Using an incandescent light bulb in series with a mains-powered device is a well known trick to limit the current used by the device. It's great for testing devices you're unsure of.

Unfortunately, incandescent light bulbs are hard to come by these days, and this solution isn't easily programmable. It requires the use of several light bulbs in series or parallel in order to roughly achieve a target current limit.

I was wondering what would be the better replacement option while remaining inexpensive. I'd like to find something easily programmable and that allows to keep the current flowing at a lower voltage when the current limit is reached instead of dropping it to zero like a breaker.

By programmable I mean: being able to set the current limit in advance. And ideally also allow transient spikes of a given intensity and duration.

A word of context FWIW: Nowadays I'm fixing my laptops power bricks. Keeping some current flowing would allow me to locate short circuits much more easily. And having the power source react in to the load in real time would be much more safe and less prone to user-error than adjusting a variac manually each time I test something.

This means that my requirement are not very demanding. I don't need an adjustable frequency and phase is irrelevant. A maximum power of 200 to 500W should be enough and more than 2kW would be useless. I don't necessarily need a great accuracy (we're replacing light bulbs here), anything more precise than 1% would be useless. Nor do I need a great stability with fast changing loads since I'll be mostly testing things manually, although for this one, the higher the better.

I was thinking of maybe rolling my own solution using a microcontroller and a PWM signal to control a switch at a high frequency. And then smooth it out with a low pass filter. But maybe that's not the best idea.

So, what would be the best way to limit mains current in a programmable and inexpensive way?

  • 4
    \$\begingroup\$ I like this question \$\endgroup\$
    – Voltage Spike
    Jan 30 at 19:56

3 Answers 3


Sounds like you need an AC benchtop power supply. Something like this is what you need: https://www.chromausa.com/product/low-power-programmable-ac-source-61600/


As a starting point (I'm not going to design the circuit right now but maybe someone else can piggy back off this idea), you can do something like this:


simulate this circuit – Schematic created using CircuitLab

This will at least reduce your problem to a DC circuit that is much easier to manage with semiconductors and provide cycle-by-cycle current limit. The current-limit block is representative and shown with arbitrary components for my convenience. A proper circuit still needs to be designed.

  • \$\begingroup\$ I'm sure the price is justified for what it does. But I was hoping to find something simpler and cheaper. For instance I only need the output to be the same frequency as the input. I also don't need great stability or high max power since it's mostly a debugging tool. \$\endgroup\$
    – Celelibi
    Jan 31 at 0:40
  • \$\begingroup\$ @Celelibi I added a concept to my answer. Considering a laptop power supply will draw at most 3A, and that would be for a large laptop, a modest circuit could be built without too much complexity with pretty cheap parts. \$\endgroup\$
    – MOSFET
    Jan 31 at 4:27
  • \$\begingroup\$ I lke the idea of rectifying to only handle DC current limiting. \$\endgroup\$
    – Celelibi
    Feb 2 at 18:35

As you describe, using an incandescent light bulb and connecting it in series with the Line conductor does limit FAULT current. However, I think you misunderstand the point of such a test. It determines if the Device Under Test has a fault before applying power. For example, say you have a WWII-era tube radio or TV. Chances are high that the old radio or TV has a line-to-neutral short circuit, and if you plug it in, BANG! It's time to change your underwear—the light bulb test for the short circuit. The light bulb's wattage does not matter; it must be rated for the correct voltage. If the light bulb lights up to full intensity, you have a short circuit, and you need to find it before applying power, or else it will ruin another pair of underwear. If the light bulb comes on dimly or dark, it is safe to apply power.

If that is what you are trying to do, you are overthinking this. Electrically, an incandescent light bulb is a very simple power resistor, and there is nothing programable about it. You can easily apply KISS (Keep It Simple Stupid) engineering techniques and achieve excellent results. All you need is a handful of power resistors configured in series and parallel to make a ballast. Twenty-five watts should be plenty. If the circuit voltage is 120 volts, you need roughly 575 Ohms of resistors with a power rating of 25 watts or more—a pair 250-Ohm, 10 watts or higher in series with 75 Ohms 5-watt or higher.

There are a couple of ways to use it since it will not light up. If it gets hot, you have a short. If It runs cool, you are safe to apply power. Or you can use a volt meter. If you see around 120 volts across the hot resistors, you have a short circuit to find. If you see a substantially lower voltage across the cool resistors, you are good to connect power.

We old-timers like to KISS. If the light is bright, have a fright.


I'm fixing my laptops power bricks. Keeping some current flowing would allow me to locate short circuits much more easily.

Every laptop power brick out there internally runs on DC. The first thing it does with AC is to rectify it to DC. So you don't need to worry about AC at all. It's just an unnecessary complication.

All you really need is a "lab" CV/CC 350V supply with 1A or 2A capacity. That's not too hard to make or buy. I have two 200V 1A supplies I connect in series for that purpose, for example.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.