Main content
Course: Electrical engineering > Unit 2
Lesson 2: Resistor circuits- Series resistors
- Series resistors
- Parallel resistors (derivation)
- Parallel resistors (derivation continued)
- Parallel resistors
- 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
© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice
Voltage divider
A voltage divider is a simple series resistor circuit. It's output voltage is a fixed fraction of its input voltage. The divide-down ratio is determined by two resistors. Written by Willy McAllister.
A very common and useful series resistor circuit goes by the nickname voltage divider. We will work out how this circuit operates, and you will see where the nickname comes from.
A voltage divider looks like this:
Our goal is to come up with an expression relating output to input . A good place to start is finding the current through and .
Assumption: Assumecurrent is flowing out of the divider. (Before we are done we will check to see what happens if this zero-current assumption doesn't hold).
With this assumption, and have the same current, and we can consider them to be in series.
To find the current, we apply Ohm's law and what we know about resistors in series, (reminder: resistors in series add),
Rearranging to solve for ,
We've solved for current in terms of and both resistors.
Next, we write an expression for using Ohm's Law,
We can substitute for in the previous equation to get,
and we have derived the voltage divider equation:
The output voltage equals the input voltage scaled by a ratio of resistors: the bottom resistor divided by the sum of the resistors.
The ratio of resistors is always less than for any values of and . This means is always less than . Input voltage is scaled down to by a fixed ratio determined by the resistor values. This is where the circuit gets its nickname: voltage divider.
Example - use the voltage divider equation to find
We want to find using the voltage divider relationship.
We insert the actual input voltage and resistor values into the equation, remembering the equation tells us the bottom resistor, , goes in the numerator.
Let's do an optional step to check the current.
Now we know the current, so we can compute how much power is being dissipated by our voltage divider,
Summary: Our voltage divider takes an input voltage (in this case , but it could be any value) and scales it down to create an output voltage that's of the input voltage. The ratio is determined by our choice of the two resistors. As long as is turned on, a current of flows down through the voltage divider, so it dissipates .
Voltage divider practice problems
All of these problems use this circuit diagram,
Review the assumption (advanced)
A voltage divider doesn't do anything useful unless its output is connected to something. You should be aware of what happens when a divider is connected to a load.
Remember we made an assumption at the beginning? We assumed the current flowing out of the output was . That let us treat and as if they were in series, and we developed the voltage divider equation. Let's check what happens if the assumption is not true.
Operating the voltage divider near its mid-range
To start this discussion, we let . With matched resistors, the expected of the voltage divider is the mid-point of the divider's range, . To cause some current to flow out of the divider, we connect a resistor . Does the divider still work? Does our voltage divider story collapse?
Resistor acts as a load on the output of the voltage divider, meaning that it causes a current to flow. The presence of means and are no longer strictly in series. Let's make fairly big, to make fairly small relative to . Let be ten times bigger than ,
This is our loaded voltage divider circuit, redrawn to show the equivalent resistance of in parallel with ,
The load resistor has the effect of reducing the resistance at the bottom of the voltage divider by roughly . What is the impact of this additional load on the divider's output voltage? Without the load, the expected output is . Now we figure out the output voltage in the presence of a load resistor.
We designed our divider with , so they cancel out,
The output voltage drops to of the input voltage. How big an error is this?
The actual output of the voltage divider is low by compared to the expected voltage. (Note the voltage error of is significantly less than the resistance change.) Does a few error matter? That's for you alone to decide. It depends on how accurate the voltage divider needs to be for your application.
The nugget to tuck away from this analysis: If the effective load resistance is greater than the bottom resistor in the voltage divider, you get roughly "one hand" of error in the output voltage. This holds when the output voltage is near the center of its range (in the neighborhood of ).
Operating the voltage divider near its extremes
If you design the voltage divider to operate near its extremes, with the output voltage close to or , the percentage error in output voltage will be different. We repeat the analysis with the output voltage set to and of the divider range. We keep the load resistor ten times the bottom resistor, so the parallel combination of and is still .
