- Series resistors
- Series resistors
- Parallel resistors (part 1)
- Parallel resistors (part 2)
- Parallel resistors (part 3)
- Parallel resistors
- Parallel conductance
- Series and parallel resistors
- Simplifying resistor networks
- Simplifying resistor networks
- Delta-Wye resistor networks
- Voltage divider
- Voltage divider
- Analyzing a resistor circuit with two batteries
The output voltage is a fixed fraction of the input voltage. The ratio is determined by two resistors. Created by Willy McAllister.
Want to join the conversation?
- What do the loose ends around the lower resistor mean? I don't really know how this is supposed to work, does it mean the circuit continues off-screen?(18 votes)
- Yes, you could think of it that way.
These connection points are for the input and output of the circuit. On the left we might find a battery or other source of power. On the right we might find a load that requires the specific voltage we designed the voltage divider to deliver.
- Conceptually, in this oversimplified case, why not just use 1 resistor? Or basically I don't know why you would want two resistors if you aren't actually using the middle wire. If you are using the middle wire, what is an example of a circuit literally requiring that complexity?(3 votes)
- Hello Horus,
Yes, at first glance it seems like a pointless exercise. However the voltage divider is a useful tool from two perspectives:
A voltage divider is a physical assemblage of resistors that allows you to lower a voltage. For example, let's assume we have a source that provided 5 VDC connected to another device that required 3.3 VDC. A properly designed voltage divider would allow us to connect the devices together. This scenario is common. The 5VDC may be an Ardunio microcontroller and the 3.3 VDC source a new sensor you would like to connect tot he Arduino...
On a big picture level a voltage divider is a mathematical model. At first it will allows us to talk about DC voltages and the relationships in series circuits. Later is will help us understand much more interesting AC circuits items like power factor and filters.
- Why we only take R2 for V out?(6 votes)
- Assigning the output voltage across R2 is the most common way to use a voltage divider. (Where R2 is the resistor connected to the reference node, ground). There's nothing keeping you from using the voltage across R1, but I have not seen that done too often.
There isn't anything special about the voltage divider circuit, other than it happens to be a very common resistor configuration. So we spend a little time looking at its properties and developing an equation for the output voltage. It's just a quick mental shortcut you will use frequently.(3 votes)
- In the first circuit, you mention over and over that the two resistors R1 and R2 are in series. But how are they in series when the middle wire is extending from the middle?(4 votes)
- Hello Sam,
The resistors are in series but we must make the assumption that that middle wire is for testing purposes only. No current can flow in this middle wire. If it did than the voltage divider calculation would be incorrect.
For more information please see the "Review the assumption (advanced)" section of this page:
- The formula is very mathematically nice but why does it work out this way? What is the intuition behind the voltage being proportional to the proportion of resistance across the two resistors?(1 vote)
- Tough question. Let me give it a try.
Draw a picture of one long resistor 10 cm tall on the page. Connect the top of the resistor to voltage Vin. Connect the bottom to 0v. The value of the long tall resistor is R.
Touch your finger to the resistor at the top. What is the voltage of your fingertip? Ans: Vin .
Touch your finger to the bottom of the resistor. What is the voltage of your fingertip? Ans: 0.
Touch your finger right in the middle of the resistor. What is the voltage of your fingertip? Ans: what does your intuition say?
Touch 1/4 the way down from the top. Touch 1/4 of the way up from the bottom. What are the voltages? Ans: (work out the proportions).
You can figure out any voltage from the physical length of resistance above and below your fingertip.
Can you come up with a formula to give the answer? Hint: Break the resistor into two parts: R1 is the portion above your fingertip, and R2 is the portion below your fingertip. The total resistance R = R1 + R2. Find Vfingertip in terms of Vin, R1, and R2.
Please let me know if this helps.(10 votes)
- Why is the voltage on the node between the resistors assumed to be very small or 0? How would you achieve this with a real circuit?(2 votes)
- The voltage divider formula works if the current leaving the second node is small (not the voltage). This can be achieved if the thing the divider is connected to has a very high resistance. Opamps and MOS transistors are often connected to voltage dividers, and they do present a very high resistance.(7 votes)
- Are those two open ends connected to something ?(2 votes)
- The pairs of little circles represent connection points (ports) to the voltage divider. In a typical circuit you will often find the two input circles facing left (the input port) are connected to a constant voltage, usually the power supply of the circuit. The output port (the two circles facing right) will be connected to something that needs the divided-down voltage, like a transistor or an operational amplifier.
Here's an example where a voltage divider provides the base voltage for a bipolar junction transistor: http://pcbheaven.com/wikipages/Transistor_theory/?p=5. Resistors R1 and R2 form a voltage divider.
