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RE: Buffer - emitter resistor value?

Posted by 13th Duke of Wymbourne (A) on December 24, 2024 at 00:35:36

In Reply to: Buffer - emitter resistor value? posted by andyr on December 23, 2024 at 21:03:36:

The pull-down resistor sets the dc current in the follower and that affects the small signal output resistance and the large signal distortion performance. Assuming the base is biased mid-rail, the voltage across Re is either (12V-Vbe)/4.7k or (12V-Vbe)/1k ~ 2.4mA or 11.3mA. The output resistance of an emitter follower is (Vt/Ie + Rbase/Beta) // RE.
Vt/Ie (what we in the trade call 'little Re') is the incremental resistance of the base-emitter diode seen from the emitter side. Vt = kT/q ~ 26mV so with the two Ie options 'little Re' will be either ~ 10.8ohms or ~ 2.3ohm. Let's assume that the circuit driving the base has an output resistance of 1kohm (including the real part of the BJT base resistance, aka 'rbb') and the follower Beta = 100, then the contribution of the resistance on the base side as seen at the emitter is ~ 1k/100 = 10ohms, which is significant. With these assumptions you only get an improvement in output resistance from ~20.8ohms to 12.3ohms for a 5x increase in current (this is shunted by the pulldown resistance, aka 'Big RE', but >1k shunting 10s of ohms does not have much effect). The small signal output resistance tells you how well the follower will drive a low load resistance &/or a capacitive load.
Considering the distortion performance of a follower, that comes from the large signal performance and depends upon the current variation in the transistor. Assuming the signal at the base is un-distorted, the output follows that voltage offset by a Vbe. If the current in the transistor changes then Vbe changes in an exponential fashion given by Vt*ln(ratio of current change), e.g. if the current doubles the change in Vbe ~ 26mV*ln(2) = 18mV and this is distortion seen at the emitter that is not present at the base. With a resistive pulldown (RE) the current in the transistor varies with the signal voltage irrespective of any current supplied to the load e.g. if the base signal voltage is 2Vpk then the emitter current changes by either 2/4.7k = 0.15mApk or 2mApk. The effect on Vbe of the two cases is Vt*ln((2.4m+0.15m)/2.4m) = 1.6mV or Vt*ln((11.3m+2m)/11.3m) = 4.2m. Relative to a 2Vpk signal those are changes of 0.08% or 0.2%, which are small but not negligible if you want lowest distortion. This distortion can be all but eliminated by using a constant current source pulldown instead of RE so that the transistor current does not vary with signal.
So far I have ignored the effect of current supplied to the load as that is supplied by the follower and will also modulate Vbe. If you are driving 10k with a 2Vpk signal then the peak load current is 0.2mA, which is slightly more than doubles the current change for the 4.7k case but is only one tenth that of the current change in the 1k case. So, a 10k load resistance will not have a big effect on distortion performance. If the load was lower, say 1kohms, then the peak load current is 2mA. The impact on Vbe is now Vt*ln((2.4m+0.15m+2m)/2.4m) = 16.6mV (0.83% of the 2Vpk signal) or Vt*ln((11.3m+2m+2m)/11.3m) = 7.9mV (0.4% of 2Vpk).

In summary, the 1khom case allows you to drive a low load resistance with lower distortion but it will, counter intuitively, produce more distortion in a high load resistance case (though that can be eliminated by using a current source pulldown). One last thing is power dissipation, that will be either ~ 12.7V*2.4mA = 30mW or ~ 12.7V*11.3mA = 143mW so check the transistor rating.

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Kimber Kable

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