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ESR of output capacitor of linear regulator

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treez

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Hello,
We wish to use the LM2936M linear regulator for vin = 24v, vout = 5v, iout = 6mA.
Page 14 says Cout must be 10uF and with ESR minimum of 0.3 Ohms.
Do you think I really need to add a 0.3R resistor to a 10uF ceramic……specially for this low load current?
LM2936 datasheet
https://www.ti.com/lit/ds/symlink/lm2936.pdf
 

Hi,

Trust the datasheet.

But you are free to use an appropriate electrolytics or tantalum...maybe without the need for a series resistor.

Klaus
 

It is written at 8.2.2.2, by the way.
 
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Basically, the ESR is an easy way of inserting a zero in the system to ensure that the crossover happens at a gain slope of -20 dB/decade and not -40 db/decade which is a recipe towards instability.
 
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With 24V available and an Iout requirement of 6mA it would be cheaper to use a Zener diode than an LDO!

Brian.
 

Ordinary voltage regulators use an emitter-follower (with its base-emitter voltage loss) as its pass transistor. An emitter-follower has no additional voltage gain and a small phase shift.
But a low dropout regulator uses a common emitter PNP pass transistor and its additonal voltage gain and phase shift requires a series RC (zobel network?) at the output. Your low dropout regulator also has a very low idle current which might be making the ESR of the output capacitor critical.
 

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