22 March 2009 Author: Giorgos Lazaridis The Voltage Divider
What is the voltage divider
The voltage divider is composed (usually) from a set of two resistors in series. The effect of the voltage divider is that it can be used to polarize other components in a circuitry such as transistors or integrated circuits, with a different voltage from the main voltage supply. Thus, a circuit may have 9 volts power supply and using a voltage divider we can supply a transistor in this circuit with 3.6 volts. A typical voltage divider is shown bellow:
How to calculate the voltage divider
The first step to calculate a voltage divider is to know the current that flows within. At first, we will suppose that the R_{LOAD} is not yet connected to the circuit. So using the ohm's law:
I = U / R_{TOTAL} => I = U / (R1+R2) (1)
The output voltage will be the drop voltage across the edges of R2. Using again the ohm's law we calculate:
V_{R2} = I x R2 (2)
(2) because of (1) becomes:
V_{R2} = V_{OUTPUT} = V1 x R2 / (R1 + R2)
This is the basic formula to calculate the output voltage of the voltage divider.
A major issue here is that R2 is connected in parallel with R_{LOAD}. When you put the voltage divider to work, you should always calculate R2 as R_{TOTAL} with the R_{LOAD} connected in parallel:
R_{2LOAD TOTAL} = (R2 x R_{LOAD}) / (R2 + R_{LOAD})
And replace R2 in the basic formula with the result of the above calculation.
Last but not least, is the max current that will be able to go within R_{LOAD}. If we consider the system R1R2R_{LOAD} as a mixed resistor connection (series and parallel connection combined), the maximum current I will be less or equal to the current that flows within R1. Thus:
I = U / R1
And
I = I_{R2} + I_{RLOAD}
An example
In the previous example, we want to supply the R_{LOAD} with 3 volts. The input voltage U is 12 Volts. We also want to limit the maximum current that will flow within R_{LOAD} to 300mA. The resistance of R_{LOAD} is 7Ohms.
At first we need to calculate the R1. The only thing that is given to us is the current limitation. We suppose that the minimum resistance of R_{LOAD} could be as low as 0 ohms. That would give us the maximum current flow. So, we will calculate R1 in a way that no more than 300mA could flow within:
R1 = U / I => R1 = 12V / 300mA => R1 = 40 Ohms
Now we need to calculate R2 in a way that the voltage across R_{LOAD} will be the desired one. Using the basic formula from previous calculations:
V_{OUTPUT} = V1 x R2 / (R1 + R2)[/B]
But R2 is connected in parallel with R_{LOAD}. So we will calculate the basic formula using R_{TOTAL} instead of R2:
V_{OUTPUT} = V1 x R_{TOTAL} / (R1 + R_{TOTAL}) =>
V_{OUTPUT} x (R1 + R_{TOTAL}) = V1 x R_{TOTAL} =>
V_{OUTPUT} x R1 + V_{OUTPUT} x R_{TOTAL} = V1 x R_{TOTAL} =>
V_{OUTPUT} x R_{TOTAL}  V1 x R_{TOTAL} = V_{OUTPUT} x R1 =>
R_{TOTAL} x (V_{OUTPUT}  V1) = V_{OUTPUT} x R1 =>
R_{TOTAL} x (V1  V_{OUTPUT}) = V_{OUTPUT} x R1 =>
R_{TOTAL} = (V_{OUTPUT} x R1) / (V1  V_{OUTPUT}) =>
R_{TOTAL} = (3 x 40) / (12  3) =>
R_{TOTAL} = 120 / 9 =>
R_{TOTAL} = 13.3 Ohms =>(approx.) R_{TOTAL} = 14 Ohms
Now we come to the final step, to calculate the R2 resistor.The total resistance R_{TOTAL} is calculated from the resistor parallel calculation as follows:
1/R_{TOTAL} = 1/R2 + 1/R_{LOAD}
solving for R2:
1/R2 = 1/R_{TOTAL}  1/R_{LOAD}=>
1/R2 = R_{LOAD}/(R_{LOAD}xR_{TOTAL})  R_{TOTAL}/(R_{TOTAL}xR_{LOAD}) =>
1/R2 = (R_{LOAD}R_{TOTAL}) / R_{LOAD}xR_{TOTAL} =>
R2 = R_{LOAD}xR_{TOTAL} / (R_{LOAD}R_{TOTAL}) =>
R2 = 7 x 14 / (14  7) =>
R2 = 14 Ohms
And these are the final values for this circuit to work!
Characteristics
Polarizing a transistor
The voltage divider is a cheap and easy solution to have different voltages within a circuit with minimal components used. Therefore, it is a very common way for polarizing for example transistors within an amplifier circuit.
Nevertheless, a voltage divider can have some major drawbacks. First of all, it is not stable. If the current drawn from R_{LOAD} changes, the voltage across R_{LOAD} will also change.
Another drawback is the current limitation. There are usually not resistors with the values required to achieve exactly the desired voltage drop. Many times, higher values of resistors must be used in order to achieve this voltage drop. Typical values are from 330 Ohms, and may go up to 22Kohms or even higher. This will dramatically decrease the maximum current flow from the R_{LOAD}. That is of course not a big deal when polarizing a transistor (as seen on the left side). That is what makes the voltage divider ideal for such kind of applications.
Relative pages Learn how dimmers workBasic transistor circuitsThe LED theoryThe Ohm lawThe resistor theoryThe transistor theory of operationDr.Calculus: Voltage divider calculatorDr.Calculus: Ohm law calculatorInternational unit converter
Comments
At 30 October 2013, 4:42:55 user John wrote: [reply @ John]Sorry but I think your calculation has an error.
