Does the AC ripple voltage at the input of a linear regulator like LM317 affect its regulation ? I had a circuit where I found that the output of the LM317 was dipping under higher loads(15V,1A).The load is a continuous load and not pulsating.I had had a 47uF cap at the input , and the problem was solved by changing it to a 1000uF cap. Is there any way to calculate the maximum input ripple allowed,or is sufficient to ensure that the ripple does not go below the drop out voltage ?
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\$\begingroup\$ On your example was the voltage dropping below the drop-out voltage? \$\endgroup\$– Andy akaJun 18, 2013 at 14:16
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\$\begingroup\$ Actually I had just replaced the cap. without checking and it worked.I will check it on a scope when I get an opportunity. \$\endgroup\$– EmbSysDevJun 18, 2013 at 17:01
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Look at the datasheet. Every decent linear regulator datasheet will specify the headroom, or minimum input voltage required for the regulator to produce the specified output voltage. In your case, the dips of the voltage between when the input cap was charged went too low and the regulator couldn't hold up the output anymore. The larger cap raised the level of the dips to above what the regulator needed.
AC on the input of a regulator even when this is above the headroom limit still matters, but there is a different spec for that. No regulator is perfect, so some amount of the input AC component will appear on the output. Sometimes this is specified as a input rejection ratio, or sometimes as a "input regulation" spec. For example, if the input rejection ratio is 60 dB and the input voltage varies over 1 V, then the output can vary by up to 1 mV from the variation of the input voltage alone.
answered Jun 18, 2013 at 14:18