So the OP's mystery resistor, (he was wondering how the designer chose it's
value) Serves what purpose? Current limiting? Seems to me there must be a
reason for it. Maybe not...
---
Here's what the OP's circuit looks like connected to a typical TTL
chip.
+5V---------------------+--->>---+
| |
[130R] |
| |
C |
B NPN |
E [1K2]R1
| |
[DIODE] |
| | R2
+--->>---+--[100R]--->LVTTL
|
C
B NPN
E
|
GND>--------------------+--->>--------------->GND
TTL, only being able to source 400µA or so and guarantee 2.4V out,
(with a 5V supply) is pretty weak, and that's what the 1.2k resistor
in parallel with it is for, to provide extra current into the load.
How much? assume the load needs to see 2V and we have:
Vcc - Vil 5V - 2.4V
I = ------------ = -------------- = 0.002 ~ 2.0mA
R1 + R2 1200R + 100R
Also, those resistors will keep any downstream circuitry from
floating if there's nothing plugged into the connector and will
provide everything downstream with known logical staes if nothing is
plugged in
The resistance of the 100 ohm resistor is only [really] important
when the load is being pulled down, and since TTL can sink 1mA and
stay below 0.1V, that current through 100 ohms will cause an
additional 0.1V drop, resulting in a total drop of 200mV with 1mA
out of the load. With a maximum current of 75µA out of the LVTTL
load and a 0.1V drop across the bottem totem-pole transistor of the
TTL sink that comes to a drop of:
Et = (Il Rs) + Vol = (7.5E-5A * 100R) + 0.1V = 0.1075V ~ 0.1V
Which is well below LVTTL Vil(min) of 0.8V.
Bottom line? the 100 ohm resistor isn't really needed but is in
there in case someone does something stupid like connect 5V with no
current limiting to the connector.
To find out, check the maximum input current allowed into LVTTL and
check if that's exceeded with raw 5VCD on its inputs when it's
running with a 3V supply.