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This question already has an answer here:

1) Why to use a LDO regulator when a simple voltage divider can do the job for stepping down voltage?

2) How do we set the current in voltage divider when use as regulator functionality ? Or

3) Can we calculate (theoretically) the current sourcing capability of voltage divider output ?

voltage divider output V(out) = Rb/(Ra+Rb) * V(in)

What is the current sourcing capability at V(out) node ?

Regards, Azlum

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marked as duplicate by PeterJ, Autistic, Daniel Grillo, Nick Alexeev Feb 16 '16 at 17:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • \$\begingroup\$ A voltage divider works as a voltage regulator when the load connected is much higher than the output impedance of the divider. \$\endgroup\$ – lucas92 Feb 15 '16 at 13:38
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A voltage divider with two resistors R1 and R2 acts like a voltage source Vin*R2/(R1+R2), in series with a source resistance of (R1*R2)/(R1+R2).

For instance, if both resistors were 1k\$\Omega\$, and the whole string was fed from 10v, the mid point would behave as a 5v source in series with a 500\$\Omega\$ resistor.

This series resistance means that the regulation is too poor in the general case, and a regulator would normally be used.

In some cases, when the output current required is very low, or the load requires some source resistance, a resistive voltage divider is adequate.

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The voltage regulator is used to get a fixed voltage whatever the input voltage is ( as far as it is within device limits). while the voltage divider is used to step down the input voltage at a specific ratio. Hence at different input, different output is generated.

The voltage regulator is comparable to a resistor + zener diode circuit , not a resistor voltage divider.

For more info check this post that discussed this issue before : When would I use a voltage regulator vs voltage divider?

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  • \$\begingroup\$ What about the current ? How we can get the sourcing current at V(out) node ? \$\endgroup\$ – Mohammed Azlum Feb 15 '16 at 13:55
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You actually confused about Linear voltage regulators and voltage divider circuit.A voltage divider bias is not a exact match for Linear voltage regulator.Okay Let me explain with a example.

    schematic

    simulate this circuit – Schematic created using CircuitLab

      Now in this diagram you can see that I gave 9v as my input and the output that you can expect is around 8.1V.When the input changes(obviously a battery wont be ideal) ,so let's sya you have 8V and at the output you'll get only 7.2V and this is not desirable voltage(when you assume that you want to have 8V at output).But in Voltage regulators,it'll deliver the 8V until the input to the Voltage reg is greater than 8V.
    The current output from the putput depends on the load.If you use a current source,you'll predict the current.

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  • \$\begingroup\$ What about the current ? How we can get the sourcing current at V(out) node ? \$\endgroup\$ – Mohammed Azlum Feb 15 '16 at 13:55
  • \$\begingroup\$ Ya,that's what I am telling that,the current at the node depends on the load you connect.For example,if you connect a 1Kohm resistor(say ,it'll take 1mA).For 10Kohm load,it wont be 1mA.It'll be lesser than 1mA.Current at the node depends on the load you connect across the resistor R2. \$\endgroup\$ – Aadarsh Feb 15 '16 at 14:04
  • \$\begingroup\$ And it's also a very bad idea that,you can use a use a voltage divider as a substitute for Voltage regulator.In fact,the voltage regulator itself is not much efficient,you can go for buck converters.for small voltages,voltages regulators are good. \$\endgroup\$ – Aadarsh Feb 15 '16 at 14:06

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