AC Step Down Voltage form 230V to 18V
- Thread starter milind
- Start date
From that level it is very hard. The safest way to detect the AC voltage it is with a half wave rectifier and an optoisolator. The problem is making a voltage divider that produces enough current to turn on the LED of the opto without creating a boatload of heat in the divider.
To actually measure the voltage, safely, without a transformer is going to be a challenge. I can think of ways to approach the problem, but I'm not sure any of them would actually work.
What is magic about 18V? Don't microcontroller ADC's work from 0V to Vdd? Last microcontroller with an 18V supply was the RCA 1802, which went to microcontroller heaven some time ago.
Hi,
As pappabravoo says, there is no safe way to do this without a transformer. If you really have to, though, you could make a voltage divider with a 112K resistor attached to the line and an 18K resistor in series attahed to neutral. This will develop 18 volts across the 18K resistor. To save yourself timing problems, use a diode to half wave rectify the AC, and a capacitor to make smooth DC. Otherwise, you have to make a peak voltage detector to trigger your A to D to get sensible data.
It's really not smart to fool around with line voltages without isolation.
As pappabravoo says, there is no safe way to do this without a transformer. If you really have to, though, you could make a voltage divider with a 112K resistor attached to the line and an 18K resistor in series attahed to neutral. This will develop 18 volts across the 18K resistor. To save yourself timing problems, use a diode to half wave rectify the AC, and a capacitor to make smooth DC. Otherwise, you have to make a peak voltage detector to trigger your A to D to get sensible data.
It's really not smart to fool around with line voltages without isolation.
The peak power disipation in the 18K will be 112 mW so a quarter watt will do.
The peak power in the 112K will be 701 mW. You need a 1 Watt minimum or 2 Watt just to be safe for this one. The current will be about 2.5 mA.
Please be careful and don't let the magic smoke out of any of your components or yourself.
windoze killa
- Joined Feb 23, 2006
- 605
Digital multimeters have the circuitry to do just that internally. If you can find the schematic for one of them you could tap of the low DC voltage and feed to your micro.
Absolutely correct. It would be pretty crazy to have no isolation. They never could stop talking about it in classes.
Honestly, I think a transformer is the easiest approach. You don't need a lot of current with an A-D convertor, your only sampling the voltages to convert them to logic, so a 230V to 18V or 12V transformer with small current shouldnt be that bulky, and that's probably what I'd use.
