Can someone tell me what is output voltage for this scheme?
Calculating output voltage
- Thread starter Weras Asdsdw
- Start date
Yeah the voltage is AC
Kind of... 
The output is a half-wave rectified sinewave. Beyond that, your question needs a lot more work.
If you are asking what the output voltage will be if measured with a digital voltmeter, it depends entirely on the load. It has a peak value, an average value, an RMS value, and an incorrect DC value on a meter - and all four of these are different numbers. If the load is a resistor, then the 1 M resistor in your drawing is one part of a voltage divider. If the load is a capacitor, then the size determines what will appear on a meter.
ak
If you are asking what the output voltage will be if measured with a digital voltmeter, it depends entirely on the load. It has a peak value, an average value, an RMS value, and an incorrect DC value on a meter - and all four of these are different numbers. If the load is a resistor, then the 1 M resistor in your drawing is one part of a voltage divider. If the load is a capacitor, then the size determines what will appear on a meter.
ak
No this is part of bigger scheme, I think it should be around 5V
What is the voltage drop of a diode? What will it do on the negative half cycle of the input?
The output voltage should activate the optocoupler
The measured voltage output was 0.5V but the optocoupler didn't activate, so I think it should be around 5V for it to work
The measured voltage output was 0.5V but the optocoupler didn't activate, so I think it should be around 5V for it to work
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How are you coming up with 5V for the output?
You said this was "sort of" homework. Is it or isn't it?
Its not homework
Then as AK said, the output will be a half wave rectified sinewave and the voltage won't be 5V (DC or AC). You need to add reverse polarity protection for the opto LED.
Thank you for answer
You're welcome.
To protect the opto LED, you can install the existing discrete diode anti-parallel to the LED.
roger finder
- Joined Nov 3, 2015
- 5
half wave rectifier voltage - diode drop voltage
Not enough current to turn on the optocoupler?
Self-assigned homework? If it is assigned homework you will get guidance, not answers.
To get the 20 mA or so the optocoupler needs you need about 11,000 ohms, but that would mean you have to be a 10 Watt resistor. The voltage across the LED sets itself. (Maybe 1.2 V?) And yes, use a protection diode.
