www.nortonkit.com 18 अक्तूबर 2013
Direct links to other DC Electronics pages:
Fundamentals of Electricity: [Introduction to DC Circuits] [What is Electricity?] [Electrons] [Static Electricity] [The Basic Circuit] [Using Schematic Diagrams] [Ohm's Law]
Basic Electronic Components and Circuits. . .
Resistors: [Resistor Construction] [The Color Code] [Resistors in Series] [Resistors in Parallel] [The Voltage Divider] [Resistance Ratio Calculator] [Three-Terminal Resistor Configurations] [Delta<==>Wye Conversions] [The Wheatstone Bridge]
Capacitors: [Capacitor Construction] [Reading Capacitor Values] [Capacitors in Series] [Capacitors in Parallel]
Inductors and Transformers: [Inductor Construction] [Inductors in Series] [Inductors in Parallel] [Transformer Concepts]
Combining Different Components: [Resistors With Capacitors] [Resistors With Inductors] [Capacitors With Inductors] [Resistors, Capacitors, and Inductors]
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 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.

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.

In this circuit, the output voltage, VOUT, can be set accurately as a fraction of the source voltage, E. Using our example above, we want to select R1 and R2 such that we will drop 4 volts across R1, leaving 5 volts across R2. Since VOUT is the voltage across R2, this will give us the voltage we want. But how do we find the correct values of resistance to do this?

The first step is to note that, with no external connection to VOUT, this is simply a series connection and the same current must flow through both resistors. (We'll deal with the load current shortly.) Therefore, in accordance with Ohm's Law, the ratio of voltage across these resistors will be equal to the ratio of the resistance values themselves. In this case, the voltage ratio we want is 4:5, or 0.8:1. Therefore, we want this resistor ratio as well.

But there are 20 different standard 5% resistance values in each decade range, so there are lots of possible resistance ratios. Most will be wrong for this purpose, of course. So how do we find two standard resistance values that will give the ratio we want? We could do it manually, testing each possible combination. But a better way is to let the computer do the tedious work and present all options to us. Then we can select the values we want from the list of valid possibilities.

To this end, remember that ratio of 0.8, and insert it into the table in the Resistance Ratio Calculator on these pages. Press <Enter> when you have entered this value, and note the result in the rest of the table.

The table shows the significant digits of standard 5% resistor values. When you type in a ratio, it calculates the corresponding significant digits that would be required to complete that ratio. All you need to do is pick out ratios of valid significant digits. In this case, the table shows that you can use resistance values of 15:12, 20:16, or 30:24 to obtain the ratio of 1:0.8. If you had specified 1:1.25 (the inverse of 0.8:1), you would have gotten the ratios of 12:15, 16:20, and 24:30. Either way, these are workable choices, while all other choices fail to match standard values.

We can get a VOUT of 5 volts, then, if we set R1 = 1.2k and R2 = 1.5k. We can also get the same VOUT if we make R1 = 12k and R2 = 15k. The exact resistor values don't matter, so long as their ratio is correct.

The one thing we haven't accounted for as yet is the current drawn by the load. This will necessarily upset the resistance balance, since any load current will flow through R1, but not through R2. As a result, the load will reduce output voltage of the voltage divider by some amount. Appropriately, this effect is called loading.

To calculate the effect of loading and its extent in any given instance, we must realize that the voltage divider circuit behaves in exactly the same way as a battery of voltage VOUT with a series resistor whose value is equal to the parallel combination of R1 and R2. The figure to the right shows the equivalent circuit for our example voltage divider.

Now, we noted earlier that our example load draws 3.5mA at 5 volts. In accordance with Ohm's Law, this current will drop a voltage of 2.33333 volts across that 667 resistor. Thus, our example voltage divider will not be able to provide +5 volts to this load.

If we reduce the resistors in the voltage divider to 120 and 150, the equivalent series resistance is only 66.7 so the voltage drop caused by this load will be 0.23333 volt. This may be a small enough loss to ignore in a practical circuit.

The drawback of this is that such low resistance values will draw a significant amount of current from the original source. This is probably acceptable if the original source is an electronic power supply, but not if it's an actual battery. Thus, this use of a voltage divider is reasonable and appropriate in some circumstances, but not in all cases.

The voltage divider is a very simple circuit that can be highly accurate if not loaded down. In many cases it cannot be used directly, as we have seen. However, in such cases it can either be adapted, or augmented with other components to preserve its operation while avoiding the problems that can occur. Thus, even in those cases where a voltage divider by itself is not sufficient to meet the need, it can serve as the basis of a circuit that will perform as required. We'll see any number of examples of this in practical examples on this Website.