Case 1: of
Let of . The expected output is .
First we design a voltage divider that gives us the desired output. Figure out in terms of for a voltage divider,
Now we load the circuit with and see how the output voltage changes. The expression we derived above for the loaded voltage divider is,
We replace with ,
All the 's cancel out, leaving,
The actual output voltage is of instead of .
The actual output voltage divided by the expected output is,
So the actual voltage is lower than the expected by only .
Case 2: of
Let of . The expected output is .
Express in terms of for a voltage divider.
Now we load the circuit with and evaluate the change in output voltage. The expression we derived above for the loaded voltage divider is,
We replace with ,
All the 's cancel out,
The actual output voltage is of instead of the expected .
The actual output voltage divided by the expected output is,
So the actual voltage differs by from the expected. This is nearly twice the error compared to the mid-range divider.
Lessons for a loaded voltage divider
With a load resistor connected to a voltage divider:
- Near mid-range, the output voltage is reduced by
. - Near the top of its range, the error goes down substantially, to around
. - Near the bottom of its range, the error roughly doubles compared to mid-range. The output voltage is
lower than expected.
Controlling error in a loaded voltage divider
If your design requires the error to be significantly smaller, the load needs to be much larger than , like an additional or more. You can get an additional two ways. Increase the load resistance. Or, redesign the voltage divider to have smaller and , (at the cost of more power dissipated by the voltage divider).
Real-world resistor tolerance also impacts accuracy
Real-world resistors always have a tolerance on their value. If the accuracy of the voltage divider is critical to your application, use resistors with tight tolerances, and check for acceptable performance by analyzing the voltage divider at the anticipated tolerance extremes.
What's in a nickname
We mentioned at the start that the nickname of this circuit is a voltage divider. In many situations, that is exactly what it does. However, we showed that under certain conditions when there is a load on the divider, the actual output voltage is slightly lower than the value predicted by the voltage divider equation. The lesson: Call a circuit by its nickname; just remember that it's only a nickname.
Summary
Voltage divider:
where is the resistor on the bottom of the divider.
Want to join the conversation?
- in problem 2 and problem 4, suggest that 20k and 50k can be seen as correct answers, since the questions already use k and μ for avoiding too many 0s.(7 votes)
- Since this is the first time an answer is used in this course where "k" is part of the answer (20k), perhaps it is worth while putting in a mention of the possible notation in the text. Anyway, nice course, I am enjoying it. Thanks!(4 votes)
- regarding problem #3 in line 4 of the solution how do you go from -R2(R2x5/2)=30k to 3/2xR2=30k
how would I know to do that?(3 votes) - What is the difference between a 'Voltage Divider' and a 'Voltage Regulator'?(3 votes)
- Good question. A Voltage Divider is the simple 2-resistor circuit we talk about here. Remember the caution that a voltage divider only behaves itself when the current leaving the center node is practically zero? That limits where you can use a voltage divider. If you need to pull a lot of current at the new lower voltage, you use a Voltage Regulator. This is an integrated circuit with 10-50 transistors specially designed to take a higher voltage and provide current at a very well-controlled lower voltage.
The original IC voltage regulator is the LM317. It's been around forever. https://en.wikipedia.org/wiki/LM317. If you scroll down a little you will see the typical circuit for using the LM317. Notice part of the circuit is a 2-resistor voltage divider that let's you adjust the output voltage. The divider determines the voltage. The regulator provides the robust current.(4 votes)
- If we put RL in the circuit, does that make any change to power dissipated by voltage divider ?(1 vote)
- Hi Mogbulumide,
This is a hard question to answer without knowing your background. I am assuming you have some knowledge of electronics but have not yet studied AC circuits.
There are two answers to this question:
1) For Direct Current circuits - with DC an inductor acts as a short circuit. In reality you will "see" the resistance of the inductor's winding. In general you will not find inductors in pure DC circuits.
