Capacitors as AC Voltage Drop Components

Ken,

If you want to a lower AC voltage , you should really use a transformer.

In this way , you can vary the load and the voltage will stay constant.

If cost is a factor , resistors are the first choice of voltage droping elements instead of capacitors.

One of the reason for that is that capacitors comes in sizes which may not lend to your effort quite easily.

If using an LED , note that an LED will not have the power handling capability to rectify and provide substantial power.

I've seen several low cost items that used capacitors as voltage dropping components (driveway automobile/people sensors, flashing or night-light L.E.D. devices, etc.) and wondered how they knew what kind of capactors would be best for such things. When I approached the manufacturers I virtually got the "we can't help you" type of response. I figured that ripple current ratings would be key but wondered if there where other important factors to consider so I went looking for application notes and just haven't stumbled upon any yet ... anywhere.

Thanks for your comments.

Ken

To start with, an LED does not have much reverse voltage blocking ability -- somewhere around 7 volts per device. That is, it won't block 170 V_peak from a 120 V source. You'll need 1N4002 at least, and you'll need a filter capacitor also.

Capacitive reactance (X_c) is 1/(2 * pi * f * C). In the case of 60 Hz voltage, Xc = 1/(377C).

If you want to limit current from a 120 VAC source to 20 mA, then you need 6000 ohms impedance. This ciphers out to be 0.44 uF.

All in all, I think I'd do it another way. You're not creating a low voltage circuit by doing it this way -- it just makes you feel like you are.

Top answer by Jack of all trades! Follow his advice and you won't go too far wrong.

I'd just add that a very common application for the capacitor-as-voltage-dropper is in the various electronic time switches that get installed in switchboards and such. These often have a small backup battery or a supercap across the rectified low-voltage dc rail which helps to shunt away those pesky transients, but even this is not sufficient to stop the timers from developing the odd spot of amnesia from time to time. Still, if you are only wanting to light up an LED this is not really a consideration.

Good Luck!

This can be done to supply small amounts of current. I have done it and it works fine. However, you must do certain things. You need a rectifier to pass the voltage onto a filter capacitor (cathode to the + of capacitor) and you also need another rectifier ahead of the pass rectifier that will discharge the capacitor on the other half cycle (cathode to the anode of the pass rectifier, anode to the negative of the capacitor). If you don't do this the capacitor will charge and remain charged and not pump any further current through to your supply. The second rectifier places a charge on the capacitor during one half cycle that gets dumped to your filter capacitor on the alternate half cycle. You should also place a nominal value of resistance in series with the capacitor to limit peak current especially in the event of transient voltage spikes on the power line. You should use a capacitor with low dielectric losses such as polypropylene to reduce heating and it should have a voltage rating considerably higher than the peak voltage of the power line on which you are running it in order to allow for transient voltages. You should test your prototype under worst case conditions to be sure you do not have excess heating. The amount of power you can get will be determined by the capacitance and so this must be selected accordingly.

This URL should give you a good starting place.

http://www.freeinfosociety.com/electronics/schemview.php?id=1491

Another good source of information is an application note from Microchip Corporation AN954. It is in PDF format and available on that companys web site.

I use capacitors rated at 400 V, NPO and have built many of these as AC power indicators. Be very careful with these circuits, as has been mentioned by all, this stuff can kill you!

Do a Google search for: AC powered LED circuit. Lots of info on the web.

As previously reported there is a safety hazard with the direct connection to the line but using the capacitor is a very inexpensive method to perform the voltage drop. In addition there is no power dissipation and heating problems caused by other low-cost methods. I have done this method of voltage drop several times. You will need the following parts:

Capacitor; Use any type of film capacitor of a high enough voltage rating. You don't have to worry about the type of film since the current required by the LED is very small. A capacitor with a high dielectric constant like polyester (mylar) will be small and compact. Calculate the value from the required impedance for the required LED current.

Select the LED: Use at least two LEDs in parallel. The reason is that you will have twice the illumination and will not require a blocking diode in series. With two LEDs in parallel each LED will conduct in alternate half cycles of the line voltage. The anode of one LEd must be connected to the cathode of the other LED.

A peak current limiting resistor: This is needed to limit any peak current from damaging the LED. The LED peak current is much higher than the average current so this will not be too much of a problem. There is a lot of noise on the AC line that will be passed by the series capacitors.

Connect the LED assembly, capacitor and resistor in series. You may have to use more than one capacitor in series or parallel to get the correct value. I'm sure this will work for you.

Let me know if additional detailed design information is necessary. I can email off-line.