Close the switch to simulate an input signal.
A previous circuit I made, had me thinking. This circuit can easily be applied to any solid state amp. When an input signal is connected, the first op amp will start charging the 330uF capacitor through the 2.2k resistor (the envelope detector). After it reaches 2V (which should happen in less than half a second) the second op amp will activate the relay driving transistor, and the relay will activate the power supply rails for the amplifier. You may need a DPST relay according to your amp schematic. A lack of input signal for more than 9minutes will disengage the power supply rails and the amp will enter standby mode. The time constant of 9mins can be varied by the 330uF/2.2Mohm RC network. I'm gonna lay down below a few values for time constants of 1min to 10min. You're going to need an autonomous power supply for the control circuit, maybe a separate winding on the transformer or something. The PNP transistor will quickly discharge the capacitor and break contact from the transformer to the amplifier if the power transformer fails or the amplifier is disengaged from the mains. The quiescent current of the amplifier in stand-by mode is around 5mA. The control circuit will draw current equivalent (±5mA) to the relay coil current when operational. Maybe around 100mA.
1min release - 2.2Mohm/33uF (55s)
2min release - 1.5Mohm/100uF (115s)
3min release - 2.2Mohm/100uF (168s)
4min release - 1.5Mohm/220uF (253s)
5 min release - 1.8Mohm/220uF (305s)
6min release - 2.2Mohm/220uF (370s)
7min release - 2.5Mohm/220uF (420s)
8min release - 2.8Mohm/220uF (470s)
9min release - 2.2Mohm/330uF (550s)
10min release - 3.6Mohm/220uF (606s)
