if your lighting is LED source that would make a lot of sense (the sine wave light) because the wild world of powering led screw in bulbs is pretty hacky.
The V = IR (the voltage and current are actually proportional) formula does state that for a GIVEN resistance, voltage and current are proportional. That is, if a current of 2A flows in in a resistance of 2 ohms, then the supposed voltage supplied to that load resistance is 4 V. But in this case, we are not really measuring the voltage or a current through a resistor. We are measuring the voltage BEFORE the resistor and AFTER the phototransistor. The way to think about this is that the phototransistor is a switch, when light falls on it, the "switch" turns on, which, by essence basically means the circuit will be closed after the switch is on. Replace the phototransistor with a regular switch, when you switch it off, the measured voltage and the current will be zero, because the circuit is open, but when it is switched on, the circuit is closed, and the voltage and current measured will be some non-zero value. So basically, the more light that falls on the phototransistor the "on-ness" of the switch wll be more, when you block off the light, the 'off-ness" of the switch will increase.
This was an excellent explanation! Thank you!
wave from your room light is the AC current fluctuating... ie hertz cycles
Hey very nice video, thanks mate !
if your lighting is LED source that would make a lot of sense (the sine wave light) because the wild world of powering led screw in bulbs is pretty hacky.
Hello, why when you block the light both current and voltage went down? Should't they be inversely proportional?
The V = IR (the voltage and current are actually proportional) formula does state that for a GIVEN resistance, voltage and current are proportional. That is, if a current of 2A flows in in a resistance of 2 ohms, then the supposed voltage supplied to that load resistance is 4 V. But in this case, we are not really measuring the voltage or a current through a resistor. We are measuring the voltage BEFORE the resistor and AFTER the phototransistor. The way to think about this is that the phototransistor is a switch, when light falls on it, the "switch" turns on, which, by essence basically means the circuit will be closed after the switch is on. Replace the phototransistor with a regular switch, when you switch it off, the measured voltage and the current will be zero, because the circuit is open, but when it is switched on, the circuit is closed, and the voltage and current measured will be some non-zero value.
So basically, the more light that falls on the phototransistor the "on-ness" of the switch wll be more, when you block off the light, the 'off-ness" of the switch will increase.
-from Iraq-Kurdistan......your visible actions are not clear well