We can calculate the voltage dropped across the resistor R 2 using this equation:. Calculate the total resistance in the circuit and just add it all up since the resistors are connected in series.
You have to know the voltage you will supply and the load resistance when creating a voltage divider for a specific load. This means that the current going through the load is ten times the current going through the bottom of the voltage divider to ground. The requirement for this voltage divider is to provide a voltage of 25V and a current of mA to the load from a source voltage of V.
This current, which does not flow through any of the load devices, is called bleeder current. In the first figure, take note that the value of resistance of a parallel network is always less than the value of the smallest resistor in the network since the load connected between point B and ground forms a parallel network of the load and resistor R1. A voltage ladder is a circuit that consists of several resistors in series with a voltage placed across the entire resistor network.
Each resistor in the network has a higher voltage drop than the one before it. Since the resistors in the ladder are in series, the current is the same all throughout. To get its value, you should divide the total voltage by the total resistance. First name. Last name. Your cart is empty. A potential divider could be thought of as a voltage divider and it is often referred to as such.
How does it work? A potential divider is a simple circuit which takes advantage of the way voltages drop across resistors in series. It is a very useful and common circuit and is widely used in our range of electronic kits. The idea is that by using two resistors in series it is possible to divide a voltage and create a different voltage between them.
In the example below two identical resistors are in series with a power supply. The amount by which the voltage drops over across each resistor depends on the relative values of each resistor and the total resistance. The formula for working out the voltage drop across two resistors in series is: One of the most useful ways to use this circuit is to replace R2 with a variable resistor. If R2 can be controlled by turning a dial then Vout can also be controlled.
Another common use of the potential divider is to replace R2 with a sensor such as an LDR. Then as the resistance of the sensor changes, Vout changes as well. This change can then be used to trigger a transistor or can be fed into the input of a microcontroller. A potentiometer schematic symbol. Pins 1 and 3 are the resistor ends.
Pin 2 connects to the wiper. If the outside pins connect to a voltage source one to ground, the other to V in , the output V out at the middle pin will mimic a voltage divider. Turn the pot all the way in one direction, and the voltage may be zero; turned to the other side the output voltage approaches the input; a wiper in the middle position means the output voltage will be half of the input.
Potentiometers come in a variety of packages, and have many applications of their own. They may be used to create a reference voltage, adjust radio stations , measure position on a joystick , or in tons of other applications which require a variable input voltage.
Many sensors in the real world are simple resistive devices. A photocell is a variable resistor, which produces a resistance proportional to the amount of light it senses.
Other devices like flex sensors , force-sensitive resistors , and thermistors , are also variable resistors. Not so much. But, by adding another resistor to the resistive sensors, we can create a voltage divider.
Once the output of the voltage divider is known, we can go back and calculate the resistance of the sensor. If we combine that with a static resistance somewhere in the middle - say 5. Photocell makes up half of this voltage divider. The voltage is measured to find the resistance of the light sensor. Many of those sensors operate at a relatively low voltage, in order to conserve power.
Unfortunately, it's not uncommon that those low-voltage sensors are ultimately interfacing with a microcontroller operating at a higher system voltage. This leads to a problem of level shifting , which has a number of solutions including voltage dividing. For example, an ADXL accelerometer allows for a maximum input voltage of 3. Voltage divider! All that's needed is a couple resistors whose ratio will divide a 5V signal to about 3.
An example of voltage dividers in a breadboard , level shifting 5V signals to 3. Click to see a larger view. Keep in mind, this solution only works in one direction. A voltage divider alone will never be able to step a lower voltage up to a higher one. As tempting as it may be to use a voltage divider to step down, say, a 12V power supply to 5V, voltage dividers should not be used to supply power to a load. Any current that the load requires is also going to have to run through R 1.
The current and voltage across R 1 produce power, which is dissipated in the form of heat. That doesn't even mention how inefficient a voltage-divider-power-supply would be.
Basically, don't use a voltage divider as a voltage supply for anything that requires even a modest amount of power. If you need to drop down a voltage to use it as a power supply, look into voltage regulators or switching supplies. If you haven't yet gotten your fill of voltage dividers, in this section we'll evaluate how Ohm's law is applied to produce the voltage divider equation.
This is a fun exercise, but not super-important to understanding what voltage dividers do. If you're interested, prepare for some fun times with Ohm's law and algebra. So, what if you wanted to measure the voltage at V out? How could Ohm's law be applied to create a formula to calculate the voltage there? Let's assume that we know the values of V in , R 1 , and R 2 , so let's get our V out equation in terms of those values. Let's start by drawing out the currents in the circuit--I 1 and I 2 --which we'll call the currents across the respective resistors.
Our goal is to calculate V out , what if we applied Ohm's law to that voltage? Easy enough, there's just one resistor and one current involved:. We know R 2 's value, but what about I 2? That's an unknown value, but we do know a little something about it.
We can assume and this turns out to be a big assumption that I 1 is equivalent to I 2. Alright, but does that help us? Hold that thought. Our circuit now looks like this, where I equals both I 1 and I 2.
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