# How this circuit works!!? Answered

I got a circuit board from an emergency light when i saw it's circuit it's not common there is a ceramic capacitor and a 330 ohms resistor in its and it give 5.0 volts at the end i don''t know how it's designed but it's giving 5.0 volts at the other end with  no transformers ! then i configured the circuit myself as shown in the given diagram but it's giving 180 volts dc instead of 5 volts so my question is  : is there anybody who can tell me whats missing in this circuit so its can also give 5volts output?

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## Comments

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There must be something you are not seeing in the original circuit. If a resistor was placed in series with R1, then it would form a voltage divider to solve the circuit. R1 would have to be a high value and R2 would have to be a low value (ohms). This would form a lower voltage but it would not have very good regulation. (the voltage would not be steady as a load is applied). Also, resistors must be chosen to with sufficient wattage to withstand the 220V input

I used the value of 220 volts and calculated it 2 ways. The first way i used the ratio method. We know we will loose 5 volts across r1, so r2 will have 5 volts across it. the ratio of VOLTS will correspond to the ratio of OHMS desired for R2. 5v divided by 220 equals 0.02272 ratio. now multiply that ratio result to the resistance value of R1 .... 550,000 times 0.2272 equals 12,700 ohms. so R2 should be about 12.7K ohms.

The other way is to just use OHMS LAW. We know R1 has 215 volts across it... and we know the value of R1 is 550,000 ohms. So what is the AMPS used by R1? I=E/R .... so 215 divided by 550000 equals 0.0003909 Amps. Now we know the AMPS will be the same through R1 and R2 because they are in "series" (on the other side of the bridge rectifier). So Resistance equals Volts divided by amps. R=E/I..... 5volts divided by 0.0003909 equals 127900 ohms or... 12.79 K ohms. We get SLIGHTLY different answers because of the long decimal answers that I shorten a little when entering them into the calculator. ALSO.... my answer will not be totally accurate because of the bridge rectifier in the middle that I did not compensate for. 12.7 K will get you close.

Remember that if you attach any LOAD to the rectifier in PARALLEL with the R2 resistor.... then the voltage will go lower because you are adding resistance (load) in parallel with R2. The LOAD ITSELF (without adding any R2 resistor)... should measure 12.7K for this to have 5 volts (thereabouts) across it.

Update.... maybe the EMERGENCY LIGHT itself is acting as an "R2" resistor. This would mean your circuit will put out 5 (or so) volts ONLY when the emergency light is attached to the bridge. The missing part is the emergency light! With NO LOAD... the voltage will be 180 volts. With the emergency light attached, the voltage should go down to 5 volts.

This is the way I would do it.... but the TOTAL circuit of the 12 volt adaptor I don't have. I would just use a little adaptor that puts out 12 volts when 220V is connected to it. Then use a 5 volt regulator to get your 5 volts. Also you could use any old computer power supply. They have 12 volt and 5volt outputs.

Warning: While a circuit like this does generate a lower voltage, that does not mean, it is safe to use!

There is no galvanic isolation between the 5V DC and main 220V AC! This means, the circuit and any load connected has to be treated as a high voltage device - double isolation, non conductive housing, protection against contact, grounding of any touchable metal parts...

And framistan is right, the resistors have to have the right wattage, but they (and the capacitor and diodes as well) have to be rated for the voltage too! This means, you can't use a SMD resistor or a 1N4148.