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If I have a circuit that requires 2V, I can use a voltage divider to get me the 2V from the 3V battery. I have been told that it is a good idea then to run that through a 2V regulator. This seems like overkill to me.

Are there any advantages to this? Is it better to drop the divider and just use the regulator?

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    \$\begingroup\$ Your circuit needs a current too at the specific voltage. \$\endgroup\$
    – Eugene Sh.
    Commented Sep 16, 2016 at 15:30
  • \$\begingroup\$ Do you need power at 2V, or do you need a reference voltage of 2V? \$\endgroup\$
    – JRE
    Commented Sep 16, 2016 at 15:33
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    \$\begingroup\$ Unfortunately you omit a crucial parameter, how much current is needed at that 2 V ? If it is 1 uA, a couple of resistors might do. For 10 mA or more, a regulator is better. Also, how stable does that 2 V need to be ? Is it OK if it becomes 1.8 V or 2.2 V ? I can think of 10 different circuits to make 2 V out of 3 V, which one is best depends on what you need. Your needs are still unclear (yes, specifying circuit requirements is difficult, it takes experience). \$\endgroup\$
    – Bimpelrekkie
    Commented Sep 16, 2016 at 15:43
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    \$\begingroup\$ The thing with a voltage divider is: the output voltage will only be as accurate as the resistors are (a few %) and the output voltage will depend on the input voltage (batteries discharge and their voltage changes). A regulator or proper reference voltage source are much more accurate. \$\endgroup\$
    – 0x6d64
    Commented Sep 16, 2016 at 15:53
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    \$\begingroup\$ @o.fithcheallaigh, when you say "reference" we think you mean a voltage level that might be compared with other voltages to measure those other voltages. A reference generally doesn't need to provide any power to the circuit it's connected to. But in another comment you say " there will be a micro-controller etc., so it will require (I am sure) a higher amount of current." If your 2 V will provide power to these components you should not be talking about a reference, you should be talking about a "power supply". \$\endgroup\$
    – The Photon
    Commented Sep 16, 2016 at 16:42

2 Answers 2

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In general, you have got three different opportunities to reduce your battery voltage from 3V to 2V. Each opportunity has got its own advantages and disadvantages and should (must) be used for different applications:

  1. Voltage divider: Voltage divider are used for applications which don't need any current but a common reference voltage. You can't drive a huge current because this would change your divide proportion. Mention that V_out depends on V_in. So if your battery voltage goes down, your reference voltage goes down too.

Typically applications: OP gain resistors, Voltage for feedback inputs

  1. Reference diode: Reference diodes are used for applications which don't need any current but a very exact reference voltage. In this case your reference voltage doesn't depend on V_in.

Typically application: ADC reference, OP circuits

  1. Voltage regulator: Voltage regulators are used for applications which need more or less load current. In this case your voltage doesn't depend on V_in. There are two different types of voltage regulators. On the one hand you can easily use a linear regulator or on the other hand you can use a clocked buck down regulator for high efficiency.

Typically applications: Supply voltage for circuit parts

Hope this help you!

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A voltage divider looks like this (roughly):

schematic

simulate this circuit – Schematic created using CircuitLab

The problem is that Vout will change if Vin changes, but also if Z_load changes:

\$ V_{out} = V_{in}\frac{R_2||Z_{load}}{R_1+(R_2||Z_{load})} \$

For \$Z_{load} \gg R_2\$, \$V_{out} = V_{in}\frac{R_2}{R_1+R_2}\$, and you can use a voltage divider. If \$Z_{load} \approx R_2\$, then you have to use the whole expression. If \$Z_{load}\$ is unknown, or changes, then you run into more problems.

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