Driving a load in parallel with a resistor internal to the source

Thread Starter

crazyjunky

Joined Oct 16, 2014
8
In the Art of Electronics, page 71, the author is discussing how to design an emitter follower circuit when it is driven by an ac-coupled signal.


At the input there is capacitor C1 (why is it needed if the input signal is already ac-coupled?). To select its capacitance, the author says "...if we assume that that the load this circuit will drive is large compared with the emitter resistor..."

However, since the load will be connected in parallel to RE, shouldn't the load be much smaller than RE, so as to have most of the current going through the load? Otherwise most of the current will flow into RE, since the load is parallel to it...

Thank you
 

Jony130

Joined Feb 17, 2009
5,176
As for the ac-coupled capacitor try read this
http://forum.allaboutcircuits.com/threads/voltage-divider-bias.71104/#post-494875 (especially panic mode post)

As for the load resistor back to this
http://forum.allaboutcircuits.com/threads/in-an-emitter-follower-the-npn-can-only-source-current.102505/#post-773485

And notice that for RE = RL output voltage can swing from 20V to 5V (15 peak to pek). But if we change RL = 0.1RE the output can only swing from 20V down to 10V*1K/(1K + 100) = 9V (11V peak to peak).
In this circuit C2 act or replace the lower 10V battery.
 

vk6zgo

Joined Jul 21, 2012
677
If we aren't looking for power gain,it is not necessary to maximise the current through the load.
The voltage gain of an emitter follower is very close to unity,& its output impedance is very low,so that if the load is a much higher impedance.the output voltage is virtually unaffected by it,compared to the unloaded case.
 
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