0
\$\begingroup\$

As far as I know, you have to put capacitors on most voltage regulators. However, the data sheet for this one specifically says "No external components required" and furthermore goes on to state "electrical characteristics" being "Vi=10V, Io=500mA, Ci=0.33uF, Co=0.1uF, Unless Otherwise Specified."

I'm fairly new to electrical engineering and I don't know what to think of this. Is it saying I should put a 0.33uF on the Vin and 0.1uF on Vout, or does it mean that it already has capacity equivalent to those values built into the IC?

As a bonus, It'd be helpful to know what it means by Vi=10V and Io=500mA because it specifically states that the output is 1A and the minimum input is 7V (dropout 2V, output 5V).

I'm thoroughly confused.

Here's a link to the datasheet: MC7805CT(TO-220)-480029.pdf

\$\endgroup\$

2 Answers 2

Reset to default
2
\$\begingroup\$

7805 does require external capacitors. The capacitances mentioned in the datasheet are external.

7805s are produced by many manufacturers. The one that you have chosen have a particularly dismal short datasheet. 7805 with better datasheet. 7805 with very thorough datasheet.

related thread: Capacitor Sizes for 7805 Regulator

\$\endgroup\$
1
  • \$\begingroup\$ Yeah, I was thinking it was poor documentation. Thanks for the links. \$\endgroup\$
    – userNo99
    Commented Jan 11, 2015 at 1:13
1
\$\begingroup\$

It will work without those capacitors (won't oscillate) unlike most more modern LDO-type regulators (it's not an LDO = Low Drop Out) type.

However it's recommended to have an input capacitor if the regulator is more than a few inches from the input filter capacitor (normally an electrolytic). It's conceivable it could oscillate if you have too long a run between.

You'd also normally have output bypass capacitors, and they do improve the transient response (meaning that if you suddenly apply or remove a load the output won't dip up or down as much). If you're using a microcontroller or a logic chip, you'll likely have 100nF (0.1uF) or 1uF across each chip rails anyway, so if you've got a 60Hz input filter capacitor nearby and a bypass cap nearby, there's no extra components.

So, they're not strictly necessary, but normally they're used. The exact value is not critical but 100nF or 1uF ceramic rated for 10V or 16V is good at the output, and I like to use a 47uF/50V aluminum electrolytic at the input if there any distance to the supply. Keep them relatively close physically, but try not to let the regulator heat up the electrolytic in particular too much (conflicting requirements, welcome to engineering).

\$\endgroup\$
4
  • \$\begingroup\$ Thanks for the detailed response! I was planning on using this to convert 9V battery power to 5V. Is there a better way to do that that would minimize power loss? \$\endgroup\$
    – userNo99
    Commented Jan 11, 2015 at 1:15
  • \$\begingroup\$ If you're talking about a PP9 battery, it's not very suitable for most things that run from 5V. The regulator will suck back a few mA and the power loss on top of that is Iout(Vin-Vout). You'd do better with AA batteries and a switching regulator if the current is more than a few tens of mA. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jan 11, 2015 at 2:09
  • \$\begingroup\$ I wouldn't know where to start. Anything you can recommend I search for more info? \$\endgroup\$
    – userNo99
    Commented Jan 12, 2015 at 2:07
  • \$\begingroup\$ Simplest switching regulator is something like an LM2576 but there are cheaper options that can use much smaller inductors if you don't mind tiny packages. TI and LTC have plenty of help available on their websites. Lesser known manufacturers, less so. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jan 12, 2015 at 2:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.