Emitter-follower analysis

Discussion in 'General Electronics Chat' started by gerases, Apr 2, 2013.

1. gerases Thread Starter Member

Oct 29, 2012
177
2
Hi all,

This circuit is driving me crazy and I'm not smart enough to figure it out on my own.

My question is: how do we get 6.42V on the base? Usually, there's a voltage divider at the base and then it's a little more simple. But here, I don't know where to start.

The whole emitter follower arrangement has a few mysteries for me. When the emitter goes straight to ground, it's clear, the base current flows, the CE path starts to pass current depending on how much current flows into the base, which in turn is determined by the base resistor.

With the emitter-follower, it's more complicated because the base current depends on Vload and the Vload depends on VCE, which in turn depends on Ib. It's like a chicken and egg problem.

Thank you!

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2. Jony130 AAC Fanatic!

Feb 17, 2009
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In this circuit you post the BJT is in a saturation.
So I don't know what you want to do with this circuit?

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3. gerases Thread Starter Member

Oct 29, 2012
177
2
Well, can you tell me how to calculate the drop over the Rin in general in a configuration like this?

4. Jony130 AAC Fanatic!

Feb 17, 2009
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Emitter follower is a very simply circuit. The output voltage is always 0.6V (Vbe) 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.

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6. gerases Thread Starter Member

Oct 29, 2012
177
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By Rin I meant the 1K resistor on the base. Sorry.

I tried to apply that method, but there, we have a voltage divider and everything is a bit simpler because we know the base voltage straight away. Here, I don't see where to begin. How come the base voltage in this case is not simply 9V?

Last edited: Apr 2, 2013
7. Jony130 AAC Fanatic!

Feb 17, 2009
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As usually we start our analysis by applying Kirchhoff's law.

Vcc = IB*Rb + Vbe + Ie*Re

And for the BJT

Ie = Ib *(β + 1)

Ib = (Vcc - Vbe)/( Rb + Re*(β+1))

and Voltage at base

Vb = Vcc - Ib*Rb or Vb = Vbe + Ie*Re

So if we assume β = 100; and Vbe = 0.65V we have

Ib = (9V - 0.65)/(1K+ 1K*101) = 8.35V/102k = 81.87μA

and
Ie = 101 * 81.87μA = 8.26mA

Now we can check whether BJT is on active region or in saturation

Collector current cannot be greater than Ic_max = (Vcc - Vce_sat)/(Rc + Re) ≈ 9V/2K = 4.5mA.

But we calculate 8.26mA.

So if Ic >> Ic_max BJT is in saturation region. And in saturation Ic = β*IB don't hold anymore.

All we can do next is to write this equation

Ie = Ib + Ic

So we have

Ie = Ve/Re

Ib = (Vcc - Vbe - Ve)/Rb

Ic = (Vcc - Vce_sat - Ve)/Rc

And using these three equations we can solve for Ve

So voltage at base is equal to

Vb = Vbe + Ve = 6.4V

I assume Vbe = 0.65V and Vce_sat = 0.1V.

This is all we need to do to find DC bias current in this circuit.

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8. gerases Thread Starter Member

Oct 29, 2012
177
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Very cool. What are the vertical bars in (RB||RC||RE)?

9. Jony130 AAC Fanatic!

Feb 17, 2009
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"||" - parallel connection.

In our case we have RC = RB = RE = 1K so equivalent resistance is equal

Req = 1K/3 = 333R

Last edited: Apr 2, 2013
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10. gerases Thread Starter Member

Oct 29, 2012
177
2
Oh, I see!

Thank you for your help. Your explanation is tremendously useful!

11. wmodavis Distinguished Member

Oct 23, 2010
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Actually the circuit you refer to is not an emitter follower. It is a common emitter BJT circuit with the emitter resistor not bypassed.

12. Jony130 AAC Fanatic!

Feb 17, 2009
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How can you know that? There is no input/output terminal on the diagram.
And what if I connect "the load" directly to Re resistor?

13. crutschow Expert

Mar 14, 2008
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Yes, if you take the output from Re it will be acting as an emitter follower. But a normal emitter follower circuit has no collector resistor since it serves no purpose and can actually slow down the stage response due to Miller negative AC feedback through the base-collector capacitance. So when a BJT stage is seen with both collector and emitter resistors, the logical assumption is that it is a CE circuit with an emitter bias stabilization resistor (which can be AC bypassed or not as required by the circuit requirements).

14. wmodavis Distinguished Member

Oct 23, 2010
739
151

I can know that because your original post was sooo incomplete wrt providing detail as to what you were asking. See Cruts post.