OK... let me try this again and see if I can do a better job explaining
the task at hand and my questions. (Thanks to the 2 folks who replied!!)
Task: Build a cable that connects my K3s to an SPE 1K-FA amp (on loan
until I can buy a KPA-1500) using the TX-INH signal from the SPE amp to
INHIBIT the K3s from transmitting ALL the time UNLESS it is the selected
exciter AND it sends the SPE amp a "key/relay" signal.
Purpose: To use the SPE as the arbiter of which INPUT/EXCITER has the
drive privileges, to future proof the cable wiring for probable
AMPLIFIER sharing between TWO exciters at some point. If BOTH exciters
are put into inhibit BY DEFAULT by the SPE, an exciter may ONLY drive
the amp IF, 1) the exciter is connected to the ACTIVE SPE exciter input,
AND 2) the exciter sends a key/relay signal to the SPE amp. There's a
lot of other logic involved, but this is the key point.... BOTH
exciters are Inhibited by default (TX light blinking on K3s), and ONLY
one may EVER drive the amp at one time. No external devices required.
Components/Connections:
K3s, using ACC port, Pin #7 (with TX-INH set to "High") (and of course
Pin # 5, GND) (There are other pins being used, but I am pairing it
down to JUST this topic)
SPE 1K-FA Amp Input #1 CAT port, Pin #13 (TX-INH) and Pin #4 GND.
The SPE Amp outputs +12VDC on the TX-INH signal line to the exciter
UNTIL the amp sends a key/relay signal to the amp, at which point the
SPE amp drops the TX-INH signal to 0 vdc.
Background: The fellow I got the cable pinout, et al. from actually had
the SPE TX-INH connected to the K3s Pin #7 (TX-Inh set "High) for over a
year with no ill effects. When he was informed that the SPE was using
12VDC, but the K3s was expecting 5VDC, he put a simple voltage divider
on the signal line and to ground to reduce the SPE's TX-INH voltage to
+5vdc, and it still works. I am simply going through this inquiry to
MAKE SURE that I am providing the K3s the specified voltages for the two
states (TX Inhibited and TX Enabled).
Questions:
1) What does the K3s expect to "see" at Pin #7 with TX-INH set to "High"
(and low for that matter) Is it, as I suspect, TX-INH set to "high" TX
inhibited = +5 VDC and TX Enabled = 0 VDC..... OR..... TX-INH set to
"Low"; TX Inhibited = 0 VDC and TX Enabled = +5 VDC, or is it more
complicated where the K3s wants to see +5VDC on BOTH sides of the TX-INH
setting, where with TX-INH set Low, it's the same" Inhibited = 0VDC and
TX enabled is +5 VDC, BUT TX-INH set to "High" it wants to see TX
Inhibited = (x volts above +5VDC) and TX Enabled = +5VDC. I would think
it is the former.... but the fact that it worked at 12 VDC for a year
without torching/letting the smoke out of anything gives me pause.
2) IF the K3s simply expects +5VDC for Inhibit and 0 VDC for Enabled on
Pin #7, and the SPE amp supplies the +12 VDC on the signal line for
INHIBIT and then grounds that output for Enabled, why can't a simple 2
resistor voltage divider (or pot) can't be used to drop the SPE output
from +12 to +5 VDC, or indeed why you couldn't simply use a single
inline resistor of the correct value to do it?
I should probably just replicate what has been proven to work and not
worry about it, but I truly want to understand why/how this is supposed
to work, so I know I am providing the signal levels intended.
Going schematic diving....
73
______________________
Clay Autery, KY5G
(318) 518-1389
On 08-Jun-18 22:12, Clay Autery wrote:
I'm going to use ACC Pin #7 as Transmit Inhibit (High), and the manual
says this:
"TX INH (Transmit Inhibit Signal)
Pin 7 of the ACC connector can be configured as a
transmit inhibit input by setting CONFIG:TX INH
to LO=Inh (or HI=Inh). Holding pin 7 low (or
high) will then prevent transmit. An external 2.2 to
10 K pull-up resistor (to 5 VDC) is required.
If TX INH is set to OFF, pin 7 reverts to its
default output function, K3S ON (see above)."
This has confused me a bit... I'm not an electronic expert. The input
signal is 12 vdc, present all the time EXCEPT when the K3s keys up...
the amp gets that message and drops the signal to ) vdc... Not sure
how to pull off reducing the input from 12 to 5vdc with a single
external resistor.
I DO know how to do it with a voltage divider.... Vout = Vin x
R2/(R1+R2) Calculate resistor values, tie them together at one end
and that end connects to Pin #7 at K3s end. R1's other end ties to
the (+) signal in from amp, and R2's other end ties to the ground
(conductor on Pin #5).
Is there a simpler way to do this with a single resistor? Am I
missing something?
And am I in fact reading the manual correctly? The K3 expects a TTL
logic level + 5 vdc on Pin #7 to activate inhibit with TX Inh set to
"high"? A straight 12vdc exceeds the spec for that pin as an input?
(When it's used as the K3s ON output to XVERTER, it OUTPUTS +5 VDC)
I'm trying to get this custom cable built for this amp this
weekend.... I'd appreciate a little help here. I don't want to let
the smoke out of anything....
73,
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