W
Winfield Hill
- Jan 1, 1970
- 0
Fred Bloggs wrote...
Right, Vout is about 10.35 volts, or 3.5% high. Of course, those
are presumed to be 5% resistors, commonplace for the 70s and 80s.
The goal is to get the maximum swing at Q2's output, while biasing
the output at 0.5 Vcc; we erred by not using 19.35k instead of 20k,
to account for 650mV of Vbe drop used in biasing on the ground side.
.. Vout = Vcc - [((Vcc - Ib1 Rb1 - Vbe1) / Rc1) - 2 Ib2] Rc2
.. = 20 - [((20 - 2.5uA * 10k - 0.65) / 19.35k) - 10uA] * 10k
.. = 20 - [((20 - 0.025 -0.65)/19.35k) -10uA] * 10k
.. = 20 - [( 19.325 / 19.35k ) - 10uA ] * 10k
.. = 20 - [ 998uA - 10uA ] * 10k
.. = 20 - 9.887
.. = 10.11 volts
which is within 1% of Vcc/2 for a beta of 200. And with a 19.1k
resistor (standard 1% value) we surreptitiously do better, now to
0.2%, more than good enough for the crude back-of-the-envelope
calculations we're illustrating. :>)
Figure 2.41B is preferred, with its two 180-ohm emitter resistors,
because the usual CE-stage distortion is reduced dramatically, and
the gain is pegged at about 49. The penalty is a modest 360mV of
output range. Surely that's a bit better use of some output-range
loss than twice that amount of loss for transistor diodes in series
with both collector resistors, and much higher distortion to boot.
Thanks,
- Win
whill_at_picovolt-dot-com
The deal was to get equal collector currents and halve the collector
resistor on the output transistor. The output transistor Vbe2
corresponds to the compensating transistor *emitter* current of
(Vcc-Ib1*Rb1-Vbe1)/Rc1-Ib2 so that the emitter current of the output
is the same making the output collector less by Ib2 or
(Vcc-Ib1*Rb1-Vbe1)/Rc1-2*Ib2, and therefore the output voltage is
Vcc-((Vcc-Ib1*Rb1-Vbe1)/Rc1-2*Ib2)*Rc2 then since Rc2/Rc1=0.5 and
Rb1=Rc2, this reduces to (Vcc+Vbe)/2+(0.5*Ib1+2*Ib2)*Rc2.
Yep- looks like you're right.
Right, Vout is about 10.35 volts, or 3.5% high. Of course, those
are presumed to be 5% resistors, commonplace for the 70s and 80s.
The goal is to get the maximum swing at Q2's output, while biasing
the output at 0.5 Vcc; we erred by not using 19.35k instead of 20k,
to account for 650mV of Vbe drop used in biasing on the ground side.
.. Vout = Vcc - [((Vcc - Ib1 Rb1 - Vbe1) / Rc1) - 2 Ib2] Rc2
.. = 20 - [((20 - 2.5uA * 10k - 0.65) / 19.35k) - 10uA] * 10k
.. = 20 - [((20 - 0.025 -0.65)/19.35k) -10uA] * 10k
.. = 20 - [( 19.325 / 19.35k ) - 10uA ] * 10k
.. = 20 - [ 998uA - 10uA ] * 10k
.. = 20 - 9.887
.. = 10.11 volts
which is within 1% of Vcc/2 for a beta of 200. And with a 19.1k
resistor (standard 1% value) we surreptitiously do better, now to
0.2%, more than good enough for the crude back-of-the-envelope
calculations we're illustrating. :>)
Figure 2.41B is preferred, with its two 180-ohm emitter resistors,
because the usual CE-stage distortion is reduced dramatically, and
the gain is pegged at about 49. The penalty is a modest 360mV of
output range. Surely that's a bit better use of some output-range
loss than twice that amount of loss for transistor diodes in series
with both collector resistors, and much higher distortion to boot.
Thanks,
- Win
whill_at_picovolt-dot-com