[aprssig] Green light labs gps

[KF4LVZ]

On 2015-08-28 04:15, Stephen H. Smith via aprssig wrote:
> On 8/27/2015 10:46 PM, Dale Urban via aprssig wrote:
>> I have a green light labs gps that I used on my Kenwood D-710. I would
>> like to use it on my D-700 so I would need to supply power through the
>> RJ-45 jack. Would anyone know what the pinout is of the RJ- 45 jack on
>> this gps and which pin is pin 1 and pin 8 Thanks Dale N0KQX
>>
>> Sent from my iPad

>>
> 
> Since the D700 and D710 use the same standard Kenwood pinout for their
> RJ-45 mic jacks, it should be trivial to power it.

The D700 and the D710 don't use the RJ45 for the same purpose.  The
control head on a D700 is a 4-conductor (4P4C) modular connector while
the D710 has the 8-pin (8P8C) RJ45.  In this case he's asking about the
RJ45 on the control head where the Green Light Labs GPS typically mounts.

Also, the voltages and supply currents are different.  The D700 supplies
8 volts on pin 7 of the microphone jack with a current limit of about
100 mA (enough to power a DTMF microphone with backlight).  The Control
head receives about 10 volts on pin 7 but with a substantially higher
current limit.  Depending on the design of the GPS power supply (e.g.
use of a low drop out voltage regulator), and the current consumption,
it might not be possible to power it from the D700 directly.

According to the schematic for the D700, the control head power supply
is 9.6 Volts with power drawn through fuse F902 (value not listed in the
parts list) and transistor Q911 (a 2SB1132 PNP power transistor with a
maximum collector current of 1 amp) and supplied by the primary input
power bus (13.7 volts).  The microphone jack has pin 7 connected to the
internal 8 Volt power bus which provides power to the audio circuits.
The 8-volts comes from a TA7808S which is an 8 volt, 1 amp linear
regulator.  However, there are additional diodes and resistors in series
with the microphone jack so the total current capacity of the jack will
not be high without risking damage to the low wattage resistors
(specifically R810 which is a 33 ohm, 1/16 watt resistor probably chosen
to act exactly like a fuse).  Maximum current through that resistor is
about 40 milliamps (P = I^2 * R, P = 1/16, R = 33, sqrt(P/R) = I =
0.0435).  That's not enough for most GPS receivers.