What is the difference between having a voltage divider vs just using a resistor in series.

So for example, I have an input voltage with 12V, and two resistors in a voltage divider, R1=10k, and R2=10k, so my voltage is evenly split to 6V. How is this different from having one resistor (R=6k, I=1mA) in series?

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    \$\begingroup\$ It's a bit unclear what you are asking here, could you clarify things a little? For example do you mean a current source with the 6k resistor? If so, do you mean in series or in parallel with the current source? If in series and no load it will assume infinite voltage (assuming ideal source) I thought you might have meant a 6V voltage source with the 6k in series... \$\endgroup\$ – Oli Glaser Jan 29 '12 at 3:46
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    \$\begingroup\$ Can you explain what it is you are trying to do? I believe you have a misunderstanding of the basic concepts, but it is difficult to explain the correct concepts with out some context. Like are you trying to cut the voltage in half to be read by an ADC? \$\endgroup\$ – Kellenjb Jan 29 '12 at 19:58

If you draw 1mA from the resistor divider circuit you mentioned, it will output one volt (the upper resistor will have 1.1mA flowing through it, thus dropping 11 volts; of that 1.1mA, 0.1mA will go through the bottom resistor while the remaining 1mA will go into your load). The 6K resistor would drop 6 volts, thus feeding 6 volts into a 100mA load.

If either the load current or the load resistance is a known constant value, one can calculate a series resistance which will convert a known input voltage into any desired known, lower, load voltage. If the load current or resistance isn't known precisely, however, deviations from the ideal will cause the load voltage to vary from what is intended. The greater the difference between the input voltage and the load voltage, the greater the variation in load voltage.

Adding a load resistor will effectively add a known fixed load in addition to the potentially-variable one. Suppose one had a 12-volt source and the intended load were 10uA +/- 5uA at 6 volts. If one just used a series resistor sized for the 10uA case (600K), it would drop only 3V at 5uA (feeding 9 volts to the load) and 9V at 15uA (feeding 3 volts to the load). Adding a 6.06K resistor in parallel with the load would cause the total current draw to be about 1.000mA+/-0.005mA, requiring the upper resistor be changed to 6K; since changes in the load current would only affect the total current by about 0.5%, they would only affect the voltage drop of the upper resistor by about 0.5%.

If the source voltage is stable, and the output current is small, a voltage divider may be a practical means of generating a stable voltage. Unfortunately, for the voltage divider to generate a stable voltage, the amount of current fed through the lower resistor (and thus wasted) must be large relative to the possible absolute variation in load current. This is usually no problem when the output current is on the order of picoamps, is sometimes acceptable when the output current is on the order of microamps, and generally becomes unacceptable when the output current is on the order of amps.

  • \$\begingroup\$ This is the best answer to this question that I have read while having searched for one hour for a good explanation. It covers all details, the example is great and your last paragraph is also very nice. Thank you! \$\endgroup\$ – IceFire Apr 27 at 10:53
  • \$\begingroup\$ but why 5.94K resistor? a voltage divider with R1=600k and R2=5.94k does yield 0.0098V, this is far from 6V? \$\endgroup\$ – IceFire Apr 28 at 11:19
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    \$\begingroup\$ @IceFire: I forgot to mention that after adding the load resistor one would have to recalculate the upper resistor as appropriate for the increased loading. I also think I probably should have said 6.06K resistor. \$\endgroup\$ – supercat Apr 28 at 16:45
  • \$\begingroup\$ Ah, ok, this makes it a bit clearer. If feasible, could you also briefly show how you come to 1mA? I tried R2/(R1+R2) and stuff, but never get to this result... \$\endgroup\$ – IceFire Apr 28 at 17:19
  • \$\begingroup\$ The current is set by the combination of the load and whatever resistance is in parallel with it. The 1mA was chosen as a value which is 100x as large as the bare load resistance (reducing 100-fold the effect of variations in load current); 6.06K was chosen to pass 990uA at 6 volts. The top side resistor would then need to be adjusted to drop 6 volts at 1mA. \$\endgroup\$ – supercat Apr 28 at 21:08

The single resistor doesn't divide the voltage.
For an ideal 12V source with 6k\$\Omega\$ in series, you get 12V with 6k (output) impedance.

The centre of two 10k resistors in series across the same source would provide 6V with an impedance of 5k\$\Omega\$.
So there is no difference between this and a 6V source with 5k in series.


If you really have the 1mA then the single resistor will do. The 1mA will flow into the input of the circuit following the resistor and this will therefore have an input resistance of 6k\$\Omega\$ (6V / 1mA). So you end up with two resistors after all: the one you placed and the input impedance.
In case you're building the divider with the two 10k\$\Omega\$ resistors keep in mind that the input impedance of the following circuit is parallel to the lower resistor. Anything but a high-impedance input (like the input of an opamp) will decrease the 6V at the node.


How is this different from

"Different from" in what way?

Two obvious differences are in the output equivalent circuit (assuming you mean that the center node of the voltage divider is the output), and in the load presented to the input voltage.

The output of the voltage divider has a Thevenin equivalent of 6V with 5K output impedance. The output of the resistor + a 1mA current source load = 6V with 6K output impedance.

The load on the supply for the voltage divider is 0.6mA via 20K load; the load on the supply for the resistor + current source is a current source load (constant current).


Some confusion here I think, and also some good thinking. The difference for the question asked is maybe none, if you are just looking to drop 6 volts down from the 12 volts. It will depend on the application. Either might be successful in the right situation.

In the divider case, the center point will be 6 volts if the load on the center point takes no current, and in the single resistor with 1 mA, the output will be 6 volts if the load takes exactly 1 mA. But they are 2 different applications and neither will be exact if the load is not infinite impedance in the first case or 6k Ohms to the 12 V return (or 1 mA current sink) in the second case.

If you are going to use a voltage divider to create one voltage level from another, then you can do that with a voltage divider, but you need to know what the impedance of the termination points are, so that so that you can calculate the complete circuit voltages and current flows and if you can treat any termination as an ideal current or voltage source (allowing you to ignore its impedance). Assuming that the divider in the initial problem is between a 12 volt source and its return and the voltage source is low enough impedance to ignore, then the answer to the question depends on what the impedance of the output point is to either the 12 volt source or the 12 volt return. To see the voltage at the divider output point, you need to consider the load impedances in parallel to the resistive divider legs and do the calculation for the divider to see the output voltage and the current flowing into or out from the load.

If the load impedances are several orders of magnitude greater than 6K then in most cases you can ignore them.


simulate this circuit – Schematic created using CircuitLab


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