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I was trying to measure the internal resistance of an LDO that will produce 2.85V. I used a Digital multimeter to do this, but I got a resistance of 810 Ohm and the LDO can support upto 450 mA. Will the internal resistance of the LDO can be that high?

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    \$\begingroup\$ how are you measuring the internal resistance? ... note: should really refer to the device by its correct name "LDO regulator" \$\endgroup\$ – jsotola Nov 29 '18 at 8:40
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    \$\begingroup\$ ...and what do you plan to do wit this value once you have it? \$\endgroup\$ – Finbarr Nov 29 '18 at 9:03
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    \$\begingroup\$ "Internal resistance" isn't a proper parameter of an LDO. You can determine output resistance though. \$\endgroup\$ – Bimpelrekkie Nov 29 '18 at 9:17
  • \$\begingroup\$ What is LDo? The 'o' in lowercase on purpose? \$\endgroup\$ – soosai steven Nov 29 '18 at 11:54
  • \$\begingroup\$ LDOs with only a few hundred milliVolts of input-output differential voltage will be operating the internal Pchannel MOSFET in "triode" mode, where the FET is nearly a pure resistor that is varied to control the I*R voltage drop. Apply a sinusoidal current from a function generator at 1,000 Hz thru 1Kohm resistor and 100uF capacitor. Watch the polarity. Compute the AC current injected into the LDO output. Use a scope to measure the AC voltage on the LDO output. You can compute the output resistance of the LDO, which depends on servo-amp bandwidth, and properties of the Pchan FET. \$\endgroup\$ – analogsystemsrf Nov 30 '18 at 3:54
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If the input supply voltage to an "ideal" 5 volt regulator is 10 volts and the output load is 10 ohms (0.5 amp output current), then the regulator will "act" like a 10 ohm resistor in order to facilitate 5 volts at the output. It's just a simple potential divider.

If the load changes rapidly between 9 ohms and 11 ohms, then the regulator (due to feedback control), will keep the output voltage at 5 volts by rapidly altering its effective series resistance to counter the changing load impedance and current. So, when the load is at 9 ohms, the regulator will act like a 9 ohm resistor in order to keep the output at 5 volts.

If the input voltage reduces to (say) 8 volts and the load is 10 ohms then the effective series resistance of the regulator has to be 3 volts/0.5 amp = 6 ohm.

The effective series resistance of an ideal voltage regulator will change slowly or rapidly to counter the load and input voltage circumstances it faces.

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  • \$\begingroup\$ Thank you, I got it the output impedance will vary with load. \$\endgroup\$ – sajin Nov 30 '18 at 9:43
  • \$\begingroup\$ The effective series-pass resistance will vary with both load resistance and input voltage (to be precise). \$\endgroup\$ – Andy aka Nov 30 '18 at 9:48

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