Here is another example where the input voltage is not a constant value. Scroll down to Figure 1.4. Can you see how Rf and Rg form a voltage divider? (Analog Devices writes excellent documentation.) http://www.analog.com/media/en/training-seminars/design-handbooks/Basic-Linear-Design/Chapter1.pdf(2 votes)
- i have a question, so for a parallel series combination circuit his formula wouldn't work right i.e the voltage divider formula works only for series circuits??(1 vote)
- The voltage divider equation works when you have two conditions: you have two resistors in series AND the current leaving from the central node is quite small.(4 votes)
- Can the V out be any one of the 2 wires that come out?(1 vote)
- Hello Ammar,
No, Vout is the potential difference between the two wires on the right.
If this were a real circuit there would likely be a ground on the lower node. Think of a ground as a common point from which all measurements are made. Vin in measured form the top wire (left side) to ground and Vout is measured from the top wire (right side) to ground.
- [Voiceover] Now I'm gonna show you what a circuit, that's called a voltage divider. This is the name we give to a simple circuit of two series resistors. So, I'm just gonna draw two series resistors here. And it's a nickname, in the sense of, it's just a pattern that we see when we look at circuits. And, I'll show you what the pattern is. The pattern is, we have two resistors in series. It's no more than that. And we assume that there's a voltage over here. We hook up a voltage over here like this. So, that's called an input voltage. We'll call it 'VI', for 'V in'. And then, the midpoint of the two resistors, and typically the bottom, that's called the 'out'. So, we basically just have a pattern here with the series resistor, driven by some voltage from the ends of the two resistors. And we're curious about the voltage across one of them. So now we're gonna develop an expression for this. Let's also label our resistors. This will be 'R1'. And this will be 'R2'. That's how we tell our resistors apart. And we're gonna develop an expression for this. So, let's first put a current through here. We'll call that current 'i'. We'll make an assumption that this current here, is zero. There's no current going out of our little circuit here. And that means, of course, that this current here is also 'i'. So, it's continuous all the way down. And now we want to develop an expression that tells us what 'V out' is, in terms of these two resistors and the input voltage. So let's go over here and do that. First thing we're gonna write is, we know that, using Ohm's law, we can write an expression for these series resistors on this side here. Ohm's law, we'll put over here. 'V' equals 'iR'. In a specific case here, 'V in' equals 'i' times what? Times the series combination of 'R1' and 'R2'. And the series combination is the sum: 'R1' plus 'R2'. I'm gonna solve this for 'i'. 'i' equals 'V in' divided by 'R1' plus 'R2'. Alright, next step is gonna be, let's solve for-- let's write an expression that's related to 'V out'. And 'V out' only depends on 'R2' and this current here. So we can write 'V out' equals 'i' times 'R2'. And I'll solve this equation for 'i' the same way. Equals 'V zero' over 'R2' And now we have two expressions for 'i' in our circuit, because we made this assumption of zero current going out, those two 'i's' are the same. So, let's set those equal to each other and see what we get. 'i' is 'V zero' over 'R2', equals 'V1' over 'R1' plus 'R2'. So now I'm gonna take 'R2' and move it over to the other side of the equation. And we get 'V out' equals 'V1'. Sorry, 'V in' times 'R2' over 'R1' plus 'R2'. And this is called, this is called the voltage divider expression. Right here. It gives us an expression for 'V out', in terms of 'V in', and the ratio of resistors. Resistors are always positive numbers. And so this fraction is always less than one. Which means that 'V out' is always somewhat less than 'V in'. And it's adjustable, by adjusting the resistor values. It's a really handy circuit to have. Let's do some examples. We'll put that up in the corner so we can see it. Then real quick, I'm gonna build a voltage divider that we can practice on. Let's make this '2k' ohms, two thousand ohms. We'll make this 6000 ohms, or '6k' ohms. And we'll hook it up to an input source that looks like, let's say it's 6 volts. Like that. And we'll take an output off of this. Right here, is where the output of our voltage divider is. And we'll say that that is 'V out'. So let's solve this using the voltage divider expression. 'V out' equals 'V in', which is 6 volts. Times the ratio of resistors. 'R2' is '6k' ohms, divided by '2k' ohms, plus '6k' ohms. And notice this always happens, the 'k's' all cancel out. That's nice. And that equals six times, six over, two plus six is eight. And if I do my calculations right, 'V out' is 4.5 volts. So that's what a voltage divider is. And if you remember at the beginning, if you remember at the beginning, we made an assumption that this current going out here, was appr-- about zero. If that current is really small, you can use this voltage divider expression. Which as, we see up here, is the ratio of the bottom resistor to both resistors. That's how I remember it. It's the bottom resistor, over the two resistors added together. If you think the current is not very small, what you do is you go back and you do this analysis. You do the same analysis again but you account for the current that's in here. So that's the story on voltage dividers