I did a performance evaluation by the load of the stabilised power supply which used an LED. Therefore, I measured the change of the voltage by the load change near it. I changed the interval of 15K-ohm from 1K-ohm with the 1K-ohm step in the load resistance.
By the graph, you find that the voltage is stable near load resistance 10K-ohm (150 ľA of electric currents). The voltage is falling when making a load resistance small. Because the voltage of the LED is about 1.7 V at the circuit this time, it drops the voltage with the resistor to make 1.5-V voltage. Therefore, the voltage in the resistor for the voltage drop becomes big when the load current increases and the load voltage falls.
At the measuring range, the voltage of the both edges of the LED is changing hardly.
The graph of the change of the voltage and the electric current by the calculation is shown below.
If making the value of resistor (R2) small, the fall of the voltage can be made little. However, R3 must be made small of course to get 1.5-V voltage. In the case, a lot of electric currents flow inside the stabilised circuit. Because it is not a big electricity consumption, it is the good way that, when wanting to make the more stable voltage, makes these resistance values small.
The graph below is the result which measured the change of the LED forward voltage when the electric current which flows through the LED changes.
When increasing the input voltage(Green line), the electric current which flows through the LED(Blue line) increases. However, the voltage of the both edges of the LED(Red line) is about 1.6 V and becomes constant. This voltage changes little with the kind of the LED.
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