Solve Emitter Follower Circuit: Get Answers Now!

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Discussion Overview

The discussion revolves around the analysis of an emitter follower circuit, focusing on input and output impedances, voltage gain, and the behavior of the circuit under different configurations, including the use of Darlington pairs. Participants are exploring theoretical calculations and practical implications, with a mix of conceptual and technical reasoning.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant outlines the calculations for input and output impedances, suggesting that the input impedance (Zin) is approximately 300k ohms and the output impedance (Zout) is around 50 ohms.
  • Another participant agrees with the input impedance calculation but challenges the output impedance calculation, proposing a different approach that includes the transistor's beta (β) factor.
  • A participant inquires about the calculations for a Darlington configuration, speculating that the input and output impedances would be calculated similarly but with a potentially lower output impedance due to a higher β.
  • Concerns are raised about the voltage gain in the circuit, with one participant suggesting that the gain is approximately 0.99 and discussing the implications of using a quasi-complementary output stage.
  • Another participant questions the clipping of the negative part of the signal, discussing the maximum output voltage swing and potential asymmetries in the circuit's behavior.
  • A participant provides a detailed formula for voltage gain, incorporating various resistances and parameters, while also addressing the limitations of the output swing based on the transistor's characteristics.

Areas of Agreement / Disagreement

Participants express differing views on the calculations for output impedance and the implications of using Darlington pairs. There is no consensus on the exact values or methods for calculating certain parameters, indicating ongoing debate and exploration of the topic.

Contextual Notes

Some calculations depend on assumptions regarding the circuit configuration and the values of components, which may not be universally agreed upon. The discussion includes various interpretations of the circuit's behavior under different conditions, leading to unresolved questions about voltage swings and gain.

JackB
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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
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JackB said:
And the actual input Z (ignoring the 5k6 source?) would become 300k || 100k || 130k = 48k ohms

That's correct.
JackB said:
Zout = 100k || 130k || 300k || 5k6 || 3k3 = 5K / 100 = 50ohms

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 . . .
 
Thanks! Ah yes I did the sum & posted it & then realized 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..
 
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.
 
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 ?
 
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
https://www.physicsforums.com/threads/emitter-follower-help.631243/page-3#post-4061388

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
 
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