# What's the difference between using a simple series resistor and a voltage divider to reduce voltage

Discussion in 'General Electronics Chat' started by jellytot, Jul 10, 2016.

1. ### jellytot Thread Starter Member

May 20, 2014
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Let's say you have 2 options to reduce voltage:
1. One resistor in series, before the load.
2. Voltage divider (2 resistors in series, with output line between them). Like this: I'm aware there are many other options to reduce voltage. But I'm just curious, between the two options above, is there any difference at all? Or is one solution "better", and why?

2. ### joeyd999 AAC Fanatic!

Jun 6, 2011
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No difference. Both have an easily calcuable Vth and Rth, and are equivalent.

3. ### #12 Expert

Nov 30, 2010
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The one resistor method works very well if the load is stable. It sucks horribly for something like a motor which has a huge starting current.
The two resistor method tries to mitigate the differences in load current by wasting a lot more current than the load in order to make the voltage at the center tap fairly stable. The other options use semiconductors like a zener diode or an amplifier stage to isolate the load changes from the voltage divider.

For any load that changes, the more sophisticated circuits would be "better".

4. ### joeyd999 AAC Fanatic!

Jun 6, 2011
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Are you suggesting that the VI curve is different in the two cases???

5. ### #12 Expert

Nov 30, 2010
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Are you drunk posting again?

6. ### WBahn Moderator

Mar 31, 2012
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They are quite different.

In the series resistor case, you only have one degree of freedom, so you HAVE to choose the particular value for the voltage dropping resistor that results in the desired voltage across the load when the load is the expected resistance (or its equivalent) and so your source has an output resistance equal to the value of the dropping resistor (plus it's own internal resistance).

In the voltage divider case, you have two degrees of freedom, so you can set the ratio of the two resistors to give the desired voltage drop across the load, but you can also scale the absolute magnitudes of the resistors to make the source have whatever output resistance you want (above it's own internal resistance, of course). Hence you can trade efficiency for stability.

7. ### joeyd999 AAC Fanatic!

Jun 6, 2011
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No! Thevinin equivalence postulates that both circuits are identical.

8. ### joeyd999 AAC Fanatic!

Jun 6, 2011
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You may argue whatever practical distinction you want, but, fundamentally, the two cases are identical (as far as the load is concerned), assuming the thevinen values are the same.

This is the way I understood the question. Perhaps I misunderstood.

9. ### WBahn Moderator

Mar 31, 2012
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Hardly.

Let's use an example. Say Vin is 12 V and we want Vout to a 10 kΩ load to be 6 V.

The first circuit requires a series resistance of 10 kΩ and the Thevenin equivalent for the source as seen by the load is a 12 V source in series with a 10 kΩ resistance.

For the second circuit, on the other hand, we have a LOT more options. Let's say that we choose to set R2 = αRload. Then R1 needs to be

R1 = [α/(α+1)]·Rload

Let's choose α = 10%. That means that

R2 = 1 kΩ and R1 = 909 Ω

So the Thevenin equivalent seen by the load is now V = 6.29 V with a source resistance of 476 Ω. That's a HUGE difference.

10. ### WBahn Moderator

Mar 31, 2012
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The ONLY way for the Thevenin values to be the same is to set R2 = ∞. But in that case, they ARE identical (your voltage divider is no longer a voltage divider -- it has become merely a series voltage dropping resistor).

11. ### joeyd999 AAC Fanatic!

Jun 6, 2011
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This assumes you restrain Vin to 12V, which was not in the specification. If you set Vin to the Vth of the second circuit, and R to the thevinen equivalent of the 2 resistors, the network performance is identical, which was my point.

12. ### #12 Expert

Nov 30, 2010
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A stable, unchanging, load current was also not specified.
If you completely ignore the fact that I explained how the circuit has different results for different loads, you can argue that the circuits are identical for identical loads. The idea that you chose to ignore the conditions I explained means...what? You assume the TS always has the same load? Or you assume I have some interest in addressing you instead of the TS?

13. ### WBahn Moderator

Mar 31, 2012
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It doesn't assume that Vin is constrained to 12 V, that was just an example in order to all the examination of actual values. But I am assuming that Vin is constrained to be Vin for both cases, otherwise we are comparing apples and oranges.

If we are free to set Vin to anything we want, just set Vin to what you want the voltage across the load to be, forget about either option, and just connect Vin to the load!

The only reasonable way to interpret the problem (in my opinion) is that, for either circuit, the value of Vin (possibly with it's own output resistance) is the same. After all, the TS is specifically asking which option is the better choice to reduce the voltage from Vin to Vout. So the only difference between the two cases being considered is which option you use to drop the voltage from the fixed input voltage to the desired output voltage.

14. ### hp1729 Well-Known Member

Nov 23, 2015
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I have to go with no, they are not. To be an effective voltage divider the current through the voltage divider should be 5 to 10 times the load current. So the voltage divider is more stable but not as efficient.

15. ### shteii01 AAC Fanatic!

Feb 19, 2010
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Ok. We need to clear up the setup.

Situation 1.
We have voltage divider composed of a resistor R1 and a load Rl.

Situation 2.
We have voltage divider composed of resistor R1 and resistor R2. Load is a resistor Rl and it is in parallel with either R1 or R2.

The way I see it.

In Situation 1 you need a somewhat precise choice of voltage source or you may need to spend some time finding R1 that would drop some odd voltage to make sure that Rl is receiving the voltage it needs. Since same current will pass through your R1-Rl voltage divider, you don't need to worry about loosing any current.

In Situation 2 you can use the R1-R2 voltage divider to produce the voltage you need for Rl, which means you don't need a carefully chosen voltage source. However, since Rl is in parallel with either R1 or R2 resistor, you are forming a current divider. This means that not all the available current will pass through Rl, so you need to be careful what values you use for R1-R2 voltage divider in order to get sufficient current to Rl.

16. ### hp1729 Well-Known Member

Nov 23, 2015
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In every instance of using a voltage divider I can think of the current through the R1, R2 is many times the load current.

17. ### jellytot Thread Starter Member

May 20, 2014
72
0
Thanks for the replies, everybody.

Related question: When using a Voltage Divider, what's the difference between using different sets of R1 and R2 values to get the same voltage drop? For example, using R1 = 50 ohms, R2 = 75 ohms versus R1 = 500 ohms and R2 = 750 ohms?

18. ### hp1729 Well-Known Member

Nov 23, 2015
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I think the question revolves around the level of current to be regulated. mA or more, just the resistor. Microamps or less the voltage divider is best for stability.

19. ### joeyd999 AAC Fanatic!

Jun 6, 2011
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If everyone would stop arguing with me, and simply elaborate on the point I was trying to make, the answer would be obvious. Case 2 degenerates into case 1 when you perform the thevinin analysis. From there, the answer to your question is elementary.

20. ### shteii01 AAC Fanatic!

Feb 19, 2010
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I am not a pro on the topic, but one example that I have come across is in electronics. Often enough you have sensor that outputs analog voltage. You want to digitize this analog voltage, but it is too high for your analog to digital converter. So you run analog voltage from sensor to the voltage divider, the output of voltage divider is then sent to ADC. However, ADC often are build using R-2R resistor network. What happens is that the resistor from voltage divider is then in parallel with a resistor in ADC, this loading of the ADC changes the resistance and in turn changes the voltage across resistance, which in turn gives you a false value for the sensor. The bigger the resistors in your voltage divider, the bigger the loading of the ADC, the bigger the change in the signal that you are trying to feed to ADC.

Last edited: Jul 10, 2016