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Once again, a small valued resistor is placed in series with the load and ground. A sampling of the voltage across this resistor reveals the current flowing into the load. So only a single feedback loop is required. The third model is the current-to-current amplifier. Like the voltage-to-current amplifier, this amplifier has an infinitely high output impedance, but differs in that its input functions as a dead short, as the input is current not voltage.
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For homework, I recommend finding a fascinating article from early 80's in the Journal of the Audio Engineering Society, JAES, written by Malcolm J Hawksford. Unfortunately, I have lost my copy. This article explains the many advantages that derive from driving loudspeaker with voltage-to-current amplifiers.
Subject: OTL output impedance I have built the driver circuit similar to your differential circuit on www.tubecad.com/april_may 2001. I used this to driver a totem pole of 2 x 6C33C (total just 2 tubes). I found that the output impedance is about 23 ohm as follow:
Load 8.7 ohms No load output 40v p-p Loaded output 11v p-p Zo = (40-11)/(11/8.7)=29/1.26=22.9 ohm
The expected impedance is that of 2 cathode follower in parallel: (Rp/1+u)/2 = (100/1+3)/2 = 12.5 ohm. Well this is what I have measured, do you think this accurate? Also, the other single tube driver on the same page is actually a Futterman variation, too claimed to have lower output impedance of (Rp/(2+2u) i.e. about 13 ohms with 2 x 6C33C, I have built this circuit before, but I unfortunately I didn't (forget to) measure the output impedance. So, would appreciate if you can please explain the difference between the two circuits, as to which one has lower output impedance. I have read this article beside your web magazine: http://members.aol.com/aria3/otlpaper/otlhist.htm
p. k. Malaysia
Two reasons for the higher output impedance present themselves. The first that the rp of a triode is not a constant. It like the
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