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enter image description here

Suppose I have an IC where I connect a series resistor with Vcc and '+Vcc' supply pin of IC. Refer to the schematic above.

Say the IC current consumption is in uA or few mA for a supply voltage of 3V, resistor value 10E/100E/1000E.

  • What effect does the resistor have on the operation?
  • Will it limit the current enough to cause malfunction in IC?
  • Will the voltage reaching the IC-pin get divided as the resistor and IC are forming a voltage divider?

Electronics engineer here, you can get technical.

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    \$\begingroup\$ What picture are you referring to? And what IC, and what circuit are you using it in? \$\endgroup\$
    – Null
    Oct 28 at 11:24
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    \$\begingroup\$ It's impossible to say. Obviously the resistor will impact the power to the IC but whether or not that is enough to affect its operation depends on parameters you have not included. \$\endgroup\$
    – jwh20
    Oct 28 at 11:31
  • \$\begingroup\$ Added the circuit. May I know which parameters? \$\endgroup\$
    – Yoyoyoyo
    Oct 28 at 11:33
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    \$\begingroup\$ We need actual info, not hand waving. What chip that is and why would it need protection and from what? Assuming 1k resistor, each milliamp drops voltage by 1V, so the chip would just not work if it needs a few milliamps. \$\endgroup\$
    – Justme
    Oct 28 at 11:40
  • \$\begingroup\$ Normally you should not add a resistor like that in series with an IC unless there is a specific reason to do so. One example of such a reason could be to limit the power dissipation in a voltage regulator. But then you know the current that is flowing and the voltage drop that is needed. Then the series resistor can be chosen such that the circuit will work and some power is dissipated in the resistor. Most ICs however expect a supply where there is (almost) no series resistance. Many logic ICs will not work if you make that resistor too high. \$\endgroup\$ Oct 28 at 11:44
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With a chip that draws a steady supply current, you can measure the pin voltage with any given resistor, and see whether it stays above the minimum voltage rating.

The problem with that is that no IC that does anything useful draws a steady current. The current can be quite spiky as it charges internal nodes or charges external capacitive loads. The first spike of current will collapse the voltage and cause it to malfunction.

If you do as you are advised in every tutorial on the net and on paper to put a power supply decoupling capacitor between VCC and ground, then this will supply current for the short duration spikes, and you may get away with it.

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Will the voltage reaching the IC-pin get divided as the resistor & IC are forming a voltage divider?

Yes. More specifically, U = RI will apply and the resistor will drop some voltage.

Will it limit the current enough to cause malfunction in IC?

Maybe! Depends on how stable a supply the IC needs, its current consumption (both average and peak), and the value of any decoupling capacitors located between the resistor and the IC pin.

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