How to use a multimeter to detect damaged electronic products or household appliances?
For electronic product maintenance, home appliance repair or computer brain maintenance, everyone has their own methods, and we cannot generalize. Generally speaking, the multimeter is just a tool with many functions (that is, it can measure many parameters). By proficiently measuring the parameters of each segment of the circuit or each component, and its synthesis, the fault can be judged. For example, by measuring the voltage of a certain circuit to determine whether it is working normally; by measuring the resistance of a certain component to determine whether it is normal; by measuring the capacitance of a certain component to determine whether it is open circuit, short circuit or deterioration . . . . etc. So the multimeter is just a tool. After you buy it back, all DIY fans should also go to various forums or participate in amateur classes to strengthen their own quality and improve the ability to use tools. Finally, to make an analogy, a multimeter is like a sword. Its power lies in its owner. As much as there is martial arts (electronic knowledge), it can exert as much effect as possible.
How is the accuracy (uncertainty) of a digital multimeter calculated?
The accuracy of the multimeter is also called uncertainty by some manufacturers. It generally says "within one year from the factory, the operating temperature is 18°C ~ 28°C (64°F ~ 82°F), and the relative humidity is less than 80%. ± (0.8% reading + 2 characters)." Many buyers or users are not very clear about this and often ask. I am assuming here that there is a meter, in a certain range, such as DC 200V, it is written like this, and the measured value shows 100.0 on the meter, so what should be the correct value at this time. I think that for general users, you can completely ignore the calculation of accuracy, and just think that it is 100V DC. Calculated according to the manufacturer's accuracy, when measuring 100V (display 100.0), the error is ±(0.8%*1000+2)=±10, that is, the error is 1.0V. When you substitute the reading, don't consider the decimal point to display The value is substituted into the calculation, and the calculated value must be added with a decimal point and then used to calculate the shipping cost. Like this example, the correct value is 100.0±1.0, which should be between DC 99.0 and 101.0V.