2) For Alternating Current circuits - yes the inductor can be part of the voltage divider. Although, we wouldn't call the circuit a voltage divider anymore. We would instead call the circuit a filter. The "voltage divider's" output would change depending on the frequency of the input signal.
This is where electronics gets interesting. You may be interested in this site:
http://sound.westhost.com/lr-passive.htm
Happy soldering,
APD(6 votes)
- When evaluating the expression
R1 || RL = R2 * 10R2 / R2 + 10R2 you get 10/11R2 = 0.91R2
Was this accomplished by factoring out R2 from the numerator and denominator?
R2(1 * 10) / R2(1 + 10) = 10/11 but wouldn't the R2/R2 just equal 1 or just cancel out? So how is it that it equals 0.91R2? We don't know the value of R2 so we just leave it in as a factor of the result?
When evaluating the expression
V-out = V-in(0.91R2/R1+0.91R2). You stated that R1 = R2 So they cancel out. leaving V-in(0.91/1+0.91) = 0.91/1.91=0.48V Assuming these computations were accomplished via factoring out the R2 this would assert the same computation as the first expression, but in this case R2 cancels out, whereas in the first expression it was left in. So how do I determine when I need to keep an a variable tied to a result, as in the first case of 0.91R2, and when to cancel it out, as in the second case 0.48v?
Thank you for your time.(2 votes)- The numerator of the resistor expression is R2 * 10R2 = 10 R2^2.
The denominator of the resistor expression is R2 + 10R2 = 11R2
When cancelling R2's you eliminate the one on the bottom but you still end up with one on top. 10R2/11.
In the voltage equation, all the R2's cancel out.(4 votes)
- Regarding Problem 4 - Challenge
Shouldn't the power be equal to P = I x Vout instead of Vin if it asks for the power dissipated by the divider? I was thinking that every element dissipates power and if it specifies the divider power it should be calculated using Vout. If not, why do we use Vin?(1 vote)- The voltage divider is made of two resistors, so the "power of the divider" has to account for both. You can compute the power a few different ways.
Assuming i_out = 0...
If you know the current i flowing through the two resistors, then p = i^2 (R1 + R2).
Or, you could compute power based on the voltages,
p_total = p_R1 + p_R2
p_R1 = (Vin - Vout) / R1
p_R2 = Vout^2 / R2
Or, you could compute power based on both i and v,
p_total = p_R1 + p_R2
p_R1 = i (Vin-Vout)
p_R2 = i Vout
Or, simplest of all,
p = i Vin(5 votes)
- So can this be used for a DC circuit? I am trying reduce the voltage of a battery from 9v to 3.7v. what is the best way to do this?(2 votes)
- Hello Swag,
It depends on your application.
I assume you are using a 9 VDC battery. If you use a voltage divider over half of the battery's energy will be lost in the resistors. It turns out that this simple circuit is very inefficient.
From an energy perspective you would be better off using a a “switching Regulator.” For an example see:
http://power.murata.com/data/power/dms-78xxsr.pdf
For an explanation of how the circuit operates see:
https://en.wikipedia.org/wiki/Buck_converter
Regards,
APD(3 votes)
- In problem 3, I don't get how you go from step 2 to step 3:
2V = 5V x ( R2 / 30kOhm + R2) -->
30kOhm + R2 = ( 5V / 2V ) x R2
It's some kind of algebra I don't understand, but I also don't know the name for it well enough to look it up.(2 votes)- First, multiply both sides of the equation by (30k + R2).
That gives you 2v * (30k + R2) = 5v * R2
Then, divide both sides of the equation by 2v.
That gives you (30k + R2) = (5v * R2)/2v which is the same as (5v/2v) * R2(2 votes)
- Problem 4
Would it be wrong to calculate the equivalent resistance (Rs=R1+ R2) of the divider using P=(Vin)^2 / Rs ?(1 vote) - In problem 2, how are we allowed to take the reciprocal of 1/10 to get 15V in?(2 votes)