140417-0856 EDT
fifty60:
Assume the TTL input is the best way to on-off control the power supply. Although you might switch the AC input power to the power supply.
There should be NO common path from the power supply 5 V common to your instrument via any cable that is to control the TTL input. This control path should be totally isolated. There should be short leads from the TTL input to where the isolation between the power supply and your instrument occurs.
I would put a 0.1 mfd 10 V disc cermanic capacitor between TTL common and the TTL input at the TTL connector, and a 100 ohm 1/4 W resistor from the TTL input to whatever is the switch that is going close to TTL common.
N1IST has suggested how to obtain a 5 V supply from the +12 V you found. When you use a different power source than Vcc for the positive logic level, then it is necessary to know its relation to the Vcc supply. I could not quickly find specs on how much + excess voltage could be applied to a gate input for a standard TTL gate.
A test with a 12 V source and a random 1N4733 5 V Zener diode produced the following voltage results across the Zener using the respective series dropping resistances:
4.4 V with 5.6 k and 1.3 mA.
4.9 V with 5.6 and 1.0 k resistors paralleled and 8.1 mA. Calculated resistance 850 ohms.
The typical minimum specified high input (minimum that a logic 1 should be) for a 74xx is 2 V, per 1976 TI TTL Data Book.
The series dropping resistor from +12 V to the Zener diode is your pullup resistor, and the Zener clamps the maximum voltage that can be applied to the gate input. The Zener cathode and series dropping resistor connect to the free end of the 100 ohm input resistor to the TTL gate. A 100 ohm resistor at the input to a standard TTL input (a 7404 gate input) when connected to common has a voltage drop of about 0.0016*100 = 0.16 V and this is sufficiently low to insure a logic 0 state.
A reed switch, low voltage drop relay, transistor, or optical coupler can be used to shunt the Zener diode for control.
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