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Emitter Follower

  1. Apr 24, 2015 #1
    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)

    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. jcsd
  3. Apr 24, 2015 #2


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    Gold Member

    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 . . .
  4. Apr 24, 2015 #3
    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..
  5. Apr 24, 2015 #4


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    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.
  6. Apr 24, 2015 #5
    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 ?
  7. Apr 25, 2015 #6
    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
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