✔ 最佳答案
Dropping Resistor
Voltage Dividers
In many circuits, it is necessary to obtain a voltage not available from the main power source. Rather than have multiple power sources for all needed voltages, we can derive other voltages from the main power source. In most cases, the needed voltage is less than the voltage from the main source, so we can use resistors in an appropriate configuration to reduce the voltage from the power source, for use in a small circuit.
If we know precisely both the voltage and current required, we can simply connect a resistor in series with the power source, with a value calculated in accordance with Ohm's Law. This resistor will drop some of the source voltage, leaving the right amount for the actual load, as shown to the right.
Usually, however, this doesn't work too well. The required value of the series dropping resistor will almost never be a standard value, and the cost of having special values manufactured for specific circuits is prohibitive. For example, suppose we have a 9 volt battery as your main power source, and want to operate a load that requires 5 volts at 3.5 milliamperes. Our series resistor, R, must drop 4 volts at 3.5 mA. Using Ohm's Law to calculate the required resistance value, we find that we need a resistance of 4/0.0035 = 1142.8571
圖片參考:
http://www.play-hookey.com/symbols/Omega.gif
or 1.1428571k. We have a choice between 1.1k and 1.2k as standard 5% values, but neither will give us what we want.
圖片參考:
http://www.play-hookey.com/dc_theory/images/voltage_divider.gif
A more practical solution to the problem is to use two resistors in series, and use the voltage appearing across one of them. This configuration is known as a voltage divider because it divides the source voltage into two parts. The basic circuit is shown to the right.