What would be the best way to limit mains (240V @ 50Hz) current? I want at least 1A of current. The output should be AC 240V at any frequency greater than or equal to 50Hz. It should be able to endure short circuit like conditions. I have tried dimmers (which instantly explode when connected) and fluorescent tube electronic ballast (too low current). I would be using this to power a hand wound toroidal transformer.
Electronic – Limiting mains current
ballastcurrent-limitingmainsshort-circuittransformer
Related Solutions
You have the right idea for a basic unregulated supply. A transformer, four diodes, and as large a cap as you can manage will serve well enough for a lot of purposes, but isn't appropriate for all.
There are two main problems with such a unregulated supply. First, the voltage is not known well. Even with ideal components, so that the AC coming out of the transformer is a fixed fraction of the AC going in, you still have variations in that AC input. Wall power can vary by around 10%, and that's without considering unusual situations like brownouts. Then you have the impedance of the transformer. As you draw current, the output voltage of the transformer will drop.
Second, there will be ripple, possibly quite significant ripple. That cap is charged twice per line cycle, or every 8.3 ms. In between the line peaks, the cap is supplying the output current. This decreases the voltage on the cap. The only way to decrease this ripple in this type of design is to use a bigger cap or draw less current.
And don't even think about power factor. The power factor a full wave bridge presents to the AC line is "not nice". The transformer will smooth that out a little, but you will still have a crappy power factor regardless of what the load does. Fortunately, power factor is of little concern for something like a bench supply. Your refrigerator probably treats the power line worse than your bench supply ever will. Don't worry about it.
Some things you can't do with this supply is run a anything that has a tight voltage tolerance. For example, many digital devices will want 5.0 V or 3.3 V ± 10%. You're supply won't be able to do that. What you should probably do is aim for 7.5 V lowest possible output under load, with the lowest valid line voltage in, and at the bottom of the ripples. If you can guarantee that, you can use a 7805 regulator to make a nice and clean 5 V suitable for digital circuits.
Note that after you account for all the reasons the supply voltage might drop, that the nominal output voltage may well be several volts higher. If so, keep the dissipation of the regulator in mind. For example, if the nominal supply output is 9 V, then the regulator will drop 4 V. That 4 V times the current is the power that will heat the regulator. For example, if this is powering a digital circuit that draws 200 mA, then the dissipation in the regulator will be 4V x 200mA = 800mW. That's will get a 7805 in free air quite hot, but it will probably still be OK. Fortunately, 7805 regulators contain a thermal shutdown circuit, so they will just shut off the output for a while instead of allowing themselves to get cooked.
It's a valid plan and I've seen others implement it before. Here's a link to a evaluation board that does AC lines monitoring and powers itself off of the AC Mains. Look at the schematic on page 10. The regulator is U2 and the rectification diode is Z1. You might be interested in the board anyways, it already does the voltage and current measurements and outputs the value over UART/I2C/SPI.
Back to your idea though. I would suggest the following modifications. We'll talk about the clamp in the next paragraph. Instead of a shunt capacitor make it a series capacitor. This provides AC coupling and blocks any DC current from forcing the diode on. The diode is a cheap form of rectification. Half wave rectification is probably good enough for your purposes. See the wiki page for some details on rectification.
Make sure that the regulator can support the output voltage of the rectifier. If you bring in the voltage directly from the line, the rectified voltage will be very large. This is one of the reasons I would use a transformer to step down the voltage. Additionally, placing a shunt zener diode can help decrease the rectifier output voltage. You might want this if the rectifier output is too larger for your regulator. See what they do in the schematic linked above. U7 is the shunt zener. Make sure that you add appropriate resistance to prevent excessive current flow if the zener conducts.
One issue I see with the plan is the current transformer that you are planning to use. Current transformers are typically used for current measurements. If you look at the I-V curve in the datasheet that you've linked, it provides mV of output voltage for tens of Amps on the primary side. This cannot supply the rectifier and regulator. Alternatives would be to use a transformer to reduce the AC voltage to something manageable and use the same rectification scheme. Another method would be to attach directly to the AC line as in the board I linked to. Be very sure that all of your parts are rated for the expected voltage or current.
Best Answer
The simplest thing to do for this sort of testing is to put an ordinary incandescent light bulb in series with your transformer primary as a "ballast" resistor. But for 1A, you'll need a 250W bulb, or maybe four 60W bulbs in parallel.
A better solution is to use a Variac (variable autotransformer) to bring up the voltage on your homemade transformer gradually. Use an ammeter to monitor the current.