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Re: [TenTec] Adding Pilot Light to RX-320D

To: Discussion of Ten-Tec Equipment <tentec@contesting.com>
Subject: Re: [TenTec] Adding Pilot Light to RX-320D
From: "Charles P. Steinmetz" <charles_steinmetz@lavabit.com>
Reply-to: Discussion of Ten-Tec Equipment <tentec@contesting.com>
Date: Sat, 29 Sep 2012 08:22:47 -0400
List-post: <tentec@contesting.com">mailto:tentec@contesting.com>
K8JHR wrote:

I would like to add a pilot light to the front panel of my TenTec RX-320D. I have SOME home brewing skills... having built several kits, including a few TT receiver kits, and having designed some small project circuits on my own. My experience suggests doing this is not as easy as merely soldering a LED into the circuit, and sticking it through a hole in the front panel.

You need a power supply, an LED, and a current limiting resistor. Typically, you use a push-in plastic bezel to mount the LED to the panel.

The schematic shows that a nominal 15 V enters the radio, is filtered with an RF choke and a 100 uF capacitor, and is available as the internal 15 V supply. I'd mount the current-limiting resistor on the PCB near C7 (one end to the +15 V PCB trace with a very short lead, one end floating). A wire (22-24 ga. stranded, insulated hook-up wire) would run from the floating end of the resistor to the anode of the LED (the one with the longer lead and without the flat on the diode body). Another wire (also 22-24 ga. stranded, insulated hook-up wire) would run from the LED cathode (shorter lead, flat on diode body) back to the PCB to ground (also preferably somewhere near C7). If you want to get fancy, make the anode wire red and the cathode wire black. Doing it this way (resistor on the PCB +15 V trace), if something goes wrong with the wiring you won't short-circuit the +15 V supply.

LEDs have forward drops between 1.5 and 3 V, depending on technology and color. Assume ~2 V, so the resistor will have ~13 V across it. Since I = E/R, a 3.3k resistor will regulate the diode current to ~4 mA. Since P = EI, the resistor will dissipate ~0.05 W (50 mW). So, a 1/4 W resistor is sufficient.

Use appropriately-sized heatshrink tubing at the resistor (run it right down to the PCB, completely covering the resistor) and on each LED lead (again, run it right to the body of the LED). Route the wires neatly and use tie wraps as appropriate.

Of course, do all the work with the radio off, and check it visually for correct polarity and for solder bridges or other shorts before you turn it on.

Note that there are several regulated voltages in the 320 (+5 digital, +5 analog, and +10). I did not suggest using them for two reasons: first, the lower voltages increase the uncertainty of the voltage across the resistor, and thus the uncertainty of the LED current. And second, there is no need to add a further current burden (with increased power dissipation) to the regulators. If you measure the drop of the particular diode you use, you can calculate the resistor for the desired current with the lower voltage, and such a small additional load will probably not adversely affect the regulator, so you could probably use a regulated voltage if you prefer. If you do, I'd use the +5 V digital supply.

Best regards,

Charles






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