I'm trying to simulate an evening lamp using a NE555 timer and a relay.

To do so, I'm stepping down the 230V AC household voltage to 16V by an RC circuit as shown.

enter image description here

Now, when I take a stand-alone circuit of the rectifier, I get this output. (Output voltage being nearly 170V)

enter image description here

What maybe the reason for this to happen?


2 Answers 2


Besides the safety issues, the problem with this sort of circuit is that the output voltage depends heavily on the load impedance. You basically have a voltage divider consisting of the impedances of the dropper circuit and the load. Capacitors C1 and C2 are meant to do most of the dropping, so their impedance is important. As shown, they have an impedance of:

$$ X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi\times 50Hz~ \times 1uF} = 3183\Omega $$

for each cap, so twice that and adding in R1 and R2 and you have roughly 6800Ω.

Now you have a 30k resistor across the output so let's take that as the minimum loading. This gives you a divider formula of: $$ V_{OUT} = 230V\times \frac{30k}{30k + 6800} = 187.5V $$ Factor in diode and other losses and there's your 170 V output.

To get the 16 V you want you have to either increase the dropper impedance (make C1 and C2 lower values) or increase the loading (in simulation 750Ω instead of 30k gives right around 16 V).

Now here's the problem, any change to the load will be reflected in the voltage, so this sort of circuit would either need to be followed by a regulator or have a nearly constant load. Imagine what would happen when your relay energizes and turns on a lamp... the voltage would drop significantly from the additional current draw. So while it's an interesting thing to play around with in simulation it has rather limited real world uses.

  • 1
    \$\begingroup\$ I was experimenting on how the voltage changes as I keep on adding my other components after the rectifier, and I found that what you said at last to be true. The 16V is reached after each of the components are implemented. If I wanted 16V just from the rectifier output, the overall output after implementing all the other components came near to 3V. So, I guess I have to implement a constant voltage source instead of this approach. Anyways, thank you so much for the explanation at the start \$\endgroup\$
    – rire
    May 16 at 6:11

First off: No. Stepping down 230VAC using your RC / diode networks there leads to a deadly situation, where, if anything fails, it's likely that you'll have 230V on various parts of the system where the user does not expect it.

In any country of whose regulations I'm aware, such power supplies are forbidden in devices that user might be able to touch any parts of, for good reasons.

Output voltage being nearly 170V

A basically unloaded divider can't divide.

What you've built is but a (bit of a strange) bridge rectifier with some smoothing and no regulation, and you load it very lightly. That's not sufficiently close to a constant voltage source. If you loaded it heavily, it would run at a lower voltage, but it would also be very hot, and if your load resistor failed (which would be likely to be the hottest, and thus most likely to fail, component), you'd be back at full 230 V. So, that approach really doesn't work.

You need a proper power supply, and it's really easy to get a safe one these days: You can probably get one for free from someone else's trash bin that gives you e.g. 12 V or 9 V DC, or you can buy (or find) a USB power supply. 5V and a few milliampere would be enough, if you didn't use the ancient NE555. Any CMOS 555 instead of the archaic bipolar NE555 would work with much lower voltages and use much less power!

  • \$\begingroup\$ I must admit I'm not sure what an "evening light" is, or what you need a 555 for light, but your question really only seemed to be about the power section, and I'm only mentioning the NE555's downsides in the answer, because if that's what's stopping you from using an off-the-shelf USB supply, then it's worth getting rid of it. \$\endgroup\$ May 15 at 9:33
  • \$\begingroup\$ Thank you for answering. I have tried heavily loading it, but still the output voltage remains at 16V. This circuit was my attempt to create a project based on my learning from my basic analog course, that's why I used the NE555 timer. I understand that I can use better equipment, but my concern is why is the following setup not working. If there is any insight on why certain stepping down actions happen due to certain parameters of the RC circuit, it would really be helpful. \$\endgroup\$
    – rire
    May 15 at 14:03
  • \$\begingroup\$ There's not that much to explain, honestly: you're relying on the complex impedance of your RC circuit, together with the load impedance and the smoothing capacitor impedance to achieve a constant voltage. This is just basic linear networks, if you will. None of this regulates the output voltage. You will have to use a different circuit than what you've learned for this, I'm afraid. And because safety needs to come first, using an existing, safe, supply is really really what you should be doing. You can still apply your 555 knowledge, just not with your power stage. \$\endgroup\$ May 15 at 15:04

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