Hello one and all,
I have an Agilent 11713A switch driver that I’m using with Agilent 8765A coaxial relays. The 8765A is a single pole double throw relay. As it turns out the 8765A relay is ideally pulse driven. That is, it only requires a pulse of current to actuate the relay and it will stay in position without any more current flow. When it is driven with continuous current, it gets very hot. The 11713A switch driver only does continuous current. My question is, what circuit could I put in between the driver and the relay to get the desired pulse drive effect from an otherwise continuous current driver? I’ve got the switch wired up with a power supply and a breadboard so I can test out some ideas.
This is essentially what the circuit looks like without any pulse driver circuit. The 11713A supplies the 24 volts which is always in contact with the relay. The 11713A completes the circuit for one of two paths (represented by the single pole double throw at the left) to energize a solenoid inside the relay and actuate the relay. The solenoids in the relay are represented by the two 200 ohm resistors.
My intention is to put a circuit here, where the question mark box is at:
This is what I have tried:
When the Agilent driver completes the circuit, current will initially surge to charge the capacitor. This allows enough current to flow for the relay to actuate. The 1k-ohm resistor is there to discharge the capacitor when the Agilent driver breaks the circuit. When the circuit is complete though, the capacitor will fully charge and do nothing until it has a chance to discharge. Meanwhile, the 1k-ohm resistor acts as a current limiting resistor, reducing what would be 120mA to ~20mA.
What’s undesirable about this solution is the current is never turned off, the 1k-ohm resistor limits the current but it’s still flowing. The 1-kohm resistor has to continuously dissipate 0.4W.
The intention is use a circuit that doesn't require a separate power source. I humbly request any assistance.