kchriste said:It is possible, but more likely because the opto doesn't have enough drive capability to activate the relay. Looking at the schematic of the opto PCB, I see some 2n2222 transistors being driven via an isolator and 10K pullup resistor. The 10K resistor means that the transistor is only getting apx 0.7ma of drive current when the battery voltage is 7.5V. The 2n2222 won't have enough beta (gain) to turn fully on (saturate) so won't apply the full voltage across the relay coil. You could test this by measuring the voltage across the relay coil with your volt meter. It would be best to have the radio turned off because the RF will probably cause the meter to give a false reading.
When I measure the voltage between OUT1 and the positive lead on the battery I get 7.4 volts (actually its a little over 8 if fully charged) and I get 10 amps. Looking here - https://www.electro-tech-online.com/custompdfs/2009/11/G5V-1.pdf I need about 20 amps at 7.4 volts. It seems clunky but I guess I could drive OUT1 and OUT2, wire them in series, boost the voltage to 15V and then the 10 amps would drive the coil. There must be a more elegant way
The radio also blanks out the GPS when transmitting; not a big deal as it only transmitts every 4 or 5 minutes. Oddly the chip on the gumstix puts off exactly the same frequency as GPS signals so you will see a second post on how to make a lightweight RF shield to keep the GPS antenna away from the gumstix 'jammer'
kchriste said:If the voltage is too low (Less than 5V) then we could add a transistor to boost the current. What types, if any, do you have on hand?
I can get whatever I need from the electronics across the road, but I think its amps I need, not volts.
kchriste said:It would go across the relay coil with the stripe facing the (+) battery side. Any of the 1N400x (Where x is a number) diodes will work. So will the common 1N4148 or 1N914.
I do have a few diodes so I'll give it a shot. To be clear, I would just wire the stripe side to the relay pin that goes to the positive terminal on the battery and the other end of the diode to the other relay pin. In essence, both the diode and the wire to the + on the battery will be connected to the same leg on the relay?