### Section 3: Voltage Dividers

**Voltage Divider Concept**

Water pressure is built up by pumping water into tall water towers. The water holds a lot of potential energy, and one can imagine if the tower had a 1" hole in it, the water would spray for quite some distance. That water pressure is akin to voltage, and the amount of water which would exit the hole is is akin to the electrical current.Instead of leaving a 1" square hole in the water tower, a hose (shown in red above) is attached to the water tower. The hose is relatively narrow and long, and will resist the flow of current. This is akin to an electrical resistance labeled as R1. The amount of pressure at the end of the hose will be less than the pressure at the end of the hose. This is the equivalent of the voltage drop.

Finally, consider that between the end of the hose and the puddle on the ground, there is one additional degree of resistance. The water trying to push into the air has to overcome 14.7 lbs per square inch at sea level. This resistance to the water current flowing out of the hose affects the pressure at the end of the hose. This is akin to the electrical resistance labels as R2.

The relationship between the voltage, current, and resistance is known as Ohms Law.

In electronics, sometimes we want to know what the voltage potential is in between two resistors, as in Figure 1.4.2. We have an input voltage potential (Vin) and an output voltage potential (Vout). Since voltage potentials are relative, we can define a zero potential (ground) wherever is convenient. Assuming we know the resistance R1 (above, the hose) and the resistance R2 (above, leaving the nozzle), then we can calculate the output voltage.

**Figure 1.4.2: Resistor divider**

#### Voltage Divider Math

Starting with Ohms Law, which relates voltage, current, and the resistance:Solve for the current through the resistors R1 and R2, given Vin:

The
current that flows through each resistor must be equal, just as the
water that enters a hose must equal the water that flows out of the
hose. Therefore, the current that flows through R1 and R2 combined must be equal to the current that flows through just R2.

Substitute equation 1.4.3 into equation 1.4.4, to get:Finally, solve for the gain. The gain is the ratio of how much voltage is on the output relative to how much voltage is on the input. For a voltage divider, this is:

(Eqn. 1.4.6)