltage any more because BJT is cut-off.

Thank you so much that makes so much sense. Continuing on your example (assuming Ve is 5v when BJT is off), if Vin is 7v then Ve will be 6.3V, is that correct? Also does it saturate at 20V?

Last edited:

- #12

- 625

- 138

Yes for Vb= 7V The output voltage Ve ≈ 6.3VThank you so much that makes so much sense. Continuing on your example, if Vin is 7v then Ve will be 6.3V, is that correct? Also does it saturate at 20V?

As for the saturation for Vb = 20V bjt is almost enter the saturation region.

To enter saturation Vb must be higher the Vcc.

Ok So now if we change the GND position as you have in Art of Electronics

I hope that now you see that without BJT we have VE = -5V. And now if you want to open the BJT the Vb voltage must we greater (lees negative) than -4.4V. If Vb is lower (more negative) than -4.4V the BJT is in cut-off mode.

Last edited:

- #13

perplexabot

Gold Member

- 329

- 5

Yes for Vb= 7V The output voltage Ve ≈ 6.3V

As for the saturation for Vb = 20V bjt is almost enter the saturation region.

To enter saturation Vb must be higher the Vcc.

Ok So now if we change the GND position as you have in Art of Electronics

I hope that now you see that without BJT we have VE = -5V. And now if you want to open the BJT the Vb voltage must we greater (lees negative) then -4.4V. If Vb is lower (more negative) the -4.4V the BJT is in cut-off mode.

Truly amazing explanation. I finally understand. Thank you so much.

Sorry for my many questions. With the BJT in cut-off mode, can it be treated as an open circuit (or infinite resistance)?

Last edited:

- #14

- 625

- 138

You can even delete BJT from the diagram.With the BJT in cut-off mode, can it be treated as an open circuit (or infinite resistance)?

- #15

perplexabot

Gold Member

- 329

- 5

That simplifies a lot.You can even delete BJT from the diagram.

Once again thank you so much, you have been a great deal of help.

- #16

perplexabot

Gold Member

- 329

- 5

How did you solve for Vb or Ib in the second circuit?

- #17

- 625

- 138

I use a circuit theory and solve the circuit.

From the II Kirchhoff's law we can write

**Vcc = I1*R1 + I2*R2** (1)

**I1 = Ib + I2 **(2)

**I2*R2 = Vbe + Ie*Re** (3)

And**Ib = Ie/(β+1)** (4)

And we can solve this for Ib.

[itex]\LARGE {I_b = \frac{R2Vcc - Vbe(R1+R2)}{(\beta +1)Re(R1+R2) + (R1R2)}}= 419.047619\mu A[/itex] I assume β = 99 and Vbe = 0.6V

Knowing Ib we can easy solve for Ie. Ve and Vb

But there is also a simpler way to solve this circuit by using thevenin's theorem.

We can replace the voltage divider (this gray rectangle) with his Thevenin's equivalent circuit

Vth = Vcc * R2/(R1+R2)

Rth = R1||R2 = (R1*R2)/(R1+R2)

And now we can solve for Ib

Vth - Ib*Rth - Vbe - Ie*Re = 0

And we also know that

**Ie = Ib*(β +1)**

so we end up with

Vth - Ib*Rth - Vbe - Ib*(β +1)*Re = 0

[itex]\LARGE Ib = \frac{Vth-Vbe}{Rth+(\beta +1)*Re}[/itex]

And this is the end.

From the II Kirchhoff's law we can write

And

And we can solve this for Ib.

[itex]\LARGE {I_b = \frac{R2Vcc - Vbe(R1+R2)}{(\beta +1)Re(R1+R2) + (R1R2)}}= 419.047619\mu A[/itex] I assume β = 99 and Vbe = 0.6V

Knowing Ib we can easy solve for Ie. Ve and Vb

But there is also a simpler way to solve this circuit by using thevenin's theorem.

