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

To: <tentec@contesting.com>
Subject: Re: [TenTec] Adding Pilot Light to RX-320D
From: James Duffer <dufferjames@hotmail.com>
Reply-to: Discussion of Ten-Tec Equipment <tentec@contesting.com>
Date: Sat, 29 Sep 2012 09:56:44 -0500
List-post: <tentec@contesting.com">mailto:tentec@contesting.com>
Very well written Charles. , Jim (wd4air)
> 
> 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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