Science Buddies: "Ask an Expert"

I tried measuring the optical output power of Led and incandescent bulb . for LED the procedure says i need to fix the clamp when the voltage across the resistor is 2.5 v to fix the clamp . is it the same for incandescent lamp too?

Also is there any way to reset the light to voltage converter. it gets saturated easily and shows 3.0 always. i tried with 2 converters.

Hi - from Step 8 in the procedure:

"Now disconnect the LED circuit and reconnect the incandescent lamp circuit. Place the incandescent lamp in exactly the same position as the LEDs. Repeat steps 2–6 for each of the incandescent lamps. Record the data in your lab notebook."

This means you should NOT move the clamp to get the reading at 2.5V for the incandescent bulb. You want to put it in the same place you had the LED.

The light to voltage converter should not need to be "reset." If you cover it or remove the light source the voltage should drop. If the voltage measurement is getting stuck at 3V no matter what then there might be something wrong with your circuit or how you are taking the measurement.

Hope that helps.

Ben

The light to voltage converter changes voltage when I cover with my hand. It gets lowered from 3.04 . But after I place it under the Led constant current or incadescent lamp circuit at any distance and change the distance .... the voltage across resistor remains 3.04 only. any help regarding this?

One more doubt... is For led circuit we provide the input current is 350mA. But for incadescent lamp the input current will be 500mA,right? Since we are not using any led drivers for incadescent lamp circuit.

How can we compare the efficiency of both.. if the input current is not the same.

Hi - if the voltage across the resistor is staying at 3 volts then the sensor is saturated and you need to keep moving the lights farther away. You may have too much ambient light in the room (ceiling lights, light from windows, etc) that is contributing to saturating the sensor.

The project instructions have equations you can use to calculate the relative efficiency of the two light sources. You know both the input current and voltage, and electrical power is equal to current times voltage, so you can calculate the input electrical power for each type of light.

Hope that helps.