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I need to use a LM7809 to reduce 12-14V to 9V , the device is using 40mA at normal state

when I search google for application circuit - I got all kind of valus for Cin , Cout. use this datasheet LM78xx datasheets

I saw in page 22 to use 0.33uF and 0.1uF - so can I use this ?

what is the thumb rule for it? and what is the formula I need to use?

Thanks ,

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    \$\begingroup\$ 2A is too much for an LM7809, it will current-limit. \$\endgroup\$
    – Spehro Pefhany
    Commented Oct 26, 2020 at 8:38
  • \$\begingroup\$ didn't read the datashet correct . it use max 100mA , normal use is ~ 40mA \$\endgroup\$
    – Afik_Android
    Commented Oct 26, 2020 at 8:42
  • \$\begingroup\$ @Afik_Android then edit your question please to correct the values. Also, the information on which in- and output capacitors to use is usually part of the LM78xx datasheet – but there's many different datasheets for these out there, so please add a link to the LM78xx datasheet you've read! \$\endgroup\$
    – Marcus Müller
    Commented Oct 26, 2020 at 9:06
  • \$\begingroup\$ You should read the datasheet. It will tell you the values or how to calculate them. \$\endgroup\$
    – Peter Karlsen
    Commented Oct 26, 2020 at 9:26
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    \$\begingroup\$ @ChristopherDyer 2.2A "typical" is at an unrealistic 25°C Tj. When it's hot as a pistol (as it surely will be under those conditions) it will typically limit at well under 2A (1.5A at max Tj) and much less than that worse-case. It might be a useful number for other purposes such as determining whether a short will damage something or for a pulsed load with low average current. \$\endgroup\$
    – Spehro Pefhany
    Commented Oct 26, 2020 at 9:53

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