Can adding a diode to a 220V/30W soldering iron as a resistive load reduce current and power
To illustrate the issue, I used an electric soldering iron labeled 220V/30W as a resistive load. Electric soldering iron is basically a resistive load, with a power factor of approximately cos φ= 1. Therefore, the power P=I x U=0.1084 x 220=23.85W. The above figure shows the measurement and calculation results of a load without a diode. The AC current can be clearly seen from the digital multimeter, and the AC voltage is AC220V.
At this point, the AC current is displayed as 62.2MA (62.2/1000=0.0622A). It should be noted that after adding a diode, the AC voltmeter actually measures the pulsating DC current after half wave rectification by the diode, which contains AC components. Therefore, the AC voltmeter displays 120V at this time. Let's calculate the power P in the current circuit regardless of its voltage condition. P=I x U=0.0622 x 220=13.648W.
Subtract the actual power after connecting the diodes in series from the power before connecting the diodes, which is 23.85-13.65=10.2W. That is to say, the actual power of a diode connected in series in the circuit is reduced by 10.2W compared to the resistive load power of the same power without a diode connected in series. It meets the 0.45/V formula requirements for half wave rectifier diodes.
After connecting diodes in series in the circuit, I used a digital multimeter to actually measure the DC voltage of the soldering iron load. At this point, the measurement is 98.5V. Using the calculation formula for half wave rectifier diodes, calculate as follows:; 220/0.45=99V, taking into account the measurement error of the multimeter.
