# Choosing wattage for current limiting filament bulb

I am building a fully featured breadboard/prototyping box for audio circuits and would like to include a current limiting light bulb that will be in circuit when I apply power to the box and then I can switch it out if everything looks OK. I am wondering how I can calculate an appropriate wattage bulb to use.

Theres four main components to my prototyping rig, two power supplies and two circuits:

• One linear +/-12V power supply that feeds the breadboard and what ever circuit is on it which will not exceed 100mA (if everything is hooked up properly)
• One linear +15V power supply that feeds a 2.5W audio amplifier

With that said, how can I calculate the wattage of a bulb that will limit enough current to prevent damage to any components but still allow everything to power up to idle state?

• I've got a wimpy power supply (I out max of 100mA) that I often use for powering a circuit the first time. (And I monitor the current so I can quickly switch it off.) I think you might want to think about a current limiting circuit rather than a light bulb (or simple resistor). The light bulb/ resistor gives a very soft limit to the current. Commented Apr 13, 2015 at 17:15

The advantage of an incandescent light bulb over a resistor is that its cold resistance is about 10 times less than when lit up to normal brightness. Therefore you can choose a wattage that limits short circuit current to a safe value, without excessive voltage drop at lower currents. Also, since it takes some time for the filament to heat up and increase resistance, short duration current surges cause much less voltage drop.

You should choose a lamp with voltage rating close to the power supply voltage (a little lower won't hurt) and wattage that gives an acceptable short circuit current. Power = Volts x Amps, so if you wanted to limit a 12V supply to 0.5A then you would need a 12V 6W lamp.

The maximum normal operating current will then be determined by how much voltage sag is acceptable. This is tricky to calculate because the lamp's change in resistance with current is very non-linear. A 12V 15W festoon lamp that I tested had a resistance of 0.9 Ohms when cold, and 9.8 Ohms at full brightness. At 0.3A (24% of the bulb's nominal current draw) the output voltage dropped from 12.0V to 11.5V. Here is a graph showing output voltage vs current for my lamp compared to using a 10 Ohm resistor.

The 'soft' current limiting that a lamp provides is good for protecting against continuous short circuits without affecting normal operation, but won't protect sensitive electronic components which cannot withstand short current surges or slightly higher than normal operating current.

If you need a fast acting current limit with sharper cutoff then you could use a bipolar transistor (PNP for positive voltage, NPN for negative) with its Emitter connected to the power supply, Collector to the load and a high value resistor going from Base to Ground. Maximum collector current will then be determined by the Base current multiplied by the transistor's current gain (which varies from unit to unit, so you may have to adjust the resistor value to get a precise current limit).

• The Incandescant lamp can make a good current limiter for some applications +1 . Bulb life will be long because most of the time the filament temp will be low. Commented Jul 24, 2016 at 0:18
• Interesting (and simple idea) to use the BJT gain and a resistor, as a simplified current source. However, it might be tricky to adjust for transistor variations and gain variations caused by the operating point. Because of these issues, I would prefer to use a smaller bipolar transistor to measure the voltage drop over a 1W-like resistor, compensating the bias of the series-pass transistor. Obviously we will start to increase the parting counting, but in terms of cost, it is still negligible.
– EJE
Commented Jun 17, 2021 at 4:46