# Check My Math

Discussion in 'Bad Dog Cafe' started by Jeru, Feb 21, 2018.

1. ### JeruTele-Holic

Posts:
734
Joined:
Nov 17, 2006
Location:
Chicago
I'm converting an old console radio into a bluetooth stereo amp
for a dear friend's brother in law. I'm ditching the tube guts and
putting in a Bluetooth-equipped stereo board. Thing is, the LEDs
that the board will power just aren't up to the task of lighting up
the radio dial -- it's got to be an incandescent bulb.

For reference - one of my builds. Scroll to near the end to see
what I'm talking about re: lighting the dial:

https://www.dropbox.com/sh/ylepfjvkqi2z6ti/AABBfSBQ1C5ZtgZYTB0_rm1Da?dl=0

So -- help me do this safely and check my math re: power resistor
needed to safely lower the voltage and use a #47 bulb to backlight
the dial. I really want this thing to be SAFE and dependable.

The wall wart for the bluetooth board puts out 19vdc. I'm going to use
a DPDT switch to connect hot/ground to 1) the BT board and 2) the
bulb circuit. #47 bulb draws .15A of current.

The Bulb circuit is: 19v+ --> resistor --> Bulb --> 19v neutral.

I figure that a 10W (ceramic white block) resistor ought to be just
fine. 75 ohm resistor gives me about 7vdc to the bulb, a voltage drop
across the resistor of ~12vdc.

Using Ohm's law to calculate the power dissipation of a 75 ohm resistor
dropping 12v, P = V(squared) ÷ R.

So, 12v squared = 144 / 75 ohms = 1.92 watts power dissipation.

So -- at five times the dissipation, even though the 10W resistor the resistor
gets warm/hot to the touch, it should be fine/safe in this circuit?

Is there anything that I'm missing?

Many thanks.

Last edited: Feb 22, 2018
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