12V Adapter for Portable Lamp

5 October 2017

My boyfriend and I often drive to Los Angeles to see friends. Since we often get there long after the sun has gone down, and I often use the time traveling to get school work done, I need a light to see what I'm doing. Using the dome light for reading impairs the driver's ability to see traffic at night. I bought a portable desk lamp with a shrouded light from Fry's Electronics for use when traveling. That was a big improvement. Unfortunately, it came with a 120 VAC-to-11 VDC adapter. Not a big deal since we have an inverter. However, since the power requirements of an LED lamp are small (0.25 Amps) and it runs off direct current, it is more elegant to bypass converting the power twice and go directly from 12 VDC to 11 VDC. While it's true that the lamp could probably handle running off 12 V, the reality is that, when the engine is running, the system puts out 14.5 V. That is because it needs to charge the battery and every 1 V the battery puts out requires the electrical system to put in 1.2 V to charge it (at least for lead-acid batteries).

There are at least a couple of ways to build this adapter. One is using a voltage regulator. Another is setup a voltage divider. I choose that latter mainly because I don't know much about voltage regulators and resistors tend to be smaller and less complicated. Since I planned to put all the electronics inside a "cigarette lighter" plug, size is a major consideration. The first thing to do was to find out how much amperage the lamp takes in. The adapter has a rating of 0.25 Amps. One could go by that since that is the maximum it could draw continuously. However, I chose to determine how much current the lamp actually draws to create a better divider. By connecting a multi-meter in series, with the assistance of my always-helpful-boyfriend, we found the current to be 0.245 Amps. Connecting the multi-meter in parallel showed the load voltage be 10.2 V. Using Ohm's Law of V = IR and solving for R, gives us 41.6 Ohms resistance for the lamp. Since the car puts out 14.5 V, we again use Ohm's Law to find the resistance required to get the same current draw as follows: R = 14.5 V / 0.245 A = 59.2 Ohms. To find the resistance required for my power adapter, we subtract 41.6 Ohms from 59.2 Ohms to get 17.6 Ohms. I wasn't able to find 18 Ohm resistors at Fry's (let alone 17.6 Ohm resistors). Getting that exact resistance would require using several resistors in series. Because of space limitations and for simplicity, I chose a 20 Ohm resistor. That brings the total resistance to 61.6 Ohms and reduces the current draw to 0.235 A = 14.5 V / 61.6 Ohms.

Besides being rated for Ohms, resistors also have a rating for how much power/heat they can dissipate measured in Watts. Power is voltage multiplied by current (P = VI). Before I pick a resistor, I need to figure out how much power it will need to dissipate from the voltage the resistor is dropping and the current going through it. To find out how much voltage the resistor is dropping, I divide the Ohms of the power supply resistor (20 Ohms) by the total Ohms of the system (61.6 Ohms = 20 Ohms for power supply + 41.6 Ohms for lamp) and that gets us 0.325. Multiplying the result by 14.5 V is 4.7 V. We find the power dissipated by the resistor as follows: 4.7 V * 0.235 A = 1.106 Watts. To have a safety factor of approximately two, I purchased a 20 Ohm resistor rated at 2 Watts. A resistor rated at a higher wattage would have been too large for the cigarette lighter plug. Since the resistor is removing 4.7 V, what is left over is 9.8 V to the lamp. That is slightly less than the voltage of 10.2 V coming out of the original power adapter.