# How this circuit works!!?

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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4 years ago
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
can to tell me what should be the value of R2 if the input ac voltage is around 230volts and the value of r1 is 550 k ??
4 years ago
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.
Thank you sir, you've helped me a lot!
Sir sorry for the mistake but the value of the r1 is around 330 kilo ohms not ohms i forgot typing ""k""
4 years ago
Multiply the ratio .02272 times any resistor for R1 gives you the resistor for R2... would be 330,000 times .02272 equals about 7.5K ohms for R2 (total load)
framistan4 years ago
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.
Sir i got a request for you !
Can you give me a circuit which gives 5 volts as output with these components actually i need a schematic of that circuit ?