Hyperbolic tangent transfer function

[Winfield Hill]

Tony Williams wrote...

In article <NKoje.1416023$35.52808274@news4.tin.it>,
Ban <bansuri@web.de> wrote:
[snip]
.............. The differential input Win proposed will IMHO
not improve the curve, but allow for a higher input voltage.

True, but an isolated signal ground allows better following
of small inputs. That's one purpose. Oft-times the common-
mode voltage can usefully be negative, that's another purpose.

Since for values over 10Vt the output is 0.99991, we can put
a limiter in front to allow for higher input voltages.

You can make the collector voltages more favourable
by swapping the collector-opamp connections. As below.

+-----+---o
| | +Vcc
.-. .-.
| | | | 10k_
| |10k| | +--|___|--+
'-' '-' | |
| | | |\ |
+-----)------------+----|-\ |
| | | >-+---o
| +------------+----|+/ Vout
| | | |/
Vin |/ \| .-.
o------|Q1 Q2|--+ | |10k
|> <| | | |
| | | '-'
+--+--+ | |
| === ===
| GND GND
.-.
( I )100uA
'-'
|
o -Vee

As Q1 takes more current, Q2 takes less current,
which raises Q2's collector voltage.

Good. Increasing the output resistors by 10x will also help
increase the common-mode range, shown above.

With respect to using this scheme as a precision nonlinear
attenuator for large signals to increase dynamic range, offset
voltage errors in the input pair can have a dramatic degrading
effect when attempting to calculate the original voltage from
the attenuated output signal - small shifts in apparent input
voltage correspond to large changes in the actual voltage for
large voltages. E.g., near zero volts a 1mV shift corresponds
to 1mV, but at 120mV it corresponds to a much higher voltage.


--
Thanks,
- Win