colombian said:
I am finishing a basic GSSL with just the turbo board, no SSC. I have an EAO MA illuminated, switch, 2NC + 2NO, (01-262.025). If I want to use the bypass and knowing I don't have a SSC, how would I wire it? have a diagram?
I need : Bypass, disable makeup gain, light the lamp (where should i take the power for the lamp?)
Welcome Lud,
for GSSL bypass you'll need 2 switch poles for the differing voltage potentials. The lamp needs another pole for a 3rd different voltage potential, so obviously one pole would be missing from your EAO switch alone. Easiest solution might be a single DPDT relais for the bypass switching, switching function
nearly as shown in Jakobs schematic. (cleanest solution probably would be bypassing the complete GSSL from XLR-in to XLR-out, but this would require 4 DPDT relais for a balanced stereo config).
For a maybe Omron-G5V-2-H1-DC24 or Multicomp-HRS2H-S-DC24V relais, connect relais pin4 to 'COM', pin6 to 'ON', pin8 to 'OFF', link pin8 to pin9, pin11 to the right side of the 'link' (further going to wiper of makeup-pot) on control pcb, pin13 to the left side of the 'link' on control pcb.
Activating this relais thru your EAO switch by connecting relais pin1 with 0V (or to the already linked pins 8/9). Connect relais pin16 to a throw position of your switch (pin1 for default activated or pin3 for default bypassed). Link the switch poles pin2 and 4 and take this junction to the raw +DC before the +15V regulator (easiest soldering spot seems to be the unused hole on GSSL main pcb between 7815 pin1 and the not mounted mid side rectifier). Add a flyback diode, maybe 1N400x, between relais pins 1/16, anode side to pin1.
Illumination with your whatever voltage (6,12,24,28,30,36,48,60V) base T 5.5 incandescent lamp might be done by connecting one lamp pin with 0V and the other lamp pin with either switch throw pin1 or pin3, depending on wanted displaying status (activated/bypassed) with a maybe needed voltage dropping resistor in between (a 24V lamp or greater obviously doesn't need a dropping resistor when fed from a ~22V raw DC source, and a 48V or 60V lamp will probably never show up to a noticable degree). Resistor in ohms, if needed and ignoring voltage drop in rectifier, would be [(transformer secondary AC voltage * root(2) - lamp voltage) / lamp current in A]. For a 12V 50mA lamp and 15VAC transformer secondary this would be (15V*1.414 - 12V)/0.05A = 184.2 ohm. If required, resistor rating would be at least [(transformer secondary AC voltage * root(2) - lamp voltage) * lamp current in A] = 0.46W. I'd pick a 200R/1W part for these example values.
Just look up the datasheets for your switch or prementioned relais types to get the idea (weblinks already posted some pages back, or google them up).
... I also want to use an EAO for Power on/off. This one is straight forward, but the lamp, where can I take the voltage needed from?
Maybe from same raw DC spot as previously described, with maybe needed voltage dropping resistor in between. Make sure, these lamp contacts never get in touch with your switched mains AC primary voltage. The LED with 1k resistor in series on control pcb probably could be left out. YMMV.
Good luck
