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  • GE VMIVME-4215 Gas turbine input card
GE VMIVME-4215 Gas turbine input card

GE VMIVME-4215 Gas turbine input card

GE VMIVME-4215 Gas turbine input card

To make sure your signal source and the 1771-VHSC module are compatibility, you need to understand the electrical characteristics of your output driver and its interaction with the 1771-VHSC input circuit. Refer to Figure C.1. The most basic circuit would consist of R1, R2, JPR4, JPR5, the photodiode and associated circuitry around half of the opto-isolator. The resistors provide first-order current limiting to the photodiodes of the dual high speed opto-isolator. With JPR4 closed, and JPR5 open, the total limiting resistance is R1 + R2 = 1150 ohms. This jumper position is designated “12 to 24 Volt Range.” Assuming a 2V drop across the photodiode and R97 and R98, you would have 8.7-19mA demanded from the driving circuit as the applied voltage ranged from 12 to 24V.

Example Circuits for 5V Differential and +12 to +24V SingleEnded

In the “5 Volt” position (JPR4 open; JPR5 closed), R1 is shorted and the limiting resistance is 250 ohms. If 5.0V was applied at the input, the current demanded would be (5.0 - 2.0)/150 = 20mA. The above type of calculation is necessary to the user since the driving device must cause a minimum of 5mA to flow through the photodiode regardless of which jumper position is selected.If the driving device is a standard 5V differential line driver, D2 and D3 provide a path for reverse current when the field wiring arm terminal 1 is logic low and terminal 2 is logic high. The combined drop is about the same at the photodiode (about 1.4V). The circuit appears more symmetrical, or balanced, to the driver as opposed to just one diode.

The optical isolator manufacturer

 recommends a maximum of 8mA to flow through the photodiode. This current could be exceeded in the 24V position. To obtain this limit, a dc shunt circuit is included, consisting of D1, Q2, R97 and R98. If the photodiode current exceeds about 8mA, the drop across R97-R98 will be sufficient to turn Q2 on, and any excess current will be shunted through D1 and Q2 instead of through the photodiode.

                                         

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