Emitter Follower

1. Apr 24, 2015

JackB

Hi, I'm currently self learning electronics and am stuck on the emitter follower..
I'm wondering if I can talk it through with someone..?

In regards to In Z and out Z, running through a document I have..
In Z = ΔVB/ΔIB
So for the ΔIB
IE = IB + IC
IC = βIB
ΔIB = ΔIE/(β+1)

Therefore ΔIB = RE(β+1)

And the ΔVB
ΔVE = ΔVB
ΔVE = RE ΔIE
ΔVB = RE ΔIE

Therefore ΔVB/ΔIB = (β+1) RE

So the input Z of the base would be around 300kohms without the two RB resistors?
And the actual input Z (ignoring the 5k6 source?) would become 300k || 100k || 130k = 48k ohms

And in regards to outZ..
= (Vin - Vout)/output current
= (Vin - ΔVB)/ΔIE
Vin = Rsource ΔIB + RE ΔIE
Vin = Rsource ΔIE/(β+1) + RE ΔIE
Vin = [Rsource /(β+1) + RE] ΔIE
Zout = Rsource /(β+1)
Rsource/(β+1)

So in my case here it would be

100k || 130k || 300k || 5k6 || 3k3 = 5K / 100 = 50ohms

Am on the right lines so far?
Any help really appreciated

2. Apr 24, 2015

Hesch

That's correct.
That's wrong:

Ignoring C1 ( you can do so by high frequency ) it will be:

Zout = ( ( 100k || 130k || 5k6 ) / (1+β) ) || 3k3 = 50Ω

Well, same result, but in principle . . .

3. Apr 24, 2015

JackB

Thanks! Ah yes I did the sum & posted it & then realised I'd forgot about the load so popped it in without thinking..

I'm also looking at Darlingtons in this configuration.. I assume the Zin & Zout would be calculated exactly the same way? Presumably the main difference would be higher β and so they'd have a lower output Z all else equal.

With regards to bootstrapping however I'm a little lost with Zin..
For output Z, I guess it's exactly the same as previous? All else equal, slightly higher perhaps as the source isn't as loaded as before.

However I'm unsure how to calculate the amplification.. So I'm stuck on input Z..
Mostly in my books it either says a little less than 1 or imagine it's 0.985 etc..

4. Apr 24, 2015

Hesch

Don't care about an amplification ≈ 0,99. A problem using darlington could be the amplitude of the voltage output ( rail to rail proporties ) in an output-stage. You could consider using a quasi-complimentary output-stage instead of a complimentary.

5. Apr 24, 2015

JackB

The negative part of the signal being clipped as voltage increases?

Yes.. that's another thing I'm not 100% sure of in this circuit..

Why does this happen?

The positive can swing to within a saturation drop of V+.. so that is 0.1V - 0.3V ?
Then it can only turn off.. so that would take it to +6V + 0.6V ? So a maximum of 17.3Vpp *0.3535 = 6.1Vrms = 17.93dBu ? That seems too high.. Have I gone wrong somewhere?

Perhaps because it's asymmetrical? In this instance before clipping it would only be 10.8Vpp? *0.3535= 3.82Vrms = 13.86dBu ?

6. Apr 25, 2015

Jony130

As for the voltage gain.

Av = V1/VB * VE/VB where

V1 - input voltage
VB - voltage at base
VE - voltage at emitter

Av = Rin/(Rs + Rin) * RE/(re + RE) = 48kΩ/( 5.6kΩ + 48k) * 1kΩ/(1k + 0.007kΩ) ≈ 0.889V/V

Also Rin = RB1||RB2||( (β+1)*(RE + re) )

Where :
Rs - source internal resistance = 5.6kΩ
RE - emitter resistance = 1kΩ
re - small signal resistance between base and emitter looking into the emitter re = Vt/Ie ≈ 26mV/Ie = 26mV/3.7mA = 7Ω

As for the output swing.
The voltage at transistor base cannot be larger than Vcc. And cannot be lower than 0.6V. So the max negative output voltage swing is equal to -Vmax = Ieq * RE||RL

Last edited: Apr 25, 2015