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Transmit Inhibit

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Clay Autery-2

Transmit Inhibit

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,

--
______________________
Clay Autery, KY5G
(318) 518-1389

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Don Wilhelm

Re: Transmit Inhibit

Clay,

The answer lies in your driving device and not the K3. Yes, a 12 volt
source may damage the K3.
A simple voltage divider will not do the whole job. That output must
swing from 5 volts to zero volts., so at the the time the output should
go to zero volts, the voltage divider will still be producing voltage.

Look at the schematic for the driving device. What is the device
providing the output? A relay, or a solid state device?
If a relay, is it a SPDT relay with the output taken from the common?
Is the NO contact connected to ground? Is the NC contact connected to
+12v - if so, remove the source of 12 volts and connect a 2.2 to 10k
resistor to that contact with its other end connected to a source of 5
volts.

If instead of a relay, the output is a solid state device (transistor or
FET) disconnect the collector (or drain) resistor and replace it with a
2.2k to 10k resistor with the other end connected to source of 5 volts
instead of 12 volts.

In other words, you will have to modify the driving device -
alternately, you could add a device to the circuit (internally or
externally) which takes the 12 volts as an input and uses a 2.2k to 10k
collector (or drain) resistor terminated in 5 volts - a 2N7000 FET would
work fine.
That output would go from 0 volts when the input is 12 volts and go to
+5v when its input is 0 volts.

73,
Don W3FPR

On 6/8/2018 11:12 PM, 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)
>

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Bob Wilson, N6TV

Re: Transmit Inhibit

In reply to this post by Clay Autery-2

What amp. are you using and why do you need to use TX INH? Perhaps it has
KEY IN / KEY OUT, and you can use that instead, to activate PTT on the
radio only when the amp. is "ready". PTT signal or footswitch keys amp.,
amp. keys K3 PTT IN, instead of PTT/footswitch keying radio, and radio
keying amp. Or you could set CONFIG:TX DLY to a large value.

If you must use the TX inhibit pin, 5V is available on K3 ACC Pin 1 (FSK
keying). I use that plus a single resistor on my Simple K3 Lockout Circuit
(see https://bit.ly/K3lockout which downloads a PDF schematic). The Y-BOX
<https://bit.ly/Y-BOX> has this lockout circuit built in, as well as a 10K
pull-up resistor for a separate 5V source if the FSK pin is already in use
for RTTY. But, the circuit uses TX INH LO and you need TX INH HIGH?

You could use the 12V from the amp. to key the base of an NPN transistor
(through a 1K resistor). Ground the emitter and and connect a 470 ohm
resistor to the collector. Connect other side of the resittor to the TX
INH pin and +5V (Pin 1), as in my lockout schematic. When your amp outputs
12V, the transistor closes the circuit, and the INH pin goes low. When
your amp. outputs 0V, that opens the circuit, and the INH line goes high
(+5V).

73,
Bob, N6TV

On Fri, Jun 8, 2018 at 8:12 PM, Clay Autery <[hidden email]> 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,
>
> --
> ______________________
> Clay Autery, KY5G
> (318) 518-1389
>
> ______________________________________________________________
> Elecraft mailing list
> Home: http://mailman.qth.net/mailman/listinfo/elecraft
> Help: http://mailman.qth.net/mmfaq.htm
> Post: mailto:[hidden email]
>
> This list hosted by: http://www.qsl.net
> Please help support this email list: http://www.qsl.net/donate.html
> Message delivered to [hidden email]

Clay Autery-2

Re: Transmit Inhibit

In reply to this post by Clay Autery-2

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,
>

Clay Autery-2

Re: Transmit Inhibit

OK.... I AM making progress.

What I have learned thusfar... (Thanks for the patience of Don, WFPR
and others.)

... and correct me if I am wrong, PLEASE.

1) Pin #7 is TTL logic... And logic pins need to NOT be left
floating. They need to see discrete logic high (+ 5vdc) or low (0 vdc)
levels at all times. (Playing with the TX-INH settings on the K3s with
nothing connected, I have seen what happens when the pin is left
floating... the MCU doesn't know what it's going to get... I've
successfully gotten it to inhibit AND enable on both settings.)

2) The pull-up resistor basically acts sort of like a current limiter on
the input pin and a load when the pin is pulled to ground.

3) Lower pull-up values improve switching speed. Smaller is faster, but
too small and you get too much current and power consumption/heating
become issues. Too high, switching too slow and voltage level may not
be high enough to clearly be taken as a "high" logic state.

4) The "switch" needs to be between Pin #7 and +5VDC.... and Pin #7

So now, it's just a matter of putting together a switching circuit
that feeds Pin #7 with +5VDC when the SPE TX-INH signal is 12 vdc, and
when the +12 vdc is removed, Pin #7 sees 0 vdc. Seems simple, but...

Here is a link to a draft circuit I've drawn up based on a suggestion as
to a particular MOSFET to use as a switching device:

http://www.montac.com/images/circuits/TX-INH_circuit.jpg

Notes/Questions:

Q1 = 2N7000G Small Signal MOSFET (N-channel)

R1 = Pull-up resistor, as per manual (2.2-10 kOhms)

?? R2 = Pull-down resistor, value unknown. Needed? I THINK it might be
needed in view of the "requirement" for R3 below, and to potentially
avoid using D1 below.

R3 = Rgs resistor, multiple references to the gate to source as a
non-polarized capacitor and the need for a bleeder resistor to ground to
aid in reliability of switch turn-off. Value undetermined as yet. My
gut tells me R3 < R2, but I can't understand why I think that yet, exactly.

?? D1 = Protection for Pin #7 from 12 Vdc via bleeder resistor path.
Yes, I realize I put it in the circuit the wrong way... <smile> I
really don't want to give up the forward voltage drop, but I can reduce
the drop with a Schottkey, et al. Reckon this extra protection is
actually necessary enough to trade the 0.3+ volts or so?

+5 VDC.... Don't know where I'm going to get this yet.... Don't really
want to steal the FSK pin... maybe the amp has a line 5 volt line I can
steal.

This device is getting bigger by the minute... the idea is to shrink it
into the cable right behind the connector housing... IDEALLY, it would
be better INSIDE the connector shell, but that's not going to happen.
May have to demo the circuit and then recreate it in SMD.

73,

______________________
Clay Autery, KY5G
(318) 518-1389

<Previous Postings Deleted for brevity>
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