Ban wrote...
Maybe the subtlety is the feedback connection between O/P and bases
of the long tailed pairs? The current into the output transistors
will regulate by itself. We do never saturate or starve the first
stage, it's a buffer, not an amplifier. Whatever current is not used
in the first stage will flow into the output transistors, it is not
controlled by a fixed bias voltage. The right current source has
10mA, half of the left one. These 10mA are consumed by the input
transistor, so there is nothing left for the power stage.
Nope, not so simple Ban. Small offsets in the input pair can easily
create say 10% current imbalance, and the resulting 1mA drive will be
magnified by the output transistor's beta, possibly leading to >100mA
of class-A rail-rail current. If each side's feedback was returned
from one of its output transistor's emitters, before the equalizing
resistor, the class-A current problem could be mitigated. But what
you really want is a way to predetermine and control the quiescent
bias current, which ideally for a high-speed amp should not be zero.
This can be done by adding current sources and resistors at the input.
.. ---+---+-----------+----+-----
.. | | | |
.. 10mA 5mA | |
.. | | _ |/ |/
.. | +------U-------| fzt2222a
.. | | | bead |> |>
.. 1mA | | | |
.. | ,---)---)----+------+ | quiescent
.. | |/ | \| | | | Iq = 10mA
.. +--| | |--' | |
.. | |> | <| 10 10
.. 100 '-+-)---' 1W 1W
.. | | | fzt2222a | |
.. o---+ | | (matched) +----+---o
.. | | | fzt2907a | |
.. 100 ,-)-+---, 10 10
.. | |< | >| 1W 1W
.. +--| | |--, | |
.. | |\ | /| | | |
.. | '-)-----)----+------+ |
.. 1mA | | | |
.. | | | _ |< |<
.. | +------U-------| fzt2907a
.. | | bead |\ |\
.. 10mA 5mA | |
.. | | | |
.. ---+-----+-----------+----+----
