Often I have needed a variable resistor with a power rating of up to 10 or more watts and a resistance of several ohms up to 100 or so ohms. These are expensive and hard to find.* I’ll explain how to make an inexpensive one, but first I want to explain what I have used them for.
I have accumulated dozens of wall wart AC adapters, scrounged, or often bought for less than a dollar each. These are often unregulated, just a transformer, rectifier and filter capacitor. The output voltage varies widely with load. The output voltage is supposed to be the rated voltage at maximum rated current, but at lower current, the voltage rises; typically to 41% higher at zero current. And this doesn’t take into account fluctuations in the AC line voltage.
I often use one of these to power a strip of LEDs, usually needing a few hundred milliamps. But in order to get the voltage and current set right, I have to put a current limiting resistor across the wall wart output. For a wall wart rated at 6VDC unregulated at a half amp and a 1 watt LED using 200 to 350 mA, the variable resistor would be under 25 ohms.
Variable resistors or potentiometers this low are difficult to find and expensive, so a good substitute is a power transistor with a higher value pot between the base and collector. This pot may be 1000 ohms in series with a 100 ohm resistor, so the resistance may be varied between 100 and 1100 ohms. The 100 ohm resistor must be there so that the pot, when turned to minimum, has at least 100 ohms resistance, to prevent excessive current. As I said, this is connected between the base and collector. The power transistor may be the common 2N3055, a TIP31, a BD433 or similar. For short periods of on time, a heat sink may not be needed. The collector is connected to the positive lead of the adapter and the emitter to the anode or positive pin of the LED. The cathode or negative LED pin then connects to the negative lead of the wall wart adapter.
I forgot to put the DMM set to the current range into this circuit. It can go between the positive lead of the adapter and the collector. The adapter is plugged in and the 1k pot is adjusted until the DMM reads the desired LED current. If the meter reads only up to 200 mA, it may have to be set on the 10 amp range. Once it is set at the desired current, the adapter is unplugged and the DMM is removed, and the collector is reconnected to positive lead. The DMM is switched to 20 volt range and connected to both leads of the adapter. The adapter is plugged in and with the LED brightly lit, the adapter’s voltage is read. Then the DMM is connected across the emitter and collector and the voltage is measured.
Assume that the variable resistor was set for 200 mA or 0.2 A, and the voltage across the emitter and collector was 4 volts. Divide 4 volts by 0.2 A and we get 20 ohms for the resistance. Using the formula I squared times R, we get 0.8 watts for the power dissipation in the 20 ohm resistance. We could use two 10 ohm, 1/2 watt resistors in series, but each of the resistors would dissipate 0.4 watts, which is too close to their 0.5 watt maximum. It would be better to use two 10 ohm, 1watt resistors or three 6.8 ohm, 1/2 watt resistors in series.
The reason we measured the adapter’s voltage with the variable resistor load is to find what its approximate voltage will be when the 1 watt LED and two 10 ohm 1 watt resistors are connected in series across the adapter leads. The LED should light up brightly, but it may get hot. The LED should be mounted on a heatsink to prevent it from overheating and shortening its lifetime.
The transistor could be used permanently, if the pot was measured and replaced with a fixed resistance. Or the 100 ohm resistor and pot could be measured and then change the 100 ohm to that value. The pot can then be used as a dimmer.
Note: I was lucky to get a few 25 ohm, 50 watt pots long ago. I used them as a load resistance for testing audio power amplifiers. But I also use them for the above setup.