The total resistance of any resistor network, in parallel, must always be lower than the lowest value.
However, your calculations show the total to be the same as one of the parallel resistors (14 ohms)
Also, the calculated values do not work correctly for the circuit.
Sorry. But many thanks for the site. Its a good educational site!
At 14 June 2013, 19:04:34 user Tim wrote: [reply @ Tim]Excuse me, but i think you have a mistake. Or i have misunderstanding
"1/RTOTAL = 1/R2 1/RLOAD
solving for R2:
1/R2 = 1/RLOAD  1/RTOTAL =>"/I THINK MISTAKE IS HERE, 1/RTOTAL=1/R2 1/RLOAD,THEN 1/R2=1/RTOTAL 1/RLOAD, THEN 1/R2=1/RTOTAL 1/RLOAD. tHE RESULT OF R2 WILL BE STE SAME BUT NEGATIVE. I BELIEVE NEGATIVE ANSWER IS UNCORRECT BUT MATH IS MATH...
At 29 July 2011, 12:23:49 user Kammenos wrote: [reply @ Kammenos]@Yoram Stein what do you mean???
At 29 July 2011, 10:28:10 user Yoram Stein wrote: [reply @ Yoram Stein]Giorgios question: How do you write R load so theat load comes lower then the R (what alt function is that)?
At 6 June 2011, 15:47:59 user Kammenos wrote: [reply @ Kammenos]@Fung in that case, you add the resistors and find the equivalent. If for example 3 resistors R1R2R3 with R1 to + and R3 to , you can find the voltage between R1 and r2 if you add r2 and r3 and calculate them as a voltage divider with 2 resistors, R1 and R2+R3 equivalent.
At 6 June 2011, 4:22:30 user Fung wrote: [reply @ Fung]How about the voltage dividers which includes more than 2 resistors, which has more than 1 loading voltage?
For example, there are 5 resistors in a voltage divider R1 to R5 with the supply voltage of 6V, the values of R1 to R5 are 82 kohms, 5 kohms, 3.3 kohms, 6.8 kohms and 68 kohms respectively; how can I calculate the voltage between the resistors (ie R1 and R2, R2 and R3, R3 and R4, R4 and R5)?
At 12 June 2010, 9:35:34 user Kammenos wrote: [reply @ Kammenos]Ed, go to the forum (http://pcbheaven.com/forum/) subscribe there and post your question with a circuit.
At 11 June 2010, 21:45:22 user Ed Edmondson wrote: [reply @ Ed Edmondson]Well I am still playing with this divider circuit. What I now have is the 3.3 megohm resistors in series with a 51 Kohm resistor to ground. The 10 Megohm resistor is still between the first two 3.3 mOhm resistors and the last 3.3 meg ohm and 51 Kohm string to ground. I calculate It is 0.0001005 A, Rt is 9,951,000 Ohms, E across each 3.3 Meg ohm resistor is 331.62 volts, E across the 51 kOhm is 5.1255 volts. The 5.1255 volts feeds the minus[] side of a comparator. The plus [+] side of the comparator is fed by a 50 K pot and a 5.6 volt zener in series with a 1K resistor. I suppose this allow a trip point to be set.
The end of the 10 meg ohm resistor feeds a 0.047 uf capacitor [400 vdc] in parallel with the primary of a trigger transformer [32.1 ohms] and a button to ground to discharge the capacitor and flash a lamp.
The question I have is what am I looking t as far as electrical specs here. I want to replace the push button with a SSR so I can connect it to a PIC micro.
Please educate me.
At 7 June 2010, 18:42:55 user Kammenos wrote: [reply @ Kammenos]So, RA = R1+R2 = 6.6M. RA is the R1 in my calculations.
RB = R3 = 3.3M. RB is the R2 in my calculations.
For the moment, forget the load (the 10M resistor and the comparator) and find the voltage from the output of the divider.
The voltage on the tap is Vt = 1000 x RB / (RA+RB) => Vt=1000x3.3/9.9 => Vt=333.33V
The max current that can go through 10M resistor in series with 333.33V, is I=U/R = 1000/10000000 =0.0001A = 0.1mA
At 7 June 2010, 15:49:45 user Ed Edmondson wrote: [reply @ Ed Edmondson]What I have is the following: 1,000 volts DC is the applied voltage. This voltage is going thru R1 [3.3 M] + R2 [3.3M] + R3 [3.3 M] to ground. There is a tap between R2 and R3 and this goes through a 10 M resistor to an comparator [Texas Instruments TL331]. The question is what voltage and current are applied to the comparator. What formulas do I need to calculate this. I have the series resistance is 9900000 ohms, the voltage is 1,000 and that gives a current through R1 through R3 as 0.000101 amps. I don't understand how to determine the rest.
Thanks,
Ed
At 7 June 2010, 4:21:03 user Kammenos wrote: [reply @ Kammenos]a circuit schematic would be easier for me to understand. You can use the forum to upload one, or upload to a photogallery site and post here the link. I am sure it is not too complicated. Send me the schematic.
At 6 June 2010, 21:10:56 user Ed Edmondson wrote: [reply @ Ed Edmondson]I have a resistive divider made up of 3 3,300,000 resistors in series with 1,000 volts applied. I have a tap between resistors 2 and 3 an it is a 10,000,000 resistor to feed a comparator. I tried to use your page to calculate the voltage coming out of the 10,000,000 resistor but your page and calculations are too confusing. What is "U"? You have no description of "U"? 

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