We can replace the voltage divider (this gray rectangle) with his Thevenin's equivalent circuit

Vth = Vcc * R2/(R1+R2)

Rth = R1||R2 = (R1*R2)/(R1+R2)

And now we can solve for Ib

Vth - Ib*Rth - Vbe - Ie*Re = 0

And we also know that

so we end up with

Vth - Ib*Rth - Vbe - Ib*(β +1)*Re = 0

[itex]\LARGE Ib = \frac{Vth-Vbe}{Rth+(\beta +1)*Re}[/itex]

And this is the end.

Last edited:

- #18

perplexabot

Gold Member

- 329

- 5

How did you get equation (1)? Is Vcc = 10V, if Yes, then I think I get it.From the II Kirchhoff's law we can write

Vcc = I1*R1 + I2*R2(1)

I1 = Ib + I2(2)

I2*R2 = Vbe + Ie*Re(3)

AndIb = Ie/(β+1)(4)

I am familiar with thevenin's theorem, but I don't know how you got Vth = Vcc * R2/(R1+R2)? Once again if Vcc = 10V, then I think I understand.But there is also a simpler way to solve this circuit by using thevenin's theorem.

We can replace the voltage divider (this gray rectangle) with his Thevenin's equivalent circuit

Vth = Vcc * R2/(R1+R2)

Rth = R1||R2 = (R1*R2)/(R1+R2)

Isn't this supposed to be: Vth - Ib*Rth - Vbe - Ie*Re = 0 ?

And now we can solve for Ib

Vth - Ib*Rth + Vbe + Ie*Re = 0

Or are you assuming the emitter current going from ground to transistor?

I am sorry if I am asking for too much. You are helping me a lot.

- #19

- 625

- 138

No I made a typing error. Point for you.Isn't this supposed to be: Vth - Ib*Rth - Vbe - Ie*Re = 0 ?

Or are you assuming the emitter current going from ground to transistor?

I edit my post and correct the error.

- #20

perplexabot

Gold Member

- 329

- 5

I have one last question for you (I hope). For the image above (top figure), did you just "copy/paste" Vin from collector to ground?

- #21

- 625

- 138

In our case Vin = Vcc = 10V. But sometimes Vin has a different value than supply voltage.

- #22

perplexabot

Gold Member

- 329

- 5

In this figure you have only one battery

In the top figure you have two batteries. How did you get the second one? Or are these two figures NOT equivelent?

- #23

- 625

- 138

And yes I just just "copy/paste" Vin from collector to ground.

But also kept in mind that this second circuit with two batters is more universal for education purpose. Because Vin need not be equal to Vcc (supply voltage).

- #24

perplexabot

Gold Member

- 329

- 5

And yes I just just "copy/paste" Vin from collector to ground.

But also kept in mind that this second circuit with two batters is more universal for education purpose. Because Vin need not be equal to Vcc (supply voltage).

Thank you very much for your help.

- #25

sophiecentaur

Science Advisor

Gold Member

2020 Award

- 25,797

- 5,197

This is basic approach to feedback theory.Thank you all for your great replies.

Why will a change in Vout cause a change in Ib? Why would Vout change in the first place? Isn't Ib the one that is being "controlled" and so is subject to change?

An emitter follower is a circuit with Feedback. That accounts for the way that the output impedance is a lot lower than the value of the emitter load Re. If you want to analyse how feedback works (this is a general principle) you introduce an error signal and see the effect of the circuit on correcting this error. A reason that Vout could change could be that Re changes (or even that the current gain changes a bit). The feedback mechanism (involving the very high current gain of the transistor) corrects this by adjusting the Emitter current and bringing it back to (very near) what it was. When an output voltage is controlled to be more or less independent of the load, this is known as a voltage source or low impedance source. This basic approach may be unfamiliar to you but it is the way that we can see how circuits can be linearised, their frequency response flattened and their input and output impedances controlled.

You can use the methods described elsewhere in this thread to analyse what the circuit will do and they will give answers but you would need to do the sums, piecewise, for a range of values of Re, say. Approaching it as a feedback problem, you get the effective output impedance (etc.) irrespective of the value of Re. The same approach is used, of course, in amplifiers using Op Amps and simplifies many amplifier circuit problems significantly by omitting many of the less relevant circuit values.

Share: